Problems & Solutions
for
Statistical Physics of Particles
Updated July 2008
by
Mehran Kardar
Department of Physics
Massachusetts Institute of Technology
Cambridge, Massachusetts 02139, USA
Table of Contents
I.
Thermodynamics
II.
Probability
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
III.
Kinetic Theory of Gases
IV.
Classical Statistical Mechanics
V. Interacting Particles
VI.
VII.
1
. . . . . . . . . . . . . . . . . . . . . . . . 38
. . . . . . . . . . . . . . . . . . . . . 72
. . . . . . . . . . . . . . . . . . . . . . . . . . 93
Quantum Statistical Mechanics
. . . . . . . . . . . . . . . . . . . .
121
Ideal Quantum Gases . . . . . . . . . . . . . . . . . . . . . . . .
138
Problems for Chapter I - Thermodynamics
1. Surface tension:
Thermodynamic properties of the interface between two phases are
described by a state function called the surface tension S. It is defined in terms of the
work required to increase the surface area by an amount dA through d̄W = SdA.
(a) By considering the work done against surface tension in an infinitesimal change in
radius, show that the pressure inside a spherical drop of water of radius R is larger than
outside pressure by 2S/R. What is the air pressure inside a soap bubble of radius R?
• The work done by a water droplet on the outside world, needed to increase the radius
from R to R + ∆R is
∆W = (P − Po ) · 4πR2 · ∆R,
where P is the pressure inside the drop and Po is the atmospheric pressure. In equilibrium,
this should be equal to the increase in the surface energy S∆A = S · 8πR · ∆R, where S
is the surface tension, and
∆Wtotal = 0,
=⇒
∆Wpressure = −∆Wsurface ,
resulting in
(P − Po ) · 4πR2 · ∆R = S · 8πR · ∆R,
=⇒
(P − Po ) =
2S
.
R
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures,
and
Pfilm − Po = Pinterior − Pfilm =
leading to
Pinterior − Po =
2S
,
R
4S
.
R
Hence, the air pressure inside the bubble is larger than atmospheric pressure by 4S/R.
(b) A water droplet condenses on a solid surface. There are three surface tensions involved
S aw , S sw , and S sa , where a, s, and w refer to air, solid and water respectively. Calculate
the angle of contact, and find the condition for the appearance of a water film (complete
wetting).
• When steam condenses on a solid surface, water either forms a droplet, or spreads on
the surface. There are two ways to consider this problem:
Method 1: Energy associated with the interfaces
1
In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore
dE = Saw dAaw + Sas dAas + Sws dAws = 0.
Since the total surface area of the solid is constant,
dAas + dAws = 0.
From geometrical considerations (see proof below), we obtain
dAws cos θ = dAaw .
From these equations, we obtain
dE = (Saw cos θ − Sas + Sws ) dAws = 0,
=⇒
cos θ =
Sas − Sws
.
Saw
Proof of dAws cos θ = dAaw : Consider a droplet which is part of a sphere of radius R,
which is cut by the substrate at an angle θ. The areas of the involved surfaces are
Aws = π(R sin θ)2 ,
and
Aaw = 2πR2 (1 − cos θ).
Let us consider a small change in shape, accompanied by changes in R and θ. These
variations should preserve the volume of water, i.e. constrained by
V =
πR3
cos3 θ − 3 cos θ + 2 .
3
Introducing x = cos θ, we can re-write the above results as

2
2
A
=
πR
1
−
x
,
ws




Aaw = 2πR2 (1 − x) ,

3


 V = πR x3 − 3x + 2 .
3
The variations of these quantities are then obtained from

dR

2

(1 − x ) − Rx dx,
dAws = 2πR


dx




dR
(1 − x) − R dx,
dAaw = 2πR 2

dx




dR

2
3
2

 dV = πR
(x − 3x + 2) + R(x − x) dx = 0.
dx
2
From the last equation, we conclude
1 dR
x2 − 1
x+1
=− 3
=−
.
R dx
x − 3x + 2
(x − 1)(x + 2)
Substituting for dR/dx gives,
dAws = 2πR2
dx
,
x+2
and
dAaw = 2πR2
x · dx
,
x+2
resulting in the required result of
dAaw = x · dAws = dAws cos θ.
Method 2: Balancing forces on the contact line
Another way to interpret the result is to consider the force balance of the equilibrium
surface tension on the contact line. There are four forces acting on the line: (1) the surface
tension at the water–gas interface, (2) the surface tension at the solid–water interface, (3)
the surface tension at the gas–solid interface, and (4) the force downward by solid–contact
line interaction. The last force ensures that the contact line stays on the solid surface, and
is downward since the contact line is allowed to move only horizontally without friction.
These forces should cancel along both the y–direction x–directions. The latter gives the
condition for the contact angle known as Young’s equation,
S as = S aw · cos θ + S ws ,
=⇒
cos θ =
S as − S ws
.
S aw
The critical condition for the complete wetting occurs when θ = 0, or cos θ = 1, i.e. for
cos θC =
S as − S ws
= 1.
S aw
Complete wetting of the substrate thus occurs whenever
S aw ≤ S as − S ws .
(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances
surface tension effects are all important. At room temperature, the surface tension of
water is S o ≈ 7 × 10−2 N m−1 . Estimate the typical length-scale that separates “large” and
“small” behaviors. Give a couple of examples for where this length-scale is important.
3
• Typical length scales at which the surface tension effects become significant are given
by the condition that the forces exerted by surface tension and relevant pressures become
comparable, or by the condition that the surface energy is comparable to the other energy
changes of interest.
Example 1: Size of water drops not much deformed on a non-wetting surface. This is given
by equalizing the surface energy and the gravitational energy,
S · 4πR2 ≈ mgR = ρV gR =
4π 4
R g,
3
leading to
R≈
s
3S
≈
ρg
s
3 · 7 × 10−2 N/m
≈ 1.5 × 10−3 m = 1.5mm.
103 kg/m3 × 10m/s2
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel
(about 1 atm) = surface tension of water, gives
πgel ≈
2S
N
kB T ≈
,
V
R
where N is the number of counter ions within the gel. Thus,
2 × 7 × 10−2 N/m
R≈
≈ 10−6 m.
105 N/m2
********
2. Surfactants:
Surfactant molecules such as those in soap or shampoo prefer to spread
on the air-water surface rather than dissolve in water. To see this, float a hair on the
surface of water and gently touch the water in its vicinity with a piece of soap. (This is
also why a piece of soap can power a toy paper boat.)
(a) The air-water surface tension S o (assumed to be temperature independent) is reduced
roughly by N kB T /A, where N is the number of surfactant particles, and A is the area.
Explain this result qualitatively.
• Typical surfactant molecules have a hydrophilic head and a hydrophobic tail, and prefer
to go to the interface between water and air, or water and oil. Some examples are,
CH3 − (CH2 )11 − SO3− · N a+ ,
4
CH3 − (CH2 )11 − N + (CH3 )3 · Cl− ,
CH3 − (CH2 )11 − O − (CH2 − CH2 − O)12 − H.
The surfactant molecules spread over the surface of water and behave as a two dimensional
gas. The gas has a pressure proportional to the density and the absolute temperature,
which comes from the two dimensional degrees of freedom of the molecules. Thus the
surfactants lower the free energy of the surface when the surface area is increased.
∆Fsurfactant =
N
kB T · ∆A = (S − So ) · ∆A,
A
S = So −
=⇒
N
kB T.
A
(Note that surface tension is defined with a sign opposite to that of hydrostatic pressure.)
(b) Place a drop of water on a clean surface. Observe what happens to the air-watersurface contact angle as you gently touch the droplet surface with a small piece of soap,
and explain the observation.
• As shown in the previous problem, the contact angle satisfies
cos θ =
S as − S ws
.
S aw
Touching the surface of the droplet with a small piece of soap reduces S aw , hence cos θ
increases, or equivalently, the angle θ decreases.
(c) More careful observations show that at higher surfactant densities
∂S
∂A
T
2a
N kB T
−
=
2
(A − N b)
A
N
A
2
,
∂T
∂S
and
A
=−
A − Nb
;
N kB
where a and b are constants. Obtain the expression for S(A, T ) and explain qualitatively
the origin of the corrections described by a and b.
• When the surfactant molecules are dense their interaction becomes important, resulting
in
∂S
∂A
T
N kB T
2a
=
−
2
(A − N b)
A
and
∂T
∂S
A
=−
N
A
2
N
A
,
A − Nb
.
N kB
Integrating the first equation, gives
N kB T
+a
S(A, t) = f (T ) −
A − Nb
5
2
,
where f (T ) is a function of only T , while integrating the second equation, yields
N kB T
,
A − Nb
S(A, T ) = g(A) −
with g(A) a function of only A. By comparing these two equations we get
N kB T
+a
S(A, T ) = S o −
A − Nb
N
A
2
,
where S o represents the surface tension in the absence of surfactants and is independent
of A and T . The equation resembles the van der Waals equation of state for gas-liquid
systems. The factor N b in the second term represents the excluded volume effect due to
the finite size of the surfactant molecules. The last term represents the binary interaction
between two surfactant molecules. If surfactant molecules attract each other the coefficient
a is positive the surface tension increases.
(d) Find an expression for CS − CA in terms of
∂E
∂T S .
∂E
∂A T ,
∂S
∂A
T
∂E
∂T
A
S,
, and
∂T
,
∂S A
• Taking A and T as independent variables, we obtain
δQ = dE − S · dA,
=⇒
and
δQ =
δQ =
∂E
∂A
T
∂E
∂A
dA +
T
− S dA +
∂E
∂T
dT − S · dA,
dT.
A
From the above result, the heat capacities are obtained as
resulting in

δQ


 CA ≡ δT

δQ

 CS ≡
δT
∂E
∂T
A
∂E
∂A
=
−S
∂A T
∂T
=
A
S
∂E
∂A
∂S
∂A
·
CS − CA =
T
−S
S
∂A
∂T
Using the chain rule relation
∂T
∂S
A
·
T
6
∂A
∂T
S
= −1,
∂E
+
∂T
.
S
,
S
for
∂E
∂T A
=
we obtain
CS − CA =
∂E
∂A
T
−S

·

−1
∂T
∂S A
·
∂S
∂A
T
.
********
3. Temperature scales:
Prove the equivalence of the ideal gas temperature scale Θ, and
the thermodynamic scale T , by performing a Carnot cycle on an ideal gas. The ideal gas
satisfies P V = N kB Θ, and its internal energy E is a function of Θ only. However, you
may not assume that E ∝ Θ. You may wish to proceed as follows:
(a) Calculate the heat exchanges QH and QC as a function of ΘH , ΘC , and the volume
expansion factors.
• The ideal gas temperature is defined through the equation of state
θ=
PV
.
N kB
The thermodynamic temperature is defined for a reversible Carnot cycle by
Qhot
Thot
=
.
Tcold
Qcold
For an ideal gas, the internal energy is a function only of θ, i.e. E = E(θ), and
d̄Q = dE − d̄W =
dE
· dθ + P dV.
dθ
adiabatics (∆Q = 0)
pressure P
1
Q hot
2
θ hot
isothermals
4
Q cold 3
volume V
7
θ cold
Consider the Carnot cycle indicated in the figure. For the segment 1 to 2, which undergoes
an isothermal expansion, we have
dθ = 0,
=⇒ d̄Qhot = P dV,
and P =
N kB θhot
.
V
Hence, the heat input of the cycle is related to the expansion factor by
Z V2
dV
V2
N kB θhot
Qhot =
.
= N kB θhot ln
V
V1
V1
A similar calculation along the low temperature isotherm yields
Z V3
V3
dV
,
= N kB θcold ln
Qcold =
N kB θcold
V
V4
V4
and thus
Qhot
θhot ln (V2 /V1 )
=
.
Qcold
θcold ln (V3 /V4 )
(b) Calculate the volume expansion factor in an adiabatic process as a function of Θ.
• Next, we calculate the volume expansion/compression ratios in the adiabatic processes.
Along an adiabatic segment
d̄Q = 0,
=⇒
0=
dE
N kB θ
· dθ +
· dV,
dθ
V
=⇒
dV
1 dE
=−
· dθ.
V
N kB θ dθ
Integrating the above between the two temperatures, we obtain
 Z θhot
1 dE
1
V3


=−
· dθ, and

 ln V
N kB θcold θ dθ
2
Z θhot

1 dE
1
V4


=−
 ln
· dθ.
V1
N kB θcold θ dθ
While we cannot explicitly evaluate the integral (since E(θ) is arbitrary), we can nonetheless conclude that
V2
V1
=
.
V4
V3
(c) Show that QH /QC = ΘH /ΘC .
• Combining the results of parts (a) and (b), we observe that
Qhot
θhot
=
.
Qcold
θcold
8
Since the thermodynamic temperature scale is defined by
Qhot
Thot
=
,
Qcold
Tcold
we conclude that θ and T are proportional. If we further define θ(triple pointH2 0 ) =
T (triple pointH2 0 ) = 273.16, θ and T become identical.
********
4. Equations of State:
The equation of state constrains the form of internal energy as
in the following examples.
(a) Starting from dE = T dS − P dV , show that the equation of state P V = N kB T , in fact
implies that E can only depend on T .
• Since there is only one form of work, we can choose any two parameters as independent
variables. For example, selecting T and V , such that E = E(T, V ), and S = S(T, V ), we
obtain
dE = T dS − P dV = T
∂S
∂T
dT + T
V
∂S
∂V
T
dV − P dV,
resulting in
∂E
∂V
=T
T
∂S
∂V
− P.
T
Using the Maxwell’s relation†
∂S
∂V
we obtain
Since T
∂P
∂T V
∂E
∂V
=
T
=T
T
∂P
∂T
∂P
∂T
B
= T Nk
= P , for an ideal gas,
V
,
V
V
∂E
∂V T
− P.
= 0. Thus E depends only on T , i.e.
E = E(T ).
(b) What is the most general equation of state consistent with an internal energy that
depends only on temperature?
• If E = E(T ),
∂E
∂V
= 0,
=⇒
T
T
∂P
∂T
= P.
V
The solution for this equation is P = f (V )T, where f (V ) is any function of only V .
†
dL = Xdx + Y dy + · · · ,
=⇒
∂X
∂y
x
=
9
∂Y
∂x y
=
∂2L
∂x·∂y .
(c) Show that for a van der Waals gas CV is a function of temperature alone.
• The van der Waals equation of state is given by
"
P −a
N
V
2 #
· (V − N b) = N kB T,
or
N kB T
P =
+a
(V − N b)
N
V
2
.
From these equations, we conclude that
∂E
CV ≡
∂T
,
=⇒
V
∂CV
∂V
T
∂ 2E
∂
=
=
∂V ∂T
∂T
∂P
T
∂T
V
−P
=T
∂ 2P
∂T 2
= 0.
V
********
5. Clausius–Clapeyron equation describes the variation of boiling point with pressure. It
is usually derived from the condition that the chemical potentials of the gas and liquid
phases are the same at coexistence.
• From the equations
µliquid (P, T ) = µgas (P, T ),
and
µliquid (P + dP, T + dT ) = µgas (P + dP, T + dT ),
we conclude that along the coexistence line
dP
dT
∂µg
∂T
=
∂µl
∂P
coX
P
T
−
∂µl
∂T
−
∂µg
∂P
P
.
T
The variations of the Gibbs free energy, G = N µ(P, T ) from the extensivity condition, are
given by
V =
∂G
∂P
,
T
S=−
∂G
∂T
.
P
In terms of intensive quantities
v=
∂µ
V
=
N
∂P
,
s=
T
10
S
∂µ
=−
N
∂T
,
P
where s and v are molar entropy and volume, respectively. Thus, the coexistence line
satisfies the condition
dP
dT
=
coX
Sg − Sl
sg − sl
=
.
Vg − Vl
vg − vl
For an alternative derivation, consider a Carnot engine using one mole of water. At the
source (P, T ) the latent heat L is supplied converting water to steam. There is a volume
increase V associated with this process. The pressure is adiabatically decreased to P − dP .
At the sink (P − dP, T − dT ) steam is condensed back to water.
(a) Show that the work output of the engine is W = V dP + O(dP 2 ). Hence obtain the
Clausius–Clapeyron equation
dP
dT
=
boiling
L
.
TV
(1)
• If we approximate the adiabatic processes as taking place at constant volume V (vertical
lines in the P − V diagram), we find
P dV = P V − (P − dP )V = V dP.
pressure
W =
I
P
liq.
1
Q hot
2
T
gas
V
P-dP 4
Q cold
T-dT
3
volume
11
Here, we have neglected the volume of liquid state, which is much smaller than that
of the gas state. As the error is of the order of
∂V
∂P
S
dP · dP = O(dP 2 ),
we have
W = V dP + O(dP 2 ).
The efficiency of any Carnot cycle is given by
η=
W
TC
=1−
,
QH
TH
and in the present case,
QH = L,
W = V dP,
TH = T,
TC = T − dT.
Substituting these values in the universal formula for efficiency, we obtain the ClausiusClapeyron equation
dT
V dP
=
,
L
T
dP
dT
or
=
coX
L
.
T ·V
(b) What is wrong with the following argument: “The heat QH supplied at the source to
convert one mole of water to steam is L(T ). At the sink L(T − dT ) is supplied to condense
one mole of steam to water. The difference dT dL/dT must equal the work W = V dP ,
equal to LdT /T from eq.(1). Hence dL/dT = L/T implying that L is proportional to T !”
• The statement “At the sink L(T − dT ) is supplied to condense one mole of water” is
incorrect. In the P − V diagram shown, the state at “1” corresponds to pure water, “2”
corresponds to pure vapor, but the states “3” and “4” have two phases coexisting. In
going from the state 3 to 4 less than one mole of steam is converted to water. Part of the
steam has already been converted into water during the adiabatic expansion 2 → 3, and
the remaining portion is converted in the adiabatic compression 4 → 1. Thus the actual
latent heat should be less than the contribution by one mole of water.
(c) Assume that L is approximately temperature independent, and that the volume change
is dominated by the volume of steam treated as an ideal gas, i.e. V = N kB T /P . Integrate
equation (1) to obtain P (T ).
12
• For an ideal gas
V =
N kB T
,
P
=⇒
dP
dT
=
coX
LP
,
N kB T 2
dP
L
=
dT.
P
N kB T 2
or
Integrating this equation, the boiling temperature is obtained as a function of the pressure
P , as
P = C · exp −
L
kB TBoiling
.
(d) A hurricane works somewhat like the engine described above. Water evaporates at the
warm surface of the ocean, steam rises up in the atmosphere, and condenses to water at
the higher and cooler altitudes. The Coriolis force converts the upwards suction of the
air to spiral motion. (Using ice and boiling water, you can create a little storm in a tea
cup.) Typical values of warm ocean surface and high altitude temperatures are 800 F and
−1200 F respectively. The warm water surface layer must be at least 200 feet thick to
provide sufficient water vapor, as the hurricane needs to condense about 90 million tons
of water vapor per hour to maintain itself. Estimate the maximum possible efficiency, and
power output, of such a hurricane. (The latent heat of vaporization of water is about
2.3 × 106 Jkg −1 .)
• For TC = −120o F = 189o K, and TH = 80o F = 300o K, the limiting efficiency, as that
of a Carnot engine, is
ηmax =
TH − TC
= 0.37.
TH
The output power, is equal to (input power) x (efficiency). The input in this case is the
energy obtained from evaporation of warm ocean temperature; hence
dQc
TH − TC
dW
=
×
dt
dt
TC
6
1hr
1000kg 2.3 × 106 J
90 × 10 tons
·
·
·
× 0.67 ≈ 4 × 1013 watts.
=
hr
3600sec
ton
kg
Power output =
(e) Due to gravity, atmospheric pressure P (h) drops with the height h. By balancing
the forces acting on a slab of air (behaving like a perfect gas) of thickness dh, show that
P (h) = P0 exp(−mgh/kT ), where m is the average mass of a molecule in air.
13
• Consider a horizontal slab of area A between heights h and h + dh. The gravitational
force due to mass of particles in the slab is
dFgravity = mg
N
P
Adh = mg
Adh,
V
kB T
where we have used the ideal gas law to relate the density (N/V ) to the pressure. The
gravitational force is balanced in equilibrium with the force due to pressure
dFpressure = A [P (h) − P (h + dh)] = −
∂P
Adh.
∂h
Equating the two forces gives
∂P
P
= −mg
,
∂h
kB T
mgh
,
P (h) = p0 exp −
kB T
=⇒
assuming that temperature does not change with height.
(f) Use the above results to estimate the boiling temperature of water on top of Mount
Everest (h ≈ 9km). The latent heat of vaporization of water is about 2.3 × 106 Jkg −1 .
• Using the results from parts (c) and (e), we conclude that
PEverest
1
mg
L
1
.
≈ exp −
(hEverest − hsea ) ≈ exp −
−
Psea
kB T
kB TEverest (boil) Tsea (boil)
Using the numbers provided, we find TEverest (boil) ≈ 346o K (74o C≈ 163o F).
********
6. Glass:
Liquid quartz, if cooled slowly, crystallizes at a temperature Tm , and releases
latent heat L. Under more rapid cooling conditions, the liquid is supercooled and becomes
glassy.
(a) As both phases of quartz are almost incompressible, there is no work input, and changes
in internal energy satisfy dE = T dS + µdN . Use the extensivity condition to obtain the
expression for µ in terms of E, T , S, and N .
• Since in the present context we are considering only chemical work, we can regard
entropy as a function of two independent variables, e.g. E, and N , which appear naturally
from dS = dE/T − µdN/T . Since entropy is an extensive variable, λS = S(λE, λN ).
Differentiating this with respect to λ and evaluating the resulting expression at λ = 1,
gives
S(E, N ) =
∂S
∂E
E+
N
∂S
∂N
14
N=
E
E
Nµ
−
,
T
T
leading to
µ=
E − TS
.
N
(b) The heat capacity of crystalline quartz is approximately CX = αT 3 , while that of
glassy quartz is roughly CG = βT , where α and β are constants.
Assuming that the third law of thermodynamics applies to both crystalline and glass
phases, calculate the entropies of the two phases at temperatures T ≤ Tm .
• Finite temperature entropies can be obtained by integrating d̄Q/T , starting from S(T =
0) = 0. Using the heat capacities to obtain the heat inputs, we find

T dScrystal

 Ccrystal = αT 3 =
,
N
dT

T dSglass

Cglass = βT =
,
N dT
N αT 3
,
3
=⇒
Scrystal =
=⇒
Sglass = βN T.
(c) At zero temperature the local bonding structure is similar in glass and crystalline
quartz, so that they have approximately the same internal energy E0 . Calculate the
internal energies of both phases at temperatures T ≤ Tm .
• Since dE = T dS + µdN , for dN = 0, we have
(
dE = T dS = αN T 3 dT
dE = T dS = βN T dT
(crystal),
(glass).
Integrating these expressions, starting with the same internal energy Eo at T = 0, yields

αN 4

 E = Eo +
T
4

 E = E + βN T 2
o
2
(crystal),
(glass).
(d) Use the condition of thermal equilibrium between two phases to compute the equilibrium melting temperature Tm in terms of α and β.
• From the condition of chemical equilibrium between the two phases, µcrystal = µglass ,
we obtain
1 1
−
3 4
1
· αT = 1 −
· βT 2 ,
2
4
15
=⇒
αT 4
βT 2
=
,
12
2
resulting in a transition temperature
Tmelt =
r
6β
.
α
(e) Compute the latent heat L in terms of α and β.
• From the assumptions of the previous parts, we obtain the latent heats for the glass to
crystal transition as
3
αTmelt
L = Tmelt (Sglass − Scrystal ) = N Tmelt βTmelt −
3
2
αTmelt
2
2
2
= N Tmelt
(β − 2β) = −N βTmelt
< 0.
= N Tmelt
β−
3
(f) Is the result in the previous part correct? If not, which of the steps leading to it is
most likely to be incorrect?
• The above result implies that the entropy of the crystal phase is larger than that of
the glass phase. This is clearly unphysical, and one of the assumptions must be wrong.
The questionable step is the assumption that the glass phase is subject to the third law
of thermodynamics, and has zero entropy at T = 0. In fact, glass is a non-ergodic state of
matter which does not have a unique ground state, and violates the third law.
********
7. Filament: For an elastic filament it is found that, at a finite range in temperature, a
displacement x requires a force
J = ax − bT + cT x,
where a, b, and c are constants. Furthermore, its heat capacity at constant displacement
is proportional to temperature, i.e. Cx = A(x)T .
(a) Use an appropriate Maxwell relation to calculate ∂S/∂x|T .
• From dF = −SdT + Jdx, we obtain
∂S
∂x
T
=−
∂J
∂T
x
= b − cx.
(b) Show that A has to in fact be independent of x, i.e. dA/dx = 0.
16
∂S
= A(x)T , where S = S(T, x). Thus
• We have Cx = T ∂T
∂ ∂S
∂ ∂S
∂A
=
=
=0
∂x
∂x ∂T
∂T ∂x
from part (a), implying that A is independent of x.
(c) Give the expression for S(T, x) assuming S(0, 0) = S0 .
• S(x, T ) can be calculated as
T ′ =T
∂S(T ′ , x = 0) ′
S(x, T ) = S(0, 0) +
dT +
∂T ′
T ′ =0
Z x
Z T
′
(b − cx′ )dx′
AdT +
= S0 +
Z
Z
x′ =x
x′ =0
∂S(T, x′ )
dx
∂x′
0
0
c
= S0 + AT + (b − x)x.
2
(d) Calculate the heat capacity at constant tension, i.e. CJ = T ∂S/∂T |J as a function of
T and J.
• Writing the entropy as S(T, x) = S(T, x(T, J)), leads to
∂S
∂T
From parts (a) and (b),
∂S
∂x T
∂x
∂x
− b + cx + cT ∂T
= 0, i.e.
a ∂T
Thus
Since x =
=
J
∂S
∂T
+
x
= b − cx and
∂S
∂x
∂S
∂T
T
∂x
∂T
.
J
x = A. Furthermore,
∂x
∂T J
is given by
b − cx
∂x
=
.
∂T
a + cT
(b − cx)2
.
CJ = T A +
(a + cT )
J+bT
a+cT
, we can rewrite the heat capacity as a function of T and J, as
"
CJ = T A +
2
(b − c J+bT
a+cT )
(a + cT )
(ab − cJ)2
.
=T A+
(a + cT )3
#
********
8.
Hard core gas:
A gas obeys the equation of state P (V − N b) = N kB T , and has a
heat capacity CV independent of temperature. (N is kept fixed in the following.)
17
(a) Find the Maxwell relation involving ∂S/∂V |T,N .
• For dN = 0,
d(E − T S) = −SdT − P dV,
∂S
∂V
=⇒
∂P
∂T
=
T,N
.
V,N
(b) By calculating dE(T, V ), show that E is a function of T (and N ) only.
• Writing dS in terms of dT and dV ,
∂S
∂T
dE = T dS − P dV = T
dT +
V,N
∂S
∂V
dV
T,N
!
− P dV.
Using the Maxwell relation from part (a), we find
∂S
∂T
dE(T, V ) = T
dT +
T
V,N
∂P
∂T
V,N
−P
!
dV.
But from the equation of state, we get
P =
N kB T
,
(V − N b)
∂P
∂T
=⇒
=
V,N
P
,
T
=⇒
dE(T, V ) = T
∂S
∂T
dT,
V,N
i.e. E(T, N, V ) = E(T, N ) does not depend on V .
(c) Show that γ ≡ CP /CV = 1 + N kB /CV (independent of T and V ).
• The hear capacity is
CP =
∂Q
∂T
=
P
∂E + P V
∂T
=
P
∂E
∂T
+P
P
∂V
∂T
.
P
But, since E = E(T ) only,
∂E
∂T
=
P
∂E
∂T
= CV ,
V
and from the equation of state we get
∂V
∂T
=
P
N kB
,
P
=⇒
CP = CV + N kB ,
=⇒
γ =1+
N kB
,
CV
which is independent of T , since CV is independent of temperature. The independence of
CV from V also follows from part (a).
18
(d) By writing an expression for E(P, V ), or otherwise, show that an adiabatic change
satisfies the equation P (V − N b)γ =constant.
• Using the equation of state, we have
dE = CV dT = CV d
P (V − N b)
N kB
=
CV
(P dV + (V − N b)dP ) .
N kB
The adiabatic condition, dQ = dE + P dV = 0, can now be written as
0 = dQ =
CV
1+
N kB
P d(V − N b) +
CV
(V − N b)dP.
N kB
Dividing by CV P (V − N b)/(N kB ) yields
d(V − N b)
dP
+γ
= 0,
P
(V − N b)
ln [P (V − N b)γ ] = constant.
=⇒
********
9. Superconducting transition: Many metals become superconductors at low temperatures
T , and magnetic fields B. The heat capacities of the two phases at zero magnetic field are
approximately given by
(
Cs (T ) = V αT 3
Cn (T ) = V βT 3 + γT
in the superconducting phase
in the normal phase
,
where V is the volume, and {α, β, γ} are constants. (There is no appreciable change in
volume at this transition, and mechanical work can be ignored throughout this problem.)
(a) Calculate the entropies Ss (T ) and Sn (T ) of the two phases at zero field, using the third
law of thermodynamics.
• Finite temperature entropies are obtained by integrating dS = d̄Q/T , starting from
S(T = 0) = 0. Using the heat capacities to obtain the heat inputs, we find




dSs
,
dT
dS


 Cn = V βT 3 + γT = T n ,
dT
Cs = V αT 3 = T
19
=⇒
=⇒
αT 3
,
3 3
.
βT
Sn = V
+ γT
3
Ss = V
(b) Experiments indicate that there is no latent heat (L = 0) for the transition between
the normal and superconducting phases at zero field. Use this information to obtain the
transition temperature Tc , as a function of α, β, and γ.
• The Latent hear for the transition is related to the difference in entropies, and thus
L = Tc (Sn (Tc ) − Ss (Tc )) = 0.
Using the entropies calculated in the previous part, we obtain
r
αTc3
3γ
βTc3
=
+ γTc , =⇒ Tc =
.
3
3
α−β
(c) At zero temperature, the electrons in the superconductor form bound Cooper pairs.
As a result, the internal energy of the superconductor is reduced by an amount V ∆, i.e.
En (T = 0) = E0 and Es (T = 0) = E0 −V ∆ for the metal and superconductor, respectively.
Calculate the internal energies of both phases at finite temperatures.
• Since dE = T dS + BdM + µdN , for dN = 0, and B = 0, we have dE = T dS = CdT .
Integrating the given expressions for heat capacity, and starting with the internal energies
E0 and E0 − V ∆ at T = 0, yields
h

α 4i

 Es (T ) = E0 + V −∆ + 4 T
β 4 γ 2 .

 En (T ) = E0 + V
T + T
4
2
(d) By comparing the Gibbs free energies (or chemical potentials) in the two phases, obtain
an expression for the energy gap ∆ in terms of α, β, and γ.
• The Gibbs free energy G = E − T S − BM = µN can be calculated for B = 0 in each
phase, using the results obtained before, as
h
h

α 3
α 4i
α 4i

T
−
T
V
T
=
E
−
V
∆
+
T
G
(T
)
=
E
+
V
−∆
+
0
0
 s
4
3
12
β 4 γ 2
β 3
β 4 γ 2 .

 Gn (T ) = E0 + V
T + T − TV
T + γT = E0 − V
T + T
4
2
3
12
2
At the transition point, the chemical potentials (and hence the Gibbs free energies) must
be equal, leading to
∆+
α 4
β 4 γ 2
Tc =
T + Tc ,
12
12 c
2
=⇒
20
∆=
γ 2 α−β 4
T −
T .
2 c
12 c
Using the value of Tc =
p
3γ/(α − β), we obtain
∆=
3 γ2
.
4α−β
(e) In the presence of a magnetic field B, inclusion of magnetic work results in dE =
T dS +BdM +µdN , where M is the magnetization. The superconducting phase is a perfect
diamagnet, expelling the magnetic field from its interior, such that Ms = −V B/(4π) in
appropriate units. The normal metal can be regarded as approximately non-magnetic,
with Mn = 0. Use this information, in conjunction with previous results, to show that the
superconducting phase becomes normal for magnetic fields larger than
T2
Bc (T ) = B0 1 − 2 ,
Tc
giving an expression for B0 .
• Since dG = −SdT − M dB + µdN , we have to add the integral of −M dB to the Gibbs
free energies calculated in the previous section for B = 0. There is no change in the
metallic phase since Mn = 0, while in the superconducting phase there is an additional
R
R
contribution of − Ms dB = (V /4π) BdB = (V /8π)B 2 . Hence the Gibbs free energies
at finite field are

h
B2
α 4i


+
V
T
G
(T,
B)
=
E
−
V
∆
+
 s
0
12
8π .
β 4 γ 2


 Gn (T, B) = E0 − V
T + T
12
2
Equating the Gibbs free energies gives a critical magnetic field
γ
α−β 4
3 γ2
γ
α−β 4
Bc2
= ∆ − T2 +
T =
− T2 +
T
8π
2
12
4α−β
2
12
"
#
2
2
α−β
3γ
6γT 2
α−β
=
−
Tc2 − T 2 ,
+ T4 =
12
α−β
α−β
12
where we have used the values of ∆ and Tc obtained before. Taking the square root of the
above expression gives
T2
Bc = B0 1 − 2 ,
Tc
where
B0 =
r
21
2π(α − β) 2
Tc =
3
s
p
6πγ 2
= Tc 2πγ.
α−β
********
10. Photon gas Carnot cycle:
The aim of this problem is to obtain the blackbody
4
radiation relation, E(T, V ) ∝ V T , starting from the equation of state, by performing an
infinitesimal Carnot cycle on the photon gas.
P
P+dP
T+dT
P
V
T
V+dV
V
(a) Express the work done, W , in the above cycle, in terms of dV and dP .
• Ignoring higher order terms, net work is the area of the cycle, given by W = dP dV .
(b) Express the heat absorbed, Q, in expanding the gas along an isotherm, in terms of P ,
dV , and an appropriate derivative of E(T, V ).
• Applying the first law, the heat absorbed is
Q = dE + P dV =
∂E
∂T
dT +
V
∂E
∂V
dV + P dV
T
=
isotherm
∂E
∂V
T
+ P dV.
(c) Using the efficiency of the Carnot cycle, relate the above expressions for W and Q to
T and dT .
• The efficiency of the Carnot cycle (η = dT /T ) is here calculated as
η=
dP
dT
W
=
=
.
Q
[(∂E/∂V )T + P ]
T
(d) Observations indicate that the pressure of the photon gas is given by P = AT 4 ,
3
4
where A = π 2 kB
/45 (h̄c) is a constant. Use this information to obtain E(T, V ), assuming
E(T, 0) = 0.
• From the result of part (c) and the relation P = AT 4 ,
4
4AT =
∂E
∂V
4
+ AT ,
T
22
or
∂E
∂V
= 3AT 4 ,
T
so that
E = 3AV T 4 .
(e) Find the relation describing the adiabatic paths in the above cycle.
• Adiabatic curves are given by dQ = 0, or
0=
∂E
∂T
dT +
V
∂E
∂V
dV + P dV = 3V dP + 4P dV,
T
i.e.
P V 4/3 = constant.
********
11. Irreversible Processes:
(a) Consider two substances, initially at temperatures T10 and T20 , coming to equilibrium
at a final temperature Tf through heat exchange. By relating the direction of heat flow
to the temperature difference, show that the change in the total entropy, which can be
written as
Z
∆S = ∆S1 + ∆S2 ≥
Tf
T10
d̄Q1
+
T1
Z
Tf
T20
d̄Q1
=
T2
Z
T1 − T2
d̄Q,
T1 T2
must be positive. This is an example of the more general condition that “in a closed
system, equilibrium is characterized by the maximum value of entropy S.”
• Defining the heat flow from substance 1 to 2 as, d̄Q1→2 , we get,
∆S = ∆S1 + ∆S2 ≥
Z
Tf
T10
d̄Q1
+
T1
Z
Tf
T20
d̄Q2
=
T2
Z
T1 − T2
d̄Q1→2 .
T1 T2
But according to Clausius’ statement of the second law d̄Q1→2 > 0, if T1 > T2 and
d̄Q1→2 < 0, if T1 < T2 . Hence, (T1 − T2 )d̄Q1→2 ≥ 0, resulting in
∆S ≥
Z
T1 − T2
d̄Q1→2 ≥ 0.
T1 T2
23
(b) Now consider a gas with adjustable volume V , and diathermal walls, embedded in a
heat bath of constant temperature T , and fixed pressure P . The change in the entropy of
the bath is given by
∆Sbath =
∆Qbath
∆Qgas
1
=−
= − (∆Egas + P ∆Vgas ) .
T
T
T
By considering the change in entropy of the combined system establish that “the equilibrium
of a gas at fixed T and P is characterized by the minimum of the Gibbs free energy G =
E + P V − T S.”
• The total change in entropy of the whole system is,
1
1
∆S = ∆Sbath + ∆Sgas = − (∆Egas + P ∆Vgas − T ∆Sgas ) = − ∆Ggas .
T
T
By the second law of thermodynamics, all processes must satisfy,
1
− ∆Ggas ≥ 0 ⇔ ∆Ggas ≤ 0,
T
that is, all processes that occur can only lower the Gibbs free energy of the gas. Therefore
the equillibrium of a gas in contact with a heat bath of constant T and P is established
at the point of minimum Gibbs free energy, i.e. when the Gibbs free energy cannot be
lowered any more.
********
12. The solar system originated from a dilute gas of particles, sufficiently separated from
other such clouds to be regarded as an isolated system. Under the action of gravity the
particles coalesced to form the sun and planets.
(a) The motion and organization of planets is much more ordered than the original dust
cloud. Why does this not violate the second law of thermodynamics?
• The formation of planets is due to the gravitational interaction. Because of the attrac-
tive nature of this interaction, the original dust cloud with uniform density has in fact
lower entropy. Clumping of the uniform density leads to higher entropy. Of course the
gravitational potential energy is converted into kinetic energy in the process. Ultimately
the kinetic energy of falling particles is released in the form of photons which carry away
a lot of entropy.
(b) The nuclear processes of the sun convert protons to heavier elements such as carbon.
Does this further organization lead to a reduction in entropy?
24
• Again the process of formation of heavier elements is accompanied by the release of large
amounts of energy which are carry away by photons. The entropy carry away by these
photons is more than enough to compensate any ordering associated with the packing of
nucleons into heavier nuclei.
(c) The evolution of life and intelligence requires even further levels of organization. How
is this achieved on earth without violating the second law?
• Once more there is usage of energy by the organisms that converts more ordered forms
of energy to less ordered ones.
********
25
Problems for Chapter II - Probability
1. Characteristic functions:
Calculate the characteristic function, the mean, and the
variance of the following probability density functions:
(a) Uniform
1
2a
p(x) =
−a < x < a , and
for
• A uniform probability distribution,

 1
p(x) = 2a

0
p(x) = 0
for − a < x < a
otherwise;
,
otherwise
for which there exist many examples, gives
1
f (k) =
2a
=
Therefore,
a
1 1
exp(−ikx)dx =
exp(−ikx)
2a −ik
−a
Z
(b) Laplace
p(x) =
• The Laplace PDF,
−a
∞
X
1
(ak)2m
sin(ka) =
(−1)m
.
ak
(2m
+
1)!
m=0
m1 = hxi = 0,
1
2a
a
exp
− |x|
a
and
m2 = hx2 i =
1 2
a .
3
;
1
|x|
p(x) =
,
exp −
2a
a
for example describing light absorption through a turbid medium, gives
Z ∞
|x|
1
dx exp −ikx −
f (k) =
2a −∞
a
Z ∞
Z 0
1
1
dx exp(−ikx − x/a) +
dx exp(−ikx + x/a)
=
2a 0
2a −∞
1
1
1
1
=
=
−
2a −ik + 1/a −ik − 1/a
1 + (ak)2
= 1 − (ak)2 + (ak)4 − · · · .
Therefore,
m1 = hxi = 0,
and
26
m2 = hx2 i = 2a2 .
(c) Cauchy
p(x) =
a
π(x2 +a2 )
.
• The Cauchy, or Lorentz PDF describes the spectrum of light scattered by diffusive
modes, and is given by
p(x) =
π(x2
a
.
+ a2 )
For this distribution,
∞
a
exp(−ikx)
dx
2
π(x + a2 )
−∞
Z ∞
1
1
1
exp(−ikx)
dx.
−
=
2πi −∞
x − ia x + ia
f (k) =
Z
The easiest method for evaluating the above integrals is to close the integration contours
in the complex plane, and evaluate the residue. The vanishing of the integrand at infinity
determines whether the contour has to be closed in the upper, or lower half of the complex
plane, and leads to
Z

1
exp(−ikx)


dx = exp(−ka)
 − 2πi
x + ia
C
f (k) =
Z

exp(−ikx)
1


dx = exp(ka)
2πi B x − ia


for k ≥ 0 

= exp(−|ka|).


for k < 0 
Note that f (k) is not an analytic function in this case, and hence does not have a Taylor expansion. The moments have to be determined by another method, e.g. by direct
evaluation, as
m1 = hxi = 0,
and
2
m2 = hx i =
Z
dx
π
x2
· 2
→ ∞.
a x + a2
The first moment vanishes by symmetry, while the second (and higher) moments diverge,
explaining the non-analytic nature of f (k).
The following two probability density functions are defined for x ≥ 0. Compute only
the mean and variance for each.
(d) Rayleigh
p(x) =
x
a2
2
x
exp(− 2a
,
2)
• The Rayleigh distribution,
x2
x
p(x) = 2 exp − 2 ,
a
2a
27
for
x ≥ 0,
can be used for the length of a random walk in two dimensions. Its characteristic function
is
Z ∞
x2
x
f (k) =
exp(−ikx) 2 exp − 2 dx
a
2a
0
Z ∞
x
x2
=
[cos(kx) − i sin(kx)] 2 exp − 2 dx.
a
2a
0
The integrals are not simple, but can be evaluated as
Z ∞
∞
X
n
(−1)n n!
x2
x
2a2 k 2 ,
cos(kx) 2 exp − 2 dx =
a
2a
(2n)!
0
n=0
and
Z
0
resulting in
∞
Z
x
x
x2
1 ∞
x2
sin(kx) 2 exp − 2 dx =
sin(kx) 2 exp − 2 dx
a
2a
2 −∞
a
2a
r
2 2
π
k a
,
ka exp −
=
2
2
r
2 2
∞
X
(−1)n n!
k a
π
2 2 n
2a k
−i
.
f (k) =
ka exp −
(2n)!
2
2
n=0
The moments can also be calculated directly, from
r
Z ∞ 2
Z ∞ 2
x
π
x
x2
x2
m1 = hxi =
dx
=
dx
=
exp
−
exp
−
a,
2
a2
2a2
2a2
2
0
−∞ 2a
2
Z ∞ 2
Z ∞ 3
x
x2
x2
x
x
2
2
exp − 2 dx = 2a
exp − 2 d
m2 = hx i =
2
2
a
2a
2a
2a
2a2
0
0
Z ∞
y exp(−y)dy = 2a2 .
= 2a2
0
q
2
2
x
(e) Maxwell p(x) = π2 xa3 exp(− 2a
.
2)
• It is difficult to calculate the characteristic function for the Maxwell distribution
r
2 x2
x2
exp − 2 ,
p(x) =
π a3
2a
say describing the speed of a gas particle. However, we can directly evaluate the mean and
variance, as
r Z ∞
x3
2
x2
exp − 2 dx
m1 = hxi =
π 0 a3
2a
r
2
Z ∞ 2
x
x2
x
2
=2
a
exp − 2 d
2
π 0 2a
2a
2a2
r
r
Z ∞
2
2
a
y exp(−y)dy = 2
a,
=2
π 0
π
28
and
2
m2 = hx i =
r
2
π
Z
∞
o
x4
x2
exp − 2 dx = 3a2 .
a3
2a
********
2. Directed random walk:
The motion of a particle in three dimensions is a series
of independent steps of length ℓ. Each step makes an angle θ with the z axis, with
a probability density p(θ) = 2 cos2 (θ/2)/π; while the angle φ is uniformly distributed
between 0 and 2π. (Note that the solid angle factor of sin θ is already included in the
definition of p(θ), which is correctly normalized to unity.) The particle (walker) starts at
the origin and makes a large number of steps N .
(a) Calculate the expectation values hzi, hxi, hyi, z 2 , x2 , and y 2 , and the covariances
hxyi, hxzi, and hyzi.
• From symmetry arguments,
hxi = hyi = 0,
while along the z-axis,
hzi =
X
i
hzi i = N hzi i = N a hcos θi i =
Na
.
2
The last equality follows from
Z
Z π
1
hcos θi i = p(θ) cos θdθ =
cos θ · (cos θ + 1)dθ
0 π
Z π
1
1
(cos 2θ + 1)dθ = .
=
2
0 2π
The second moment of z is given by
X
XX
X
zi2
hzi zj i =
hzi zj i +
z2 =
i
i,j
=
XX
i
i6=j
hzi i hzj i +
2
i
i6=j
X
zi2
i
= N (N − 1) hzi i + N zi2 .
Noting that
zi2
=
a2
Z
0
π
1
cos2 θ(cos θ + 1)dθ =
π
29
Z
0
π
1
1
(cos 2θ + 1)dθ = ,
2π
2
we find
z
2
= N (N − 1)
a 2
2
+N
a2
a2
= N (N + 1) .
2
4
The second moments in the x and y directions are equal, and given by
x2 =
X
i,j
hxi xj i =
XX
i
i6=j
hxi xj i +
X
x2i = N x2i .
i
Using the result
x2i
= sin2 θ cos2 φ
a2
Z π
Z 2π
1
1
2
dθ sin2 θ(cos θ + 1) = ,
dφ cos φ
=
2
2π 0
4
0
we obtain
x2 = y 2 =
N a2
.
4
While the variables x, y, and z are not independent because of the constraint of unit
length, simple symmetry considerations suffice to show that the three covariances are in
fact zero, i.e.
hxyi = hxzi = hyzi = 0.
(b) Use the central limit theorem to estimate the probability density p(x, y, z) for the
particle to end up at the point (x, y, z).
• From the Central limit theorem, the probability density should be Gaussian. However,
for correlated random variable we may expect cross terms that describe their covariance.
However, we showed above that the covarainces between x, y, and z are all zero. Hence
we can treat them as three independent Gaussian variables, and write
(x − hxi)2
(y − hyi)2
(z − hzi)2
p(x, y, z) ∝ exp −
−
−
.
2σx2
2σy2
2σz2
(There will be correlations between x, y, and z appearing in higher cumulants, but all such
cumulants become irrelevant in the N → ∞ limit.) Using the moments
hxi = hyi = 0,
and
2
σx2 = x2 − hxi = N
30
a
hzi = N ,
2
a2
= σy2 ,
4
σz2
and
= z
2
a2
− hzi = N (N + 1) −
4
2
we obtain
p(x, y, z) =
2
πN a2
3/2
"
Na
2
2
=N
2
x2 + y 2 + (z − N a/2)
exp −
N a2 /2
a2
,
4
#
.
********
Consider any probability density p(x) for (−∞ < x < ∞),
3. Tchebycheff inequality:
with mean λ, and variance σ 2 . Show that the total probability of outcomes that are more
than nσ away from λ is less than 1/n2 , i.e.
Z
|x−λ|≥nσ
dxp(x) ≤
1
.
n2
Hint: Start with the integral defining σ 2 , and break it up into parts corresponding to
|x − λ| > nσ, and |x − λ| < nσ.
• By definition, for a system with a PDF p(x), and average λ, the variance is
Z
2
σ = (x − λ)2 p(x)dx.
Let us break the integral into two parts as
Z
Z
2
2
(x − λ) p(x)dx +
σ =
|x−λ|≥nσ
|x−λ|<nσ
(x − λ)2 p(x)dx,
resulting in
2
σ −
Now since
Z
2
|x−λ|<nσ
Z
(x − λ) p(x)dx =
2
|x−λ|≥nσ
(x − λ) p(x)dx ≥
Z
|x−λ|≥nσ
Z
(x − λ)2 p(x)dx.
(nσ)2 p(x)dx,
|x−λ|≥nσ
we obtain
Z
|x−λ|≥nσ
and
2
2
(nσ) p(x)dx ≤ σ −
Z
|x−λ|≥nσ
Z
|x−λ|<nσ
p(x)dx ≤
31
(x − λ)2 p(x)dx ≤ σ 2 ,
1
.
n2
********
4. Optimal selection:
In many specialized populations, there is little variability among
the members. Is this a natural consequence of optimal selection?
(a) Let {rα } be n random numbers, each independently chosen from a probability density
p(r), with r ∈ [0, 1]. Calculate the probability density pn (x) for the largest value of this
set, i.e. for x = max{r1 , · · · , rn }.
• The probability that the maximum of n random numbers falls between x and x + dx
is equal to the probability that one outcome is in this interval, while all the others are
smaller than x, i.e.
n
pn (x) = p(r1 = x, r2 < x, r3 < x, · · · , rn < x) ×
1
!
,
where the second factor corresponds to the number of ways of choosing which rα = x. As
these events are independent
n
pn (x) = p(r1 = x) · p(r2 < x) · p(r3 < x) · · · p(rn < x) ×
1
n
n−1
.
= p(r = x) [p(r < x)]
×
1
The probability of r < x is just a cumulative probability function, and
pn (x) = n · p(x) ·
x
Z
p(r)dr
0
n−1
.
(b) If each rα is uniformly distributed between 0 and 1, calculate the mean and variance
of x as a function of n, and comment on their behavior at large n.
R1
R1
• If each rα is uniformly distributed between 0 and 1, p(r) = 1 ( 0 p(r)dr = 0 dr = 1).
With this PDF, we find
pn (x) = n · p(x) ·
Z
x
p(r)dr
0
n−1
=n
Z
x
dr
0
n−1
= nxn−1 ,
and the mean is now given by
hxi =
Z
1
xpn (x)dx = n
0
Z
0
32
1
xn dx =
n
.
n+1
The second moment of the maximum is
Z
2
x =n
1
xn+1 dx =
0
n
,
n+2
resulting in a variance
2
σ = x
2
n
− hxi =
−
n+2
2
n
n+1
2
=
n
.
(n + 1)2 (n + 2)
Note that for large n the mean approaches the limiting value of unity, while the variance
vanishes as 1/n2 . There is too little space at the top of the distribution for a wide variance.
********
5. Benford’s law describes the observed probabilities of the first digit in a great variety
of data sets, such as stock prices. Rather counter-intuitively, the digits 1 through 9 occur
with probabilities 0.301, .176, .125, .097, .079, .067, .058, .051, .046 respectively. The key
observation is that this distribution is invariant under a change of scale, e.g. if the stock
prices were converted from dollars to persian rials! Find a formula that fits the above
probabilities on the basis of this observation.
• Let us consider the observation that the probability distribution for first integers is
unchanged under multiplication by any (i.e. a random) number. Presumably we can
repeat such multiplications many times, and it is thus suggestive that we should consider
the properties of the product of random numbers. (Why this should be a good model for
stock prices is not entirely clear, but it seems to be as good an explanation as anything
else!)
Consider the x =
QN
i=1 ri ,
where ri are positive, random variables taken from some reaPN
sonably well behaved probability distribution. The random variable ℓ ≡ ln x = i=1 ln ri
is the sum of many random contributions, and according to the central limit theorem
should have a Gaussian distribution in the limit of large N , i.e.
"
2 #
ℓ − Nℓ
1
√
lim p(ℓ) = exp −
,
2
N→∞
2N σ
2πN σ 2
where ℓ and σ 2 are the mean and variance of ln r respectively. The product x is distributed
according to the log-normal distribution
"
2 #
ln(x) − N ℓ
dℓ
1
1
√
p(x) = p(ℓ)
.
= exp −
2
dx
x
2N σ
2πN σ 2
33
The probability that the first integer of x in a decimal representation is i is now obtained
approximately as follows:
pi =
XZ
10q (i+1)
dxp(x).
10q i
q
The integral considers cases in which x is a number of magnitude 10q (i.e. has q + 1 digits
before the decimal point). Since the number is quite widely distributed, we then have to
sum over possible magnitudes q. The range of the sum actually need not be specified! The
next stage is to change variables from x to ℓ = ln x, leading to
pi =
XZ
q+ln(i+1)
dℓp(ℓ) =
q+ln i
q
XZ
q
q+ln(i+1)
q+ln i
"
ℓ − Nℓ
dℓ exp −
2N σ 2
2 #
√
1
2πN σ 2
.
We shall now make the approximation that over the range of integration (q + ln i to q +
ln(i + 1)), the integrand is approximately constant. (The approximation works best for
q ≈ N ℓ where the integral is largest.) This leads to
pi ≈
X
q
"
q − Nℓ
exp −
2N σ 2
2 #
1
√
,
[ln(i + 1) − ln i] ∝ ln 1 +
i
2πN σ 2
1
where we have ignored the constants of proportionality which come from the sum over
q. We thus find that the distribution of the first digit is not uniform, and the properly
normalized proportions of ln(1+1/i) indeed reproduce the probabilities p1 , · · · , p9 of 0.301,
.176, .125, .097, .079, .067, .058, .051, .046 according to Benford’s law. (For further
information check http://www.treasure-troves.com/math/BenfordsLaw.html.)
********
6. Information:
Consider the velocity of a gas particle in one dimension (−∞ < v < ∞).
(a) Find the unbiased probability density p1 (v), subject only to the constraint that the
average speed is c, i.e. h|v|i = c.
• For an unbiased probability estimation, we need to maximize entropy subject to the
two constraints of normalization, and of given average speed (h|v|i = c.). Using Lagrange
multipliers α and β to impose these constraints, we need to maximize
S = − hln pi = −
Z
∞
−∞
p(v) ln p(v)dv + α 1 −
34
Z
∞
−∞
Z
p(v)dv + β c −
∞
−∞
p(v)|v|dv .
Extremizing the above expression yields
∂S
= − ln p(v) − 1 − α − β|v| = 0,
∂p(v)
which is solved for
ln p(v) = −1 − α − β|v|,
or
p(v) = Ce−β|v| ,
with C = e−1−α .
The constraints can now be used to fix the parameters C and β:
Z ∞
Z ∞
Z ∞
1 −βv
−β|v|
p(v)dv =
Ce
dv = 2C
e−βv dv = 2C
1=
e
−β
−∞
−∞
0
which implies
C=
−∞
∞
2C
,
β
e−βv vdv,
0
which when integrated by parts yields
Z
∞
1 ∞ −βv
1
1 −βv
+
e
dv = − e−βv
c = β − ve
β
β 0
β
0
β=
0
=
β
.
2
From the second constraint, we have
Z
Z ∞
−β|v|
Ce
|v|dv = β
c=
or,
∞
∞
0
=
1
.
β
1
.
c
The unbiased PDF is then given by
−β|v|
p(v) = Ce
1
|v|
.
exp −
=
2c
c
(b) Now find the probability density p2 (v), given only the constraint of average kinetic
energy, mv 2 /2 = mc2 /2.
• When the second constraint is on the average kinetic energy, mv 2 /2 = mc2 /2, we
have
S=−
Z
∞
−∞
Z
p(v) ln p(v)dv + α 1 −
∞
p(v)dv + β
−∞
35
mc2
−
2
Z
∞
mv 2
p(v)
dv .
2
−∞
The corresponding extremization,
mv 2
∂S
= − ln p(v) − 1 − α − β
= 0,
∂p(v)
2
results in
βmv 2
.
p(v) = C exp −
2
The normalization constraint implies
Z ∞
Z
1=
p(v)dv = C
−∞
∞
e−βmv
2
−∞
or
C=
The second constraint,
p
/2
=C
p
2π/βm,
βm/2π.
r
Z
βmv 2
mv 2
m βm ∞
exp −
p(v)
v 2 dv
dv =
2
2
2π
2
−∞
−∞
"√ r
3/2 #
π
1
2
m βm
=
,
=
2
2π
2
βm
2β
mc2
=
2
Z
∞
gives
β=
1
,
mc2
for a full PDF of
βmv 2
v2
1
p(v) = C exp −
exp − 2 .
=√
2
2c
2πc2
(c) Which of the above statements provides more information on the velocity? Quantify
the difference in information in terms of I2 − I1 ≡ (hln p2 i − hln p1 i) / ln 2.
• The entropy of the first PDF is given by
Z ∞
1
|v|
−|v|
− ln(2c) −
dv
S1 = − hln p1 i = −
exp
c
c
−∞ 2c
Z
Z
v
v v
ln(2c) ∞
1 ∞
=
exp −
exp −
dv +
dv
c
c
c 0
c c
0
1
∞
∞
= − ln(2c) exp(−v/c)|0 + [−c exp(−v/c)|0 ]
c
= ln(2c) + 1 = 1 + ln 2 + ln c.
36
For the second distribution, we obtain
∞
1
v2
v2
2
− ln 2πc − 2
exp − 2
S2 = − hln p2 i = − √
2c
2
2c
2πc2 −∞
Z ∞ 2
2 Z ∞
ln 2πc
v
1
2
2
exp −v 2 /2c2 dv + √
= √
exp
−v
/2c
dv
2 2πc2 −∞
2πc2 −∞ 2c2
#
"√
2 c2
2πc
1
1
= ln 2πc2 + √
2
2
c2 2πc2
1
1
1 1
= ln 2πc2 + = + ln(2π) + ln c.
2
2
2 2
1
Z
For a discrete probability, the information content is
Iα = ln2 M − Sα / ln 2,
where M denotes the number of possible outcomes. While M , and also the proper measure
of probability are not well defined for a continuous PDF, the ambiguities disappear when
we consider the difference
I2 − I1 = (−S2 + S1 ) / ln 2
= − (S2 − S1 ) / ln 2
(ln π − ln 2 − 1)
=−
≈ 0.3956.
2 ln 2
Hence the constraint of constant energy provides 0.3956 more bits of information. (This
is partly due to the larger variance of the distribution with constant speed.)
********
7. Dice: A dice is loaded such that 6 occurs twice as often as 1.
(a) Calculate the unbiased probabilities for the 6 faces of the dice.
P
• We want to find pi that maximizes, S = − pi ln pi with the constraints, p6 = 2p1
P
P
P
and
pi = 1. Since for a fixed sum,
pi = 1 − 3p1 , − i=2,···,5 pi ln pi is maximized by
p2 = p3 = p4 = p5 , our problem reduces to maximizing,
S(p1 ) = −p1 ln p1 − 2p1 ln 2p1 − (1 − 3p1 ) ln
Differentiating this with respect to p1 , we get,
1 − 3p1
dS
= 3 ln
.
dp1
28/3 p1
37
1 − 3p1
.
4
Note that this has only one zero and its sign changes from + to − at that zero. The
maximum value of S is obtained for at that zero which is,
p1 =
1
.
28/3 + 3
Hence we obtain,
p1 =
1
22/3
1 − 3p1
2
=
,
p
=
p
=
p
=
p
=
, p6 = 2p1 = 8/3
.
2
3
4
5
8/3
8/3
4
2 +3
2 +3
2 +3
(b) What is the information content (in bits) of the above statement regarding the dice?
• By definitiion of the information content,
I = ln2 6 +
6
X
pi ln2 pi = ln2 6 + ln2
i=1
5
3
22/3
=
,
+
ln
2
3
3 + 28/3
3 + 28/3
which is about 0.03 bits.
********
8. Random matrices:
As a model for energy levels of complex nuclei, Wigner considered
N × N symmetric matrices whose elements are random. Let us assume that each element
Mij (for i ≥ j) is an independent random variable taken from the probability density
function
p(Mij ) =
1
2a
− a < Mij < a ,
for
and p(Mij ) = 0
otherwise.
(a) Calculate the characteristic function for each element Mij .
• Since each element is uniformly distributed in the interval (−a, a), the characteristic
function is
p̃ij (k) =
Z
a
−ikx
dx e
−a
eika − e−ika
sin ak
1
=
=
.
2a
2aik
ak
(b) Calculate the characteristic function for the trace of the matrix, T ≡ tr M =
38
P
i
Mii .
• The trace of the matrix is the sum of the N diagonal elements which are independent
random variables. The characteristic function for the sum of independent variables is
simply the product of the corresponding characteristic functions, and thus
p̃T (k) =
N
Y
p̃ii (k) =
i=1
sin ak
ak
N
.
(c) What does the central limit theorem imply about the probability density function of
the trace at large N ?
• Since the trace is the sum of N ≫ 1 independent random variables, its cumulants are
simply N times those of a single element. The leading cumulants are
hT ic = N hMij i = 0,
and
T
2
c
=N
2
Mij
=N
Z
a
dx
−a
√
x2
a2
=N .
2a
3
For the qunatity t = T / N , higher order cumulants vanish in the limit of N → ∞, and
thus
r
tr M
3t2
3
lim p t = √
= exp − 2
.
N→∞
2a
2πa2
N
(d) For large N , each eigenvalue λα (α = 1, 2, · · · , N ) of the matrix M is distributed
according to a probability density function
s
λ2
2
1 − 2 for − λ0 < λ < λ0 ,
p(λ) =
πλ0
λ0
and
p(λ) = 0
otherwise,
(known as the Wigner semi-circle rule). Find the variance of λ.
(Hint: Changing variables to λ = λ0 sin θ simplifies the integrals.)
• The mean value of λ is zero by symmetry, and hence its variance is given by
s
Z λ0
2
λ2
dλ λ2
λ2 c =
1 − 2.
πλ0
λ0
−λ0
In the integral, change variables to λ = λ0 sin θ and dλ = λ0 cos θdθ, to get
2
λ
c
2λ2
= 0
π
Z
π/2
λ2
dθ cos θ sin θ = 0
2π
−π/2
2
2
Z
π/2
λ2
dθ sin 2θ = 0
4π
−π/2
39
2
Z
π/2
−π/2
dθ (1 − cos 4θ) =
λ20
.
4
(e) If in the previous result, we have λ20 = 4N a2 /3, can the eigenvalues be independent of
each other?
• The trace of a matrix is related to its eigenvalues by
T =
N
X
λα ,
=⇒
T
2
c
=
α=1
N
X
α=1
λ2α +
X
α6=β
hλα λβ i .
The cross-correlations of eigenvalues thus satisfy
X
α6=β
hλα λβ i = T
2
−
c
N
X
λ2α = N
α=1
4N a2
a2
−N ×
6= 0.
3
3
Clearly, this is inconsistent with independent eigenvalues. In fact, the well known result
that eigenvalues do not cross implies a repulsion between eigenvalues which leads to a
much wider distribution than would result from independent eigenvalues.
********
9. Random deposition: A mirror is plated by evaporating a gold electrode in vaccum by
passing an electric current. The gold atoms fly off in all directions, and a portion of them
sticks to the glass (or to other gold atoms already on the glass plate). Assume that each
column of deposited atoms is independent of neighboring columns, and that the average
deposition rate is d layers per second.
(a) What is the probability of m atoms deposited at a site after a time t? What fraction
of the glass is not covered by any gold atoms?
• This is an example of a Poisson process, for which the probability of m deposition events
at average rate d is
(dt)m e−dt
.
p(m) =
m!
The fraction of the glass that is not covered by gold atoms can be obtained by p(0), that
is,
p(0) = e−dt .
An explicit derivation of this follows: First note that in a given time interval ∆t, the
probability for an atom to stick to a site is ∆p. We will eventually send both to zero. In
this case, the probability for a column at a site to have m atoms in time t would be,
p(m) =(t/∆t) Cm (∆p)m (1 − ∆p)t/∆t−m .
40
So the average number of layers made in time t would be for N ≡ t/∆t,
N
X
n=0
n ·N Cn (∆p)n (1 − ∆p)N−n = N ∆p =
∆p
t = dt.
∆t
Hence the relation, ∆p = d∆t. Note that this relation was obtained without any assumptions on ∆t, ∆p. Now let’s obtain p(m) as we send ∆t, ∆p → 0
(t/∆t)(t/∆t − 1) · · · (t/∆t − m + 1)
(∆p)m (1 − ∆p)t/∆t−m .
∆t→0
m!
p(m) = lim
Rearranging the terms we get,
p(m) = lim
∆t→0
(t∆p/∆t) · · · (t∆p/∆t − (m − 1)∆p)
(1 − ∆p)t/∆t−m .
m!
Inserting, ∆t = d∆p we get,
(dt)m e−dt
(dt)m
(1 − ∆p)dt/∆p =
.
∆p→0 m!
m!
p(m) = lim
(b) What is the variance in the thickness?
• We need hm2 i, which is,
2
hm i =
∞
X
m=0
m2 p(m) =
X m(m − 1)
m!
Since,
hmi =
the variance in the thickness is,
(dt)m e−td +
X m
(dt)m e−td = (dt)2 + dt
m!
∞
X
m
(dt)m e−td = dt.
m!
m=0
hm2 i − hmi2 = dt.
Note that all cumulants of this Poisson process are equal to dt.
********
10. Diode:
The current I across a diode is related to the applied voltage V via I =
I0 [exp(eV /kT ) − 1]. The diode is subject to a random potential V of zero mean and
variance σ 2 which is Gaussian distributed. Find the probability density p(I) for the current
41
I flowing through the diode. Find the most probable value for I, the mean value of I, and
indicate them on a sketch of p(I).
********
11. Mutual information:
Consider random variables x and y, distributed according to a
joint probability p(x, y). The mutual information between the two variables is defined by
M (x, y) ≡
X
x,y
p(x, y)
p(x, y) ln
px (x)py (y)
,
where px and py denote the unconditional probabilities for x and y.
(a) Relate M (x, y) to the entropies S(x, y), S(x), and S(y) obtained from the corresponding
probabilities.
• Expanding the expression for M (x, y), we get
M (x, y) =
X
x,y
p(x, y) ln p(x, y) −
X
x,y
p(x, y) ln px (x) −
X
p(x, y) ln py (y).
x,y
Since
X
p(x, y) = px (x),
X
p(x, y) = py (y),
y
and
x
we obtain
M (x, y) = −S(x, y) + S(x) + S(y).
(b) Calculate the mutual information for the joint Gaussian form
by 2
ax2
−
− cxy .
p(x, y) ∝ exp −
2
2
• First, we calculate the normalization factor for the joint probability distribution
1 = Axy
Z
∞
∞
a 2 b 2
dx
dy exp − x − y − cxy
2
2
−∞
−∞
Z
42
calculating the integral, we get
Axy =
1 p
ab − c2 .
2π
Next, we have to find p(x) and p(y). For that purpose, we “integrate out” y and x
respectively:
px (x) =
Z
∞
s
dyp(x, y) =
−∞
and
py (y) =
s
1
2π
1
2π
c2
b−
a
c2
a−
b
a c2
exp −
x2 ,
−
2 2b
b
c2
exp −
y2 .
−
2 2a
Thus,
c2 2 c2 2
Axy
M (x, y) =
p(x, y) ln
− cxy − x − y
A
A
2b
2a
x
y
x,y
Z
Z
Z
Z
Axy
c2
c2
2
2
= ln
−
x px (x)dx −
y py (y)dy − c dx dyxy p(x, y),
Ax Ay
2b
2a
X
where Ax,y are the normalization factors for px (x) and py (y). This expression finally leads
to
M (x, y) = ln
Axy
Ax Ay
1
c2
= − ln 1 −
.
2
ab
********
12. Semi-flexible polymer in two dimensions Configurations of a model polymer can be
described by either a set of vectors {ti } of length a in two dimensions (for i = 1, · · · , N ),
or alternatively by the angles {φi } between successive vectors, as indicated in the figure
below.
t3
t 2ø
t N-1
1
t N ø N-1
t1
R
43
The polymer is at a temperature T , and subject to an energy
H = −κ
N−1
X
i=1
ti · ti+1 = −κa2
N−1
X
cos φi
,
i=1
where κ is related to the bending rigidity, such the probability of any configuration is
proportional to exp (−H/kB T ).
(a) Show that htm · tn i ∝ exp (−|n − m|/ξ), and obtain an expression for the persistence
length ℓp = aξ. (You can leave the answer as the ratio of simple integrals.)
• In terms of the angles, the dot product can be written as
tm · tn = a2 cos (φm + φm+1 + · · · + φn−1 ) = a2 ℜei(φm +φm+1 +···+φn−1 ) .
Note that the angles {φn } are independent variables, distributed according to the Boltz-
mann weight
p [{φn }] ∝
N−1
Y
n=1
exp
κa2
cos φn
kB T
.
Hence the average of the product is the product of averages, and
htm · tn i = a2 ℜ
n−1
Y
k=m

eiφn = a2 
The persistence length is thus given by
ℓp =
ln
R
dφ e
κa2
kB T
cos φ
/
a
R
R
κa2
kB T
dφ cos φ e
R
κa2
dφ e kB T
dφ cos φ e
cos φ

cos φ
κa2
kB T
cos φ
|n−m|
.
.
(b) Consider the end–to–end distance R as illustrated in the figure. Obtain an expression
for R2 in the limit of N ≫ 1.
PN−1
• Using R = n=1 tn , we obtain
R2 =
X
m,n
htm · tn i =
X
a2 e−|n−m|/ξ .
m,n
The above sum decays exponentially around each point. Ignoring corrections from end
effects, which are asymptotically negligible for N → ∞, we obtain
e−1/ξ
1
2
2
R ≃a N 1+2
= a2 N coth .
−1/ξ
2ξ
1−e
44
(c) Find the probability p(R) in the limit of N ≫ 1.
PN−1
• Since R = n=1 tn , we can use the central limit theorem to conclude that in the limit
of N → ∞, the probability distribution p(R) approaches a Gaussian form. Thus, we just
need to evaluate the mean and variance of R. Since the mean hRi = 0 by symmetry,
the variance is equal to R2 calculated in part (b). Noting that in two dimensions,
Rx2 = Ry2 = R2 /2, the properly normalized Gaussian form is
1
R·R
p(R) =
.
exp −
π hR2 i
hR2 i
(d) If the ends of the polymer are pulled apart by a force F, the probabilities
for poly
mer configurations are modified by the Boltzmann weight exp kF·R
. By expanding this
BT
weight, or otherwise, show that
hRi = K −1 F + O(F 3 ) ,
and give an expression for the Hookian constant K in terms of quantities calculated before.
• Let us indicate by hi0 averages taken for F = 0. Then the average of R at finite F is
given by
ReβF·R 0
hRi =
heβF·R i0
,
where β = 1/(kB T ). The exponential factors can now be expanded in powers of F . Noting
that zero-force averages of odd powers of R are zero by symmetry, we obtain
hRµ i = βFν hRν Rµ i0 + O(F 3 )
By symmetry hRν Rµ i0 = δµν R2
0
/2, where R2
0
.
is the zero-force expression calculated
in the previous part. Thus,
hRi = K −1 F + O(F 3 )
, with K =
2kB T
.
hR2 i0
********
13. The book of records: Consider a sequence of random numbers {x1 , x2 , · · · , xn , · · ·}; the
entry xn is a record if it is larger than all numbers before it, i.e. if xn > {x1 , x2 , · · · , xn−1 }.
45
We can then define an associated sequence of indicators {R1 , R2 , · · · , Rn , · · ·} in which
Rn = 1 if xn is a record, and Rn = 0 if it is not (clearly R1 = 1).
(a) Assume that each entry xn is taken independently from the same probability distribution p(x). [In other words, {xn } are IIDs (independent identically distributed).] Show
that, irrespective of the form of p(x), there is a very simple expression for the probability
Pn that the entry xn is a record.
• Consider the n–entries {x1 , x2 , · · · , xn }. Each one of them has the same probability to
be the largest one. Thus the probability that xn is the largest, and hence a record, is
Pn = 1/n.
(b) The records are entered in the Guinness Book of Records. What is the average number
hSN i of records after N attempts, and how does it grow for, N ≫ 1? If the number of
trials, e.g. the number of participants in a sporting event, doubles every year, how does
the number of entries asymptotically grow with time.
•
N
X
N
X
1
1
hSN i =
Pn =
≈ ln N + γ + O
n
N
n=1
n=1
for
N ≫ 1,
where γ ≈ 0.5772 . . . is the Euler number. Clearly if N ∝ 2t , where t is the number of
years,
hSt i ≈ ln N (t) = t ln 2.
(c) Prove that the record indicators {Rn } are independent random variables (though not
identical), in that hRn Rm ic = 0 for m 6= n.
• hRn Rm i = 1 · Pn · Pm while hRn i = 1 · Pn , hence yielding,
hRn Rm ic = Pn Pm − Pn · Pm = 0.
This is by itself not sufficient to prove that the random variables Rn and Rm are independent variables. The correct proof is to show that the joint probability factorizes,
i.e. p(Rn , Rm ) = p(Rn )p(Rm ). Let us suppose that m > n, the probability that xm is the
largest of the random variables up to m is 1/m, irrespective of whether some other random
number in the set was itself a record (the largest of the random numbers up to n). The
equality of the conditional and unconditional probabilities is a proof of independence.
(d) Compute all moments, and the first three cumulants of the total number of records
SN after N entries. Does the central limit theorem apply to SN ?
46
• We want to compute he−ikSN i. This satisfies,
he−ikSN i =
e−ik −ikSN −1
N − 1 −ikSN −1
e−ik + N − 1 −ikSN −1
he
i+
he
i=
he
i,
N
N
N
since the probability for aquiring an additional phase −ik is PN = 1/N . Since he−ikS1 i =
e−ik , by recursion we obtain,
he−ikSN i =
N
N
1 X
1 Y −ik
(e
+ n − 1) =
S1 (N, n)e−ikn ,
N ! n=1
N ! n=0
where S1 (n, m) denotes the unsigned Stirling number of the first kind. Expanding the
exponent, we obtain,
−ikSN
he
N
∞
∞
X
X
1 X
nm (−ik)m
(−ik)m
i=
S1 (N, n)
=
N ! n=0
m!
m!
m=0
m=0
"
#
N
1 X
m
S1 (N, n)n .
N ! n=0
Hence we obtain the mth moment,
m
hSN
i
N
1 X
S1 (N, n)nm .
=
N ! n=0
The first three cumulants can be obtained using the moments, but the easier way to obtain
them is by using the fact that SN = R1 + · · · + RN and that Rn are independent variables.
Due to the independence of Rn ,
m
hSN
ic
=
N
X
n=1
hRnm ic .
From this, and the fact that for any m, hRnm i = 1/n, we easily obtain,
N
X
N
X
1
hSN ic =
hRn ic =
,
n
n=1
n=1
2
hSN
ic
3
hSN
ic
=
=
N
X
n=1
N
X
n=1
hRn2 ic
hRn3 ic
N
X
1
1
( − 2 ),
=
n n
n=1
N
X
3
2
1
=
( − 2 + 3 ).
n n
n
n=1
47
The central limit theorem applies to SN since Rn are independent variables. Seeing this
in another way is to observe that for large N ,
m
hSN
ic =
N
X
n=1
hRnm ic <
N
X
n=1
hRnm i =
N
X
1
≈ ln N + O (1) .
n
n=1
Since the variance is of order ln N , all the higher moments can be disregarded.
(e) The first record, of course occurs for n1 = 1. If the third record occurs at trial number
n3 = 9, what is the mean value hn2 i for the position of the second record? What is the
mean value hn4 i for the position of the fourth record?
• The probability for n2 = n and n3 = 9 is,
(1 − R2 ) · · · (1 − Rn−1 )Rn (1 − Rn+1 ) · · · (1 − R8 )R9 =
1
.
72(n − 1)
Hence hn2 in3 =9 can be obtained by,
hn2 i =
P8
1
n2 =2 n2 · 72(n2 −1)
P8
1
n2 =2 72(n2 −1)
As for hn4 i, for given value of x3 , define, P (x3 ) =
value of x3 ,
hn4 ix3 =a =
∞
X
n=10
n · (1 − P (a))n−10P (a) =
Hence we obtain hn4 i by,
R∞
hn4 i =
−∞
1343
≈ 3.70.
363
=
R∞
x3
dxp(x). Then we get for a given
P (a)
1
9P (a)
+
= 9+
.
2
1 − (1 − P (a)) (1 − (1 − P (a)))
P (a)
dap(a)(9 + 1/P (a))
R∞
=
dap(a)
−∞
Z
0
1
dP (9 +
1
) → ∞,
P
where we used the fact that p(a) = −dP (a)/da.
********
14. Jarzynski equality: In equilibrium at a temperature T , the probability that a macroscopic system is in a microstate µ is p(µ) = exp [−βH(µ)] /Z, where H(µ) is the energy of
the microstate, β = 1/(kB T ), and the normalization factor is related to the free energy by
−βF = ln Z. We now change the macroscopic state of the system by performing external
work W , such that the new state is also in equilibrium at temperature T . For example,
imagine that the volume of a gas in changed by moving a piston as L(t) = L1 +(L2 −L1 )t/τ .
48
Depending on the protocol (e.g. the speed u = (L2 −L1 )/τ ), the process may be close to or
far from reversible. Nonetheless, the Jarzynski equality relates the probability distribution
for the work W to the equilibrium change in free energy!
(a) Assume that the process by which work is performed is fully deterministic, in the sense
that for a given protocol, any initial microstate µ will evolve to a specific final microstate
µ′ . The amount of work performed W (µ) will vary with the initial microstate, and there is
thus a probability distribution pf (W ) which can be related to the equilibrium p(µ). The
energy of the final microstate, however, is precisely H′ (µ′ ) = H(µ) + W (µ). Time reversal
symmetry implies that if we now instantaneously reverse all the momenta, and proceed
according to the reversed protocol, the time-reversed microstate µ′ will deterministically
evolve back to microstate µ, and the work −W (µ) is recovered. However, rather than
doing so, we first allow the system to equilibrate into its new macrostate at temperature
T , before reversing the protocol to recover the work. The recovered work −W will now be
a function of the selected microstate, and distributed according to a different probability
pb (−W ), related to p′ (µ′ ) = exp −βH′ (µ′ ) /Z ′ . It is in general not possible to find pf (W )
or pb (−W ). However, by noting that the probabilities of a pair of time-reversed microstates
are exactly equal, show that their ratio is given by
pf (W )
= exp [β(W + F − F ′ )] .
pb (−W )
While you were guided to prove the above result with specific assumptions, it is in fact
more generally valid, and known as the work–fluctuation theorem.
• pf (W ) can be defined such that,
pf (W )dW =
X
p(µ)dµ.
W (µ)=W
Also,
pb (−W )dW =
X
p′ (µ′ )dµ′ .
W ′ (µ′ )=−W
We used W (µ) to denote the work done on the microstate µ in the forward process, and
W ′ (µ′ ) to denote the work done on the microstate µ′ in the backward process. Note that
there is a one to one correspondence between initial microstates µ such that W (µ) = W
and final microstates µ′ with W ′ (µ′ ) = −W , since each µ is mapped to a µ′ by a given
49
protocol. Also, since the density of microstates do not change by thermodynamic processes,
it is safe to say that at a given microstate µ, dµ = dµ′ . Hence we may write,
pb (−W )dW =
X
W ′ (µ′ )=−W
Therefore,
Now since for each µ,
X
p′ (µ′ )dµ′ =
p′ (µ′ (µ))dµ.
W (µ)=W
P
pf (W )
W (µ)=W p(µ)dµ
.
=P
′ ′
pb (−W )
W (µ)=W p (µ (µ))dµ
p(µ)dµ
Z′
=
exp(−β(H(µ) − H′ (µ′ ))) = exp(β(W + F − F ′ )),
p′ (µ′ (µ))dµ
Z
we finally obtain,
pf (W )
= exp(β(W + F − F ′ )).
pb (−W )
(b) Prove the Jarzynski equality
′
−βW
∆F ≡ F − F = −kB T ln e
≡ −kB T ln
Z
−βW
dW pf (W )e
.
This result can in principle be used to compute equilibrium free energy differences from
an ensemble of non-equilibrium measurements of the work. For example, in Liphardt, et.
al., Science 296, 1832 (2002), the work needed to stretch a single RNA molecule was
calculated and related to the free energy change. However, the number of trials must be
large enough to ensure that the averaged exponential, which is dominated by rare events,
is accurately obtained.
• From the previous problem,
pf (W )e−βW = pb (−W )eβ(F −F
′
)
Integrating both sides with respect to W , and using
= pb (−W )e−β∆F .
R
dW pb (−W ) = 1 we get,
he−βW i = e−β∆F .
Taking the logarithm of both sides and sorting out the terms we get,
∆F = −kB T lnhe−βW i.
50
(c) Use the Jarzynski equality to prove the familiar thermodynamic inequality
hW i ≤ ∆F
.
• Since exp(−βW ) is a convex function of W , for any distribution p(W ) with p(W ) ≥ 0
R
and dW p(W ) = 1,
Z
Z
−βW
he
i = dW p(W ) exp(−βW ) ≥ exp(−β dW p(W )W ) = e−βhW i .
Hence we obtain,
∆F = −
1
1
lnhe−βW i ≤ − ln e−βhW i = hW i.
β
β
(d) Consider a cycle in which a work W + ω is performed in the first stage, and work
−W is returned in the reversed process. According to the second law of thermodynamics,
the net gain ω must be negative. However, within statistical physics, it is always possible
that this condition is violated. Use the above results to conclude that the probability of
violating the second law decays with the degree of violation according to
Pviolating
second law (ω)
≤ e−βω .
• Let us define p(ω) as the probability density to violate the second law by ω > 0. Since
the same ω can be obtained for different choices of W , we have
Z
p(ω) = dW pf (W − ω)pb (−W ).
By the relation obtained in (a), we get,
pf (W − ω)pb (−W )
exp(β(W − ω + F − F ′ ))
=
,
pb (−W + ω)pf (W )
exp(β(W + F − F ′ ))
which yields,
pf (W − ω)pb (−W ) = pb (−W + ω)pf (W ) exp(−βω),
hence,
p(ω) =
Z
dW pb (−W + ω)pf (W ) exp(−βω) = p(−ω) exp(−βω).
The cumulative probability of violating the second law by ω or more is now obtained as
Z ∞
Z ∞
Z ∞
′
′
′
′
′
P (ω) =
ω p(ω ) =
ω p(−ω ) exp(−βω ) < exp(−βω)
ω ′ p(−ω ′ ) < exp(−βω).
ω
ω
ω
′
The first inequality is because for all ω in the integrand, exp(−βω ′ ) < exp(−βω), and the
second since any cumulative probability is less than one.
********
51
Problems for Chapter III - Kinetic Theory of Gases
1. One dimensional gas:
A thermalized gas particle is suddenly confined to a one–
dimensional trap. The corresponding mixed state is described by an initial density function
√
ρ(q, p, t = 0) = δ(q)f (p), where f (p) = exp(−p2 /2mkB T )/ 2πmkB T .
(a) Starting from Liouville’s equation, derive ρ(q, p, t) and sketch it in the (q, p) plane.
• Liouville’s equation, describing the incompressible nature of phase space density, is
∂ρ
∂ρ
∂ρ
∂H ∂ρ ∂H ∂ρ
= −q̇
− ṗ
=−
+
≡ −{ρ, H}.
∂t
∂q
∂p
∂p ∂q
∂q ∂p
For the gas particle confined to a 1–dimensional trap, the Hamiltonian can be written as
H=
p2
p2
+ V (qx ) =
,
2m
2m
since Vqx = 0, and there is no motion in the y and z directions. With this Hamiltonian,
Liouville’s equation becomes
p ∂ρ
∂ρ
=−
,
∂t
m ∂q
whose solution, subject to the specified initial conditions, is
p p
ρ(q, p, t) = ρ q − t, p, 0 = δ q − t f (p).
m
m
p
t=0
q
slope m/ t
52
(b) Derive the expressions for the averages q 2 and p2 at t > 0.
• The expectation value for any observable O is
Z
hOi = dΓOρ(Γ, t),
and hence
2
p
Z
p = p f (p)δ q − t dp dq = p2 f (p)dp
m
Z ∞
1
p2
2
dp p √
exp −
=
= mkB T.
2mkB T
2πmkB T
−∞
Z
2
Likewise, we obtain
2 Z
Z 2
Z
kB T 2
p
t
p 2
2
p2 f (p)dp =
t f (p)dp =
t .
q = q f (p)δ q − t dp dq =
m
m
m
m
(c) Suppose that hard walls are placed at q = ±Q. Describe ρ(q, p, t ≫ τ ), where τ is an
appropriately large relaxation time.
• Now suppose that hard walls are placed at q = ±Q. The appropriate relaxation time
τ , is the characteristic length between the containing walls divided by the characteristic
velocity of the particle, i.e.
r
2Q
2Qm
m
= 2Q
τ∼
=p
.
|q̇|
kB T
hp2 i
Initially ρ(q, p, t) resembles the distribution shown in part (a), but each time the particle
hits the barrier, reflection changes p to −p. As time goes on, the slopes become less, and
ρ(q, p, t) becomes a set of closely spaced lines whose separation vanishes as 2mQ/t.
p
q
-Q
+Q
53
(d) A “coarse–grained” density ρ̃, is obtained by ignoring variations of ρ below some small
resolution in the (q, p) plane; e.g., by averaging ρ over cells of the resolution area. Find
ρ̃(q, p) for the situation in part (c), and show that it is stationary.
• We can choose any resolution ε in (p, q) space, subdividing the plane into an array of
pixels of this area. For any ε, after sufficiently long time many lines will pass through this
area. Averaging over them leads to
ρ̃(q, p, t ≫ τ ) =
1
f (p),
2Q
as (i) the density f (p) at each p is always the same, and (ii) all points along q ∈ [−Q, +Q]
are equally likely. For the time variation of this coarse-grained density, we find
∂ ρ̃
p ∂ ρ̃
=−
= 0, i.e. ρ̃ is stationary.
∂t
m ∂q
********
2. Evolution of entropy:
The normalized ensemble density is a probability in the phase
R
space Γ. This probability has an associated entropy S(t) = − dΓρ(Γ, t) ln ρ(Γ, t).
(a) Show that if ρ(Γ, t) satisfies Liouville’s equation for a Hamiltonian H, dS/dt = 0.
• A candidate “entropy” is defined by
S(t) = −
Z
dΓρ(Γ, t) ln ρ(Γ, t) = − hln ρ(Γ, t)i .
Taking the derivative with respect to time gives
dS
=−
dt
Z
dΓ
∂ρ
1 ∂ρ
ln ρ + ρ
∂t
ρ ∂t
=−
Z
dΓ
∂ρ
(ln ρ + 1) .
∂t
Substituting the expression for ∂ρ/∂t obtained from Liouville’s theorem gives
dS
=−
dt
Z
3N X
∂ρ ∂H
∂ρ ∂H
−
(ln ρ + 1) .
dΓ
∂pi ∂qi
∂qi ∂pi
i=1
54
(Here the index i is used to label the 3 coordinates, as well as the N particles, and hence
runs from 1 to 3N .) Integrating the above expression by parts yields†
3N X
∂
∂H
∂
∂H
ρ
(ln ρ + 1) − ρ
(ln ρ + 1)
dΓ
∂pi ∂qi
∂qi ∂pi
i=1
Z
3N X
∂2H
∂H 1 ∂ρ
∂2H
∂H 1 ∂ρ
ρ
= dΓ
(ln ρ + 1) + ρ
−ρ
(ln ρ + 1) − ρ
∂p
∂q
∂q
ρ
∂p
∂q
∂p
∂pi ρ ∂qi
i
i
i
i
i
i
i=1
Z
3N X
∂H ∂ρ
∂H ∂ρ
= dΓ
−
.
∂qi ∂pi
∂pi ∂qi
i=1
dS
=
dt
Z
Integrating the final expression by parts gives
dS
=−
dt
Z
3N X
∂ 2H
∂ 2H
−ρ
+ρ
= 0.
dΓ
∂pi ∂qi
∂qi ∂pi
i=1
(b) Using the method of Lagrange multipliers, find the function ρmax (Γ) which maximizes
R
the functional S[ρ], subject to the constraint of fixed average energy, hHi = dΓρH = E.
• There are two constraints, normalization and constant average energy, written respectively as
Z
†
dΓρ(Γ) = 1,
hHi =
and
This is standard integration by parts, i.e.
Ra
b
Z
dΓρ(Γ)H = E.
a
F dG = F G|b −
Ra
b
GdF . Looking
explicitly at one term in the expression to be integrated in this problem,
Z Y
3N
∂ρ ∂H
=
dVi
∂qi ∂pi
i=1
Z
dq1 dp1 · · · dqi dpi · · · dq3N dp3N
∂ρ ∂H
,
∂qi ∂pi
∂ρ
we identify dG = dqi ∂q
, and F with the remainder of the expression. Noting that
i
ρ(qi ) = 0 at the boundaries of the box, we get
Z Y
3N
∂ρ ∂H
=−
dVi
∂qi ∂pi
i=1
55
Z Y
3N
i=1
dVi ρ
∂ ∂H
.
∂qi ∂pi
Rewriting the expression for entropy,
S(t) =
Z
dΓρ(Γ) [− ln ρ(Γ) − α − βH] + α + βE,
where α and β are Lagrange multipliers used to enforce the two constraints. Extremizing
the above expression with respect to the function ρ(Γ), results in
∂S
∂ρ(Γ)
ρ=ρmax
= − ln ρmax (Γ) − α − βH(Γ) − 1 = 0.
The solution to this equation is
ln ρmax = −(α + 1) − βH,
which can be rewritten as
ρmax = C exp (−βH) ,
where
C = e−(α+1) .
(c) Show that the solution to part (b) is stationary, i.e. ∂ρmax /∂t = 0.
• The density obtained in part (b) is stationary, as can be easily checked from
n
o
∂ρmax
−β H
= − {ρmax , H} = − Ce
,H
∂t
∂H
∂H −β H ∂H
∂H −β H
=
C(−β)
e
−
C(−β)
e
= 0.
∂p
∂q
∂q
∂p
(d) How can one reconcile the result in (a), with the observed increase in entropy as
the system approaches the equilibrium density in (b)? (Hint: Think of the situation
encountered in the previous problem.)
• Liouville’s equation preserves the information content of the PDF ρ(Γ, t), and hence S(t)
does not increase in time. However, as illustrated in the example in problem 1, the density
becomes more finely dispersed in phase space. In the presence of any coarse-graining of
phase space, information disappears. The maximum entropy, corresponding to ρ̃, describes
equilibrium in this sense.
********
56
3. The Vlasov equation is obtained in the limit of high particle density n = N/V , or large
inter-particle interaction range λ, such that nλ3 ≫ 1. In this limit, the collision terms are
dropped from the left hand side of the equations in the BBGKY hierarchy.
The BBGKY hierarchy
"
s
s
X
X
∂
~pn
∂
−
+
·
∂t n=1 m ∂~qn n=1
!
#
X ∂V(~qn − ~ql )
∂
∂U
+
·
fs
∂~qn
∂~qn
∂~pn
l
s Z
X
∂V(~qn − ~qs+1 ) ∂fs+1
·
,
=
dVs+1
∂~qn
∂~pn
n=1
has the characteristic time scales

∂U ∂
v
1

∼
·
∼ ,



τU
∂~q ∂~p
L


 1
∂V ∂
v
∼
·
∼ ,

τc
∂~q ∂~p
λ


Z



 1 ∼ dx ∂V · ∂ fs+1 ∼ 1 · nλ3 ,
τX
∂~q ∂~p fs
τc
where nλ3 is the number of particles within the interaction range λ, and v is a typical
velocity. The Boltzmann equation is obtained in the dilute limit, nλ3 ≪ 1, by disregarding
terms of order 1/τX ≪ 1/τc . The Vlasov equation is obtained in the dense limit of nλ3 ≫ 1
by ignoring terms of order 1/τc ≪ 1/τX .
(a) Assume that the N body density is a product of one particle densities, i.e. ρ =
QN
pi , ~qi ). Calculate the densities fs , and their normalizations.
i=1 ρ1 (xi , t), where xi ≡ (~
• Let bf xi denote the coordinates and momenta for particle i. Starting from the joint
QN
probability ρN = i=1 ρ1 (xi , t), for independent particles, we find
N!
fs =
(N − s)!
Z
N
Y
α=s+1
The normalizations follow from
Z
dΓρ = 1,
and
Z Y
s
n=1
dVα ρN =
dVn fs =
=⇒
Z
dV1 ρ1 (x, t) = 1,
N!
≈ Ns
(N − s)!
57
s
Y
N!
ρ1 (xn , t).
(N − s)! n=1
for
s ≪ N.
(b) Show that once the collision terms are eliminated, all the equations in the BBGKY
hierarchy are equivalent to the single equation
∂
~p ∂
∂Ueff ∂
f1 (~
p, ~q, t) = 0,
+
·
−
·
∂t m ∂~q
∂~q
∂~p
where
Ueff (~q, t) = U (~q ) +
• Noting that
Z
dx′ V(~q − ~q ′ )f1 (x′ , t).
(N − s)!
fs+1
=
ρ1 (xs+1 ),
fs
(N − s − 1)!
the reduced BBGKY hierarchy is
"
≈
s Z
X
s
X
∂
+
∂t n=1
∂U
∂
p~ n
∂
−
·
·
m ∂~qn
∂~qn ∂~pn
#
fs
∂V(~qn − ~qs+1 ) ∂
·
[(N − s)fs ρ1 (xs+1 )]
∂~qn
∂~pn
n=1
Z
s
X
∂
∂
≈
dVs+1 ρ1 (xs+1 )V(~qn − ~qs+1 ) · N
fs ,
∂~qn
∂~pn
n=1
dVs+1
where we have used the approximation (N − s) ≈ N for N ≫ s. Rewriting the above
expression,
"
where
s
X
∂
+
∂t n=1
p~n
∂
∂Uef f
∂
·
−
·
m ∂~qn
∂~qn
∂~pn
Uef f = U (~q ) + N
Z
#
fs = 0,
dV ′ V(~q − ~q ′ )ρ1 (x′ , t).
(c) Now consider N particles confined to a box of volume V , with no additional potential.
Show that f1 (~q, p~ ) = g(~
p )/V is a stationary solution to the Vlasov equation for any g(~
p ).
Why is there no relaxation towards equilibrium for g(~
p )?
• Starting from
ρ1 = g(~
p )/V,
we obtain
Hef f =
N X
p~i 2
i=1
2m
58
+ Uef f (~qi ) ,
with
Z
N
1
d3 qV(~q ).
p) =
Uef f = 0 + N dV V(~q − ~q ) g(~
V
V
R 3
(We have taken advantage of the normalization d pg(~
p ) = 1.) Substituting into the
Z
′
′
Vlasov equation yields
∂
p~ ∂
+
·
∂t m ∂~q
ρ1 = 0.
There is no relaxation towards equilibrium because there are no collisions which allow
g(~
p ) to relax. The momentum of each particle is conserved by Hef f ; i.e. {ρ1 , Hef f } = 0,
preventing its change.
********
4. Two component plasma:
Consider a neutral mixture of N ions of charge +e and mass
m+ , and N electrons of charge −e and mass m− , in a volume V = N/n0 .
(a) Show that the Vlasov equations for this two component system are

p~
∂
∂Φeff ∂
∂


p, ~q, t) = 0
 ∂t + m · ∂~q + e ∂~q · ∂~p f+ (~
+

~p
∂
∂Φeff ∂
∂


f− (~
p, ~q, t) = 0
+
·
−e
·
∂t m− ∂~q
∂~q
∂~p
,
where the effective Coulomb potential is given by
Φeff (~q, t) = Φext (~q ) + e
Z
dx′ C(~q − ~q ′ ) [f+ (x′ , t) − f− (x′ , t)] .
Here, Φext is the potential set up by the external charges, and the Coulomb potential C(~
q)
satisfies the differential equation ∇2 C = 4πδ 3 (~q ).
• The Hamiltonian for the two component mixture is
X
N 2N
2N
X
X
p~i 2
1
~pi 2
H=
ei ej
ei Φext (~qi ),
+
+
+
2m
2m
|~
q
−
~
q
|
+
−
i
j
i=1
i,j=1
i=1
where C(~qi − ~qj ) = 1/|~qi − ~qj |, resulting in
X
∂Φext
∂
∂H
= ei
+ ei
ej
C(~qi − ~qj ).
∂~qi
∂~qi
∂~qi
j6=i
59
Substituting this into the Vlasov equation, we obtain

p
~
∂
∂Φ
∂
∂
ef
f


f+ (~
p, ~q, t) = 0,
+
·
+e
·

∂t m+ ∂~q
∂~q
∂~p

∂
p~
∂
∂Φef f ∂


f− (~
p, ~q, t) = 0.
+
·
−e
·
∂t m− ∂~q
∂~q
∂~p
(b) Assume that the one particle densities have the stationary forms f± = g± (~
p )n± (~
q ).
Show that the effective potential satisfies the equation
∇2 Φeff = 4πρext + 4πe (n+ (~q ) − n− (~q )) ,
where ρext is the external charge density.
• Setting f± (~
p, ~q ) = g± (~
p )n± (~q ), and using
potential simplify to
Φef f (~q, t) = Φext (~q ) + e
Z
R
d3 pg± (~
p ) = 1, the integrals in the effective
d3 q ′ C(~q − ~q ′ ) [n+ (~q ′ ) − n− (~q ′ )] .
Apply ∇2 to the above equation, and use ∇2 Φext = 4πρext and ∇2 C(~q−~q ′ ) = 4πδ 3 (~q−~
q ′ ),
to obtain
∇2 φef f = 4πρext + 4πe [n+ (~q ) − n− (~q )] .
(c) Further assuming that the densities relax to the equilibrium Boltzmann weights
n± (~q ) = n0 exp [±βeΦeff (~q)], leads to the self-consistency condition
∇2 Φeff = 4π ρext + n0 e eβeΦeff − e−βeΦeff ,
known as the Poisson–Boltzmann equation. Due to its nonlinear form, it is generally not
possible to solve the Poisson–Boltzmann equation. By linearizing the exponentials, one
obtains the simpler Debye equation
∇2 Φeff = 4πρext + Φeff /λ2 .
Give the expression for the Debye screening length λ.
• Linearizing the Boltzmann weights gives
n± = no exp[∓βeΦef f (~q )] ≈ no [1 ∓ βeΦef f ] ,
60
resulting in
∇2 Φef f = 4πρext +
1
Φef f ,
λ2
with the screening length given by
λ2 =
kB T
.
8πno e2
(d) Show that the Debye equation has the general solution
Z
Φeff (~q ) = d3 ~q′ G(~q − ~q ′ )ρext (~q ′ ),
where G(~q ) = exp(−|~q |/λ)/|~q | is the screened Coulomb potential.
• We want to show that the Debye equation has the general solution
Z
Φef f (~q ) = d3 ~qG(~q − ~q ′ )ρext (~q ′ ),
where
G(~q ) =
exp(−|q|/λ)
.
|q|
Effectively, we want to show that ∇2 G = G/λ2 for ~q 6= 0. In spherical coordinates,
G = exp(−r/λ)/r. Evaluating ∇2 in spherical coordinates gives
1 e−r/λ
1 ∂ 2
1 ∂
e−r/λ
2 ∂G
r
= 2 r −
∇ G= 2
−
r ∂r
∂r
r ∂r
λ r
r2
1 1 −r/λ 1 −r/λ 1 −r/λ
1 e−r/λ
G
= 2
=
re
−
e
+
e
= 2.
2
2
r λ
λ
λ
λ
r
λ
2
(e) Give the condition for the self-consistency of the Vlasov approximation, and interpret
it in terms of the inter-particle spacing?
• The Vlasov equation assumes the limit no λ3 ≫ 1, which requires that
(kB T )3/2
1/2
no e3
≫ 1,
where ℓ is the interparticle spacing.
consistency condition is
=⇒
e2
≪ no−1/3 ∼ ℓ,
kB T
In terms of the interparticle spacing, the self–
e2
≪ kB T,
ℓ
61
i.e. the interaction energy is much less than the kinetic (thermal) energy.
(f) Show that the characteristic relaxation time (τ ≈ λ/c) is temperature independent.
What property of the plasma is it related to?
• A characteristic time is obtained from
r
r
r
λ
1
kB T
m
m
τ∼ ∼
·
∼
∼
,
2
2
c
no e
kB T
no e
ωp
where ωp is the plasma frequency.
********
5. Two dimensional electron gas in a magnetic field: When donor atoms (such as P or As)
are added to a semiconductor (e.g. Si or Ge), their conduction electrons can be thermally
excited to move freely in the host lattice. By growing layers of different materials, it is
possible to generate a spatially varying potential (work–function) which traps electrons at
the boundaries between layers. In the following, we shall treat the trapped electrons as a
gas of classical particles in two dimensions.
If the layer of electrons is sufficiently separated from the donors, the main source of
scattering is from electron–electron collisions.
(a) The Hamiltonian for non–interacting free electrons in a magnetic field has the form

~
X  ~pi − eA
H=

2m
i
2

~ 
± µB |B|
.
(The two signs correspond to electron spins parallel or anti-parallel to the field.) The
~=B
~ × ~q/2 describes a uniform magnetic field B.
~ Obtain the classical
vector potential A
equations of motion, and show that they describe rotation of electrons in cyclotron orbits
~
in a plane orthogonal to B.
• The Hamiltonian for non–interacting free electrons in a magnetic field has the form

~
X  ~pi + eA
H=

2m
i
or in expanded form
H=
2

~ 
± µB |B|
,
2
e
p2
~ 2 ± µB |B|.
~
~+ e A
+ p~ · A
2m m
2m
62
~=B
~ × ~q/2, results in
Substituting A
e
p2
~ × ~q +
+
~p · B
2m 2m
p2
e
~ · ~q +
=
+
~ ×B
p
2m 2m
H=
2
e2 ~
~
B × ~q ± µB |B|
8m
e2 2 2
~ · ~q )2 ± µB |B|.
~
B q − (B
8m
Using the canonical equations, ~q˙ = ∂H/~
p and ~p˙ = −∂H/~q , we find

∂H
e~
~p
e ~

˙

× ~q , =⇒ p~ = m~q˙ − B
× ~q ,
 ~q = ∂~p = m + 2m B
2
2
2 
e
∂H

~ · ~q B.
~
~ − e B 2 ~q + e
 ~p˙ = −
B
=−
p~ × B
∂~q
2m
4m
4m
Differentiating the first expression obtained for p~, and setting it equal to the second expression for p~˙ above, gives
e~ ˙
e ˙ e~
e2 ~ 2
e2 ~
¨
~
~
m~q − B × ~q = −
m~q − B × ~q × B −
B · ~q B.
|B| ~q +
2
2m
2
4m
4m
~ × ~q × B
~ = B 2 ~q − B
~ · ~q B,
~ leads to
Simplifying the above expression, using B
~ × ~q˙ .
m¨~q = eB
This describes the rotation of electrons in cyclotron orbits,
¨~q = ω
~ c × ~q˙ ,
~
~
where ~ωc = eB/m;
i.e. rotations are in the plane perpendicular to B.
(b) Write down heuristically (i.e. not through a step by step derivation), the Boltzmann
equations for the densities f↑ (~
p, ~q, t) and f↓ (~
p, ~q, t) of electrons with up and down spins, in
terms of the two cross-sections σ ≡ σ↑↑ = σ↓↓ , and σ× ≡ σ↑↓ , of spin conserving collisions.
• Consider the classes of collisions described by cross-sections σ ≡ σ↑↑ = σ↓↓ , and σ× ≡
σ↑↓ . We can write the Boltzmann equations for the densities as
∂f↑
− {H↑ , f↑ } =
∂t
Z
2
d p2 dΩ|v1 − v2 |
dσ
[f↑ (p~1 ′ )f↑ (p~2 ′ ) − f↑ (p~1 )f↑ (p~2 )] +
dΩ
dσ×
′
′
[f↑ (p~1 )f↓ (p~2 ) − f↑ (p~1 )f↓ (p~2 )] ,
dΩ
63
and
∂f↓
− {H↓ , f↓ } =
∂t
dσ
[f↓ (p~1 ′ )f↓ (p~2 ′ ) − f↓ (p~1 )f↓ (p~2 )] +
dΩ
dσ×
′
′
[f↓ (p~1 )f↑ (p~2 ) − f↓ (p~1 )f↑ (p~2 )] .
dΩ
Z
2
d p2 dΩ|v1 − v2 |
(c) Show that dH/dt ≤ 0, where H = H↑ + H↓ is the sum of the corresponding H functions.
• The usual Boltzmann H–Theorem states that dH/dt ≤ 0, where
Z
H = d2 qd2 pf (~q, ~p, t) ln f (~q, ~p, t).
For the electron gas in a magnetic field, the H function can be generalized to
Z
H = d2 qd2 p [f↑ ln f↑ + f↓ ln f↓ ] ,
where the condition dH/dt ≤ 0 is proved as follows:
Z
∂f↑
∂f↓
dH
2
2
= d qd p
(ln f↑ + 1) +
(ln f↓ + 1)
dt
∂t
∂t
Z
= d2 qd2 p [(ln f↑ + 1) ({f↑ , H↑ } + C↑↑ + C↑↓ ) + (ln f↓ + 1) ({f↓ , H↓ } + C↓↓ + C↓↑ )] ,
with C↑↑ , etc., defined via the right hand side of the equations in part (b). Hence
Z
dH
= d2 qd2 p (ln f↑ + 1) (C↑↑ + C↑↓ ) + (ln f↓ + 1) (C↓↓ + C↓↑ )
dt
Z
= d2 qd2 p (ln f↑ + 1) C↑↑ + (ln f↓ + 1) C↓↓ + (ln f↑ + 1) C↑↓ + (ln f↓ + 1) C↓↑
≡
d
dH↑↑ dH↓↓
+
+
(H↑↓ + H↓↑ ) ,
dt
dt
dt
where the H′ s are in correspondence to the integrals for the collisions. We have also made
R
R
use of the fact that d2 pd2 q {f↑ , H↑ } = d2 pd2 q {f↓ , H↓ } = 0. Dealing with each of the
terms in the final equation individually,
Z
dσ
dH↑↑
= d2 qd2 p1 d2 p2 dΩ|v1 − v2 | (ln f↑ + 1)
[f↑ (~
p1 ′ )f↑ (~
p2 ′ ) − f↑ (~
p1 )f↑ (~
p2 )] .
dt
dΩ
After symmetrizing this equation, as done in the text,
Z
dσ
1
dH↑↑
d2 qd2 p1 d2 p2 dΩ|v1 − v2 |
=−
[ln f↑ (~
p1 )f↑ (~
p2 ) − ln f↑ (~
p1 ′ )f↑ (~
p2 ′ )]
dt
4
dΩ
64
· [f↑ (~
p1 )f↑ (~
p2 ) − f↑ (~
p1 ′ )f↑ (~
p2 ′ )] ≤ 0.
Similarly, dH↓↓ /dt ≤ 0. Dealing with the two remaining terms,
dH↑↓
=
dt
dσ×
[f↑ (~
p1 ′ )f↓ (~
p2 ′ ) − f↑ (~
p1 )f↓ (~
p2 )]
dΩ
Z
dσ×
[f↑ (~
p1 )f↓ (~
p2 ) − f↑ (~
p1 ′ )f↓ (~
p2 ′ )] ,
= d2 qd2 p1 d2 p2 dΩ|v1 − v2 | [ln f↑ (~
p1 ′ ) + 1]
dΩ
Z
d2 qd2 p1 d2 p2 dΩ|v1 − v2 | [ln f↑ (~
p1 ) + 1]
where we have exchanged (~
p1 , ~p2 ↔ ~p1 ′ , ~p2 ′ ). Averaging these two expressions together,
dH↑↓
1
=−
dt
2
Z
d2 qd2 p1 d2 p2 dΩ|v1 − v2 |
dH↓↑
1
=−
dt
2
Z
d2 qd2 p1 d2 p2 dΩ|v1 − v2 |
dσ×
[ln f↑ (~
p1 ) − ln f↑ (~
p1 ′ )]
dΩ
· [f↑ (~
p1 )f↓ (~
p2 ) − f↑ (~
p1 ′ )f↓ (~
p2 ′ )] .
Likewise
dσ×
[ln f↓ (~
p2 ) − ln f↓ (~
p2 ′ )]
dΩ
· [f↓ (~
p2 )f↑ (~
p1 ) − f↓ (~
p2 ′ )f↑ (~
p1 ′ )] .
Combining these two expressions,
Z
dσ×
1
d
d2 qd2 p1 d2 p2 dΩ|v1 − v2 |
(H↑↓ + H↓↑ ) = −
dt
4
dΩ
′
′
[ln f↑ (~
p1 )f↓ (~
p2 ) − ln f↑ (~
p1 )f↓ (~
p2 )] [f↑ (~
p1 )f↓ (~
p2 ) − f↑ (~
p1 ′ )f↓ (~
p2 ′ )] ≤ 0.
Since each contribution is separately negative, we have
dH
dH↑↑
dH↓↓
d
=
+
+
(H↑↓ + H↓↑ ) ≤ 0.
dt
dt
dt
dt
(d) Show that dH/dt = 0 for any ln f which is, at each location, a linear combination of
quantities conserved in the collisions.
• For dH/dt = 0 we need each of the three square brackets in the previous derivation to be
zero. The first two contributions, from dH↑↓ /dt and dH↓↓ /dt, are similar to those discussed
in the notes for a single particle, and vanish for any ln f which is a linear combination of
quantities conserved in collisions
ln fα =
X
aα
q )χi (~
p),
i (~
i
65
where α = (↑ or ↓) . Clearly at each location ~q, for such fα ,
ln fα (~
p1 ) + ln fα (~
p2 ) = ln fα (~
p1 ′ ) + ln fα (~
p2 ′ ).
If we consider only the first two terms of dH/dt = 0, the coefficients aα
q) can vary
i (~
with both ~q and α = (↑ or ↓).
This changes when we consider the third term
d (H↑↓ + H↓↑ ) /dt. The conservations of momentum and kinetic energy constrain the corresponding four functions to be the same, i.e. they require a↑i (~q) = a↓i (~q). There is,
however, no similar constraint for the overall constant that comes from particle number
conservation, as the numbers of spin-up and spin-down particles is separately conserved, i.e.
a↑0 (~q) = a↓0 (~q). This implies that the densities of up and down spins can be different in the
final equilibrium, while the two systems must share the same velocity and temperature.
(e) Show that the streaming terms in the Boltzmann equation are zero for any function
that depends only on the quantities conserved by the one body Hamiltonians.
• The Boltzmann equation is
∂fα
= − {fα , Hα } + Cαα + Cαβ ,
∂t
where the right hand side consists of streaming terms {fα , Hα }, and collision terms C.
Let Ii denote any quantity conserved by the one body Hamiltonian, i.e. {Ii , Hα } = 0.
Consider fα which is a function only of the Ii′ s
fα ≡ fα (I1 , I2 , · · ·) .
Then
{fα , Hα } =
X ∂fα
j
∂Ij
{Ij , Hα } = 0.
~ = ~q × p~, is conserved during, and away from collisions.
(f) Show that angular momentum L
• Conservation of momentum for a collision at ~q
(~
p1 + p~2 ) = (~
p1 ′ + p~2 ′ ),
implies
~q × (~
p1 + p~2 ) = ~q × (~
p1 ′ + p~2 ′ ),
or
~1 + L
~2 = L
~1 ′ + L
~ 2 ′,
L
66
~ i = ~q × p~i . Hence angular momentum L
~ is conserved during colliwhere we have used L
sions. Note that only the z–component Lz is present for electrons moving in 2–dimensions,
~q ≡ (x1 , x2 ), as is the case for the electron gas studied in this problem. Consider the Hamiltonian discussed in (a)
e
e2 2 2
p2
2
~
~
~
B q − (B · ~q ) ± µB |B|.
+
p~ × B · ~q +
H=
2m 2m
8m
Let us evaluate the Poisson brackets of the individual terms with Lz = ~q × p~ |z . The first
term is
2
|~
p | , ~q × ~p
= εijk {pl pl , xj pk } = εijk 2pl
∂
(xj pk ) = 2εilk pl pk = 0,
∂xl
where we have used εijk pj pk = 0 since pj pk = pk pj is symmetric. The second term is
proportional to Lz ,
n
~ · ~q, Lz
p~ × B
o
= {Bz Lz , Lz } = 0.
The final terms are proportional to q 2 , and q 2 , ~q × p~ = 0 for the same reason that
2
p , ~q × p~ = 0, leading to
{H, ~q × p~ } = 0.
Hence angular momentum is conserved away from collisions as well.
(g) Write down the most general form for the equilibrium distribution functions for particles
confined to a circularly symmetric potential.
• The most general form of the equilibrium distribution functions must set both the
collision terms, and the streaming terms to zero. Based on the results of the previous
parts, we thus obtain
fα = Aα exp [−βHα − γLz ] .
The collision terms allow for the possibility of a term −~u · p~ in the exponent, corresponding
to an average velocity. Such a term will not commute with the potential set up by a
stationary box, and is thus ruled out by the streaming terms. On the other hand, the
angular momentum does commute with a circular potential {V (~q), L} = 0, and is allowed
by the streaming terms. A non–zero γ describes the electron gas rotating in a circular
box.
(h) How is the result in part (g) modified by including scattering from magnetic and
non-magnetic impurities?
67
~ in collisions.
• Scattering from any impurity removes the conservation of p~, and hence L,
The γ term will no longer be needed. Scattering from magnetic impurities mixes populations of up and down spins, necessitating A↑ = A↓ ; non–magnetic impurities do not have
this effect.
(i) Do conservation of spin and angular momentum lead to new hydrodynamic equations?
• Conservation of angular momentum is related to conservation of p~, as shown in (f), and
hence does not lead to any new equation. In contrast, conservation of spin leads to an
additional hydrodynamic equation involving the magnetization, which is proportional to
(n↑ − n↓ ).
********
6. The Lorentz gas describes non-interacting particles colliding with a fixed set of scatterers. It is a good model for scattering of electrons from donor impurities. Consider a
uniform two dimensional density n0 of fixed impurities, which are hard circles of radius a.
(a) Show that the differential cross section of a hard circle scattering through an angle θ is
dσ =
θ
a
sin dθ,
2
2
and calculate the total cross section.
• Let b denote the impact parameter, which (see figure) is related to the angle θ between
p~ ′ and ~p by
68
π−θ
θ
= a cos .
2
2
The differential cross section is then given by
b(θ) = a sin
θ
dσ = 2|db| = a sin dθ.
2
Hence the total cross section
σtot =
Z
π
0
π
θ
θ
= 2a.
dθa sin = 2a − cos
2
2 0
(b) Write down the Boltzmann equation for the one particle density f (~q, ~p, t) of the Lorentz
gas (including only collisions with the fixed impurities). (Ignore the electron spin.)
• The corresponding Boltzmann equation is
∂f
p~ ∂f
~ · ∂f
+
·
+F
∂t
m ∂~q
∂~p
Z
Z
dσ
p|
no |~
p|
dσ |~
′
dθ
n0 [−f (~
p ) ) + f (~
p )] =
[f (~
p ′ ) − f (~
p )] ≡ C [f (~
p )] .
= dθ
dθ m
m
dθ
(c) Find the eigenfunctions and eigenvalues of the collision operator.
~ ≡ −∂U/∂~q, and
(d) Using the definitions F
Z
n(~q, t) = d2 ~pf (~q, ~p, t),
and
hg(~q, t)i =
1
n(~q, t)
Z
d2 p~f (~q, ~p, t)g(~q, t),
show that for any function χ(|~
p|), we have
∂
∂
p~
∂χ
~
.
(n hχi) +
· n
χ
=F · n
∂t
∂~q
m
∂~p
• Using the definitions F~ ≡ −∂U/∂~q,
Z
n(~q, t) = d2 ~pf (~q, ~p, t),
and
1
hg(~q, t)i =
n(~q, t)
Z
d2 p~f (~q, ~p, t)g(~q, t),
we can write
Z
p~ ∂f
∂f
dσ
|~
p
|
′
~·
d pχ(|~
p |) − ·
−F
+ dθ
no (f (~
p) − f (~
p ))
m ∂~q
∂~p
dθ m
∂
p~
∂χ
.
=−
· n
χ
+ F~ · n
∂~q
m
∂~p
d
(n hχ(|~
p |)i) =
dt
Z
2
69
Rewriting this final expression gives the hydrodynamic equation
∂
p~
∂χ
∂
~
.
(n hχi) +
· n
χ
=F · n
∂t
∂~q
m
∂~p
(e) Derive the conservation equation for local density ρ ≡ mn(~q, t), in terms of the local
velocity ~u ≡ h~p/mi.
• Using χ = 1 in the above expression
∂
∂
p~
= 0.
n+
· n
∂t
∂~q
m
In terms of the local density ρ = mn, and velocity ~u ≡ h~
p/mi, we have
∂
∂
ρ+
· (ρ~u ) = 0.
∂t
∂~q
(f) Since the magnitude of particle momentum is unchanged by impurity scattering, the
Lorentz gas has an infinity of conserved quantities |~
p|m . This unrealistic feature is removed
upon inclusion of particle–particle collisions. For the rest of this problem focus only on
p2 /2m as a conserved quantity. Derive the conservation equation for the energy density
ǫ(~q, t) ≡
ρ 2
c ,
2
where
~c ≡
p~
− ~u,
m
in terms of the energy flux ~h ≡ ρ ~c c2 /2, and the pressure tensor Pαβ ≡ ρ hcα cβ i.
• With the kinetic energy χ = p2 /2m as a conserved quantity, the equation found in (c)
gives
∂ n
∂
h~
p
i
n p~ p2
2
~· n
+
=F
.
|~
p|
·
∂t 2m
∂~q
2 mm
m
Substituting ~p/m = ~u + ~c , where h~c i = 0, and using ρ = nm,
i
∂ hρ
ρ
∂ hρ 2 ρ 2 i
2
2
+
c
(~u + ~c )(u + c + 2~u · ~c ) = F~ · ~u.
u +
·
∂t 2
2
∂~q 2
m
From the definition ε = ρ c2 /2, we have
i
i
∂ hρ 2
ρ
∂ hρ
~uu2 + ~u c2 + ~cc2 + 2~u · h~c ~c i = F~ · ~u.
u +ε +
·
∂t 2
∂~q 2
m
70
Finally, by substituting ~h ≡ ρ ~c c2 /2 and Pαβ = ρ hcα cβ i , we get
i
i
∂ hρ 2
∂ h ρ 2
∂
ρ
u +ε +
· ~u
u + ε + ~h +
(uβ Pαβ ) = F~ · ~u.
∂t 2
∂~q
2
∂qα
m
(g) Starting with a one particle density
1
p2
,
f (~
p, ~q, t) = n(~q, t) exp −
2mkB T (~q, t) 2πmkB T (~q, t)
0
reflecting local equilibrium conditions, calculate ~u, ~h, and Pαβ . Hence obtain the zeroth
order hydrodynamic equations.
• There are only two quantities, 1 and p2 /2m, conserved in collisions. Let us start with
the one particle density
p2
1
f (~
p, ~q, t) = n(~q, t) exp −
.
2mkB T (~q, t) 2πmkB T (~q, t)
0
Then
~u =
p~
m
= 0,
and
0
~h =
p~ p2
mm
0
ρ
= 0,
2
o
since both are odd functions of p~, while f is an even function of p~, while
Pαβ = ρ hcα cβ i =
n
n
hpα pβ i = δαβ · mkB T = nkB T δαβ .
m
m
Substituting these expressions into the results for (c) and (d), we obtain the zeroth–order
hydrodynamic equations

∂ρ


= 0,
∂t

 ∂ ε = ∂ ρ c2 = 0.
∂t
∂t 2
The above equations imply that ρ and ε are independent of time, i.e.
ρ = n(~q ),
or
and
ε = kB T (~q ),
p2
n(~q )
.
exp −
f =
2πmkB T (~q )
2mkB T (~q )
0
71
(h) Show that in the single collision time approximation to the collision term in the Bolzmann equation, the first order solution is
"
!#
2
~
p~
p
∂ln ρ ∂ln T
∂T
F
f 1 (~
p, ~q, t) = f 0 (~
p, ~q, t) 1 − τ ·
.
−
+
−
2
m
∂~q
∂~q
2mkB T ∂~q
kB T
• The single collision time approximation is
f0 − f
.
C [f ] =
τ
The first order solution to Boltzmann equation
f = f 0 (1 + g) ,
is obtained from
as
Noting that
f 0g
L f0 = −
,
τ
∂
p~ ∂
∂
1 0
0
0
0
~
ln f +
·
ln f + F ·
ln f .
g = −τ 0 L f = −τ
f
∂t
m ∂~q
∂~p
ln f 0 = −
p2
+ ln n − ln T − ln(2πmkB ),
2mkB T
where n and T are independent of t, we have ∂ ln f 0 /∂t = 0, and
2
p
~
∂n
1
∂T
p
1
∂T
−~
p
+
·
−
+
g = −τ F~ ·
mkB T
m n ∂~q
T ∂~q
2mkB T 2 ∂~q
!)
(
~
F
1 ∂T
p2
1 ∂ρ
∂T
p~
.
·
−
+
−
= −τ
m
ρ ∂~q
T ∂~q
2mkB T 2 ∂~q
kB T
(i) Show that using the first order expression for f , we obtain
h
i
ρ~u = nτ F~ − kB T ∇ ln (ρT ) .
R
d2 qf 0 (1 + g) = d2 qf 0 = n, and
Dp E 1 Z
pα
α
=
d2 p f 0 (1 + g)
uα =
m
n
m
Z
pβ ∂ ln ρ ∂ ln T
1
Fβ
p2
∂T
2 pα
d p
−τ
f 0.
=
−
−
+
2
n
m
m
∂qβ
∂qβ
kB T
2mkB T ∂qβ
• Clearly
R
72
Wick’s theorem can be used to check that
hpα pβ i0 = δαβ mkB T,
p2 pα pβ
0
= (mkB T )2 [2δαβ + 2δαβ ] = 4δαβ (mkB T )2 ,
resulting in
ρ
∂
1
∂T
nτ
δαβ kB T
ln
−
δαβ .
Fβ + 2kB
uα = −
ρ
∂qβ
T
kB T
∂qβ
Rearranging these terms yields
∂
ln (ρT ) .
ρuα = nτ Fα − kB T
∂qα
(j) From the above equation, calculate the velocity response function χαβ = ∂uα /∂Fβ .
• The velocity response function is now calculated easily as
χαβ =
∂uα
nτ
=
δαβ .
∂Fβ
ρ
(k) Calculate Pαβ , and ~h, and hence write down the first order hydrodynamic equations.
• The first order expressions for pressure tensor and heat flux are
ρ
hpα pβ i = δαβ nkB T,
and
δ 1 Pαβ = 0,
m2
E
τρ D
ρ
2 pi
2
2
ai + bi p
pα p = − 3 pα p
.
hα =
2m3
2m
m
0
Pαβ =
The latter is calculated from Wick’s theorem results
2
pi pα p2 = 4δαi (mkB T ) ,
and
3
pi pα p2 p2 = (mkB T ) [δαi (4 + 4) + 4 × 2δαi + 4 × 2δαi ] = 24δαi ,
as
∂
ρ
24(mkB T )3 ∂
ρτ
2
~
ln − F +
ln T
hα = − 3 (mkB T )
2m
∂qα T
2mkb T ∂qα
∂T
2
= −12nkB
Tτ
.
∂qα
Substitute these expressions for Pαβ and hα into the equation obtained in (e)
i
∂ hρ 2
∂
∂
ρ 2
∂T
ρ
2
+
u +ǫ +
· ~u
u + ǫ − 11nkB T τ
(~u nkB T ) = F~ · ~u.
∂t 2
∂~q
2
∂~q
∂~q
m
73
********
7. Thermal conductivity:
Consider a classical gas between two plates separated by a
distance w. One plate at y = 0 is maintained at a temperature T1 , while the other plate at
y = w is at a different temperature T2 . The gas velocity is zero, so that the initial zeroth
order approximation to the one particle density is,
f10 (~
p, x, y, z)
=
p~ · p~
.
exp −
2mkB T (y)
n(y)
3/2
[2πmkB T (y)]
(a) What is the necessary relation between n(y) and T (y) to ensure that the gas velocity ~u
remains zero? (Use this relation between n(y) and T (y) in the remainder of this problem.)
• Since there is no external force acting on the gas between plates, the gas can only flow
locally if there are variations in pressure. Since the local pressure is P (y) = n(y)kB T (y),
the condition for the fluid to be stationary is
n(y)T (y) = constant.
(b) Using Wick’s theorem, or otherwise, show that
p2
0
0
≡ hpα pα i = 3 (mkB T ) ,
0
p4
and
0
2
≡ hpα pα pβ pβ i = 15 (mkB T ) ,
0
where hOi indicates local averages with the Gaussian weight f10 . Use the result p6
0
=
105(mkB T )3 in conjunction with symmetry arguments to conclude
p2y p4
0
3
= 35 (mkB T ) .
0
• The Gaussian weight has a covariance hpα pβ i = δαβ (mkB T ). Using Wick’s theorem
gives
p2
0
0
= hpα pα i = (mkB T ) δαα = 3 (mkB T ) .
Similarly
p4
0
0
2
2
= hpα pα pβ pβ i = (mkB T ) (δαα + 2δαβ δαβ ) = 15 (mkB T ) .
74
The symmetry along the three directions implies
p2x p4
0
= p2y p4
0
= p2z p4
0
=
1 2 4
p p
3
0
=
1
3
3
× 105 (mkB T ) = 35 (mkB T ) .
3
(c) The zeroth order approximation does not lead to relaxation of temperature/density
variations related as in part (a). Find a better (time independent) approximation f11 (~
p, y),
by linearizing the Boltzmann equation in the single collision time approximation, to
L f11 ≈
∂
f 1 − f10
py ∂
f10 ≈ − 1
,
+
∂t
m ∂y
τK
where τK is of the order of the mean time between collisions.
• Since there are only variations in y, we have
∂
py ∂
3
p2
3
0
0 py
0
0 py
f = f1 ∂y ln f1 = f1 ∂y ln n − ln T −
+
− ln (2πmkB )
∂t
m ∂y 1
m
m
2
2mkB T
2
2
5
∂y T
3 ∂y T
p
p2
∂T
0 py
0 py ∂y n
= f1
− +
−
+
,
= f1
m
n
2 T
2mkB T T
m
2 2mkB T
T
where in the last equality we have used nT = constant to get ∂y n/n = −∂y T /T . Hence
the first order result is
f11 (~
p, y)
=
f10 (~
p, y)
py
p2
5 ∂y T
1 − τK
.
−
m 2mkB T
2
T
(d) Use f11 , along with the averages obtained in part (b), to calculate hy , the y component
of the heat transfer vector, and hence find K, the coefficient of thermal conductivity.
• Since the velocity ~u is zero, the heat transfer vector is
1
n
mc2
=
py p2
hy = n cy
2
2m2
1
.
In the zeroth order Gaussian weight all odd moments of p have zero average. The corrections in f11 , however, give a non-zero heat transfer
n ∂y T
hy = −τK
2m2 T
py
m
75
p2
5
−
2mkB T
2
2
py p
0
.
Note that we need the Gaussian averages of p2y p4
0
0
and p2y p2 . From the results of part
(b), these averages are equal to 35(mkB T )3 and 5(mkB T )2 , respectively. Hence
n ∂y T
2
hy = −τK
(mkB T )
3
2m T
35 5 × 5
−
2
2
=−
2
5 nτK kB
T
∂y T.
2
m
The coefficient of thermal conductivity relates the heat transferred to the temperature
gradient by ~h = −K∇T , and hence we can identify
K=
2
5 nτK kB
T
.
2
m
(e) What is the temperature profile, T (y), of the gas in steady state?
• Since ∂t T is proportional to −∂y hy , there will be no time variation if hy is a constant.
But hy = −K∂y T , where K, which is proportional to the product nT , is a constant in
the situation under investigation. Hence ∂y T must be constant, and T (y) varies linearly
between the two plates. Subject to the boundary conditions of T (0) = T1 , and T (w) = T2 ,
this gives
T (y) = T1 +
T2 − T1
y.
w
********
8. Zeroth-order hydrodynamics: The hydrodynamic equations resulting from the conservation of particle number, momentum, and energy in collisions are (in a uniform box):

∂t n + ∂α (nuα ) = 0





1
∂β Pαβ
∂ t uα + uβ ∂ β uα = −
,
mn




 ∂t ε + uα ∂α ε = − 1 ∂α hα − 1 Pαβ uαβ
n
n
where n is the local density, ~u = h~p/mi, uαβ = (∂α uβ + ∂β uα ) /2, and ε = mc2 /2 , with
~c = p~/m − ~u.
(a) For the zeroth order density
f10 (~
p, ~q, t) =
n(~q, t)
3/2
(2πmkB T (~q, t))
0
"
2
(~
p − m~u(~q, t))
exp −
2mkB T (~q, t)
#
,
0
0
calculate the pressure tensor Pαβ
= mn hcα cβ i , and the heat flux h0α = nm cα c2 /2 .
76
• The PDF for ~c is proportional to the Gaussian exp −mc2 /(2kB T ) , from which we
immediately get
0
hcα cβ i =
kB T
δαβ
m
=⇒
0
Pαβ
= nkB T δαβ ,
and ε =
3
kB T.
2
All odd expectation values of the symmetric weight are zero, specifically h~ci = 0, and
D E
~h0 = 0.
(b) Obtain the zeroth order hydrodynamic equations governing the evolution of n(~
q, t),
~u(~q, t), and T (~q, t).
•
Substituting
0
Pαβ
equations gives:
D E
= nkB T δαβ , ε = 3kB T /2, and ~h0 = 0 in the hydrodynamic

∂t n + uα ∂α n = −n∂α uα





kB
∂α (nT )
∂ t uα + uβ ∂ β uα = −
mn




 3 (∂t T + uα ∂α T ) = −T ∂α uα
2
.
(c) Show that the above equations imply Dt ln nT −3/2 = 0, where Dt = ∂t + uβ ∂β is the
material derivative along streamlines.
• Using Dt = ∂t + uβ ∂β , the above equations can be written as

Dt ln n = −∂α uα





kB
Dt uα = −
∂α (nT )
mn




 3 Dt ln T = −∂α uα
2
.
Eliminating ∂α uα between the first and third equations gives the required result of
Dt ln nT −3/2 = 0.
R
(d) Write down the expression for the function H0 (t) = d3 ~qd3 p~f10 (~
p, ~q, t) ln f10 (~
p, ~q, t),
after performing the integrations over p~, in terms of n(~q, t), ~u(~q, t), and T (~q, t).
• Using the expression for f10 ,
H0 (t) =
Z
d3 ~qd3 p~
n
3/2
"
exp −
(~
p − m~u)
2mkB T
2
#
(2πmkB T )
"
#.
2
3
(~
p
−
m~
u
)
× ln nT −3/2 − ln (2πmkB ) −
2
2mkB T
77
The Gaussian averages over ~p are easily performed to yield
Z
3
3
0
3
−3/2
− ln (2πmkB ) −
H (t) = d ~q n ln nT
.
2
2
(e) Using the hydrodynamic equations in (b) calculate dH0 /dt.
• Taking the time derivative inside the integral gives
Z
h
i
dH0
3
−3/2
−3/2
= d ~q ∂t n ln nT
+ n∂t ln nT
.
dT
Use the results of parts (b) and (c) to substitute for ∂t n and ∂t ln nT −3/2 , to get
dH0
=−
dT
=−
Z
Z
i
h −3/2
−3/2
∂α (nuα ) + nuα ∂α ln nT
d ~q ln nT
3
h
i
d3 ~q ∂α nuα ln nT −3/2 = 0,
since the integral of a complete derivative is zero.
(f) Discuss the implications of the result in (e) for approach to equilibrium.
• The expression for −H0 is related to the entropy of the gas. The result in (f) implies
that the entropy of the gas does not change if its n ~u, and T vary according to the zeroth
order equations. The corrections due to first order hydrodynamics are necessary in order
to describe the increase in entropy.
********
9. Viscosity:
Consider a classical gas between two plates separated by a distance w.
One plate at y = 0 is stationary, while the other at y = w moves with a constant velocity
vx = u. A zeroth order approximation to the one particle density is,
1
n
0
2
2
2
exp −
f1 (~
p, ~q) =
(px − mαy) + py + pz ,
3/2
2mkB T
(2πmkB T )
obtained from the uniform Maxwell–Boltzmann distribution by substituting the average
value of the velocity at each point. (α = u/w is the velocity gradient.)
(a) The above approximation does not satisfy the Boltzmann equation as the collision
term vanishes, while df10 /dt 6= 0. Find a better approximation, f11 (~
p), by linearizing the
Boltzmann equation, in the single collision time approximation, to
1
f 1 − f10
p~ ∂
∂
f10 ≈ − 1
,
+
·
L f1 ≈
∂t m ∂~q
τ×
78
where τ× is a characteristic mean time between collisions.
• We have
p~ ∂
∂
+
·
∂t m ∂~q
whence
f11
=
f10
f10 =
1 − τx
α
py (px − mαy)f10 ,
mkB T
α
py (px − mαy) .
mkB T
(b) Calculate the net transfer Πxy of the x component of the momentum, of particles
passing through a plane at y, per unit area and in unit time.
• The transfer of x-momentum in the y direction, across a plane at y, per unit area and
per unit time, is calculated as
Z
Z
(−py )
3 py
1
Πxy =
d p px f1 (y) −
d3 p
px f11 (y)
m
m
p >0
py <0
Z y
py
= d3 p px f11 (y)
m
Z
τx α
3 py
py (px − mαy)f10
= d p px −
m
mkB T
 


p2y
(px −mαy)2

Z
Z
 

exp − 2mkB T
exp − 2mkB T 
τx αn 
2
2
√
dpx (px − mαy)
=− 2
·
dpy py √
 
m kB T 
2πmkB T
2πmkB T 


=−
τx αn
(mkB T )2 = −αnτx kB T.
m2 kB T
(c) Note that the answer to (b) is independent of y, indicating a uniform transverse force
Fx = −Πxy , exerted by the gas on each plate. Find the coefficient of viscosity, defined by
η = Fx /α.
• From part (b),
η=
Fx
= nτx kB T.
a
********
10. Light and matter:
In this problem we use kinetic theory to explore the equilibrium
between atoms and radiation.
(a) The atoms are assumed to be either in their ground state a0 , or in an excited state a1 ,
which has a higher energy ε. By considering the atoms as a collection of N fixed two-state
79
systems of energy E (i.e. ignoring their coordinates and momenta), calculate the ratio
n1 /n0 of densities of atoms in the two states as a function of temperature T .
• The energy and temperature of a two-state system are related by
E=
Nǫ
,
1 + exp (ǫ/kB T )
leading to
n0 =
N exp (ǫ/kB T )
N − E/ǫ
=
,
V
V 1 + exp (ǫ/kB T )
so that
and
n1 =
E/ǫ
N
1
=
,
V
V 1 + exp (ǫ/kB T )
n1
ǫ
= exp −
.
n0
kB T
Consider photons γ of frequency ω = ε/h̄ and momentum |~
p | = h̄ω/c, which can
interact with the atoms through the following processes:
(i) Spontaneous emission: a1 → a0 + γ.
(ii) Adsorption: a0 + γ → a1 .
(iii) Stimulated emission: a1 + γ → a0 + γ + γ.
Assume that spontaneous emission occurs with a probability σsp , and that adsorption
and stimulated emission have constant (angle-independent) differential cross-sections of
σad /4π and σst /4π, respectively.
(b) Write down the Boltzmann equation governing the density f of the photon gas, treating
the atoms as fixed scatterers of densities n0 and n1 .
• The Boltzmann equation for photons in the presence of fixed scatterers reads
∂f
∂f
+ p~ ·
= −σad n0 cf + σst n1 cf + σsp n1 .
∂t
∂~q
(c) Find the equilibrium density feq. for the photons of the above frequency.
• In uniform equilibrium, the left-hand side vanishes, leaving
−σad n0 cfeq. + σst n1 cfeq. + σsp n1 = 0,
i.e.
feq. =
1
σsp
σsp
1
=
.
c σad n0 /n1 − σst
c σad exp (ǫ/kB T ) − σst
80
(d) According to Planck’s law, the density of photons at a temperature T depends on their
frequency ω as feq. = [exp (h̄ω/kB T ) − 1]
−1
/h3 . What does this imply about the above
cross sections?
• The result of part (c) agrees with Planck’s law if
σad = σst ,
and σsp =
c
σst ,
h3
a conclusion first reached by Einstein, and verified later with explicit quantum mechanical
calculations of cross-sections.
(e) Consider a situation in which light shines along the x axis on a collection of atoms
whose boundary coincides with the x = 0 plane, as illustrated in the figure.
γ
x
matter (n0 , n1 )
vacuum
Clearly, f will depend on x (and px ), but will be independent of y and z. Adapt the
Boltzmann equation you propose in part (b) to the case of a uniform incoming flux of
photons with momentum p~ = h̄ω x̂/c. What is the penetration length across which the
incoming flux decays?
• In this situation, the Boltzmann equation reduces to
h
∂f
n1 i
px
= σst c (n1 − n0 ) f + 3 θ (x) .
∂x
h
To the uniform solution obtained before, one can add an exponentially decaying term for
x > 0, i.e.
f (px , x > 0) = A (px ) e−ax/px + feq. (px ) .
The constant A (px ) can be determined by matching to solution for x < 0 at x = 0, and is
related to the incoming flux. The penetration depth d is the inverse of the decay parameter,
and given by
d=
px
,
a
with
a = σst c (n0 − n1 ) > 0.
81
********
11. Equilibrium density: Consider a gas of N particles of mass m, in an external potential
U (~q ). Assume that the one body density ρ1 (~p, ~q, t), satisfies the Boltzmann equation. For
a stationary solution, ∂ρ1 /∂t = 0, it is sufficient from Liouville’s theorem for ρ1 to satisfy
ρ1 ∝ exp −β p2 /2m + U (~q ) . Prove that this condition is also necessary by using the
H-theorem as follows.
(a) Find ρ1 (~
p, ~q ) that minimizes H = N
R
d3 p~d3 ~qρ1 (~
p, ~q ) ln ρ1 (~
p, ~q), subject to the con-
straint that the total energy E = hHi is constant. (Hint: Use the method of Lagrange
multipliers to impose the constraint.)
• Using Lagrange multipliers to impose the constraints, hHi = E and
minimizing H with the given constraints reduces to minimizing,
N
Z
R
d3 p~d3 ~q ρ1 = 1,
d3 ~pd3 ~q(ρ1 ln ρ1 + βρ1 H + αρ1 ) − βE − αN.
Differentiating with respect to α, β, and the function ρ1 we get,
N
Z
Z
3
3
d p~d ~q ρ1 = N
3
3
d ~pd ~q ρ1 H = E
→
→
Z
Z
d3 p~d3 ~q ρ1 = 1,
d3 p~d3 ~q ρ1 H = E/N,
ln ρ1 + βH + α = 0 → ρ1 = exp(−βH − α),
respectively. Hence we conclude,
ρ1 = R
where β is determined by,
R
exp(−βH)
d3 p~d3 ~q exp(−βH)
,
d3 ~pd3 ~q H exp(−βH)
E
R
= .
3
3
N
d ~pd ~q exp(−βH)
(a)
(b) For a mixture of two gases (particles of masses ma and mb ) find the distributions ρ1
(b)
and ρ1 that minimize H = H(a) + H(b) subject to the constraint of constant total energy.
Hence show that the kinetic energy per particle can serve as an empirical temperature.
82
• If we have Na and Nb of each particle type with total energy E, then H is minimized
with the total energy constraint by extremizing,
Z
(a)
(a)
(b)
(b)
(a)
(b)
d3 p~d3 ~q(Na ρ1 ln ρ1 + Nb ρ1 ln ρ1 + β(Na Ha ρ1 + Nb Hb ρ1 )
.
(a)
′ (b)
′
+Na αρ1 + Nb α ρ1 ) − βE − αNa − α Nb
(a)
(b)
Differentiating this expression with respect to α, α′ , β, ρ1 , and ρ1 , we get,
Z
(a)
d3 p~d3 ~q ρ1 = 1,
Z
(b)
d3 p~d3 ~q ρ1 = 1,
Z
(a)
(b)
3
3
d p~d ~q Na Ha ρ1 + Nb Hb ρ1 = E,
(a)
ln ρ1 + βHa + α = 0,
(b)
ln ρ1 + βHb + α′ = 0.
So we get,
(a)
ρ1 = R
(a)
ρ2 = R
exp(−βHa )
3
d ~pd3 ~q exp(−βH
a)
,
exp(−βHb )
.
d3 ~pd3 ~q exp(−βHb )
where β is obtained by,
R 3 3
R 3 3
d p~d ~qHa exp(−βHa )
d p~d ~qHb exp(−βHb )
Na R 3 3
+ Nb R 3 3
= E.
d p~d ~q exp(−βHa )
d p~d ~q exp(−βHb )
Note that β is a value defined for both gases a and b, and hence can serve as an empirical
temperature.
For the specific case of
Ha =
p2
+ Ua (~q),
2ma
Hb =
p2
+ Ub (~q),
2ma
the kinetic energy per particle in a distribution with equal β is also equal, since
R 3 3 p2
R 3
R
p2
d p~d ~q 2ma exp −β(p2 /2ma + Ua (~q))
exp(−βp2 /2ma )
d ~q exp(−βUa ) d3 p~ 2m
a
R
R
R
=
d3 p~d3 ~q exp(−βHa )
d3 ~q exp(−βUa ) d3 p~ exp(−βp2 /2ma )
R∞
p4
4π 0 dp 2m
exp(−βp2 /2ma )
a
.
R
=
∞
4π 0 dp p2 exp(−βp2 /2ma )
R∞
2
3
1 0 dt t4 e−t
=
= R∞
2
β 0 dt t2 e−t
2β
83
So we see that the kinetic energy per particle for the gas can also serve as an empirical
temperature in this case.
********
12. Moments of momentum: Consider a gas of N classical particles of mass m in thermal
equilibrium at a temperature T , in a box of volume V .
(a) Write down the equilibrium one particle density feq. (~p, ~q ), for coordinate ~q, and momentum p~.
• The equilibrium Maxwell-Boltzmann distribution reads
f (~
p, ~q ) =
n
(2πmkB T )
3/2
exp −
D
p2
2mkB T
(b) Calculate the joint characteristic function, exp −i~k · p~
• Performing the Gaussian average yields
E
.
, for momentum.
D
E
mkB T 2
−i~
k·~
p
~
k .
p̃ k = e
= exp −
2
n
(c) Find all the joint cumulants pℓx pm
y pz c .
• The cumulants are calculated from the characteristic function, as
n
pℓx pm
y pz c
=
∂
∂ (−ikx )
ℓ ∂
∂ (−iky )
m ∂
∂ (−ikz )
n
ln p̃ ~k
= mkB T (δℓ2 δm0 δn0 + δℓ0 δm2 δn0 + δℓ0 δm0 δn2 ) ,
i.e., there are only second cumulants; all other cumulants are zero.
(d) Calculate the joint moment hpα pβ (~
p · p~ )i.
• Using Wick’s theorem
hpα pβ (~p · p~ )i = hpα pβ pγ pγ i
= hpα pβ i hpγ pγ i + 2 hpα pγ i hpβ pγ i
2
2
= (mkB T ) δαβ δγγ + 2 (mkB T ) δαγ δβγ
2
= 5 (mkB T ) δαβ .
84
~
k=0
Alternatively, directly from the characteristic function,
∂
∂
∂
∂
p̃ ~k
∂ (−ikα ) ∂ (−ikβ ) ∂ (−ikγ ) ∂ (−ikγ )
~
k=0
i
h
mkB T ~ 2
∂
∂
2
3mkB − (mkB T ) ~k 2 e− 2 k
=
∂ (−ikα ) ∂ (−ikβ )
hpα pβ (~
p · p~ )i =
~
k=0
2
= 5 (mkB T ) δαβ .
********
13. Generalized ideal gas: Consider a gas of N particles confined to a box of volume V
in d-dimensions. The energy, ǫ, and momentum, p, of each particle are related by ǫ = ps ,
where p = |p|. (For classical particles s = 2, while for highly relativistic ones s = 1.) Let
f (v)dv denote the probability of finding particles with speeds between v and v + dv, and
n = N/V .
(a) Calculate the number of impacts of gas molecules per unit area of the wall of the box,
and per unit time as follows:
(i) Show that the number of particles hitting area A in a time dt arriving from a
~ with a speed v, is proportional to A · vdt cos θ · nf (v)dv ,
specific direction Ω,
~ and the normal to the wall.
where θ is the angle between the direction Ω
~ with the same polar angle θ, demonstrate that
(ii) Summing over all directions Ω
dN (θ, v) = A · vdt cos θ · nf (v)dv ·
Sd−1 sind−2 θ dθ
Sd
,
where Sd = 2π d/2 /(d/2 − 1)! is the total solid angle in d dimensions.
(iii) By averaging over v and θ show that
Sd−1
N
· nv ,
=
A dt
(d − 1)Sd
where
v=
Z
vf (v)dv
is the average speed.
(b) Each (elastic) collision transfers a momentum 2p cos θ to the wall. By examining the
net force on an element of area prove that the pressure P equals
s
d
.
E
V ,
where E is the
average (kinetic) energy. (Note that the velocity v is not p/m but ∂ǫ/∂p.) Hint: Clearly
upon averaging over all directions cos2 θ = 1/d.
(c) Using thermodynamics and the result in (b) show that along an adiabatic curve P V γ
is constant, and calculate γ.
85
(d) According to the equipartition theorem, each degree of freedom which appears quadratically in the energy has an energy kB T /2. Calculate the value of γ if each gas particle has
ℓ such quadratic degrees of freedom in addition to its translational motion. What values
of γ are expected for helium and hydrogen at room temperature?
(e) Consider the following experiment to test whether the motion of ants is random. 250
ants are placed inside a 10cm × 10cm box. They cannot climb the wall, but can escape
through an opening of size 5mm in the wall. If the motion of ants is indeed random, and
they move with an average speed of 2mms−1 , how may are expected to escape the box in
the first 30 seconds?
********
14. Effusion:
A box contains a perfect gas at temperature T and density n.
(a) What is the one-particle density, ρ1 (~v), for particles with velocity ~v ?
A small hole is opened in the wall of the box for a short time to allow some particles
to escape into a previously empty container.
(b) During the time that the hole is open what is the flux (number of particles per unit time
and per unit area) of particles into the container? (Ignore the possibility of any particles
returning to the box.)
(c) Show that the average kinetic energy of escaping particles is 2kB T . (Hint: calculate
contributions to kinetic energy of velocity components parallel and perpendicular to the
wall separately.)
(d) The hole is closed and the container (now thermally insulated) is allowed to reach
equilibrium. What is the final temperature of the gas in the container?
(e) A vessel partially filled with mercury (atomic weight 201), and closed except for a hole
of area 0.1mm2 above the liquid level, is kept at 00 C in a continuously evacuated enclosure.
After 30 days it is found that 24mg of mercury has been lost. What is the vapor pressure
of mercury at 00 C?
********
15. Adsorbed particles: Consider a gas of classical particles of mass m in thermal equilibrium at a temperature T , and with a density n. A clean metal surface is introduced into
the gas. Particles hitting this surface with normal velocity less than vt are reflected back
into the gas, while particles with normal velocity greater than vt are absorbed by it.
86
(a) Find the average number of particles hitting one side of the surface per unit area and
per unit time.
(b) Find the average number of particles absorbed by one side of the surface per unit area
and per unit time.
********
16. Electron emission:
When a metal is heated in vacuum, electrons are emitted from
its surface. The metal is modeled as a classical gas of noninteracting electrons held in the
solid by an abrupt potential well of depth φ (the work function) relative to the vacuum.
(a) What is the relationship between the initial and final velocities of an escaping electron?
(b) In thermal equilibrium at temperature T , what is the probability density function for
the velocity of electrons?
(c) If the number density of electrons is n, calculate the current density of thermally
emitted electrons.
********
87
Problems for Chapter IV - Classical Statistical Mechanics
1. Classical harmonic oscillators:
Consider N harmonic oscillators with coordinates and
momenta {qi , pi }, and subject to a Hamiltonian
N 2
X
pi
mω 2 qi2
.
+
H({qi , pi }) =
2m
2
i=1
(a) Calculate the entropy S, as a function of the total energy E.
(Hint: By appropriate change of scale, the surface of constant energy can be deformed
into a sphere. You may then ignore the difference between the surface area and volume
for N ≫ 1. A more elegant method is to implement this deformation through a canonical
transformation.)
• The volume of accessible phase space for a given total energy is proportional to
Z
1
dq1 dq2 · · · dqN dp1 dp2 · · · dpN ,
Ω= N
h
H=E
where the integration is carried out under the condition of constant energy,
N 2
X
pi
mω 2 qi2
.
E = H ({qi , pi }) =
+
2m
2
i=1
Note that Planck’s constant h, is included as a measure of phase space volume, so as to
make the final result dimensionless.
The surface of constant energy is an ellipsoid in 2N dimensions, whose area is difficult
to calculate. However, for N → ∞ the difference between volume and area is subleading in
N , and we shall instead calculate the volume of the ellipsoid, replacing the constraint H =
E by H ≤ E. The ellipsoid can be distorted into a sphere by a canonical transformation,
changing coordinates to
qi′ ≡
√
mωqi ,
and
pi
.
p′i ≡ √
mω
The Hamiltonian in this coordinate system is
E=
H ({qi′ , p′i })
N
ω X ′2
pi + qi′2 .
=
2 i=1
88
Since the canonical transformation preserves volume in phase space (the Jacobian is unity),
we have
Z
1
′
dq1′ · · · dqN
dp′1 · · · dp′N ,
Ω≈ N
h
H≤E
where the integral is now over the 2N -dimensional (hyper-) sphere of radius R =
As the volume of a d-dimensional sphere of radius R is Sd Rd /d, we obtain
N
=
S ≡ kB ln Ω ≈ N kB ln
2πeE
N hω
2π N
1
Ω≈
·
(N − 1)! 2N
2E
hω
2πE
hω
N
p
2E/ω.
1
.
N!
The entropy is now given by
.
(b) Calculate the energy E, and heat capacity C, as functions of temperature T , and N .
• From the expression for temperature,
1
∂S
≡
T
∂E
N
≈
N kB
,
E
we obtain the energy
E = N kB T,
and the heat capacity
C = N kB .
(c) Find the joint probability density P (p, q) for a single oscillator. Hence calculate the
mean kinetic energy, and mean potential energy for each oscillator.
• The single particle distribution function is calculated by summing over the undesired
coordinates and momenta of the other N − 1 particles. Keeping track of the units of h
used to make phase space dimensionless, gives
p(p1 , q1 )dp1 dq1 =
1
dq2 · · · dqN dp2 · · · dpN
(H≤EN −1 ) hN −1
R
1
dq1 dq2 · · · dqN dp1 dp2 · · · dpN
(H≤E) hN
R
89
×
dp1 dq1
,
h
where EN−1 = E − p21 /2m − mω 2 q12 /2. Using the results from part (a),
p(p1 , q1 ) =
=
=
Ω (N − 1, EN−1 )
hΩ(N, E)
2 N−1 π N −1
hω
ω N
2π E
p2
2
1
E − 2m
− mω
q12
(N−1)!
2
2 N πN N
h hω
N! E
!N−1
p21
mω 2 2
+
q
1
2
.
1 − 2m
E
N−1
Using the approximation (N − 1) ∼ N , for N ≫ 1, and setting E = N kB T , we have
ω N
p(p1 , q1 ) =
2π N kB T
1−
!N
2
2
+ mω
q
1
2
N kB T
!
2
2
+ mω
q
1
2
.
kB T
p21
2m
p2
1
ω
≈
exp − 2m
2πkB T
Let us denote (p1 , q1 ) by (p, q), then
ω
p2
mω 2 q 2
p(p, q) =
,
exp −
−
2πkB T
2mkB T
2kB T
is a properly normalized product of two Gaussians. The mean kinetic energy is
p2
2m
=
Z
p(p, q)
p2
kB T
dqdp =
,
2m
2
p(p, q)
mω 2 q 2
kB T
dqdp =
.
2
2
while the mean potential energy is also
mω 2 q 2
2
=
Z
********
2. Quantum harmonic oscillators: Consider N independent quantum oscillators subject
to a Hamiltonian
N
X
1
h̄ω ni +
H ({ni }) =
,
2
i=1
where ni = 0, 1, 2, · · ·, is the quantum occupation number for the ith oscillator.
90
(a) Calculate the entropy S, as a function of the total energy E.
(Hint: Ω(E) can be regarded as the number of ways of rearranging M =
N − 1 partitions along a line.)
P
i
ni balls, and
• The total energy of the set of oscillators is
N
X
N
ni +
2
i=1
E = h̄ω
!
.
Let us set the sum over the individual quantum numbers to
M≡
N
X
ni =
i=1
E
N
− .
h̄ω
2
The number of configurations {ni } for a given energy (thus for a given value of M ) is
equal to the possible number of ways of distributing M energy units into N slots, or of
partitioning M particles using N − 1 walls. This argument gives to the number of states
as
Ω=
(M + N − 1)!
,
M ! (N − 1)!
and a corresponding entropy
M
M +N −1
N −1
− M ln
.
S = kB ln Ω ≈ kB (M + N − 1) ln
− (N − 1) ln
e
e
e
(b) Calculate the energy E, and heat capacity C, as functions of temperature T , and N .
• The temperature is calculated by
1
∂S
≡
T
∂E
N
kB
≈
ln
h̄ω
M +N −1
M
kB
=
ln
h̄ω
E
h̄ω
+
E
h̄ω
N
2
−
−1
N
2
!
kB
ln
≈
h̄ω
E+
E−
By inverting this equation, we get the energy
N
exp (h̄ω/kB T ) + 1
1
1
E = h̄ω
,
= N h̄ω
+
2
exp (h̄ω/kB T ) − 1
2 exp (h̄ω/kB T ) − 1
and a corresponding heat capacity
∂E
C≡
∂T
= N kB
N
h̄ω
kB T
2
91
exp (h̄ω/kB T )
2.
[exp (h̄ω/kB T ) − 1]
N
h̄ω
2
N
h̄ω
2
!
.
(c) Find the probability p(n) that a particular oscillator is in its nth quantum level.
• The probability that a particular oscillator is in its nth quantum level is given by
summing the joint probability over states for all the other oscillators, i.e.
P
X
Ω N − 1, E − (n + 12 )h̄ω
(E−(n+1/2)h̄ω) 1
P
=
p(n) =
p(ni ) =
Ω(N, E)
(E) 1
{ni6=1 }
[(M − n) + N − 2]! M ! · (N − 1)!
·
(M − n)! · (N − 2)! (M + N − 1)!
M (M − 1) · · · (M − n + 1) · N
≈
(M + N − 1)(M + N − 2) · · · (M + N − n − 1)
=
≈ N (M + N )−n−1 M n ,
where the approximations used are of the from (I − 1) ≈ I, for I ≫ 1. Hence,
p(n) = N
E
N
−
+N
h̄ω
2
N
= E
h̄ω +
N
2
E
h̄ω
E
h̄ω
−n−1 N
2
N
2
−
+
!n
N
E
−
h̄ω
2
n
,
which using
1
kB
=
ln
T
h̄ω
E+
E−
Nh̄ω
2
Nh̄ω
2
!
,
E
h̄ω
E
h̄ω
=⇒
+
−
N
2
N
2
!n
nh̄ω
,
= exp −
kB T
leads to the probability
h̄ω
h̄ω
p(n) = exp −n
1 − exp −
.
kB T
kB T
(d) Comment on the difference between heat capacities for classical and quantum oscillators.
• As found in part (b),
Cquantum = N kB
h̄ω
kB T
2
h
h̄ω
kB T
exp
i2 .
h̄ω
exp kB T − 1
In the high temperature limit, h̄ω/(kB T ) ≪ 1, using the approximation ex ≈ 1 + x for
x ≪ 1, gives
Cquantum = N kB = Cclassical .
92
At low temperatures, the quantized nature of the energy levels of the quantum oscillators
becomes noticeable. In the limit T → 0, there is an energy gap between the ground state
and the first excited state. This results in a heat capacity that goes to zero exponentially,
as can be seen from the limit h̄ω/(kB T ) ≫ 1,
Cquantum = N kB
h̄ω
kB T
2
h̄ω
.
exp −
kB T
********
3. Relativistic particles: N indistinguishable relativistic particles move in one dimension
subject to a Hamiltonian
H ({pi , qi }) =
N
X
i=1
[c|pi | + U (qi )] ,
with U (qi ) = 0 for 0 ≤ qi ≤ L, and U (qi ) = ∞ otherwise. Consider a microcanonical
ensemble of total energy E.
(a) Compute the contribution of the coordinates qi to the available volume in phase space
Ω(E, L, N ).
•
Each of N coordinates explores a length L, for an overall contribution of LN /N !.
Division by N ! ensures no over-counting of phase space for indistinguishable particles.
(b) Compute the contribution of the momenta pi to Ω(E, L, N ).
Pd
(Hint: The volume of the hyper-pyramid defined by
i=1 xi ≤ R, and xi ≥ 0, in d
dimensions is Rd /d! .)
• The N momenta satisfy the constraint
PN
i=1
|pi | = E/c. For a particular choice of the
signs of {pi }, this constraint describes the surface of a hyper-pyramid in N dimensions. If
we ignore the difference between the surface area and volume in the large N limit, we can
calculate the volume in momentum space from the expression given in the hint as
1
·
Ωp = 2 ·
N!
N
E
c
N
.
The factor of 2N takes into account the two possible signs for each pi . The surface area
√
√
of the pyramid is given by dRd−1 /(d − 1)!; the additional factor of d with respect to
93
dvolume/dR is the ratio of the normal to the base to the side of the pyramid. Thus, the
volume of a shell of energy uncertainly ∆E , is
Ω′p
√
N
·
=2 ·
(N − 1)!
N
E
c
N−1
·
∆E
.
c
We can use the two expressions interchangeably, as their difference is subleading in N .
(c) Compute the entropy S(E, L, N ).
• Taking into account quantum modifications due to anisotropy, and phase space measure,
we have
√
N−1
1 LN N
N
E
∆E
Ω(E, L, N ) = N ·
·2 ·
·
·
.
h
N!
(N − 1)!
c
c
Ignoring subleading terms in the large N limit, the entropy is given by
S(E, L, N ) = N kB ln
2e2 L E
·
·
hc N N
.
(d) Calculate the one dimensional pressure P .
• From dE = T dS − P dV + µdN , the pressure is given by
P =T
∂S
∂L
=
E,N
N kB T
.
L
(e) Obtain the heat capacities CL and CP .
• Temperature and energy are related by
∂S
1
=
T
∂E
=
L,N
N kB
,
E
=⇒
E = N kB T,
=⇒
CL =
∂E
∂T
= N kB .
L,N
Including the work done against external pressure, and using the equation of state,
CP =
∂E
∂T
+P
P,N
∂L
∂T
= 2N kB .
P,N
(f) What is the probability p(p1 ) of finding a particle with momentum p1 ?
94
• Having fixed p1 for the first particle, the remaining N − 1 particles are left to share
an energy of (E − c|p1 |). Since we are not interested in the coordinates, we can get the
probability from the ratio of phase spaces for the momenta, i.e.
Ωp (E − c|p1 |, N − 1)
Ωp (E, N )
"
N−1 # N ! c N
E − c|p1 |
2N−1
× N ·
=
·
(N − 1)!
c
2
E
N
cN
c|p1 |
cN |p1 |
cN
≈
.
· 1−
· exp −
≈
2E
E
2E
E
p(p1 ) =
Substituting E = N kB T , we obtain the (property normalized) Boltzmann weight
c|p1 |
c
.
· exp −
p(p1 ) =
2kB T
kB T
********
4. Hard sphere gas: Consider a gas of N hard spheres in a box. A single sphere occupies
volume ω, while its center of mass can explore a volume V (if the box is otherwise empty).
There are no other interactions between the spheres, except for the constraints of hard-core
exclusion.
(a) Calculate the entropy S, as a function of the total energy E.
2
(Hint: V − aω V − (N − a)ω ≈ V − N ω/2 .)
• The available phase space for N identical particles is given by
Z
1
Ω=
d3 ~q1 · · · d3 ~qN d3 p~1 · · · d3 ~pN ,
N !h3N H=E
where the integration is carried out under the condition,
N
X
p2i
E = H(~qi , ~pi ) =
,
2m
i=1
N
X
or
p2i = 2mE.
i=1
The momentum integrals are now performed as in an ideal gas, yielding
(2mE)3N/2−1
·
Ω=
N !h3N
2π 3N
·
3N
2 −1 !
95
Z
d3 ~q1 · · · d3 ~qN .
The joint integral over the spacial coordinates with excluded volume constraints is best
performed by introducing particles one at a time. The first particle can explore a volume
V , the second V − ω, the third V − 2ω, etc., resulting in
Z
d3 ~q1 · · · d3 ~qN = V (V − ω)(V − 2ω) · · · (V − (N − 1)ω).
Using the approximation (V − aω)(V − (N − a)ω) ≈ (V − N ω/2)2 , we obtain
Z
N
Nω
d ~q1 · · · d q~N ≈ V −
.
2
3
3
Thus the entropy of the system is
"
e
S = kB ln Ω ≈ N kB ln
N
3/2 #
Nω
4πmEe
V −
.
2
3N h2
(b) Calculate the equation of state of this gas.
• We can obtain the equation of state by calculating the expression for the pressure of
the gas,
∂S
P
=
T
∂V
E,N
≈
N kB
,
V − Nω
2
which is easily re-arranged to,
P
Nω
V −
2
= N kB T.
Note that the joint effective excluded volume that appears in the above expressions is one
half of the total volume excluded by N particles.
(c) Show that the isothermal compressibility, κT = −V −1 ∂V /∂P |T , is always positive.
• The isothermal compressibility is calculated from
κT = −
1 ∂V
V ∂P
=
T,N
N kB T
> 0,
P 2V
and is explicitly positive, as required by stability constraints.
********
96
5. Non-harmonic gas:
Let us reexamine the generalized ideal gas introduced in the
previous section, using statistical mechanics rather than kinetic theory. Consider a gas of
N non-interacting atoms in a d-dimensional box of “volume” V , with a kinetic energy
H=
N
X
i=1
s
A |~
pi | ,
where p~i is the momentum of the ith particle.
(a) Calculate the classical partition function Z(N, T ) at a temperature T . (You don’t have
to keep track of numerical constants in the integration.)
• The partition function is given by
1
Z(N, T, V ) =
N !hdN
1
=
N !hdN
Z
···
Z Z
Z
"
dd ~q1 · · · dd ~qN dd p~1 · · · dd ~pN exp −β
d
s
d
d ~qd p~ exp (−βA |~
p| )
N
N
X
i=1
A |~
pi |
s
#
.
Ignoring hard core exclusion, each atom contributes a d–dimensional volume V to the
integral over the spatial degrees of freedom, and
Z
N
VN
s
d
d p~ exp (−βA |~p | ) .
Z(N, T, V ) =
N !hdN
Observing that the integrand depends only on the magnitude |~
p | = p, we can evaluate the
R d
R
d−1
integral in spherical coordinates using d p~ = Sd dpp , where Sd denotes the surface
area of a unit sphere in d–dimensions, as
N
Z ∞
VN
d−1
s
dpp
exp (−βAp ) .
Z(N, T, V ) =
Sd
N !hdN
0
Introducing the variable x ≡ (βA)1/s p, we have
−dN /s Z ∞
N
A
V N SdN
d−1
s
dxx
exp(−x )
Z(N, T, V ) =
N !hdN kB T
0
−dN /s
N 1 V Sd
A
N
= C (d, s)
,
N!
hd
kB T
where C denotes the numerical value of the integral. (We assume that A and s are both
real and positive. These conditions ensure that energy increases with increasing |~
p |.) The
integral is in fact equal to
C(d, s) =
Z
∞
0
dxx
d−1
1
exp(−x )dx = Γ
s
s
97
d
,
s
and the partition function is
1
Z=
N!
V Sd
hd s
N A
kB T
−dN /s N
d
Γ
.
s
(b) Calculate the pressure and the internal energy of this gas. (Note how the usual equipartition theorem is modified for non-quadratic degrees of freedom.)
• To calculate the pressure and internal energy, note that the Helmholtz free energy is
F = E − T S = −kB T ln Z,
and that
P =−
∂F
∂V
,
E=−
while
T
∂ ln Z
∂β
.
V
First calculating the pressure:
P =−
∂F
∂V
= kB T
T
∂ ln Z
∂V
=
T
N kB T
.
V
Now calculating the internal energy:
∂ ln Z
E=−
∂β
V
∂
dN
d
A
=−
−
= N kB T.
ln
∂β
s
kB T
s
Note that for each degree of freedom with energy A|~
pi |s , we have the average value,
hA|~
pi |s i = ds kB T. This evaluates to 32 kB T for the 3–dimensional ideal gas.
(c) Now consider a diatomic gas of N molecules, each with energy
Hi = A
p~i
(1)
s
+ p~i
s
(2)
+ K ~qi
(1)
− ~qi
(2)
t
,
where the superscripts refer to the two particles in the molecule. (Note that this unrealistic
potential
allowsthe two atoms to occupy the same point.) Calculate the expectation value
t
(1)
(2)
~qi − ~qi
, at temperature T .
• Now consider N diatomic molecules, with
H=
N
X
i=1
Hi ,
where
Hi = A
(1)
~p i
98
s
+
(2)
p~ i
s
(1)
+ K ~q i
(2)
− ~q i
t
.
The expectation value
(1)
~q i
(2)
− ~q i
t
1
N!
=
t
P
(1)
(2)
d (1) d (2) d (1) d (2)
d
q
~
d
q
~
d
p
~
d
p
~
~
q
−
q
~
exp
[−β
i
i
i
i
i
i
i=1
i Hi ]
,
R
Q
P
(1) d (2) d (1) d (2)
N
1
d~
d
q
d
~
q
d
p
~
d
p
~
exp
[−β
H
]
i
i
i
i
i
i=1
i
N!
R QN
is easily calculated by changing variables to
~x ≡ ~q (1) − ~q (2) ,
q~ (1) + ~q (2)
,
2
~y ≡
and
as (note that the Jacobian of the transformation is unity)
~q (1) − ~q (2)
t
=
=
R
R
dd ~xdd ~y · |~x |t · exp [−βK|~x |t ]
R
dd ~xdd ~y · exp [−βK|~x |t ]
dd ~x · |~x |t · exp [−βK|~x |t ]
R
.
dd ~x · exp [−βK|~x |t ]
Further simplifying the algebra by introducing the variable ~z ≡ (βK)1/t~x, leads to
~q
(1)
− ~q
(2)
t
(βK)−t/t
R
=
dd ~z · |~z |t · exp [−|~z |t ]
d kB T
.
= ·
d
t
t
K
d ~z · exp [−|~z | ]
R
Here we have assumed that the volume is large enough, so that the range of integration
over the relative coordinate can be extended from 0 to ∞.
Alternatively, note that for the degree of freedom ~x = ~q (1) − ~q (2) , the energy is K|~x |t .
Thus, from part (b) we know that
d kB T
·
,
t
K
K|~x |t =
i.e.
|~x |
t
=
~q
(1)
− ~q
(2)
t
=
d kB T
·
.
t
K
And yet another way of calculating the expectation value is from
N X
i=1
(1)
~q i
−
(2)
~q i
t
=−
1 ∂ ln Z
N d kB T
=
·
,
β ∂K
t
K
(note that the relevant part of Z is calculated in part (d) below).
(d) Calculate the heat capacity ratio ratio γ = CP /CV , for the above diatomic gas.
99
• For the ideal gas, the internal energy depends only on temperature T . The gas in part
(c) is ideal in the sense that there are no molecule–molecule interactions. Therefore,
d̄Q
dT
CV =
and
d̄Q
dT
CP =
dE + P dV
dT
=
V
dE + P dV
dT
=
P
Since P V = N kB T,
CP =
=
P
=
V
dE(T )
,
dT
dE(T )
∂V (T )
+P
dT
∂T
.
P
dE(T )
+ N kB .
dT
We now calculate the partition function
Z=
=
1
N !hdN
Z Y
N
i=1
(1)
(2)
(1)
(2)
"
dd ~q i dd ~q i dd p~ i dd p~ i exp −β
1
z1N ,
dN
N !h
X
i
Hi
#
where
z1 =
=
Z
Z
d (1) d (2) d
d ~q
d ~q
d p~
d (1) d (2)
d ~q
d ~q
(1) d
d p~
(2)
exp −β · A p
~ (1)
s
+ A p~
(2)
s
+ K ~q
(1)
− ~q
(2)
Z
t
s 2
(1)
(2)
d (1)
(1)
exp −βK ~q − ~q
·
d p~ exp −βA p~
.
Introducing the variables, ~x, ~y , and ~z, as in part (c),
N Z ∞
2N
Z ∞
VN
−d/t
d−1
t
d−1
s
(βK)
z
exp(−z )dz
·
p
exp(−βAp )dp
Z∝
N!
0
0
N 2N
1
d
d
VN
−d/t 1
−d/s
(βK)
· (βA)
Γ
Γ
=
N!
t
t
s
s
∝
t
VN
−dN/t
−2Nd/s
(βK)
(βA)
.
N!
Now we can calculate the internal energy as
∂ ln Z
d
2d
hEi = −
= N kB T + N kB T = dN kB T
∂β
t
s
1 2
+
t
s
From this result, the heat capacities are obtained as
∂E
CP =
∂T
∂E
CV =
∂T
P
V
∂V
+P
∂T
= N kB
2 1
= dN kB
.
+
s
t
P
100
2d d
+ +1 ,
s
t
.
resulting in the ratio
γ=
CP
2d/s + d/t + 1
st
=
=1+
.
CV
2d/s + d/t
d(2t + s)
********
6. Surfactant adsorption:
A dilute solution of surfactants can be regarded as an ideal
three dimensional gas. As surfactant molecules can reduce their energy by contact with air,
a fraction of them migrate to the surface where they can be treated as a two dimensional
ideal gas. Surfactants are similarly adsorbed by other porous media such as polymers and
gels with an affinity for them.
(a) Consider an ideal gas of classical particles of mass m in d dimensions, moving in
a uniform attractive potential of strength εd . By calculating the partition function, or
otherwise, show that the chemical potential at a temperature T and particle density nd , is
given by
µd = −εd + kB T ln nd λ(T )d ,
where
λ(T ) = √
h
.
2πmkB T
• The partition function of a d-dimensional ideal gas is given by
(
"
#)
Z
Z Y
Nd
Nd 2
X
p
~
1
i
···
dd ~qi dd ~p i exp −β Nd εd +
Zd =
Nd !hdNd
2m
i=1
i=1
Nd
1
Vd
=
e−βNd εd ,
Nd ! λd
where
λ≡ √
h
.
2πmkB T
The chemical potential is calculated from the Helmholtz free energy as
∂ ln Zd
∂Nd
V,T
Vd
.
= −εd + kB T ln
Nd λd
µd =
∂F
∂N
= −kB T
101
V,T
(b) If a surfactant lowers its energy by ε0 in moving from the solution to the surface, calculate the concentration of floating surfactants as a function of the solution concentration
n (= n3 ), at a temperature T .
•
The density of particles can also be calculated from the grand canonical partition
function, which for particles in a d–dimensional space is
Ξ(µ, Vd , T ) =
=
∞
X
Nd =0
∞
X
Nd =0
Z(Nd , Vd , T )eβNd µ
1
Nd !
Vd
λd
Nd
−βNd εd βNd µ
e
e
= exp
Vd
λd
β(µ−εd )
·e
.
The average number of particles absorbed in the space is
1 ∂
1 ∂
hNd i =
ln Ξ =
β ∂µ
β ∂µ
Vd
λd
β(µ−εd )
·e
=
Vd
λd
· eβ(µ−εd ) .
We are interested in the coexistence of surfactants between a d = 3 dimensional solution,
and its d = 2 dimensional surface. Dividing the expressions for hN3 i and hN2 i, and taking
into account ε0 = ε3 − ε2 , gives
Aλ βε0
hN2 i
=
e ,
hN3 i
V
which implies that
n2 =
hN2 i
= nλeβε0 .
A
(c) Gels are formed by cross-linking linear polymers. It has been suggested that the porous
gel should be regarded as fractal, and the surfactants adsorbed on its surface treated as
a gas in df dimensional space, with a non-integer df . Can this assertion be tested by
comparing the relative adsorption of surfactants to a gel, and to the individual polymers
(presumably one dimensional) before cross-linking, as a function of temperature?
• Using the result found in part (b), but regarding the gel as a df –dimensional container,
the adsorbed particle density is
hngel i = nλ3−df exp [β(ε3 − εgel )] .
Thus by studying the adsorption of particles as a function of temperature one can determine
the fractal dimensionality, df , of the surface. The largest contribution comes from the
102
difference in energies. If this leading part is accurately determined, there is a subleading
dependence via λ3−df which depends on df .
********
7. Molecular adsorption:
N diatomic molecules are stuck on a metal surface of square
symmetry. Each molecule can either lie flat on the surface in which case it must be aligned
to one of two directions, x and y, or it can stand up along the z direction. There is an
energy cost of ε > 0 associated with a molecule standing up, and zero energy for molecules
lying flat along x or y directions.
(a) How many microstates have the smallest value of energy? What is the largest microstate
energy?
• The ground state energy of E = Emin = 0 is obtained for 2N configurations. The largest
microstate energy is N ǫ is unique.
(b) For microcanonical macrostates of energy E, calculate the number of states Ω(E, N ),
and the entropy S(E, N ).
• Let Nz = E/ǫ. Expressing Ω as the number of ways to choose the Ng excited molecules,
multiplied by the number of possible configurations
Ω(E, N ) =
N!
· 2N−Nz ,
Nz !(N − Nz )!
and
S(E, N ) = Stwo−level system + kb (N − Nz ) ln 2
E
E
E
E
E
= −N kB
ln
+ (1 −
) ln(1 −
) + kb (N − ) ln 2.
Nǫ Nǫ
Nǫ
Nǫ
ǫ
(c) Calculate the heat capacity C(T ) and sketch it.
• The temperature dependence of the energy is obtained from the relation
∂S
1
=
T
∂E
N
whence
E=
and
exp( kǫB ( T1
=−
kB
E
kB
ln(
)−
ln 2,
ǫ
Nǫ − E
ǫ
Nǫ
Nǫ
=
,
kB
2 exp( kBǫ T ) + 1
+ ǫ ln 2)) + 1
2 exp( kBǫ T )
ǫ 2
dE
= N kB (
)
.
C=
dT
kB T (1 + 2 exp( kBǫ T ))2
103
C/kBN
(kεT)2e- ε/k T
B
1 ε 2
4 kB T
( )
B
T
(d) What is the probability that a specific molecule is standing up?
• The probability that a specific molecule is standing up is
Ω(E − ǫ, N − 1)
Ω(E, N )
(N − 1)!
Nz !(N − Nz )! 1
=
2(N−1)−(Nz −1)
(Nz − 1)!((N − 1) − (Nz − 1))!
N!
2N−Nz
E
Nz
=
=
N
Nǫ
1
.
=
2 exp kBǫ T + 1
p(~r1 = ẑ) =
(e) What is the largest possible value of the internal energy at any positive temperature?
• Since
dE
dT
> 0 for all T > 0, the energy is largest for → ∞, i.e.
Emax =
Nǫ
.
3
********
8. Curie susceptibility: Consider N non-interacting quantized spins in a magnetic field
~ = B ẑ, and at a temperature T . The work done by the field is given by BMz , with a
B
PN
magnetization Mz = µ i=1 mi . For each spin, mi takes only the 2s + 1 values −s, −s +
1, · · · , s − 1, s.
(a) Calculate the Gibbs partition function Z(T, B). (Note that the ensemble corresponding
to the macrostate (T, B) includes magnetic work.)
• The Gibbs partition function is
Z=
X
{mi }
~ ·M
~ =
exp β B
X
exp βBµ
N
X
i=1
{mi }
104
mi
!
=
"
m
i =s
X
mi =−s
exp(βµB · mi )
#N
.
Thus we obtain the series
N
Z = [exp(−βBµs) + exp(−βBµ(s − 1)) + · · · + exp(βBµ(s − 1)) + exp(βBµs)] .
In general, to evaluate a geometrical series of the form
S = x−s + x−(s−1) + · · · + xs−1 + xs ,
increase the order of the series by one,
Sx = x−s+1 + · · · + xs + xs+1 ,
and subtract from the original series:
(1 − x)S = x−s − xs+1 ,
=⇒
S=
x−s − xs+1
.
1−x
(Note that the same result is obtained whether s is an integer or half–integer quantity.)
Using this expression, we get
Z=
=
exp(−βBµs) − exp(βBµ(s + 1))
1 − exp(βBµ)
N
exp(−βBµ(s + 1/2)) − exp(−βBµ(s + 1/2))
exp(−βBµ/2) − exp(−βBµ/2)
N
.
Substituting in the proper trigonometric identity,
sinh (βµB(s + 1/2))
Z=
sinh(βµB/2)
N
.
(b) Calculate the Gibbs free energy G(T, B), and show that for small B,
G(B) = G(0) −
N µ2 s(s + 1)B 2
+ O(B 4 ).
6kB T
• The Gibbs free energy is
G = E − BM = −kB T ln Z
= −N kB T ln[sinh(βµB(s + 1/2))] + N kB T ln[sinh(βµB/2)].
105
Using an approximation of sinh θ for small θ,
1
1 θ
e − e−θ ≈
sinh θ =
2
2
θ3
2θ + 2
+ O(θ 5 ),
3!
for
θ ≪ 1,
we find (setting α = βµB),
G ≈ −N kB T
(
" 2 !#
1
α2
1
α
α2
4
ln α s +
1+
s+
− ln
1+
+ O(α ) .
2
6
2
2
24
Using the expansion ln(1 + x) = x − x2 /2 + x3 /3 − · · ·, we find
1
1
2
2
G ≈ −N kB T ln(2s + 1) + (α(s + 1/2)) − (α/2) + O(α4 )
6
6
2
(s + s)
≈ −N kB T ln(2s + 1) − N kB T α2
6
2 2
N µ B s(s + 1)
= G0 −
+ O(B 4 ).
6kB T
(c) Calculate the zero field susceptibility χ = ∂Mz /∂B|B=0 , and show that is satisfies
Curie’s law
χ = c/T.
• The magnetic susceptibility, χ = ∂Mz /∂B, is obtained by noting that the average
magnetization is
hMz i = kB T
Thus
χ=
∂G
∂ ln Z
=−
.
∂B
∂B
∂ hMz i
∂ ∂G
N µ2 ts(s + 1)
=−
=
,
∂B
∂B ∂B
3kB T
which obeys Curie’s law, χ = c/T , with c = N µ2 s(s + 1)/3kB .
(d) Show that CB − CM = cB 2 /T 2 where CB and CM are heat capacities at constant B
and M respectively.
********
9. Langmuir isotherms:
An ideal gas of particles is in contact with the surface of a
catalyst.
106
(a) Show that the chemical potential of the gas particles is related to their temperature
and pressure via µ = kB T ln P/T 5/2 + A0 , where A0 is a constant.
• For convenience, we begin by defining the characteristic length,
λ= √
h
,
2πmkB T
in terms of which the free energy of ideal gas is given by,
F = −N kB T ln
Ve
N λ3
.
Then using the equation of state of ideal gas we get,
∂F
µ=
= −kB T ln
∂N
"
V
N λ3
= kB T ln(P T −5/2 ) + ln
= −kB T ln
h3
kB T
P λ3
5/2
kB 23/2 π 3/2 m3/2
!#
.
(b) If there are N distinct adsorption sites on the surface, and each adsorbed particle gains
an energy ǫ upon adsorption, calculate the grand partition function for the two dimensional
gas with a chemical potential µ.
• We have,
Q(T, µ) =
N X
N
n=0
n
N
enβµ e−nβǫ = 1 + eβ(µ−ǫ)
.
(c) In equilibrium, the gas and surface particles are at the same temperature and chemical
potential. Show that the fraction of occupied surface sites is then given by f (T, P ) =
P/ P + P0 (T ) . Find P0 (T ).
• Since the average number of absorption sites occupied is,
hN i = kB T
eβ(µ−ǫ)
∂ ln Q
=N
,
∂µ
1 + eβ(µ−ǫ)
and the fraction of occupied sites is given by,
f (T, P ) =
eβ(µ−ǫ)
.
1 + eβ(µ−ǫ)
107
Since the gas and surface particles have the same temperature and chemical potential, the
realtion in (a), namely,
P λ3
kB T
eβµ =
holds. Plugging this into the formula for f we obtain,
f (T, P ) =
P
,
P + P0 (T )
with P0 (T ) =
kB T βǫ
e .
λ3
(d) In the grand canonical ensemble, the particle number N is a random variable. Calculate
its characteristic function hexp(−ikN )i in terms of Q(βµ), and hence show that
hN m ic = −(kB T )m−1
∂ mG
∂µm
,
T
where G is the grand potential.
• Note that in calculating Q(βµ) the term for N particles is proportional to eβµN . In
calculating the average of e−ikN , we just replace the initial factor with e(βµ−ik)N , and
hence
e−ikN =
Q(βµ − ik)
.
Q(βµ)
For the cumulant generating function, we get
ln e−ikN = ln Q(βµ − ik) − ln Q(βµ) = −βG(βµ − ik) + βG(βµ).
Hence,
∂m
(−βG(βµ − ik))|T
∂(−ik)m
m
∂ mG
1−m ∂ G
= −β
=
−β
∂(βµ)m T
∂µm
∂ mG
.
= −(kB T )m−1
∂µm T
hN m ic =
(e) Using the characteristic function, show that
N2
c
= kB T
108
∂hN i
∂µ
.
T
T
•
∂ 2G
∂µ2
hN 2 ic = −kB T
= kB T
T
∂hN i
∂µ
.
T
(f) Show that fluctuations in the number of adsorbed particles satisfy
N2
c
2
hN ic
• By definition of f ,
hN ic = −
and since,
∂f
∂µ
=
T
e−βǫ
∂
∂µ e−βµ + e−βǫ
we get,
hN 2 ic =
1−f
.
Nf
=
∂G
∂µ
= N f,
T
βe−βµ e−βǫ
=
(e−βµ + e−βǫ )
T
2
= βf (1 − f ),
1 ∂f
1 ∂N
= N
= N f (1 − f ),
β ∂µ
β ∂µ
leading to the equality,
hN 2 ic
2
hN ic
=
1−f
.
Nf
********
~ of unity, i.e. S z is quantized to -1, 0,
10. Molecular oxygen has a net magnetic spin, S,
~ k ẑ is
or +1. The Hamiltonian for an ideal gas of N such molecules in a magnetic field B
H=
N X
p~i
i=1
2
2m
−
µBSiz
,
where {~
pi } are the center of mass momenta of the molecules. The corresponding coordi-
nates {~qi } are confined to a volume V . (Ignore all other degrees of freedom.)
(a) Treating {~
pi , ~qi } classically, but the spin degrees of freedom as quantized, calculate the
partition function, Z̃(T, N, V, B).
•
Z=
X
µs
−βH(µs )
e
=
X
µs
−β
e
P
(~
p2i /2m−µBSiz )
109
1
=
N!
V βµB
(e
+ 1 + e−βµB )
λ3
N
,
where,
λ= √
h
.
2πmkB T
(b) What are the probabilities for Siz of a specific molecule to take on values of -1, 0, +1
at a temperature T ?
• The probabilities for Siz of a given molecule to take values -1, 0, +1, are,
e−βµB
,
2 cosh βµB + 1
eβµB
,
2 cosh βµB + 1
1
,
2 cosh βµB + 1
respectively.
(c) Find the average magnetic dipole moment, hM i /V , where M = µ
•
1 ∂ ln Z
1 ∂N ln(2 cosh βµB + 1)
=
β ∂B
β
∂B
2 sinh βµB
.
= Nµ
2 cosh βµB + 1
PN
i=1
Siz .
hM i =
(d) Calculate the zero field susceptibility χ = ∂ < M > /∂B|B=0 .
•
χ = N βµ2
2 cosh βµB(2 cosh βµB + 1) − 4 sinh βµB sinh βµB
(2 cosh βµB + 1)2
=
B=0
2
N βµ2 .
3
********
~ of unity, i.e. S z is quantized to -1, 0,
11. Molecular oxygen has a net magnetic spin, S,
~ k ẑ is
or +1. The Hamiltonian for an ideal gas of N such molecules in a magnetic field B
H=
N X
p~i
i=1
2
2m
−
µBSiz
,
where {~
pi } are the center of mass momenta of the molecules. The corresponding coordi-
nates {~qi } are confined to a volume V . (Ignore all other degrees of freedom.)
(a) Treating {~
pi , ~qi } classically, but the spin degrees of freedom as quantized, calculate the
partition function, Z̃(T, N, V, B).
•
Z=
X
µs
−βH(µs )
e
=
X
µs
−β
e
P
(~
p2i /2m−µBSiz )
110
1
=
N!
V βµB
(e
+ 1 + e−βµB )
λ3
N
,
where,
λ= √
h
.
2πmkB T
(b) What are the probabilities for Siz of a specific molecule to take on values of -1, 0, +1
at a temperature T ?
• The probabilities for Siz of a given molecule to take values -1, 0, +1, are,
e−βµB
,
2 cosh βµB + 1
1
,
2 cosh βµB + 1
eβµB
,
2 cosh βµB + 1
respectively.
(c) Find the average magnetic dipole moment, hM i /V , where M = µ
•
1 ∂N ln(2 cosh βµB + 1)
1 ∂ ln Z
=
β ∂B
β
∂B
2 sinh βµB
.
= Nµ
2 cosh βµB + 1
PN
i=1
Siz .
hM i =
(d) Calculate the zero field susceptibility χ = ∂ < M > /∂B|B=0 .
•
χ = N βµ2
2 cosh βµB(2 cosh βµB + 1) − 4 sinh βµB sinh βµB
(2 cosh βµB + 1)2
=
B=0
2
N βµ2 .
3
********
12. Polar rods:
Consider rod shaped molecules with moment of inertia I, and a dipole
moment µ. The contribution of the rotational degrees of freedom to the Hamiltonian is
given by
Hrot.
1
=
2I
p2θ +
p2φ
sin2 θ
!
− µE cos θ
,
where E is an external electric field. (φ ∈ [0, 2π], θ ∈ [0, π] are the azimuthal and polar
angles, and pφ , pθ are their conjugate momenta.)
(a) Calculate the contribution of the rotational degrees of freedom of each dipole to the
classical partition function.
111
• The classical partition function is obtained by integrating over the angles θ and φ, and
the corresponding momenta as
"
Z π Z 2π
Z ∞
1
β
Zrot = 2
dθ
dφ
dpθ dpφ exp −
h 0
2I
0
−∞
p2θ +
p2φ
sin2 θ
!
#
+ βµ cos θ .
The Gaussian integrations over momenta are easily carried out, and after the change of
variables to x = cos θ, we find
2 Z +1
2πI
8π I sinh(βµE)
βµEx
Zrot =
dxe
=
(2π)
.
2
βh
βh2
βµE
−1
(b) Obtain the mean polarization P = hµ cos θi, of each dipole.
• From the form of the partition function, we note that
∂ ln Zrot
1
P = hµ cos θi =
.
= µ coth(βµE) −
∂βE
βµE
(c) Find the zero–field polarizability
χT =
∂P
∂E
.
E=0
• The susceptibility is
∂P
1
1 µ2
1
2
χT =
=
+
= βµ −
∂E
3 kB T
sinh2 (βµE) (βµE)2
for E = 0 .
(d) Calculate the rotational energy per particle (at finite E), and comment on its high and
low temperature limits.
• The energy stored in this degree of freedom is
hErot i = −
∂ ln Zrot
= 2kB T − µE coth(βµE).
∂β
At high temperature hErot i ≈ kB T coming entirely from the kinetic energies. At temper-
atures kB T ≪ µE, the polarization is saturated, and hErot i ≈ −µE + 2kB T .
(e) Sketch the rotational heat capacity per dipole.
• The classical heat capacity is 2kB at low temperatures, and decreases to 1kB at high
temperatures, with a ‘knee’ at kB T ≈ µE.
********
112
Problems for Chapter V - Interacting Particles
1. Debye–Hückel theory and ring diagrams:
The virial expansion gives the gas pressure
as an analytic expansion in the density n = N/V . Long range interactions can result
in non-analytic corrections to the ideal gas equation of state. A classic example is the
Coulomb interaction in plasmas, whose treatment by Debye–Hückel theory is equivalent
to summing all the ring diagrams in a cumulant expansion.
For simplicity consider a gas of N electrons moving in a uniform background of positive
charge density N e/V to ensure overall charge neutrality. The Coulomb interaction takes
the form
UQ =
X
i<j
V(~qi − ~qj ) ,
with V(~q ) =
e2
−c.
4π|~q |
The constant c results from the background and ensures that the first order correction
R
vanishes, i.e. d3 ~q V(~q ) = 0.
(a) Show that the Fourier transform of V(~q ) takes the form
2
2
Ṽ (~ω ) = e /ω for ~ω 6= 0 .
0
for ~ω = 0
• The Fourier transform of V(~q) is singular at the origin, and can be defined explicitly as
Z
Ṽ(~ω ) = lim d3 ~qV(~q)ei~ω·~q−εq .
ε→0
The result at ~ω = 0 follows immediately from the definition of c. For ~ω 6= 0,
2 2
Z
Z
e
e
i~
ω ·~
q −εq
3
3
ei~ω·~q−εq
−c e
= lim d ~q
Ṽ(~ω ) = lim d ~q
ε→0
ε→0
4πq
4πq
Z π
2 Z ∞
e
iωq cos θ−εq
2
e
= lim 2π
sin θdθ
q dq
ε→0
4πq
0
0
Z
e2 ∞ eiωq−εq − e−iωq−εq
=−
dq
2 0
iω
e2
1
1
e2
e2
= lim
= lim
+
=
.
ε→0 2iω
ε→0 ω 2 + ε2
iω − ε iω + ε
ω2
(b) In the cumulant expansion for
D
U ℓQ
E0
c
, we shall retain only the diagrams forming a
ring; which are proportional to
Z 3
d3 ~qℓ
d ~q1
···
V(~q1 − ~q2 )V(~q2 − ~q3 ) · · · V(~qℓ − ~q1 ).
Rℓ =
V
V
113
Use properties of Fourier transforms to show that
Z
1
d3 ~ω
Rℓ = ℓ−1
Ṽ(~ω )ℓ .
3
V
(2π)
D E0
• In the cumulant expansion for U ℓQ , we retain only the diagrams forming a ring. The
c
contribution of these diagrams to the partition function is
Z 3
d q~ 1 d3 ~q 2
d3 ~q ℓ
Rℓ =
···
V(~q 1 − ~q 2 )V(~q 2 − ~q 3 ) · · · V(~q ℓ − ~q 1 )
V
V
V
Z
Z
1
· · · d3 ~x1 d3 ~x2 · · · d3 ~xℓ−1 d3 ~q ℓ V(~x1 )V(~x2 ) · · · V(~xℓ−1 )V
= ℓ
V
−
ℓ−1
X
i=1
~xi
!
,
where we introduced the new set of variables {~xi ≡ ~q i − ~q i+1 }, for i = 1, 2, · · · , ℓ − 1. Note
that since the integrand is independent of ~q ℓ ,
Rℓ =
1
V ℓ−1
Z
···
Z
d3 ~x1 d3 ~x2 · · · d3 ~xℓ−1 V(~x1 )V(~x2 ) · · · V
−
ℓ−1
X
~xi
i=1
!
.
Using the inverse Fourier transform
1
V(~q ) =
(2π)3
Z
d3 ~ω · Ṽ(~ω )e−i~q ·~ω ,
the integral becomes
1
Rℓ =
3ℓ
(2π) V ℓ−1
Z
Z
d3 ~x1 · · · d3 ~xℓ−1 Ṽ(~ω1 )e−i~ω1 ·~x1 Ṽ(~ω2 )e−i~ω2 ·~x2
!
ℓ−1
X
· · · Ṽ(~ωℓ ) exp −i
~ωℓ · ~xk d3 ~ω1 · · · d3 ~ωℓ .
···
k=1
Since
Z
d3 ~q −i~ω ·~q
e
= δ 3 (~ω ),
(2π)3
we have
1
Rℓ =
3
(2π) V ℓ−1
Z
···
ℓ−1
Y
Z
k=1
resulting finally in
Rℓ =
1
V ℓ−1
Z
δ(~ωk − ~ωℓ )Ṽ(~ωk )d3 ~ωk
d3 ω
~
Ṽ(~ω )ℓ .
3
(2π)
114
!
d3 ~ωℓ ,
D E0
(c) Show that the number of ring graphs generated in U ℓQ is
c
Sℓ =
N!
(ℓ − 1)!
(ℓ − 1)! ℓ
×
≈
N .
(N − ℓ)!
2
2
D E0
• The number of rings graphs generated in U ℓQ is given by the product of the number
of ways to choose ℓ particles out of a total of N ,
c
N!
.
(N − ℓ)!
multiplied by the number of ways to arrange the ℓ particles in a ring
ℓ!
.
2ℓ
The numerator is the number of ways of distributing the ℓ labels on the ℓ points of the
ring. This overcounts by the number of equivalent arrangements that appear in the denominator. The factor of 1/2 comes from the equivalence of clockwise and counterclockwise
arrangements (reflection), and there are ℓ equivalent choices for the starting point of the
ring (rotations). Hence
Sℓ =
ℓ!
N!
(ℓ − 1)!
N!
×
=
×
.
(N − ℓ)! 2ℓ
(N − ℓ)!
2
For N ≫ ℓ, we can approximate N (N − 1) · · · (N − ℓ + 1) ≈ N ℓ , and
Sℓ ≈
(ℓ − 1)! ℓ
N .
2
Another way to justify this result is by induction: A ring of length ℓ + 1 can be created
from a ring of ℓ links by inserting an additional point in between any of the existing ℓ
nodes. Hence Sℓ+1 = Sℓ × (N − ℓ − 1) × ℓ, leading to the above result, when starting with
S2 = N (N − 1)/2.
(d) Show that the contribution of the ring diagrams can be summed as
ln Zrings = ln Z0 +
∞
X
(−β)ℓ
ℓ!
ℓ=2
V
≈ ln Z0 +
2
Z
0
∞
Sℓ Rℓ
4πω 2 dω
(2π)3
115
κ 2
ω
κ2
− ln 1 + 2
,
ω
p
βe2 N/V is the inverse Debye screening length.
P∞
(Hint: Use ln(1 + x) = − ℓ=1 (−x)ℓ /ℓ.)
where κ =
• The contribution of the ring diagrams is summed as
ln Zrings = ln Z0 +
∞
X
(−β)ℓ
ℓ!
ℓ=2
∞
X
Sℓ Rℓ
Z
d3 ~ω
(−β)ℓ (ℓ − 1)! ℓ 1
N ℓ−1
Ṽ(~ω )ℓ
= ln Z0 +
3
ℓ!
2
V
(2π)
ℓ=2
ℓ
Z ∞
∞
V
4πω 2 dω X 1
βN
= ln Z0 +
−
Ṽ(ω)
2 0
(2π)3
ℓ
V
ℓ=2
ℓ
Z ∞
∞
4πω 2 dω X 1
βN e2
V
−
= ln Z0 +
2 0
(2π)3
ℓ
V ω2
ℓ=2
Z ∞
V
4πω 2 dω βN e2
βN e2
= ln Z0 +
,
− ln 1 +
2 0
(2π)3
V ω2
V ω2
where we have used ln(1 + x) = −
leads to
ln Zrings
V
= ln Z0 +
2
P∞
ℓ=1 (−x)
Z
∞
0
ℓ
/ℓ. Finally, substituting κ =
4πω 2 dω κ 2
κ2
− ln 1 + 2
.
(2π)3
ω
ω
p
βe2 N/V ,
(e) The integral in the previous part can be simplified by changing variables to x = κ/ω,
and performing integration by parts. Show that the final result is
ln Zrings = ln Z0 +
V 3
κ
12π
.
• Changing variables to x = κ/ω, and integrating the integrand by parts, gives
Z ∞
Z ∞
κ 2
dx 2
κ2
2
3
2
ω dω
− ln 1 + 2
=κ
x
−
ln(1
+
x
)
ω
ω
x4
0
0
Z
Z
2x
2κ3 ∞ dx
πκ3
κ3 ∞ dx
2x
−
=
=
,
=
3 0 x3
1 + x2
3 0 1 + x2
3
resulting in
ln Zrings = ln Z0 +
V
πκ3
V 3
·
= ln Z0 +
κ .
2
4π
3
12π
(f) Calculate the correction to pressure from the above ring diagrams.
116
• The correction to the ideal gas pressure due to the Debye–Hückel approximation is
!
∂ ln Zrings
P = kB T
∂V
T,N
kB T 3
V κ3
∂
= P0 −
κ
= P0 + k B T
∂V 12π T,N
24π
2
3/2
kB T
e N
= P0 −
.
24π kB T V
Note that the correction to the ideal gas behavior is non-analytic, and cannot be expressed
by a virial series. This is due to the long range nature of the Coulomb interaction.
(g) We can introduce an effective potential V (~q − ~q ′ ) between two particles by integrating
over the coordinates of all the other particles. This is equivalent to an expectation value
that can be calculated perturbatively in a cumulant expansion. If we include only the
loop-less diagrams (the analog of the rings) between the particles, we have
′
′
V (~q − ~q ) = V (~q − ~q ) +
∞
X
(−βN )
ℓ=1
ℓ
Z
d3 ~qℓ
d3 ~q1
···
V(~q − ~q1 )V(~q1 − ~q2 ) · · · V(~qℓ − ~
q ′ ).
V
V
Show that this sum leads to the screened Coulomb interaction V (~q ) = exp(−κ|~q |)/(4π|~
q |).
• Introducing the effective potential V(~q − ~q ′ ), and summing over the loop–less diagrams
gives
′
′
V(~q − ~q ) = V(~q − ~q ) +
∞
X
(−βN )
ℓ=1
ℓ
Z
d3 ~q 1
d3 ~q ℓ
···
V(~q − ~q 1 )V(~q 1 − ~q 2 ) · · · V(~q ℓ − ~
q ′)
V
V
d3 ~q 1
V(~q − ~q 1 )V(~q 1 − ~q ′ )
= V(~q − ~q ) − βN
V
Z 3
d ~q 1 d3 ~q 2
2
V(~q − ~q 1 )V(~q 1 − ~q 2 )V(~q 2 − ~q ′ ) − · · · .
+ (βN )
V
V
′
Z
Using the changes of notation
~x1 ≡ ~q ,
~x2 ≡ ~q ′ ,
V 12 ≡ V(~x1 − ~x2 ),
~x3 ≡ ~q 1 ,
~x4 ≡ ~q 2 ,
and
n ≡ N/V,
···
~xℓ ≡ ~q ℓ ,
we can write
V 12 = V 12 − βn
Z
3
d ~x3 V 13 V 32 + (βn)
117
2
Z
d3 ~x3 d3 ~x4 V 13 V 34 V 42 − · · · .
Using the inverse Fourier transform (as in part (a)), and the notation ~xij ≡ ~xi − ~xj ,
V 12 = V 12 − βn
d3 ~x3
Ṽ(~ω13 )Ṽ(~ω32 )e−i(~x13 ·~ω13 +~x32 ·~ω32 ) d3 ~ω13 d3 ~ω32 + · · · ,
(2π)6
Z
and employing the delta function, as in part (a)
d3 ~ω13 d3 ω
~ 32 3
δ (~ω13 − ~ω32 ) Ṽ(~ω13 )Ṽ(~ω32 ) exp[~x1 · ~ω13 − ~x2 · ~ω32 ] + · · ·
(2π)3
Z
i2
d3 ~ω h
Ṽ(~ω ) exp[~ω · ~x12 ] + · · · .
= V 12 − βn
(2π)3
V 12 = V 12 − βn
Z
Generalizing this result and dropping the subscript such that ~x ≡ ~x12 ,
V 12
Z h
∞
iℓ+1
X
(−βn)ℓ
Ṽ(~
ω
)
ei~x·~ω d3 ~ω .
= V 12 +
(2π)3
ℓ=1
Finally, including the Fourier transform of the direct potential (first term), gives
V 12
∞ Z
2ℓ+2
X
d3 ~ω (−1)ℓ e2 κ2ℓ ixω cos θ
d3 ~ω
ℓe
i~
x·~
ω
(−βn)
e
=
e
=
(2π)3
ω 2ℓ+2
(2π)3 ω 2ℓ+2
ℓ=0
ℓ=0
Z ∞
∞
ℓ 2 2ℓ Z 1
X
κ
(−1) e
eixω cos θ d cos θ
=
dω
2
2π
ω
−1
0
ℓ=0
Z ∞
∞
κ 2ℓ
e2 2 sin xω X
(−1)ℓ
.
=
dω 2
2π
xω
ω
0
∞ Z
X
ℓ=0
Setting y ≡ ω/κ, gives
V 12
1
=
2
Z
∞
−∞
e2 eixκy − e−ixκy −1
κ
dy.
2π 2
2ixκy
y2 + 1
Intergrating in the complex plane, via the residue theorem, gives
−κx
e2 e−κx
e−κx
e
e2
·
π
=
+
.
V 12 =
4π 2
2x
2x
4πx
Recalling our original notation, x = |~q − q~ ′ | ≡ |~q |, we obtain the screened Coulomb
potential
V(~q ) =
e2 e−κ|~q |
.
4π |~q |
********
118
2. Virial coefficients:
Consider a gas of particles in d-dimensional space interacting
through a pair-wise central potential, V(r), where
(
V(r) =
+∞ for 0 < r < a,
−ε
for a < r < b,
0
for b < r < ∞.
(a) Calculate the second virial coefficient B2 (T ), and comment on its high and low temperature behaviors.
• The second virial coefficient is obtained from
Z
1
B2 ≡ −
dd r12 {exp[−βV(r12 )] − 1} ,
2
where r12 ≡ |~r1 − ~r2 |, as
1
B2 = −
2
"Z
a
dd r12 (−1) +
Z
a
0
b
dd r12 eβε − 1
#
1
= − {Vd (a)(−1) + [Vd (b) − Vd (a)] · [exp(βε) − 1]} ,
2
where
Vd (r) =
Sd d
2π d/2
r =
rd ,
d
d (d/2 − 1)!
is the volume of a d–dimensional sphere of radius r. Thus,
B2 (T ) =
1
1
Vd (b) − exp(βε) [Vd (b) − Vd (a)] .
2
2
For high temperatures exp(βε) ≈ 1 + βε, and
B2 (T ) ≈
1
βε
Vd (a) −
[Vd (b) − Vd (a)] .
2
2
At the highest temperatures, βε ≪ 1, the hard-core part of the potential is dominant, and
B2 (T ) ≈
1
Vd (a).
2
For low temperatures β ≫ 1, the attractive component takes over, and
1
B2 = − {Vd (a)(−1) + [Vd (b) − Vd (a)] · [exp(βε) − 1]}
2
1
≈ − [Vd (b) − Vd (a)] exp(βε),
2
119
resulting in B2 < 0.
(b) Calculate the first correction to isothermal compressibility
κT = −
1 ∂V
V ∂P
.
T,N
• The isothermal compressibility is defined by
κT ≡ −
From the expansion
1 ∂V
V ∂P
.
T,N
P
N
N2
=
+ 2 B2 ,
kB T
V
V
for constant temperature and particle number, we get
N
N2
1
dP = − 2 dV − 2B2 3 dV.
kB T
V
V
Thus
and
∂V
∂P
T,N
1
V2
1
=−
=
−
kB T N/V 2 + 2B2 N 2 /V 3
N kB T
V
κT =
N kB T
1
1 + 2B2 N/V
V
≈
N kB T
1
1 + 2B2 N/V
,
N
1 − 2B2
.
V
(c) In the high temperature limit, reorganize the equation of state into the van der Waals
form, and identify the van der Waals parameters.
• Including the correction introduced by the second virial coefficient, the equation of state
becomes
PV
N
= 1 + B2 (T ).
N kB T
V
Using the expression for B2 in the high temperature limit,
N
PV
=1+
{Vd (a) − βε[Vd (b) − Vd (a)]} ,
N kB T
2V
and
N
N2
ε[Vd (b) − Vd (a)] = kB T
P+
2
2V
V
120
N
1+
Vd (a) .
2V
Using the variable n = N/V , and noting that for low concentrations
−1
−1
n
n
N
1 + Vd (a) ≈ 1 − Vd (a)
= V V − Vd (a)
,
2
2
2
the equation of state becomes
N
n2 ε
[Vd (b) − Vd (a)] · V − Vd (a) = N kB T.
P+
2
2
This can be recast in the usual van der Waals form
(P − an2 ) · (V − N b) = N kB T,
with
a=
ε
[Vd (b) − Vd (a)] ,
2
and
b=
1
Vd (a).
2
(d) For b = a (a hard sphere), and d = 1, calculate the third virial coefficient B3 (T ).
• By definition, the third virial coefficient is
Z
1
dd rdd r ′ f (r)f (r ′ )f (r − r ′ ),
B3 = −
3
where, for a hard core gas
(
−1
for
0 < r < a,
V(r)
f (r) ≡ exp −
−1=
.
kB T
0
for
a<r<∞
In one–dimension, the only contributions come from 0 < r, and r ′ < a, where f (r) =
f (r ′ ) = −1. Using the notations |x| ≡ r, |y| ≡ r ′ (i.e. −a < x, and y < a),
Z
Z a
Z Z
1
16
1 a
dx
dy · f (x − y) =
(−1) =
(2a)2 = a2 ,
B3 = −
3 −a
3
3
8
−a
−a<x,y<a,−a<x−y<a
where the relevant integration area is plotted below.
y
a
-a
a
-a
121
x
********
3. Dieterici’s equation: A gas obeys Dieterici’s equation of state:
P (v − b) = kB T exp −
a
kB T v
,
where v = V /N .
(a) Find the ratio P v/kB T at the critical point.
• The critical point is the point of inflection, described by
∂P
∂v
= 0,
∂ 2P
∂v 2
and
Tc ,N
= 0.
Tc ,N
T>Tc
Pc
T=Tc
pressure P
T<Tc
vc
specific volume v
The first derivative of P is
kB T
a
∂P
a
a
1
∂ kB T
=
exp −
exp −
−
=
∂v Tc ,N ∂v v − b
kB T v
v−b
kB T v
kB T v 2
v−b
1
a
,
−
=P
kB T v 2
v−b
while a second derivative gives
∂ 2P
∂v 2
Tc ,N
∂
a
1
=
P
−
∂v
kB T v 2
v−b
a
2a
∂P
1
1
=
−P
−
−
.
∂v kB T v 2
v−b
kB T v 3
(v − b)2
122
Therefore vc and Tc are determined by
a
1
−
= 0,
2
kB Tc vc
vc − b
and
1
2a
−
= 0,
3
kB Tc vc
(vc − b)2
and
kB Tc =
with the solutions
vc = 2b,
a
.
4b
The critical pressure is
kB Tc
a
a
Pc =
= 2 e−2 ,
exp −
vc − b
kB Tc vc
4b
resulting in the ratio
Pc vc
= 2e−2 ≈ 0.27.
kB Tc
Note that for the van der Waals gas
3
Pc vc
= = 0.375,
kB Tc
8
while for some actual gases
Pc vc
= 0.230,
kB Tc water
and
Pc vc
kB Tc
= 0.291.
Argon
(b) Calculate the isothermal compressibility κT for v = vc as a function of T − Tc .
• The isothermal compressibility is defined by
κT ≡ −
1 ∂v
v ∂P
,
T,N
and from part (a), given by
∂P
∂v
=P
Tc ,N
a
1
−
2
kB T v
v−b
.
Expanding this expression, at v = vc , in terms of t ≡ kB T − kB Tc (for T > Tc ), yields
Pc 4bt
∂P
1
2Pc
a
≈−
−
=−
t,
≈ Pc
2
∂v Tc ,N
(a/4b + t) 4b
b
b a
vc kB Tc
and thus
κT =
kB Tc 1
be2
=
.
2Pc t
2kB (T − Tc )
123
Note that expanding any analytic equation of state will yield the same simple pole for the
divergence of the compressibility.
(c) On the critical isotherm expand the pressure to the lowest non-zero order in (v − vc ).
• Perform a Taylor–series expansion along the critical isotherm T = Tc , as
P (v, Tc ) = Pc +
∂P
∂v
Tc ,vc
(v − vc ) +
1 ∂ 2P
2! ∂v 2
Tc ,vc
(v − vc )2 +
1 ∂ 3P
3! ∂v 3
Tc ,vc
(v − vc )3 + · · · .
The first two terms are zero at the critical point, and
∂ 3P
∂v 3
Tc ,vc
∂
2a
1
= −Pc
−
∂v kB Tc v 3
(v − b)2
6a
2
= −Pc
−
kB Tc vc4
(vc − b)3
Pc
= − 3.
2b
Substituting this into the Taylor expansion for P (v, Tc ), results in
(v − vc )3
,
P (v, Tc ) = Pc 1 −
12b3
which is equivalent to
P
2
−1=−
Pc
3
v
−1
vc
3
.
********
4. Two dimensional Coulomb gas:
Consider a classical mixture of N positive, and N
negative charged particles in a two dimensional box of area A = L × L. The Hamiltonian
is
H=
2N
X
p~i 2
i=1
2m
−
2N
X
i<j
ci cj ln |~qi − ~qj |
,
where ci = +c0 for i = 1, · · · N , and ci = −c0 for i = N + 1, · · · 2N , denote the charges of
the particles; {~qi } and {~pi } their coordinates and momenta respectively.
(a) Note that in the interaction term each pair appears only once, and that there is no
self interaction i = j. How many pairs have repulsive interactions, and how many have
attractive interactions?
124
• There are N positively charged particles, and N negatively charged particles. Hence
there are N · N = N 2 pairs of opposite charges, and nattractive = N 2 . For like charges,
we can choose pairs from the N particles of positive charge, or from the N particles with
negative charges. Hence the number of pairs of like pairs is
!
N
N!
nrepulsive = 2 ×
=2×
= N (N − 1).
2!(N − 2)!
2
(b) Write down the expression for the partition function Z(N, T, A) in terms of integrals
over {~qi } and {~pi }. Perform the integrals over the momenta, and rewrite the contribution
of the coordinates as a product involving powers of {~qi }, using the identity eln x = x.
• The partition function is


Z Y
2N
2N
2
X
X p
1
i
2
2
−β
Z(N, T, A) =
d
q
~
d
p
~
exp
+
β
ci cj ln |~q i − ~q j |
i
i
(N !)2 h4N
2m
i<j
i=1
i=1
1
= 4N
λ (N !)2
√
Z Y
2N
i=1
d2 ~q i exp [β ln |~q i − ~q j |ci cj ] ,
where λ = h/ 2πmkB T . Further simplifying the expression for the partition function
1
Z(N, T, A) = 4N
λ (N !)2
Z Y
2N
d2 ~q i
i=1
2N
Y
i<j
|~q i − ~q j |βci cj ,
where we have used the fact that eln x = x.
(c) Although it is not possible to perform the integrals over {~qi } exactly, the dependence
of Z on A can be obtained by the simple rescaling of coordinates, ~qi ′ = ~qi /L. Use the
2
results in parts (a) and (b) to show that Z ∝ A2N−βc0 N/2 .
• The only length scale appearing in the problem is set by the system size L. Rescaling
the expression using ~q i ′ = ~q i /L, then yields
1
Z(N, T, A) = 4N
λ (N !)2
Z Y
2N
L2 d2 ~q i ′
2N
Y
i<j
i=1
Lβci cj |~q i ′ − ~q j ′ |βci cj .
Note that there are N 2 terms for which the interaction is attractive (βci cj = −βc20 ), and
N (N − 1) terms for which the interaction is repulsive (βci cj = βc20 ). Thus
Z(N, T, A) = L
=L
4N
·L
βc20 N(N−1)
4N−βNc20
·L
−βc20 N 2
1
4N
λ (N !)2
Z0 (N, T, A′ = L′2 = 1) ∝ A
125
Z Y
2N
2
d ~q i
2N
Y
i<j
i=1
2N−βc20 N/2
′
,
|~q i ′ − ~q j ′ |βci cj
since A = L2 .
(d) Calculate the two dimensional pressure of this gas, and comment on its behavior at
high and low temperatures.
• The pressure is then calculated from
P =−
1 ∂ ln Z
β ∂A
= kB T
N,T
= kB T 2N − βc20 N/2
At high temperatures,
2
∂
ln A2N−βc0 N/2 Z0
∂A
∂
2N kB T
N c20
ln A =
−
.
∂A
A
2A
P =
2N kB T
,
A
which is the ideal gas behavior for 2N particles. The pressure becomes negative at temperature below
Tc0 =
c20
,
4kB
which is unphysical, indicating the collapse of the particles due to their attractions.
(e) The unphysical behavior at low temperatures is avoided by adding a hard–core which
prevents the coordinates of any two particles from coming closer than a distance a. The
appearance of two length scales a and L, makes the scaling analysis of part (c) questionable.
By examining the partition function for N = 1, obtain an estimate for the temperature Tc
at which the short distance scale a becomes important in calculating the partition function,
invalidating the result of the previous part. What are the phases of this system at low and
high temperatures?
• A complete collapse of the system (to a single point) can be avoided by adding a hard
core repulsion which prevents any two particles from coming closer than a distance a. The
partition function for two particles (i.e. N = 1) is now given by
1
Z(N = 1, T, A) = 4
λ
Z
2
d2 ~q 1 d2 ~q 2 · |~q 1 − ~q 2 |−βc0 .
To evaluate this integral, first change to center of mass and relative coordinates

Q
~ = 1 (~q 1 + ~q 2 ),
2

~q =~q 1 − ~q 2 .
126
Integrating over the center of mass gives
A
Z(N = 1, T, A) = 4
λ
Z
2
d ~q q
−βc20
2−βc20
2πA q
= 4
λ 2 − βc20
If 2 − βc20 < 0, as L → ∞,
L
a
2πA
≈ 4
λ
Z
L
a
2
dq · q 1−βc0
2
2
2πA L2−βc0 − a2−βc0
= 4
.
λ
2 − βc20
2
2πA a2−βc0
,
Z≈ 4
λ 2 − βc20
is controlled by the short distance cutoff a; while if 2−βc20 > 0, the integral is controlled by
the system size L, as assumed in part (c). Hence the critical temperature can be estimated
by βc20 = 2, giving
Tc =
c20
,
2kB
which is larger than Tc0 by a factor of 2. Thus the unphysical collapse at low temperatures
is preempted at the higher temperature where the hard cores become important. The high
temperature phase (T > Tc ) is a dissociated plasma; while the low temperature phase is a
gas of paired dipoles.
********
5. Exact solutions for a one dimensional gas:
In statistical mechanics, there are very
few systems of interacting particles that can be solved exactly. Such exact solutions are
very important as they provide a check for the reliability of various approximations. A
one dimensional gas with short-range interactions is one such solvable case.
(a) Show that for a potential with a hard core that screens the interactions from further
neighbors, the Hamiltonian for N particles can be written as
N
N
X
X
p2i
+
V(xi − xi−1 ).
H=
2m
i=2
i=1
The (indistinguishable) particles are labelled with coordinates {xi } such that
0 ≤ x1 ≤ x2 ≤ · · · ≤ xN ≤ L,
where L is the length of the box confining the particles.
127
• Each particle i interacts only with adjacent particles i − 1 and i + 1, as the hard cores
from these nearest neighbors screen the interactions with any other particle. Thus we
need only consider nearest neighbor interactions, and, included the kinetic energies, the
Hamiltonian is given by
N
N
X
X
p2i
H=
+
V(xi − xi−1 ),
2m i=2
i=1
0 ≤ x1 ≤ x2 ≤ · · · xN ≤ L.
for
(b) Write the expression for the partition function Z(T, N, L). Change variables to δ1 =
x1 , δ2 = x2 − x1 , · · · , δN = xN − xN−1 , and carefully indicate the allowed ranges of
integration and the constraints.
• The partition function is
1
Z(T, N, L) = N
h
Z
L
·
Z
∞
1
= N
λ
Z
L
dx1
Z
L
x1
0
−∞
dp1 · · ·
dx1
0
Z
dx2 · · ·
Z
∞
−∞
"
L
xN −1
dxN exp −β
"
dpN exp −β
L
x1
Z
dx2 · · ·
Z
L
xN −1
X
N
i=1
N
X
i=2
p2i
2m
"
dxN exp −β
V(xi − xi−1 )
#
#
N
X
i=2
#
V(xi − xi−1 ) ,
√
where λ = h/ 2πmkB T . (Note that there is no N ! factor, as the ordering of the particles
is specified.) Introducing a new set of variables
δ2 = x2 − x1 ,
δ1 = x1 ,
···
δn = xN − xN−1 ,
or equivalently
x1 = δ1 ,
x2 = δ1 + δ2 ,
···
xN =
1
Z(T, N, L) = N
λ
L
dδ1
0
Z
L−δ1
dδ2
0
Z
L−(δ1 +δ2 )
dδ3 · · ·
0
δi ,
i=1
the integration becomes
Z
N
X
Z
PN
L−
0
i=1
δi
−β
dδN e
This integration can also be expressed as
1
Z(T, N, L) = N
λ
Z
dδ1 dδ2 · · · dδN
128
′
"
exp −β
N
X
i=2
#
V(δi ) ,
PN
i=2
V (δi ) .
with the constraint
0≤
N
X
i=1
δi ≤ L.
This constraint can be put into the equation explicitly with the use of the step function
Θ(x) =
(
0
for
x<0
1
for
x≥0
,
as
1
Z(T, N, L) = N
λ
Z
∞
0
dδ1
Z
∞
0
dδ2 · · ·
Z
∞
0
"
dδN exp −β
N
X
i=2
#
V(δi ) Θ L −
N
X
i=1
δi
!
.
(c) Consider the Gibbs partition function obtained from the Laplace transformation
Z(T, N, P ) =
Z
∞
dL exp(−βP L)Z(T, N, L),
0
and by extremizing the integrand find the standard formula for P in the canonical ensemble.
• The Gibbs partition function is
Z(T, N, P ) =
Z
∞
dL exp(−βP L)Z(T, N, L).
0
The saddle point is obtained by extremizing the integrand with respect to L,
∂
exp(−βP L)Z(T, N, L)
∂L
= 0,
T,N
which implies that
βP =
∂
ln Z(T, N, L)
∂L
,
=⇒
P = kB T
T,N
∂ ln Z
∂L
From thermodynamics, for a one–dimensional gas we have
dF = −SdT − P dL,
=⇒
P =−
Further noting that
F = −kB T ln Z,
129
∂F
∂L
.
T,N
.
T,N
again results in
Pcanonical = kB T
(d) Change variables from L to δN+1 = L−
∂ ln Z
∂L
PN
.
T,N
and find the expression for Z(T, N, P )
i=1 δi ,
as a product over one-dimensional integrals over each δi .
• The expression for the partition function given above is
1
Z(T, N, L) = N
λ
Z
∞
dδ1
0
Z
∞
0
dδ2 · · ·
∞
Z
0
"
dδN exp −β
N
X
i=2
#
V(δi ) Θ L −
N
X
δi
i=1
!
.
The Laplace transform of this equation is
1
Z(T, N, P ) = N
λ
∞
Z
Z
∞
dδ2 · · ·
dδ1
dL exp(−βP L)
0
0
!
#
"
N
N
X
X
δi
V(δi ) Θ L −
· exp −β
0
1
= N
λ βP
=
∞
dδ1
0
Z
∞
0
1
N
λ (βP )2
Z
dδ2 · · ·
∞
0
∞
Z
dδN
0
i=1
i=2
Z
∞
Z
Z
dδ2 · · ·
∞
0
Z
"
dδN exp −β
∞
0
(
N
X
i=2
dδN exp −
#
"
V(δi ) exp −βP
N
X
N
X
δ1
i=1
)
!#
[βV(δi ) + βP δi ] .
i=2
Since the integrals for different δi′ s are equivalent, we obtain
1
Z(T, N, P ) = N
λ (βP )2
∞
Z
dδ exp [−β (V(δ) + P δ)]
0
N−1
.
This expression can also be obtained directly, without use of the step function as follows.
1
Z(T, N, P ) = N
λ
Z
·
Z
1
= N
λ
Z
L
dδ1
Z
L−δ1
dδ2
0
0
∞
0
"
Z
L−(δ1 +δ2 )
0
N
X
dL exp −βP L − β
L
dδ1
Z
L−δ1
0
0
(
· exp −βP
Z
dδ2 · · ·
"N
X
i=1
δi +
i=2
dδ3 · · ·
!#
0
L−
130
PN
L−
0
i=1
δi
dδN
V(δ)
PN
L−
Z
i=1
N
X
i=1
δi
dδN
Z
∞
−
δi
!#
−β
PN
i=1
N
X
i=2
δi
d L−
V(δi )
!)
N
X
i=1
.
δi
!
Change variables to δN+1 ≡ L −
PN
and note that each of the δ ′ s indicates the
i=1 δi ,
distance between neighboring particles. The size of the gas L, has been extended to any
value, hence each δ can be varied independently from 0 to ∞. Thus the Gibbs partition
function is
1
Z(T, N, P ) = N
λ
Z
∞
0
dδ1
"
Z
∞
0
· exp −βP
1
= N
λ
·
1
= N
λ (βP )2
∞
Z
Z
0
∞
dδ2 · · ·
N+1
X
Z
δi
i=1
∞
dδN
0
!
−β
∞
Z
0
N
X
i=2
dδ · exp [−βV(δ) − βP δ]
dδN+1
!#
V(δi )
N−1 Z
∞
dδ1 exp(−βP δ1 )
0
dδN+1 exp(−βP δN+1 )
0
Z
∞
dδ exp [−β (V(δ) + P δ)]
0
N−1
.
(e) At a fixed pressure P , find expressions for the mean length L(T, N, P ), and the density
n = N/L(T, N, P ) (involving ratios of integrals which should be easy to interpret).
• The mean length is
∂
ln Z(T, N, P )
∂(βP )
T,N
R∞
dδ · δ · exp [−βV(δ) − βP δ]
2
+ (N − 1) 0R ∞
,
=
βP
dδ · exp [−βV(δ) − βP δ]
0
L(T, N, P ) = −kB T
and the density n is given by
n=
N
=N
L(T, N, P )
(
)−1
R∞
dδ
·
δ
·
exp
[−β
(V(δ)
−
P
δ)]
2kB T
+ (N − 1) 0R ∞
.
P
dδ · exp [−β (V(δ) − P δ)]
0
Note that for an ideal gas V i.g. (δ) = 0, and
Li.g. (T, N, P ) =
leading to
n(p)i.g. =
(N + 1)kB T
,
P
N
P
.
N + 1 kB T
131
Since the expression for n(T, P ) in part (e) is continuous and non-singular for any
choice of potential, there is in fact no condensation transition for the one-dimensional
gas. By contrast, the approximate van der Waals equation (or the mean-field treatment)
incorrectly predicts such a transition.
(f) For a hard sphere gas, with minimum separation a between particles, calculate the
equation of state P (T, n). Compare the excluded volume factor with the approximate
result obtained in earlier problems, and also obtain the general virial coefficient Bℓ (T ).
• For a hard sphere gas
δi ≥ a,
i = 2, 3, · · · , N,
for
the Gibbs partition function is
1
Z(T, N, P ) = N
λ (βP )2
Z
∞
a
dδ exp (−βV(δ) − βP δ)
Z ∞
N−1
1
dδ exp (−βP δ)
= N
λ (βP )2 a
N+1
1
1
N−1
= N
exp (−βP a)
.
λ
βP
N−1
From the partition function, we can calculate the mean length
L = −kB T
∂ ln Z
∂P
=
T,N
(N + 1)kB T
+ (N − 1)a,
P
which after rearrangement yields
βP =
(N + 1)
n + 1/L
=
≈ (n + 1/l)(1 + (n − 1/L)a + (n − 1/L)2 a2 + · · ·).
L − (N − 1)a
1 − (n − 1/L)a
For N ≫ 1, n ≫ 1/L, and
βP ≈ n(1 + na + n2 a2 + · · ·) = n + an2 + a2 n3 + · · · ,
which gives the virial coefficients
Bℓ (T ) = aℓ−1 .
The value of B3 = a2 agrees with the result obtained in an earlier problem. Also note
that the exact ‘excluded volume’ is (N − 1)a, as opposed to the estimate of N a/2 used in
deriving the van der Waals equation.
132
********
6. The Manning transition:
When ionic polymers (polyelectrolytes) such as DNA are
immersed in water, the negatively charged counter-ions go into solution, leaving behind
a positively charged polymer. Because of the electrostatic repulsion of the charges left
behind, the polymer stretches out into a cylinder of radius a, as illustrated in the figure.
While thermal fluctuations tend to make the ions wander about in the solvent, electrostatic
attractions favor their return and condensation on the polymer. If the number of counterions is N , they interact with the N positive charges left behind on the rod through the
potential φ (r) = −2 (N e/L) ln (r/L), where r is the radial coordinate in a cylindrical
geometry. If we ignore the Coulomb repulsion between counter-ions, they can be described
by the classical Hamiltonian
N 2
r X
pi
2
,
+ 2e n ln
H=
2m
L
i=1
where n = N/L.
2a
R
-
z
+
z
+
h
z
+
z
+
r
-
L
z
+
(a) For a cylindrical container of radius R, calculate the canonical partition function Z in
terms of temperature T , density n, and radii R and a.
133
• The canonical partition function is
)
(
Z Q 3 3
N 2
r X
d
p
d
q
p
i
i
i
i
exp −β
+ 2e2 n ln
Z=
N !h3N
2m
L
i=1
#N
"Z
N
R
2
2
2πLe
rdr · r −2e n/kB T
=
LN·β2e n
N λ3
a
#N
"
N
2
2(1−e2 n/kB T )
2πe
− a2(1−e n/kB T )
2Ne2 nβ R
=
L
.
nλ3
2 (1 − e2 n/kB T )
(b) Calculate the probability distribution function p (r) for the radial position of a counterion, and its first moment hri, the average radial position of a counter-ion.
• Integrating out the unspecified N momenta and N − 1 positions from the canonical
distribution, one obtains the distribution function
2
2
re−(2e n/kB T ) ln(r/L)
e2 n
r 1−2e n/kB T
p (r) = R R
.
=2 1−
−(2e2 n/kB T ) ln(r/L)
kB T R2(1−e2 n/kB T ) − a2(1−e2 n/kB T )
drre
a
RR
(Note the normalization condition a drp(r) = 1.) The average position is then
!
Z R
2
2
2kB T − 2e2 n
R3−2e n/kB T − a3−2e n/kB T
hri =
rp (r) dr =
.
3kB T − 2e2 n
R2−2e2 n/kB T − a2−2e2 n/kB T
a
(c) The behavior of the results calculated above in the limit R ≫ a is very different at high
and low temperatures. Identify the transition temperature, and characterize the nature of
the two phases. In particular, how does hri depend on R and a in each case?
• Consider first low temperatures, such that e2 n/kB T > 1. In the R ≫ a limit, the
distribution function becomes
1−2e2 n/kB T
e2 n
r
p (r) = 2 1 −
,
kB T a2(1−e2 n/kB T )
and hri ∝ a. To see this, either examine the above calculated average hri in the R ≫ a
limit, or notice that
2
e2 n
p (r) dr = 2 1 −
x1−2e n/kB T dx,
kB T
R∞
2
where x = r/a, immediately implying hri ∝ a (as 1 dxx1−2e n/kB T < ∞ if e2 n/kB T > 1).
On the other hand, at high temperatures (e2 n/kB T < 1), the distribution function reduces
to
1−2e2 n/kB T
r
e2 n
,
p (r) = 2 1 −
kB T R2(1−e2 n/kB T )
134
and hri ∝ R, from similar arguments. Thus, at temperature Tc = e2 n/kB there is a
transition from a “condensed” phase, in which the counter-ions are stuck on the polymer,
to a “gas” phase, in which the counter-ions fluctuate in water at typical distances from
the polymer which are determined by the container size.
(d) Calculate the pressure exerted by the counter-ions on the wall of the container, at
r = R, in the limit R ≫ a, at all temperatures.
• The work done by the counter-ions to expand the container from a radius R to a radius
R + dR is
dW = dF = (force) dR = −P (2πRL) dR,
leading to
P =−
1 ∂F
kB T ∂lnZ
=
.
2πRL ∂R
2πRL ∂R
At low temperatures, T < Tc , the pressure vanishes, since the partition function is independent of R in the limit R ≫ a. At T > Tc , the above expression results in
kB T
e2 n 1
P =
2N 1 −
,
2πRL
kB T R
i.e.
e2 n
.
P V = N kB T 1 −
kB T
(e) The character of the transition examined in part (d) is modified if the Coulomb interactions between counter-ions are taken into account. An approximate approach to the
interacting problem is to allow a fraction N1 of counter-ions to condense along the polymer
rod, while the remaining N2 = N − N1 fluctuate in the solvent. The free counter-ions are
again treated as non-interacting particles, governed by the Hamiltonian
N 2
r X
pi
2
H=
,
+ 2e n2 ln
2m
L
i=1
where n2 = N2 /L. Guess the equilibrium number of non-interacting ions, N2∗ , and justify
your guess by discussing the response of the system to slight deviations from N2∗ . (This is
a qualitative question for which no new calculations are needed.)
• Consider a deviation (n2 ) from n∗2 ≡ N2∗ /V ≡ kB T /e2 , occuring at a temperature lower
than Tc (i.e. e2 n/kB T > 1). If n2 > n∗2 , the counter-ions have a tendency to condensate
(since e2 n/kB T > 1), thus decreasing n2 . On the other hand, if n2 > n∗2 , the counter-ions
tend to “evaporate” (since e2 n/kB T < 1). In both cases, the system drives the density n2
135
to the (equilibrium) value of n∗2 = kB T /e2 . If the temperature is higher than Tc , clearly
n∗2 = n and there is no condensation.
********
7. Hard rods:
A collection of N asymmetric molecules in two dimensions may be modeled
as a gas of rods, each of length 2l and lying in a plane. A rod can move by translation of
its center of mass and rotation about latter, as long as it does not encounter another rod.
Without treating the hard-core interaction exactly, we can incorporate it approximately
by assuming that the rotational motion of each rod is restricted (by the other rods) to an
angle θ, which in turn introduces an excluded volume Ω (θ) (associated with each rod).
The value of θ is then calculated self consistently by maximizing the entropy at a given
density n = N/V , where V is the total accessible area.
θ
excluded
volume
2l
(a) Write down the entropy of such a collection of rods in terms of N , n, Ω, and A (θ), the
phase space volume associated to the rotational freedom of a single rod. (You may ignore
the momentum contributions throughout, and consider the large N limit.)
• Including both forms of entropy, translational and rotational, leads to
"
1
S = kB ln
N!
N Ω(θ)
V −
2
N
A(θ)
N
#
≈ N kB
Ω(θ)
−1
ln n −
+ 1 + ln A(θ) .
2
(b) Extremizing the entropy as a function of θ, relate the density to Ω, A, and their
derivatives Ω′ , A′ ; express your result in the form n = f (Ω, A, Ω′ , A′ ).
136
• The extremum condition ∂S/∂θ = 0 is equivalent to
Ω′
A′
=
,
2n−1 − Ω
A
where primes indicate derivatives with respect to θ. Solving for the density gives
n=
2A′
.
ΩA′ + Ω′ A
(c) Express the excluded volume Ω in terms of θ and sketch f as a function of θ ∈ [0, π],
assuming A ∝ θ.
• Elementary geometry yields
Ω = l2 (θ + sin θ) ,
so that the equilibrium condition becomes
2
−1
[θ (2 + cos θ) + sin θ] ,
2
l
n = f (θ) =
with the function f (θ) plotted below:
f(θ)
n
nc
0
θc
θ
π
(d) Describe the equilibrium state at high densities. Can you identify a phase transition
as the density is decreased? Draw the corresponding critical density nc on your sketch.
What is the critical angle θc at the transition? You don’t need to calculate θc explicitly,
but give an (implicit) relation defining it. What value does θ adopt at n < nc ?
• At high densities, θ ≪ 1 and the equilibrium condition reduces to
N≈
V
;
2θl2
137
the angle θ is as open as allowed by the close packing. The equilibrium value of θ increases
as the density is decreased, up to its “optimal” value θc at nc , and θ (n < nc ) = θc . The
transition occurs at the minimum of f (θ), whence θc satisfies
d
[θ (2 + cos θ) + sin θ] = 0,
dθ
i.e.
2 (1 + cos θc ) = θc sin θc .
Actually, the above argument tracks the stability of a local maximum in entropy (as density
is varied) which becomes unstable at θc . There is another entropy maximum at θ = π,
corresponding to freely rotating rods, which becomes more advantageous (i.e. the global
equilibrium state) at a density slightly below θc .
********
8. Surfactant condensation:
N surfactant molecules are added to the surface of water
over an area A. They are subject to a Hamiltonian
H=
N
X
p~i 2
i=1
2m
+
1X
V(~ri − ~rj ),
2 i,j
where ~ri and p~i are two dimensional vectors indicating the position and momentum of
particle i. (This simple form ignores the couplings to the fluid itself. The actual kientic
and potential energies are more complicated.)
(a) Write down the expression for the partition function Z(N, T, A) in terms of integrals
over ~ri and p~i , and perform the integrals over the momenta.
• The partition function is obtained by integrating the Boltzmann weight over phase space,
as
Z(N, T, A) =
Z QN
2

2
N
X

p2i
X
d p~i d ~qi
−β
V(~qi − ~qj ) ,
exp
−
β
N !h2N
2m
i<j
i=1
i=1
with β = 1/(kB T ). The integrals over momenta are simple Gaussians, yielding
Z(N, T, A) =
Z Y
N
1
1
N ! λ2N
i=1

d2 ~qi exp −β
X
i<j
V(~qi − ~qj ) ,
√
where as usual λ = h/ 2πmkB T denotes the thermal wavelength.
138

The inter–particle potential V(~r) is infinite for separations |~r | < a, and attractive for
R∞
|~r | > a such that a 2πrdrV(r) = −u0 .
(b) Estimate the total non–excluded area available in the positional phase space of the
system of N particles.
• To estimate the joint phase space of particles with excluded areas, add them to the
system one by one. The first one can occupy the whole area A, while the second can
explore only A − 2Ω, where Ω = πa2 . Neglecting three body effects (i.e. in the dilute
limit), the area available to the third particle is (A − 2Ω), and similarly (A − nΩ) for the
n-th particle. Hence the joint excluded volume in this dilute limit is
A(A − Ω)(A − 2Ω) · · · (A − (N − 1)Ω) ≈ (A − N Ω/2)N ,
where the last approximation is obtained by pairing terms m and (N − m), and ignoring
order of Ω2 contributions to their product.
(c) Estimate the total potential energy of the system, within a uniform density approximation n = N/A. Using this potential energy for all configurations allowed in the previous
part, write down an approximation for Z.
• Assuming a uniform density n = N/A, an average attractive potential energy, Ū , is
estimated as
Z
1
1X
V attr. (~qi − ~qj ) =
Ū =
d2~r1 d2~r2 n(~r1 )n(~r2 )V attr. (~r1 − ~r2 )
2 i,j
2
Z
N2
n2
u0 .
≈ A d2~r V attr. (~r ) ≡ −
2
2A
Combining the previous results gives
1 1
βu0 N 2
N
Z(N, T, A) ≈
.
(A − N Ω/2) exp
N ! λ2N
2A
(d) The surface tension of water without surfactants is σ0 , approximately independent of
temperature. Calculate the surface tension σ(n, T ) in the presence of surfactants.
• Since the work done is changing the surface area is dW = σdA, we have dF = −T dS +
σdA + µdN , where F = −kB T ln Z is the free energy. Hence, the contribution of the
surfactants to the surface tension of the film is
σs = −
∂ ln Z
∂A
T,N
=−
N kB T
u0 N 2
+
,
A − N Ω/2
2A2
139
which is a two-dimensional variant of the familiar van der Waals equation. Adding the
(constant) contribution in the absence of surfactants gives
∂ ln Z
∂A
σ(n, T ) = σ0 −
T,N
=−
N kB T
u0 N 2
+
.
A − N Ω/2
2A2
(e) Show that below a certain temperature, Tc , the expression for σ is manifestly incorrect.
What do you think happens at low temperatures?
• Thermodynamic stability requires δσδA ≥ 0, i.e. σ must be a monotonically increasing
function of A at any temperature. This is the case at high temperatures where the first
term in the equation for σs dominates, but breaks down at low temperatures when the term
from the attractive interactions becomes significant. The critical temperature is obtained
by the usual conditions of ∂σs /∂A = ∂ 2 σs /∂A2 = 0, i.e. from

∂σs



 ∂A

∂ 2 σs



∂A2
=
T
T
N kB T
u0 N 2
−
=0
(A − N Ω/2)2
A3
2N kB T
3u0 N 2
=−
+
=0
(A − N Ω/2)3
A4
,
The two equations are simultaneously satisfied for Ac = 3N Ω/2, at a temperature
Tc =
8u0
.
27kB Ω
As in the van der Waals gas, at temperatures below Tc , the surfactants separate into a
high density (liquid) and a low density (gas) phase.
(f) Compute the heat capacities, CA and write down an expression for Cσ without explicit
evaluation, due to the surfactants.
• The contribution of the surfactants to the energy of the film is given by
Es = −
∂ ln Z
kB T
u0 N 2
= 2N ×
−
.
∂β
2
2A
The first term is due to the kinetic energy of the surfactants, while the second arises from
their (mean-field) attraction. The heat capacities are then calculated as
CA =
dQ
dT
=
A
∂E
∂T
140
= N kB ,
A
and
Cσ =
dQ
dT
=
σ
∂E
∂T
σ
−σ
∂A
∂T
.
σ
********
9. Critical point behavior:
The pressure P of a gas is related to its density n = N/V ,
and temperature T by the truncated expansion
b
c
P = kB T n − n 2 + n 3
2
6
,
where b and c are assumed to be positive temperature independent constants.
(a) Locate the critical temperature Tc below which this equation must be invalid, and
the corresponding density nc and pressure Pc of the critical point. Hence find the ratio
kB Tc nc /Pc .
• Mechanical stability of the gas requires that any spontaneous change in volume should be
opposed by a compensating change in pressure. This corresponds to δP δV < 0, and since
δn = −(N/V 2 )δV , any equation of state must have a pressure that is an increasing func-
tion of density. The transition point between pressure isotherms that are monotonically
increasing functions of n, and those that are not (hence manifestly incorrect) is obtained
by the usual conditions of dP/dn = 0 and d2 P/dn2 = 0. Starting from the cubic equation
of state, we thus obtain
dP
c
=kB Tc − bnc + n2c = 0
dn
2
.
d2 P
= − b + cnc = 0
dn2
From the second equation we obtain nc = b/c, which substituted in the first equation
gives kB Tc = b2 /(2c). From the equation of state we then find Pc = b3 /(6c2 ), and the
dimensionless ratio of
kB Tc nc
= 3.
Pc
(b) Calculate the isothermal compressibility κT = − V1
∂V
∂P T ,
and sketch its behavior as a
function of T for n = nc .
• Using V = N/n, we get
1 ∂V
κT (n) = −
V ∂P
=
T
1 ∂P
n ∂n
−1
T
−1
= n kB T − bn + cn2 /2
.
141
For n = nc , κT (nc ) ∝ (T − Tc )−1 , and diverges at Tc .
(c) On the critical isotherm give an expression for (P − Pc ) as a function of (n − nc ).
• Using the coordinates of the critical point computed above, we find
b3
b2
b
c
+
n − n2 + n3
2
6c
2c
2
6
2
b 2
b
b3
c
3
n − 3 n + 3 2n − 3
=
6
c
c
c
c
3
= (n − nc ) .
6
P − Pc = −
(d) The instability in the isotherms for T < Tc is avoided by phase separation into a liquid
of density n+ and gas of density n− . For temperatures close to Tc , these densities behave
as n± ≈ nc (1 ± δ). Using a Maxwell construction, or otherwise, find an implicit equation
for δ(T ), and indicate its behavior for (Tc − T ) → 0. (Hint: Along an isotherm, variations
of chemical potential obey dµ = dP/n.)
• According to the Gibbs–Duhem relation, the variations of the intensive variables are
related by SdT − V dP + N dµ = 0, and thus along an isotherm (dT = 0) dµ = dP/n =
∂P/∂n|T dn/n. Since the liquid and gas states are in coexistence they should have the
same chemical potential. Integrating the above expression for dµ from n− to n+ leads to
the so-called Maxwell construction, which reads
0 = µ(n+ ) − µ(n− ) =
Z
n+
n−
dP
=
n
Z
nc (1+δ)
dn
nc (1−δ)
kB T − bn + cn2 /2
n
.
Performing the integrals gives the equation
c 2
1+δ
1+δ
2
2
0 = kB T ln
− bnc (2δ) + nc (1 + δ) − (1 − δ) = kB T ln
− 2kB Tc δ,
1−δ
4
1−δ
where for the find expression, we have used nc = b/c and kB Tc = b2 /(2c). The implicit
equation for δ is thus
T
δ=
ln
2Tc
1+δ
1−δ
≈
T
δ − δ3 + · · · .
Tc
The leading behavior as (Tc − T ) → 0 is obtained by keeping up to the cubic term, and
given by
δ≈
r
1−
142
Tc
.
T
********
10. The binary alloy: A binary alloy (as in β brass) consists of NA atoms of type A, and
NB atoms of type B. The atoms form a simple cubic lattice, each interacting only with its
six nearest neighbors. Assume an attractive energy of −J (J > 0) between like neighbors
A − A and B − B, but a repulsive energy of +J for an A − B pair.
(a) What is the minimum energy configuration, or the state of the system at zero temperature?
• The minimum energy configuration has as little A-B bonds as possible. Thus, at zero
temperature atoms A and B phase separate, e.g. as indicated below.
A
B
(b) Estimate the total interaction energy assuming that the atoms are randomly distributed
among the N sites; i.e. each site is occupied independently with probabilities pA = NA /N
and pB = NB /N .
• In a mixed state, the average energy is obtained from
E = (number of bonds) × (average bond energy)
= 3N · −Jp2A − Jp2B + JpA pB
2
NA − NB
= −3JN
.
N
(c) Estimate the mixing entropy of the alloy with the same approximation. Assume
NA , NB ≫ 1.
• From the number of ways of randomly mixing NA and NB particles, we obtain the
mixing entropy of
S = kB ln
N!
NA !NB !
143
.
Using Stirling’s approximation for large N (ln N ! ≈ N ln N − N ), the above expression can
be written as
S ≈ kB (N ln N − NA ln NA − NB ln NB ) = −N kB (pA ln pA + pB ln pB ) .
(d) Using the above, obtain a free energy function F (x), where x = (NA −NB )/N . Expand
F (x) to the fourth order in x, and show that the requirement of convexity of F breaks
down below a critical temperature Tc . For the remainder of this problem use the expansion
obtained in (d) in place of the full function F (x).
• In terms of x = pA − pB , the free energy can be written as
F = E − TS
2
= −3JN x + kB T
1+x
2
ln
1+x
2
+
1−x
2
ln
1−x
2
.
Expanding about x = 0 to fourth order, gives
F ≃ −N kB T ln 2 + N
kB T
− 3J
2
x2 +
N kB T 4
x .
12
Clearly, the second derivative of F ,
∂ 2F
= N (kB T − 6J) + N kB T x2 ,
2
∂x
becomes negative for T small enough. Upon decreasing the temperature, F becomes
concave first at x = 0, at a critical temperature Tc = 6J/kB .
(e) Sketch F (x) for T > Tc , T = Tc , and T < Tc . For T < Tc there is a range of
compositions x < |xsp (T )| where F (x) is not convex and hence the composition is locally
unstable. Find xsp (T ).
• The function F (x) is concave if ∂ 2 F /∂x2 < 0, i.e. if
2
x <
6J
−1 .
kB T
This occurs for T < Tc , at the spinodal line given by
xsp (T ) =
r
6J
− 1,
kB T
144
F(x)/NJ
T=0
x sp(T)
T<Tc
T=Tc
T>Tc
+1
-1
x
as indicated by the dashed line in the figure below.
(f) The alloy globally minimizes its free energy by separating into A rich and B rich phases
of compositions ±xeq (T ), where xeq (T ) minimizes the function F (x). Find xeq (T ).
• Setting the first derivative of dF (x) /dx = N x (kB T − 6J) + kB T x2 /3 , to zero yields
the equilibrium value of

r

6J
 ±√3
− 1 for T < Tc
.
xeq (T ) =
kB T


0
for T > Tc
(g) In the (T, x) plane sketch the phase separation boundary ±xeq (T ); and the so called
spinodal line ±xsp (T ). (The spinodal line indicates onset of metastability and hysteresis
effects.)
• The spinodal and equilibrium curves are indicated in the figure above. In the interval
between the two curves, the system is locally stable, but globally unstable. The formation
of ordered regions in this regime requires nucleation, and is very slow. The dashed area is
locally unstable, and the system easily phase separates to regions rich in A and B.
********
145
T
Tc
x eq(T)
x sp(T)
unstable
metastable
metastable
1
1
x
Problems for Chapter VI - Quantum Statistical Mechanics
1. One dimensional chain: A chain of N +1 particles of mass m is connected by N massless
springs of spring constant K and relaxed length a. The first and last particles are held
fixed at the equilibrium separation of N a. Let us denote the longitudinal displacements
of the particles from their equilibrium positions by {ui }, with u0 = uN = 0 since the end
particles are fixed. The Hamiltonian governing {ui }, and the conjugate momenta {pi }, is
H=
N−1
X
i=1
"
#
N−2
X
p2i
K
2
u21 +
(ui+1 − ui ) + u2N−1 .
+
2m
2
i=1
(a) Using the appropriate (sine) Fourier transforms, find the normal modes {ũk }, and the
corresponding frequencies {ωk }.
• From the Hamiltonian
H=
N−1
X
i=1
#
"
N−1
X
K
p2i
2
(ui − ui−1 ) + u2N−1 ,
u21 +
+
2m
2
i=2
the classical equations of motion are obtained as
m
d 2 uj
= −K(uj − uj−1 ) − K(uj − uj+1 ) = K(uj−1 − 2uj + uj+1 ),
dt2
146
for j = 1, 2, · · · , N − 1, and with u0 = uN = 0. In a normal mode, the particles oscillate
in phase. The usual procedure is to obtain the modes, and corresponding frequencies,
by diagonalizing the matrix of coefficeints coupling the displacements on the right hand
side of the equation of motion. For any linear system, we have md2 ui /dt2 = Kij uj , and
we must diagonalize Kij . In the above example, Kij is only a function of the difference
i − j. This is a consequence of translational symmetry, and allows us to diagonalize the
matrix using Fourier modes. Due to the boundary conditions in this case, the appropriate
transformation involves the sine, and the motion of the j-th particle in a normal mode is
given by
r
2 ±iωn t
e
sin (k(n) · j) .
N
The origin of time is arbitrary, but to ensure that uN = 0, we must set
ũk(n) (j) =
k(n) ≡
nπ
,
N
n = 1, 2, · · · , N − 1.
for
Larger values of n give wave-vectors that are simply shifted by a multiple of π, and hence
coincide with one of the above normal modes. The number of normal modes thus equals
the number of original displacement variables, as required. Furthermore, the amplitudes
are chosen such that the normal modes are also orthonormal, i.e.
N−1
X
j=1
ũk(n) (j) · ũk(m) (j) = δn,m .
By substituting the normal modes into the equations of motion we obtain the dispersion
relation
where ω0 ≡
nπ h
nπ i
= ω02 sin2
,
ωn2 = 2ω02 1 − cos
N
2N
p
K/m.
The potential energy for each normal mode is given by
N
N
h nπ
io2
KX
K X n nπ 2
Un =
|ui − ui−1 | =
sin
i − sin
(i − 1)
2 i=1
N i=1
N
N
N
X
1
4K
2 nπ
2 nπ
i−
.
cos
sin
=
N
2N i=1
N
2
Noting that
N
X
i=1
cos
2
nπ
N
1
i−
2
N
h nπ
io N
1 Xn
1 + cos
=
(2i − 1) = ,
2 i=1
N
2
147
we have
2
Uk(n) = 2K sin
nπ 2N
.
(b) Express the Hamiltonian in terms of the amplitudes of normal modes {ũk }, and evaluate
the classical partition function. (You may integrate the {ui } from −∞ to +∞).
• Before evaluating the classical partition function, lets evaluate the potential energy by
first expanding the displacement using the basis of normal modes, as
uj =
N−1
X
n=1
an · ũk(n) (j).
The expression for the total potential energy is
(N−1
)2
N
N
X
X
K
KX
.
(ui − ui−1 )2 =
an ũk(n) (j) − ũk(n) (j − 1)
U=
2 i=1
2 i=1 n=1
Since
N−1
X
j=1
ũk(n) (j) · ũk(m) (j − 1) =
N−1
X
1
δn,m
{− cos [k(n)(2j − 1)] + cos k(n)} = δn,m cos k(n),
N
j=1
the total potential energy has the equivalent forms
U=
=
N−1
N
X
KX
2
(ui − ui−1 ) = K
a2n (1 − cos k(n)) ,
2 i=1
n=1
N−1
X
i=1
a2k(n) ε2k(n)
= 2K
N−1
X
i=1
a2k(n)
2
sin
nπ 2N
.
The next step is to change the coordinates of phase space from uj to an . The Jacobian
associated with this change of variables is unity, and the classical partition function is now
obtained from
#
"
Z ∞
Z ∞
N−1
X
1
2 nπ
2
,
da1 · · ·
daN−1 exp −2βK
an sin
Z = N−1
λ
2N
−∞
−∞
n=1
√
where λ = h/ 2πmkB T corresponds to the contribution to the partition function from
each momentum coordinate. Performing the Gaussian integrals, we obtain
N−1 Z ∞
i
h
1 Y
2 nπ
2
,
dan exp −2βKan sin
Z = N−1
λ
2N
−∞
n=1
N2−1 N−1
Y h nπ i−1
1
πkB T
= N−1
.
sin
λ
2K
2N
n=1
148
D
2
(c) First evaluate |ũk |
E
, and use the result to calculate u2i . Plot the resulting squared
displacement of each particle as a function of its equilibrium position.
• The average squared amplitude of each normal mode is
a2n
R∞
2 nπ
2
2
da
(a
)
exp
−2βKa
sin
n
n
n
2N
R∞
= −∞
2 sin2 nπ
da
exp
−2βKa
n
n
2N
−∞
h
nπ i−1
kB T
1
= 4βK sin2
=
2 nπ .
2N
4K sin 2N
The variation of the displacement is then given by
u2j
=
*"N−1
X
an ũn (j)
n=1
2
=
N
N−1
X
n=1
a2n
#2 +
=
N−1
X
a2n ũ2n (j)
n=1
N−1
nπ kB T X sin2
j =
sin
N
2KN n=1 sin2
2
nπ
j
N nπ
2N
.
The evaluation of the above sum is considerably simplified by considering the combination
u2j+1 + u2j−1 − 2 u2j
where we have used
N−1
kB T X 2 cos
=
2KN n=1
2nπ 2nπ
2nπ
j
−
cos
(j
+
1)
−
cos
(j
−
1)
N
N
N
1 − cos nπ
N
N−1
1 − cos nπ
kB T X 2 cos 2nπ
kB T
N j
N
=
,
=−
nπ
2KN n=1
KN
1 − cos N
PN−1
boundary conditions of
n=1
u20
cos(πn/N ) = −1. It is easy to check that subject to the
= u2N = 0, the solution to the above recursion relation is
u2j =
kB T j(N − j)
.
K
N
(d) How are the results modified if only the first particle is fixed (u0 = 0), while the other
end is free (uN 6= 0)? (Note that this is a much simpler problem as the partition function
can be evaluated by changing variables to the N − 1 spring extensions.)
• When the last particle is free, the overall potential energy is the sum of the contributions
PN−1
of each spring, i.e. U = K j=1 (uj − uj−1 )2 /2. Thus each extension can be treated
independently, and we introduce a new set of independent variables ∆uj ≡ uj − uj−1 . (In
149
Amplitude squared
NkT/K
B
free end
fixed end
0
N/2
Position j
0
N
the previous case, where the two ends are fixed, these variables were not independent.)
The partition function can be calculated separately for each spring as


Z ∞
Z ∞
N−1
X
K
1
(uj − uj−1 )2 
du1 · · ·
duN−1 exp −
Z = N−1
λ
2k
T
B
−∞
−∞
j=1


(N−1)/2
Z ∞
Z ∞
N−1
X
K
2πk
T
1
B
2
d∆uN−1 exp −
.
d∆u1 · · ·
∆uj  =
= N−1
λ
2kB T
λ2 K
−∞
−∞
j=1
For each spring extension, we have
∆u2j = (uj − uj−1 )2 =
kB T
.
K
The displacement
uj =
j
X
∆ui ,
i=1
is a sum of independent random variables, leading to the variance
* j
!2 +
j
X
X
kB T
2
2
uj =
(∆ui ) =
∆ui
=
j.
K
i=1
i=1
The results for displacements of open and closed chains are compared in the above figure.
********
2. Black hole thermodynamics: According to Bekenstein and Hawking, the entropy of a
black hole is proportional to its area A, and given by
S=
kB c3
A .
4Gh̄
150
(a) Calculate the escape velocity at a radius R from a mass M using classical mechanics.
Find the relationship between the radius and mass of a black hole by setting this escape
velocity to the speed of light c. (Relativistic calculations do not modify this result which
was originally obtained by Laplace.)
• The classical escape velocity is obtained by equating the gravitational energy and the
kinetic energy on the surface as,
2
Mm
mvE
G
=
,
R
2
leading to
vE =
r
2GM
.
r
Setting the escape velocity to the speed of light, we find
R=
2G
M.
c2
For a mass larger than given by this ratio (i.e. M > c2 R/2G), nothing will escape from
distances closer than R.
(b) Does entropy increase or decrease when two black holes collapse into one? What is the
entropy change for the universe (in equivalent number of bits of information), when two
solar mass black holes (M⊙ ≈ 2 × 1030 kg) coalesce?
• When two black holes of mass M collapse into one, the entropy change is
kB c3
kB c3
∆S = S2 − 2S1 =
(A2 − 2A1 ) =
4π R22 − 2R12
4Gh̄
" 4Gh̄ 2 #
2
3
πkB c
2G
8πGkB M 2
2G
=
2M
M
> 0.
−
2
=
Gh̄
c2
c2
ch̄
Thus the merging of black holes increases the entropy of the universe.
Consider the coalescence of two solar mass black holes. The entropy change is
2
8πGkB M⊙
ch̄
8π · 6.7 × 10−11 (N · m2 /kg 2 ) · 1.38 × 10−23 (J/K) · (2 × 1030 )2 kg 2
≈
3 × 108 (m/s) · 1.05 × 10−34 (J · s)
∆S =
≈ 3 × 1054 (J/K).
151
In units of bits, the information lost is
NI =
∆S ln 2
= 1.5 × 1077 .
kB
(c) The internal energy of the black hole is given by the Einstein relation, E = M c2 . Find
the temperature of the black hole in terms of its mass.
• Using the thermodynamic definition of temperature
1
T
=
∂S
∂E ,
and the Einstein relation
2
E = Mc ,
1 ∂
1
= 2
T
c ∂M
"
3
kB c
4π
4Gh̄
2G
M
c2
2 #
8πkB G
=
M,
h̄c3
=⇒
h̄c3 1
T =
.
8πkB G M
(d) A “black hole” actually emits thermal radiation due to pair creation processes on its
event horizon. Find the rate of energy loss due to such radiation.
• The (quantum) vacuum undergoes fluctuations in which particle–antiparticle pairs are
constantly created and destroyed. Near the boundary of a black hole, sometimes one
member of a pair falls into the black hole while the other escapes. This is a hand-waving
explanation for the emission of radiation from black holes. The decrease in energy E of a
black body of area A at temperature T is given by the Stefan-Boltzmann law,
1 ∂E
= −σT 4 ,
A ∂t
where
4
π 2 kB
σ=
.
60h̄3 c2
(e) Find the amount of time it takes an isolated black hole to evaporate. How long is this
time for a black hole of solar mass?
• Using the result in part (d) we can calculate the time it takes a black hole to evaporate.
For a black hole
2
A = 4πR = 4π
Hence
which implies that
2G
M
c2
2
=
16πG2 2
M ,
c4
4
d
π 2 kB
M c2 = −
dt
60h̄3 c2
M2
E = M c2 ,
16πG2 2
M
c4
h̄c3 1
8πkB G M
dM
h̄c4
=−
≡ −b.
dt
15360G2
152
and T =
4
,
h̄c3 1
.
8πkB G M
This can be solved to give
M (t) = M03 − 3bt
1/3
.
The mass goes to zero, and the black hole evaporates after a time
τ=
M03
5120G2 M ⊙3
=
≈ 2.2 × 1074 s,
3b
h̄c4
which is considerably longer than the current age of the universe (approximately ×1018 s).
(f) What is the mass of a black hole that is in thermal equilibrium with the current cosmic
background radiation at T = 2.7K?
• The temperature and mass of a black hole are related by M = h̄c3 /(8πkB GT ). For a
black hole in thermal equilibrium with the current cosmic background radiation at T =
2.7◦ K,
M≈
1.05 × 10−34 (J · s)(3 × 108 )3 (m/s)3
≈ 4.5 × 1022 kg.
8π · 1.38 × 10−23 (J/K) · 6.7 × 10−11 (N · m2 /kg 2 ) · 2.7◦ K
(g) Consider a spherical volume of space of radius R. According to the recently formulated
Holographic Principle there is a maximum to the amount of entropy that this volume of
space can have, independent of its contents! What is this maximal entropy?
• The mass inside the spherical volume of radius R must be less than the mass that would
make a black hole that fills this volume. Bring in additional mass (from infinity) inside
the volume, so as to make a volume-filling balck hole. Clearly the entropy of the system
will increase in the process, and the final entropy, which is the entropy of the black hole is
larger than the initial entropy in the volume, leading to the inequality
S ≤ SBH =
kB c3
A,
4Gh̄
where A = 4πR2 is the area enclosing the volume. The surprising observation is that the
upper bound on the entropy is proportional to area, whereas for any system of particles
we expect the entropy to be proportional to N . This should remain valid even at very
high temperatures when interactions are unimportant. The ‘holographic principle’ is an
allusion to the observation that it appears as if the degrees of freedom are living on the
surface of the system, rather than its volume. It was formulated in the context of string
theory which attempts to construct a consistent theory of quantum gravity, which replaces
particles as degrees of freedom, with strings.
153
********
3. Quantum harmonic oscillator: Consider a single harmonic oscillator with the Hamiltonian
H=
p2
mω 2 q 2
+
,
2m
2
with p =
h̄ d
i dq
.
(a) Find the partition function Z, at a temperature T , and calculate the energy hHi.
• The partition function Z, at a temperature T , is given by
Z = tr ρ =
X
e−βEn .
n
As the energy levels for a harmonic oscillator are given by
1
,
ǫn = h̄ω n +
2
the partition function is
Z=
X
n
=
1
exp −βh̄ω n +
= e−βh̄ω/2 + e−3βh̄ω/2 + · · ·
2
1
1
.
=
2 sinh (βh̄ω/2)
eβh̄ω/2 − e−βh̄ω/2
The expectation value of the energy is
∂ ln Z
1
h̄ω cosh(βh̄ω/2)
h̄ω
hHi = −
=
=
.
∂β
2
sinh(βh̄ω/2)
2
tanh(βh̄ω/2)
(b) Write down the formal expression for the canonical density matrix ρ in terms of the
eigenstates ({|ni}), and energy levels ({ǫn }) of H.
• Using the formal representation of the energy eigenstates, the density matrix ρ is
!
X
1
βh̄ω
< n| .
|n > exp −βh̄ω n +
ρ = 2 sinh
2
2
n
In the coordinate representation, the eigenfunctions are in fact given by
2
mω 1/4 H (ξ)
ξ
n
√
hn|qi =
exp −
,
n
πh̄
2
2 n!
154
where
ξ≡
with
r
mω
q,
h̄
n
d
dξ
exp(−ξ 2 )
Hn (ξ) = (−1) exp(ξ )
Z
exp(ξ 2 ) ∞
(−2iu)n exp(−u2 + 2iξu)du.
=
π
−∞
n
2
For example,
H0 (ξ) = 1,
and
H1 (ξ) = − exp(ξ 2 )
d
exp(−ξ 2 ) = 2ξ,
dξ
result in the eigenstates
h0|qi =
and
h1|qi =
mω 1/4
πh̄
mω exp −
q2 ,
2h̄
mω 1/4 r 2mω
mω q · exp −
q2 .
h̄
2h̄
πh̄
Using the above expressions, the matrix elements are obtained as
1
· hq ′ |ni hn|qi
exp
−βh̄ω
n
+
2
hq ′ |ρ|qi =
hq ′ |n′ i hn′ |ρ|ni hn|qi = n P
1
exp
−βh̄ω
n
+
′
n
2
n,n
X
βh̄ω
1
= 2 sinh
·
exp −βh̄ω n +
· hq ′ |ni hn|qi .
2
2
n
P
X
(c) Show that for a general operator A(x),
∂A
∂
exp [A(x)] 6=
exp [A(x)] ,
∂x
∂x
while in all cases
∂
tr {exp [A(x)]} = tr
∂x
• By definition
unless
155
∂A
= 0,
A,
∂x
∂A
exp [A(x)] .
∂x
∞
X
1 n
A ,
e =
n!
n=0
A
and
But for a product of n operators,
The
∂A
∂x
∞
X
∂eA
1 ∂An
=
.
∂x
n!
∂x
n=0
∂A
∂A
∂A
∂
(A · A · · · A) =
· A···A + A ·
···A + ··· + A · A···
.
∂x
∂x
∂x
∂x
can be moved through the A′ s surrounding it only if A, ∂A
∂x = 0, in which case
∂A n−1
∂A
=n
A
,
∂x
∂x
and
∂eA
∂A A
=
e .
∂x
∂x
However, as we can always reorder operators inside a trace, i.e. tr(BC) = tr(CB),
and
∂A
∂A
n−1
,
tr A · · · A · · ·
· · · A = tr
·A
∂x
∂x
and the identity
∂A A
,
·e
∂x
can always be satisfied, independent of any constraint on A, ∂A
.
∂x
∂
tr eA = tr
∂x
(d) Note that the partition function calculated in part (a) does not depend on the mass
m, i.e. ∂Z/∂m = 0. Use this information, along with the result in part (c), to show that
2 p
mω 2 q 2
=
.
2m
2
• The expectation values of the kinetic and potential energy are given by
2 2
p
mω 2 q 2
mω 2 q 2
p
= tr
= tr
ρ ,
and
ρ .
2m
2m
2
2
Noting that the expression for the partition function derived
in
part (a) is independent of
mass, we know that ∂Z/∂m = 0. Starting with Z = tr e−β H , and differentiating
∂Z
∂
∂
−β H
−β H
= 0,
= tr
=
tr e
(−βH)e
∂m
∂m
∂m
where we have used the result in part (c). Differentiating the Hamiltonian, we find that
mω 2 q 2 −β H
p2 −β H
+ tr −β
= 0.
e
e
tr β
2m2
2
156
Equivalently,
p2 −β H
mω 2 q 2 −β H
tr
= tr
,
e
e
2m
2
which shows that the expectation values of kinetic and potential energies are equal.
(e) Using the results in parts (d) and (a), or otherwise, calculate q 2 . How are the results
in Problem #1 modified at low temperatures by inclusion of quantum mechanical effects.
•
−1
In part (a) it was found that hHi = (h̄ω/2) (tanh(βh̄ω/2))
. Note that hHi =
p2 /2m + mω 2 q 2 /2 , and that in part (d) it was determined that the contribution from
the kinetic and potential energy terms are equal. Hence,
mω 2 q 2 /2 =
1
−1
(h̄ω/2) (tanh(βh̄ω/2)) .
2
Solving for q 2 ,
q2 =
h̄
h̄
−1
(tanh(βh̄ω/2)) =
coth(βh̄ω/2).
2mω
2mω
While the classical result q 2 = kB T /mω 2 , vanishes as T → 0, the quantum result satu-
rates at T = 0 to a constant value of q 2 = h̄/(2mω). The amplitude of the displacement
curves in Problem #1 are effected by exactly the same saturation factors.
(f) In a coordinate representation, calculate hq ′ |ρ|qi in the high temperature limit. One
approach is to use the result
exp(βA) exp(βB) = exp β(A + B) + β 2 [A, B]/2 + O(β 3 ) .
• Using the general operator identity
exp(βA) exp(βB) = exp β(A + B) + β 2 [A, B]/2 + O(β 3 ) ,
the Boltzmann operator can be decomposed in the high temperature limit into those for
kinetic and potential energy; to the lowest order as
mω 2 q 2
p2
≈ exp(−βp2 /2m) · exp(−βmω 2 q 2 /2).
−β
exp −β
2m
2
157
The first term is the Boltzmann operator for an ideal gas. The second term contains an
operator diagonalized by |q >. The density matrix element
< q ′ |ρ|q > =< q ′ | exp(−βp2 /2m) exp(−βmω 2 q 2 /2)|q >
Z
= dp′ < q ′ | exp(−βp2 /2m)|p′ >< p′ | exp(−βmω 2 q 2 /2)|q >
Z
= dp′ < q ′ |p′ >< p′ |q > exp(−βp′2 /2m) exp(−βq 2 mω 2 /2).
Using the free particle basis < q ′ |p′ >=
√ 1 e−iq·p/h̄ ,
2πh̄
Z
′
′
′2
2
2
1
dp′ eip (q−q )/h̄ e−βp /2m e−βq mω /2
< q |ρ|q >=
2πh̄

!2 
r
r
Z
β
2m
i
1
2m
−βq 2 mω 2 /2 1
′
′
′
′
2
=e
dp exp − p
+
(q − q )  exp −
(q − q ) ,
2πh̄
2m 2h̄
β
4 βh̄2
′
where we completed the square. Hence
1 −βq 2 mω2 /2 p
mkB T
′ 2
2πmkB T exp −
< q |ρ|q >=
e
(q − q ) .
2πh̄
2h̄2
′
The proper normalization in the high temperature limit is
Z
2
2 2
Z = dq < q|e−βp /2m · e−βmω q /2 |q >
Z
Z
2
2 2
= dq dp′ < q|e−βp /2m |p′ >< p′ |e−βmω q /2 |q >
Z
Z
′2
2 2
kB T
2
= dq dp |< q|p >| e−βp /2m e−βmω q /2 =
.
h̄ω
Hence the properly normalized matrix element in the high temperature limit is
s
mω 2
mkB T
mω 2 2
′
′ 2
< q |ρ|q >lim T →∞ =
exp −
q exp −
(q − q ) .
2πkB T
2kB T
2h̄2
(g) At low temperatures, ρ is dominated by low energy states. Use the ground state
wave-function to evaluate the limiting behavior of hq ′ |ρ|qi as T → 0.
• In the low temperature limit, we retain only the first terms in the summation
ρlim T →0 ≈
|0 > e−βh̄ω/2 < 0| + |1 > e−3βh̄ω/2 < 1| + · · ·
.
e−βh̄ω/2 + e−3βh̄ω/2
158
Retaining only the term for the ground state in the numerator, but evaluating the geometric
series in the denominator,
< q ′ |ρ|q >lim T →0 ≈< q ′ |0 >< 0|q > e−βh̄ω/2 · eβh̄ω/2 − e−βh̄ω/2 .
Using the expression for < q|0 > given in part (b),
′
< q |ρ|q >lim T →0 ≈
r
h mω
i
mω
q 2 + q ′2
1 − e−βh̄ω .
exp −
πh̄
2h̄
(h) Calculate the exact expression for hq ′ |ρ|qi.
********
4. Relativistic Coulomb gas:
Consider a quantum system of N positive, and N negative
charged relativistic particles in box of volume V = L3 . The Hamiltonian is
H=
2N
X
i=1
c|~
pi | +
2N
X
i<j
ei ej
|~ri − ~rj |
,
where ei = +e0 for i = 1, · · · N , and ei = −e0 for i = N + 1, · · · 2N , denote the charges of
the particles; {~ri } and {~
pi } their coordinates and momenta respectively. While this is too
complicated a system to solve, we can nonetheless obtain some exact results.
(a) Write down the Schrödinger equation for the eigenvalues εn (L), and (in coordinate
space) eigenfunctions Ψn ({~ri }). State the constraints imposed on Ψn ({~ri }) if the particles
are bosons or fermions?
• In the coordinate representation ~pi is replaced by −ih̄∇i , leading to the Schrödinger
equation


2N
X
i=1
c| − ih̄∇i | +
2N
X
i<j

ei ej 
Ψn ({~ri }) = εn (L)Ψn ({~ri }).
|~ri − ~rj |
There are N identical particles of charge +e0 , and N identical particles of charge −e0 .
We can examine the effect of permutation operators P+ and P− on these two sets. The
symmetry constraints can be written compactly as
P
P
P+ P− Ψn ({~ri }) = η++ · η−− Ψn ({~ri }),
159
where η = +1 for bosons, η = −1 for fermions, and (−1)P denotes the parity of the
permutation. Note that there is no constraint associated with exchange of particles with
opposite charge.
(b) By a change of scale ~ri ′ = ~ri /L, show that the eigenvalues satisfy a scaling relation
εn (L) = εn (1)/L.
•
After the change of scale ~ri ′ = ~ri /L (and corresponding change in the derivative
∇i ′ = L∇i ), the above Schrödinger equation becomes


′ ′ 2N
2N
′
X
X
e
e
∇
~ri
~
r
i
j
i
i

 Ψn
c −ih̄
+
=
ε
(L)Ψ
.
n
n
′ −~
′|
L
L
|~
r
r
L
L
i
j
i=1
i<j
The coordinates in the above equation are confined to a box of unit size. We can regard it
as the Schrödinger equation in such a box, with wave-functions Ψ′n ({~ri }) = Ψn ({~ri ′ /L}).
The corresponding eigenvalues are εn (1) = Lεn (L) (obtained after multiplying both sides
of the above equation by L We thus obtain the scaling relation
εn (L) =
εn (1)
.
L
(c) Using the formal expression for the partition function Z(N, V, T ), in terms of the
eigenvalues {εn (L)}, show that Z does not depend on T and V separately, but only on a
specific scaling combination of them.
• The formal expression for the partition function is
εn (L)
exp −
=
Z(N, V, T ) = tr e
kB T
n
X
εn (1)
,
=
exp −
k
T
L
B
n
−β H
X
where we have used the scaling form of the energy levels. Clearly, in the above sum T and
L always occur in the combination T L. Since V = L3 , the appropriate scaling variable is
V T 3 , and
Z(N, V, T ) = Z(N, V T 3 ).
(d) Relate the energy E, and pressure P of the gas to variations of the partition function.
Prove the exact result E = 3P V .
160
• The average energy in the canonical ensemble is given by
E=−
∂ ln Z
∂ ln Z
∂ ln Z
4 ∂ ln Z
= kB T 2
= kB T 2 (3V T 2 )
=
3k
V
T
.
B
∂β
∂T
∂(V T 3 )
∂(V T 3 )
The free energy is F = −kB T ln Z, and its variations are dF = −SdT − P dV + µdN .
Hence the gas pressure is given by
P =−
∂F
∂ ln Z
∂ ln Z
= kB T
= kB T 4
.
∂V
∂V
∂(V T 3 )
The ratio of the above expressions gives the exact identity E = 3P V .
(e) The Coulomb interaction between charges in in d-dimensional space falls off with separation as ei ej / |~ri − ~rj |
d−2
. (In d = 2 there is a logarithmic interaction.) In what dimension
d can you construct an exact relation between E and P for non-relativistic particles (kiP
netic energy i p~2i /2m)? What is the corresponding exact relation between energy and
pressure?
• The above exact result is a consequence of the simple scaling law relating the energy
eigenvalues εn (L) to the system size. We could obtain the scaling form in part (b) since
the kinetic and potential energies scaled in the same way. The kinetic energy for nonP
P
relativistic particles i ~p2i /2m = − i h̄2 ∇2i /2m, scales as 1/L2 under the change of scale
P2N
d−2
~ri ′ = ~ri /L, while the interaction energy
ri − ~rj |
in d space dimensions
i<j ei ej / |~
scales as 1/Ld−2 . The two forms will scale the same way in d = 4 dimensions, leading to
εn (L) =
εn (1)
.
L2
The partition function now has the scaling form
Z(N, V = L4 , T ) = Z N, (T L2 )2 = Z N, V T 2 .
Following steps in the previous part, we obtain the exact relationship E = 2P V .
(f) Why are the above ‘exact’ scaling laws not expected to hold in dense (liquid or solid)
Coulomb mixtures?
• The scaling results were obtained based on the assumption of the existence of a single
scaling length L, relevant to the statistical mechanics of the problem. This is a good
approximation in a gas phase. In a dense (liquid or solid) phase, the short-range repulsion
between particles is expected to be important, and the particle size a is another relevant
161
length scale which will enter in the solution to the Schrödinger equation, and invalidate
the scaling results.
********
5. The virial theorem is a consequence of the invariance of the phase space for a system
of N (classical or quantum) particles under canonical transformations, such as a change of
scale. In the following, consider N particles with coordinates {~qi }, and conjugate momenta
{~
pi } (with i = 1, · · · , N ), and subject to a Hamiltonian H ({~
pi } , {~qi }).
(a) Classical version: Write down the expression for the classical partition function, Z ≡
Z [H]. Show that it is invariant under the rescaling ~q1 → λ~q1 , p~1 → p~1 /λ of a pair of
conjugate variables, i.e. Z [Hλ ] is independent of λ, where Hλ is the Hamiltonian obtained
after the above rescaling.
• The classical partition function is obtained by appropriate integrations over phase space
as
1
Z=
N !h3N
Z
Y
3
3
d pi d qi
i
!
e−β H .
The rescaled Hamiltonian Hλ = H (~
p1 /λ, {~pi6=1 } , λ~q1 , {~qi6=1 }) leads to a rescaled partition
function
1
Z [Hλ ] =
N !h3N
which reduces to
1
Z [Hλ ] =
N !h3N
Z
Z
Y
d3 pi d3 qi
i
!
e−β Hλ ,
Y 3 3
d pi d qi
λ3 d3 p′1 λ−3 d3 q1′
i
!
e−β H = Z,
under the change of variables ~q1 ′ = λ~q1 , p~1 ′ = p~1 /λ.
(b) Quantum mechanical version: Write down the expression for the quantum partition
function. Show that it is also invariant under the rescalings ~q1 → λ~q1 , p~1 → ~p1 /λ, where
p~i and ~qi are now quantum mechanical operators. (Hint: start with the time-independent
Schrödinger equation.)
• Using the energy basis
X
Z = tr e−β H =
e−βEn ,
n
where En are the energy eigenstates of the system, obtained from the Schrödinger equation
H ({~pi } , {~qi }) |ψn i = En |ψn i ,
162
where |ψn i are the eigenstates. After the rescaling transformation, the corresponding
equation is
E
E
H (~
p1 /λ, {~
pi6=1 } , λ~q1 , {~qi6=1 }) ψn(λ) = En(λ) ψn(λ) .
In the coordinate representation, the momentum operator is p~i = −ih̄∂/∂~qi , and therefore
(λ)
ψλ ({~qi }) = ψ ({λ~qi }) is a solution of the rescaled equation with eigenvalue En
= En .
Since the eigen-energies are invariant under the transformation, so is the partition function
which is simply the sum of corresponding exponentials.
(c) Now assume a Hamiltonian of the form
H=
X p~i 2
i
2m
+ V ({~qi }) .
Use the result that Z [Hλ ] is independent of λ to prove the virial relation
2 p~1
∂V
=
· ~q1 ,
m
∂~q1
where the brackets denote thermal averages.
• Differentiating the free energy with respect to λ at λ = 1, we obtain
p~1 2
∂ ln Zλ
∂Hλ
∂V
= −β
= −β −
0=
+
· ~q1 ,
∂λ λ=1
∂λ λ=1
m
∂~q1
i.e.,
p~1 2
m
=
∂V
· ~q1 .
∂~q1
(d) The above relation is sometimes used to estimate the mass of distant galaxies. The
stars on the outer boundary of the G-8.333 galaxy have been measured to move with
velocity v ≈ 200 km/s. Give a numerical estimate of the ratio of the G-8.333’s mass to its
size.
• The virial relation applied to a gravitational system gives
GM m
2
mv =
.
R
Assuming that the kinetic and potential energies of the starts in the galaxy have reached
some form of equilibrium gives
v2
M
≈
≈ 6 × 1020 kg/m.
R
G
163
********
~ is
6. Electron spin: The Hamiltonian for an electron in a magnetic field B
0 1
0 −i
1 0
~
H = −µB ~σ .B, where σx =
, σy =
, and σz =
,
1 0
i 0
0 −1
are the Pauli spin operators, and µB is the Bohr magneton.
~ is along the z
(a) In the quantum canonical ensemble evaluate the density matrix if B
direction.
• By definition,
ρ=
exp(−βH)
.
tr exp(−βH)
Noting that for any i = 1, 2, 3, σi2 = I, we get,
exp(aσi ) = (1 +
1 2
1
1
1
a + a4 + · · ·)I + (a + a3 + a5 + · · ·)σi = cosh aI + sinh aσi ,
2!
4!
3!
5!
and thus
tr exp(aσi ) = tr(cosh aI + sinh aσi ) = 2 cosh a,
~ = B ẑ,
as the Pauli matrices are traceless. Hence in the case B
exp(βµB σz B)
2 cosh(βµB B)
βµ B
cosh(βµB B)I + sinh(βµB B)σz
1
e B
=
=
0
2 cosh(βµB B)
2 cosh(βµB B)
ρ=
0
e−βµB B
.
~ points along the x-direction.
(b) Repeat the calculation assuming that B
~ = B x̂,
• For B
exp(βµB σx B)
exp(−βH)
=
tr exp(−βH)
2 cosh(βµB B)
cosh(βµB B)I + sinh(βµB B)σx
1/2
tanh(βµB B)/2
=
=
.
tanh(βµB B)/2
1/2
2 cosh(βµB B)
ρ=
(c) Calculate the average energy in each of the above cases.
• Note that
tr σi exp(aσi ) = tr(cosh aσi + sinh aI) = 2 sinh a,
164
so that for H = −µB Bσi ,
tr[−µB Bσi exp(βµB σi B)]
tr Hρ
=
tr ρ
tr[exp(βµB σi B)]
−2µB B sinh(βµB B)
=
= −µB B tanh(βµB B),
2 cosh(βµB B)
~
independent of the direction of B.
********
7. Quantum rotor: Consider a rotor in two dimensions with
H=−
h̄2 d2
,
2I dθ 2
and
0 ≤ θ < 2π.
(a) Find the eigenstates and energy levels of the system.
• Solving,
−
h̄2 d2
ψ(θ) = Eψ(θ)
2I dθ 2
we get,
!
r
2IE
1
θ .
ψ(θ) = √ exp ±i
h̄2
2π
Imposing ψ(θ) = ψ(θ + 2π), we finally obtain,
1
h̄2 n2
ψn (θ) = √ exp(inθ), En =
, n: integer.
2I
2π
(b) Write the expression for the density matrix hθ ′ |ρ|θi in a canonical ensemble of temper-
ature T , and evaluate its low and high temperature limits.
• Defining, Z = tr e−β H ,
ρ=
2 2
1 X
|nie−βh̄ n /2I hn|
Z n
,
and,
1
hθ |ρ|θi =
2π
′
P
′
n
2
2
ein(θ−θ ) e−βh̄ n
P −βh̄2 n2 /2I
ne
165
/2I
.
In the low temperature limit (h̄2 β/2I ≫ 1) the ground state dominates. In other words,
since e−βh̄
2
/2I
≪ 1, to leading order the contributions come from the states with n = 0
and n = ±1, resulting in
h
i
−2h̄2 /2I
′
′
1
+
2e
cos(θ
−
θ
)
+
·
·
·
2
1
1
2 θ−θ
′
−2h̄
/2I
hθ |ρ|θi ≈
1 − 4e
sin
+··· .
≈
2
2π
2π
2
1 + 2e−2h̄ /2I + · · ·
q
In the high temperature limit (h̄ β/2I ≪ 1), by taking x = n h̄2 β/2I we may write,
2
X
2
2
2
2
−βh̄ n /2I
e
n
X
′
in(θ−θ ) −βh̄ n /2I
e
e
n
→
s
2I
h̄2 β
Z
dxe−x ,
→
s
2I
h̄2 β
Z
dxe−x e
=
s
and,
hθ ′ |ρ|θi ≈
2
2
p
i
2I −I(θ−θ′ )2 /2βh̄2
e
h̄2 β
2I
(θ−θ′ )x
h̄2 β
Z
2
dxe−x ,
′ )2
1 − I(θ−θ
e 2βh̄2 .
2π
Note that this expression is valid as long as (θ − θ ′ ) ∼
q
h̄2 β/2I ≪ 1, since this is our
scale of coarse graining. That’s why the ρ matrix element for θ − θ ′ = 2π is different from
that for θ − θ ′ = 0 in this approximation; it simply doesn’t apply for ∆θ ∼ 1.
********
8. Quantum mechanical entropy: A quantum mechanical system (defined by a Hamiltonian H), at temperature T , is described by a density matrix ρ(t), which has an associated
entropy S(t) = −trρ(t) ln ρ(t).
(a) Write down the time evolution equation for the density matrix, and calculate dS/dt.
• The time evolution of the density operator is given by the Heisenberg equation
∂ρ
= ih̄ [ρ, H] ,
∂t
leading to
dS
= ih̄tr {(1 + ρ) [ρ, H]} = ih̄ tr(ρH) − tr(Hρ) + tr(ρ∈ H) − tr(ρHρ) = 0,
dt
since the trace of a product is invariant under cyclic permutations.
166
(b) Using the method of Lagrange multipliers, find the density operator ρmax which maximizes the functional S[ρ], subject to the constraint of fixed average energy hHi = trρH =
E.
• In order to satisfy the two constraints, normalization and constant average energy,
written respectively as
trρ = 1
hHi = trρH = E,
and
we construct the functional
S(t) = trρ [− ln ρ − α − βH] + α + βE,
where α and β are Lagrange multipliers. Extremizing the above expression with respect
to ρ results in
∂S
∂ρ
ρ=ρmax
= − ln ρmax − α − βH − 1 = 0.
The solution to this equation is
ln ρmax = −(α + 1) − βH,
which can be rewritten as
ρmax = C exp (−βH) ,
where
C = e−(α+1) .
(c) Show that the solution to part (b) is stationary, i.e. ∂ρmax /∂t = 0.
• The density obtained in part (b) is stationary, as the commutator of H with any function
of H vanishes, implying, in particular,
∂ρmax
= 0.
∂t
********
9. Ortho/para–hydrogen: Hydrogen molecules can exist in ortho and para states.
167
(a) The two electrons of H2 in para-hydrogen form a singlet (antisymmetric) state. The
orbital angular momentum can thus only take even values; i.e.
Hp =
h̄2
ℓ(ℓ + 1),
2I
where ℓ = 0, 2, 4, · · ·. Calculate the rotational partition function of para-hydrogen, and
evaluate its low and high temperature limits.
• Since there are 2ℓ + 1 states for each ℓ,
X
Zp =
−βh̄2 ℓ(ℓ+1)/2I
(2ℓ + 1)e
=
∞
X
2
(4n + 1)e−βh̄
n(2n+1)/I
.
n=0
ℓ : even
In the low temperature limit, only the first few terms matter, and to leading order
Zp ≈ 1 + 5e−3βh̄
2
/I
.
In the high temperature limit,
Zp =
∞
X
(4n + 1)e−βh̄
2
n(2n+1)/I
n=0
I
=
βh̄2
Z
∞
2
4xe−2x dx =
0
∞
I X −2
e
≈
βh̄2 n=0
p
2
β
I h̄n
4
r
β
h̄n
I
! r
!
β
h̄
I
I
.
βh̄2
(b) In ortho-hydrogen the electrons are in a triply degenerate symmetric state, hence
Ho =
h̄2
ℓ(ℓ + 1),
2I
with ℓ = 1, 3, 5, · · ·. Calculate the rotational partition function of ortho-hydrogen, and
evaluate its low and high temperature limits.
• Since there are 2ℓ + 1 states for each ℓ, and a 3-fold degeneracy in the spin state,
Zo = 3
X
−βh̄2 ℓ(ℓ+1)/2I
(2ℓ + 1)e
∞
X
=3
(4n + 3)e−βh̄
2
(n+1)(2n+1)/I
.
n=0
ℓ : odd
In the low temperature limit, only the first few terms matter, and to leading order
2
Zo ≈ 9e−βh̄
168
/I
.
In the high temperature limit,
Zo =
∞
X
2
−βh̄ (n+1)(2n+1)/I
(4n + 3)e
n=0
3I
=
βh̄2
Z
∞
2
4xe−2x dx =
0
3I
.
βh̄2
∞
3I X −2
≈
e
βh̄2 n=0
p
2
β
I h̄n
4
r
β
h̄n
I
! r
β
h̄
I
!
(c) For an equilibrium gas of N hydrogen molecules calculate the partition function.
(Hint: Sum over contributions from mixtures of Np para- and No = N −Np ortho-hydrogen
particles. Ignore vibrational degrees of freedom.)
•
Z=
X
Np
1
1
ZpNp ZoN−Np =
(Zo + Zp )N ,
(N − Np )!Np !
N!
which is what we should expect.
(d) Write down the expression for the rotational contribution to the internal energy hErot. i,
and comment on its low and high temperature limits.
Actually, due to small transition rates between ortho- and para-hydrogen, in most
circumstances the mixture is not in equilibrium.
•
hErot i = −
∂ ln(Z0 + Zp )
∂ ln Z
= −N
.
∂β
∂β
In the low energy limit,
2
Z0 + Zp ≈ 1 + 9e−βh̄
/I
,
giving up to leading order,
hErot i ≈
9N h̄2 −βh̄2 /I
e
.
I
In the high energy limit,
Z0 + Zp ≈
4I
,
βh̄2
giving,
hErot i ≈ N kB T,
which is what we expect since we have 2 rotational degrees of freedom.
********
169
10. van Leeuwen’s theorem:
Consider a gas of charged particles subject to a general
Hamiltonian of the form
H=
N
X
p~i
i=1
2
2m
+ U (~q1 , · · · , ~qN ).
~ the canonical momenta, ~pn , are replaced with p~n − eA,
~
In an external magnetic field, B,
~ is the vector potential, B
~ =∇
~ × A.
~ Show that if quantum effects are ignored, the
where A
~
thermodynamics of the problem is independent of B.
• Ignoring quantum effects and taking the coarse graining scale to be h,
Z=
X
−β H(µs )
e
µs
Z Y
N
d3 ~pn d3 ~qn −β[PN
i=1
=
e
h3
n=1
p
~i 2
2m
+U(~
q1 ,···,~
qN )]
.
Now with a magnetic field,
~ =
Z(B)
X
µs
−β H(µs )
e
Z Y
N
d3 ~pn d3 ~qn −β[PN
i=1
=
e
3
h
n=1
Z Y
N
d3 ~p′n d3 ~qn −β[PN
i=1
e
=
h3
n=1
~ 2
(~
pi −eA)
2m
p
~′ 2
i
2m
+U(~
q1 ,···,~
qN )]
+U(~
q1 ,···,~
qN )]
,
~ =∇
~ × A,
~ where we have defined p~′ = p~ − eA.
~ Hence we see that classically,
for B
~ = Z(~0),
Z(B)
~
so the thermodynamics of the problem is independent of B.
********
170
Problems for Chapter VII - Ideal Quantum Gases
1. Identical particle pair: Let Z1 (m) denote the partition function for a single quantum
particle of mass m in a volume V .
(a) Calculate the partition function of two such particles, if they are bosons, and also if
they are (spinless) fermions.
• A two particle wave function is constructed from one particle states k1 , and k2 . Depending on the statistics of the two particles, we have
Bosons:
|k1 , k2 iB =
Fermions:
|k1 , k2 iF =

√
 (|k1 i|k2 i + |k2 i|k1 i) / 2 for k1 6= k2
,

√
 (|k1 i|k2 i − |k2 i|k1 i) / 2 for k1 6= k2
.


|k1 i|k2 i
for k1 = k2
no state
for k1 = k2
For the Boson system, the partition function is given by
X
Z2B = tr e−βH =
hk1 , k2 |e−βH |k1 , k2 iB
k1 ,k2
X hk1 |hk2 | + hk2 |hk1 |
|k1 i|k2 i + |k2 i|k1 i X
√
√
e−βH
+
hk|hk|e−βH |ki|ki
=
2
2
k
k1 >k2
2
2
X
X
βh̄
2βh̄ k 2
=
exp −
(k12 + k22 ) +
exp −
2m
2m
k1 >k2
k
1 X
βh̄2 2
βh̄2 k 2
1X
2
=
exp −
exp −
,
(k + k2 ) +
2
2m 1
2
m
k1 ,k2
k
and thus
Z2B
For the Fermion system,
m i
1h 2
=
Z (m) + Z1
.
2 1
2
X
Z2F = tr e−βH =
hk1 , k2 |e−βH |k1 , k2 iF
k1 ,k2
X hk1 |hk2 | − hk2 |hk1 |
|k1 i|k2 i − |k2 i|k1 i
√
√
=
e−βH
2
2
k1 >k2
βh̄2 2
βh̄2 k 2
1X
1 X
2
exp −
exp −
,
(k + k2 ) −
=
2
2m 1
2
m
k1 ,k2
k
171
and thus
Z2F =
m i
1h 2
Z1 (m) − Z1
.
2
2
Note that classical Boltzmann particles have a partition function
Z2classical =
1
Z1 (m)2 .
2
√
(b) Use the classical approximation Z1 (m) = V /λ3 with λ = h/ 2πmkB T . Calculate the
corrections to the energy E, and the heat capacity C, due to Bose or Fermi statistics.
• If the system is non-degenerate, the correction term is much smaller than the classical
term, and
ln Z2± = ln
Z1 (m)2 ± Z1 (m/2) /2
Z1 (m/2)
− ln 2
= 2 ln Z1 (m) + ln 1 ±
Z1 (m)2
Z1 (m/2)
≈ 2 ln Z1 (m) ±
− ln 2.
Z1 (m)2
Using
Z1 (m) =
V
,
λ(m)3
where
λ(m) ≡ √
h
,
2πmkB T
we can write
λ(m)3
±
V
ln Z2±
≈ ln Z2Classical
=⇒
∆ ln Z2± = ± 2−3/2
λ(m)
λ(m/2)
3
= ln Z2Classical ± 2−3/2
λ(m)3
,
V
h3 β 3/2
h3
−3/2
=
±
2
.
V (2πmkB T )3/2
V (2πm)3/2
Thus the energy differences are
∆E
±
3
∂
h3 β 3/2
3
= − ∆ ln Z2 = ∓ 5/2
= ∓ 5/2 kB T
3/2
∂β
2 V (2πm)
2
λ3
V
,
resulting in heat capacity differences of
∆CV±
∂∆E
=
∂T
h3 kB
3
= ± 7/2 kB
= ± 7/2
3/2
2 V (2πmkB T )
2
3
V
λ3
V
.
This approximation holds only when the thermal volume is much smaller than the volume
of the gas where the ratio constitutes a small correction.
(c) At what temperature does the approximation used above break down?
172
• The approximation is invalid if the correction terms are comparable to the first term.
This occurs when the thermal wavelength becomes comparable to the size of the box, i.e.
for
λ= √
h
≥ L ∼ V 1/3 ,
2πmkB T
or T ≤
h2
.
2πmkB L2
********
2. Generalized ideal gas: Consider a gas of non-interacting identical (spinless) quantum
particles with an energy spectrum ε = |~
p/h̄ |s , contained in a box of “volume” V in ddimensions.
(a) Calculate the grand potential G η = −kB T ln Qη , and the density n = N/V , at a
η
chemical potential µ. Express your answers in terms of s, d, and fm
(z), where z = eβµ ,
and
η
fm
(z)
1
=
Γ(m)
Z
∞
0
dxxm−1
.
z −1 ex − η
(Hint: Use integration by parts on the expression for ln Qη .)
• The grand partition function is given by
Fermions:
QF =
∞
X
βµN
e
N=0
X
−β
e
{nν }
P
ν
nν εν
1
Y X
=
eβ(µ−εν )nν =
ν nν =0
i
Yh
1 + eβ(µ−εν ) .
ν
Bosons:
QB =
∞
X
eβµN
N=0
X
{nν }
e−β
P
ν
nν εν
YX
=
eβ(µ−εν )nν =
ν {nν }
i−1
Yh
1 − eβ(µ−εν )
.
ν
Hence we obtain
ln Qη = −η
X
ν
i
h
β(µ−εν )
,
ln 1 − ηe
where η = +1 for bosons, and η = −1 for fermions.
Changing the summation into an integration in d-dimensions yields
Z
Z
Z
X
V Sd
V
d
d
d k =
k d−1 dk ,
→
d n →
d
d
(2π)
(2π)
ν
where
Sd =
2π d/2
.
(d/2 − 1)!
173
Then
Z
h
i
V Sd
d−1
β(µ−εν )
k
dk
ln
1
−
ηe
ln Qη = −η
(2π)d
Z
i
h
V Sd
d−1
−βks
= −η
,
k
dk ln 1 − ηze
(2π)d
where
z ≡ eβµ .
Introducing a new variable
s
x ≡ βk ,
results in
1/s
x
,
k=
β
→
V Sd β −d/s
ln Qη = −η
(2π)d s
Integrating by parts,
V Sd β −d/s
ln Qη =
(2π)d d
Z
dx x
d/s
Z
1
and dk =
s
1/s−1
x
dx
,
β
β
dx xd/s−1 ln 1 − ηze−x .
ze−x
V Sd β −d/s
=
1 − ηze−x
(2π)d d
Z
dx
xd/s
.
z −1 ex − η
Finally,
V Sd −d/s
d
η
ln Qη =
(z),
β
Γ
+ 1 · fd/s+1
d
(2π)
s
where
fnη (z)
1
≡
Γ(n)
Z
xn−1
dx −1 x
.
z e −η
Using similar notation, the total particle number is obtained as
Z
∂
V Sd β d/s ∂
d/s−1
−x
N=
dx
x
ln
1
−
ηze
ln Qη = −η
∂(βµ)
(2π)d s ∂(βµ)
β
Z
Z
xd/s−1 ze−x
V Sd β d/s
xd/s−1
V Sd β d/s
dx
=
z
dx
,
=
(2π)d s
1 − ηze−x
(2π)d s
z −1 e−x − η
and thus
N
Sd
d
η
d/s
n=
(z).
· fd/s
=
(kB T ) Γ
d
V
(2π) s
s
(b) Find the ratio P V /E, and compare it to the classical result obtained previously.
• The gas pressure is given by
d
V Sd (kB T )d/s+1
Γ
+ 1 · fd/s+1 (z).
P V = −kB T ln Qη =
(2π)d
d
s
174
Note that since ln Qη ∝ β −d/s ,
E=−
∂
ln Qη
∂β
z
=−
d ln Qη
d
= − kB T ln Qη ,
s β
s
=⇒
PV
s
= ,
E
d
which is the same result obtained classically.
(c) For fermions, calculate the dependence of E/N , and P , on the density n = N/V , at
zero temperature. (Hint: fm (z) → (ln z)m /m! as z → ∞.)
• For fermions at low T ,
1
eβ(ε−εF ) + 1
≈ θ(ε − εF ),
=⇒
fn (z) ≈
(βεF )n
,
nΓ(n)
resulting in
d/s+1
E ≈
V Sd εF
(2π)d d
d/s+1
,
PV ≈
V Sd εF
(2π)d s
d/s
,
n ≈
Sd εF
.
(2π)d s
Hence, we obtain
d/s+1
1
Sd εF
E
≈
·
N
n (2π)d d
d/s+1
P ∝ εF
≈
s
εF ∝ ns/d ,
d
and
∝ ns/d+1 .
(d) For bosons, find the dimension dℓ (s), below which there is no Bose condensation. Is
there condensation for s = 2 at d = 2?
• To see if Bose-Einstein condensation occurs, check whether a density n can be found
+
(z) is a monotonic function of z for 0 ≤ z ≤ 1, the
such that z = zmax = 1. Since fd/s
maximum value for the right hand side of the equation for n = N/V , given in part (a) is
Sd
(kB T )d/s Γ
(2π)d s
d
+
(1).
fd/s
s
If this value is larger than n, we can always find z < 1 that satisfies the above equation. If
this value is smaller than n, then the remaining portion of the particles should condense
+
(1) diverges at z = 1, where
into the ground state. Thus the question is whether fd/s
+
(zmax
fd/s
1
= 1) =
Γ (d/s)
175
Z
dx
xd/s−1
.
ex − 1
The integrand may diverge near x = 0; the contribution for x ∼ 0 is
Z ε
Z ε
xd/s−1
dx x
dx xd/s−2 ,
≃
e −1
0
0
which converges for d/s − 2 > −1, or d > s. Therefore, Bose-Einstein condensation occurs for d > s. For a two dimensional gas, d = s = 2, the integral diverges logarithmically,
and hence Bose-Einstein condensation does not occur.
********
3. Pauli paramagnetism: Calculate the contribution of electron spin to its magnetic
susceptibility as follows. Consider non-interacting electrons, each subject to a Hamiltonian
p~ 2
~ ,
− µ0 ~σ · B
2m
~ are ±B.
where µ0 = eh̄/2mc, and the eigenvalues of ~σ · B
~ has been ignored.)
(The orbital effect, p~ → ~p − eA,
H1 =
(a) Calculate the grand potential G − = −kB T ln Q− , at a chemical potential µ.
• The energy of the electron gas is given by
X
−
E≡
Ep (n+
p , np ),
p
where n±
p (= 0 or 1), denote the number of particles having ± spins and momentum p,
and
2
2
p
p
+
−
+
Ep (np , np ) ≡
− µ0 B np +
+ µ0 B n−
p
2m
2m
2
−
− p
− (n+
= (n+
+
n
)
p − np )µ0 B.
p
p
2m
The grand partition function of the system is
P + −
N=
(n +n )
∞
Xp p
X
−
exp −βEp (n+
Q=
exp(−βµN )
p , np )
−
{n+
p ,np }
N=0
X
=
−
{n+
p ,np }
=
−
+
−
exp βµ n+
p + np − βEp np , np
X
Y
p2
p2
· 1 + exp β µ + µ0 B −
1 + exp β µ − µ0 B −
2m
2m
p {n+ ,n− }
p
p
=
p2
p2
+
exp β µ − µ0 B −
np + µ + µ0 B −
n−
p
2m
2m
Y
p
= Q0 (µ + µ0 B) · Q0 (µ − µ0 B) ,
176
where
Y
p2
Q0 (µ) ≡
1 + exp β µ −
.
2m
p
Thus
ln Q = ln Q0 (µ + µ0 B) + ln Q0 (µ − µ0 B) .
Each contribution is given by
Z
p2 V
p2 3
−β 2m
) =
d
p
ln
1
+
ze
ln Q0 (µ) =
ln 1 + exp β(µ −
2m
(2πh̄)3
p
r
Z
√
V 4πm 2m
dx x ln(1 + ze−x ), where z ≡ eβµ ,
= 3
h β
β
X
and integrating by parts yields
ln Q0 (µ) = V
2πmkB T
h2
3/2
2 2
√
π3
Z
dx
x3/2
V −
=
f (z).
z −1 ex + 1
λ3 5/2
The total grand free energy is obtained from
ln Q(µ) =
as
i
V h −
−
−βµ0 B
βµ0 B
,
ze
ze
+
f
f
5/2
λ3 5/2
G = −kB T ln Q(µ) = −kB T
i
V h −
−
−βµ0 B
βµ0 B
.
ze
ze
+
f
f
5/2
λ3 5/2
(b) Calculate the densities n+ = N+ /V , and n− = N− /V , of electrons pointing parallel
and antiparallel to the field.
• The number densities of electrons with up or down spins is given by
where we used
N±
∂
V −
ze±βµ0 B ,
=z
ln Q± = 3 f3/2
V
∂z
λ
z
∂ −
−
f (z) = fn−1
(z).
∂z n
The total number of electrons is the sum of these, i.e.
V
−
−
−βµ0 B
βµ0 B
.
+ f3/2 ze
N = N+ + N− = 3 f3/2 ze
λ
177
(c) Obtain the expression for the magnetization M = µ0 (N+ − N− ), and expand the result
for small B.
• The magnetization is related to the difference between numbers of spin up and down
electrons as
i
V h −
−
βµ0 B
−βµ0 B
− f3/2 ze
.
M = µ0 (N+ − N− ) = µ0 3 f3/2 ze
λ
Expanding the results for small B, gives
∂ −
−
−
−
(z) ± z · βµ0 B f3/2
[z (1 ± βµ0 B)] ≈ f3/2
ze±βµ0 B ≈ f3/2
f3/2
(z),
∂z
which results in
M = µ0
V
2µ20 V
−
−
(z)
=
(z).
(2βµ
B)
·
f
· B · f1/2
0
1/2
3
3
λ
kB T λ
(d) Sketch the zero field susceptibility χ(T ) = ∂M/∂B|B=0 , and indicate its behavior at
low and high temperatures.
• The magnetic susceptibility is
χ≡
∂M
∂B
=
B=0
with z given by,
N =2
2µ20 V
· f − (z),
kB T λ3 1/2
V
−
· f3/2
(z).
3
λ
In the low temperature limit, (ln z = βµ → ∞)
Z ln(z)
n
1
[ln(z)]
n−1
dx x
=
,
Γ(n) 0
nΓ(n)
T →0
0
2/3
√
3N π 3
V 4(ln z)3/2
√
,
λ
, =⇒ ln z =
N =2 3 ·
λ
8V
3 π
1/3
1/3
1/3
√
2µ2o V
4µ20 V
4πmµ20 V
3N
3N
3N π 3
=
χ= √
=
.
λ
·
·
·
πkB T λ3
8V
kB T λ2
πV
h2
πV
fn− (z)
1
=
Γ(n)
Z
∞
xn−1
dx
1 + ex−ln(z)
≈
Their ratio of the last two expressions gives
χ
N
T →0
−
µ20 f1/2
3µ20 1
3µ20
1
3µ20
=
=
=
.
=
−
kB T f3/2
2kB T ln(z)
2kB T βεF
2kB TF
178
In the high temperature limit (z → 0),
→
z
fn (z)
z→0 Γ(n)
Z
∞
dx xn−1 e−x = z,
0
and thus
→ 2V
N
· z,
3
β→0 λ
=⇒
N
N
z≈
· λ3 =
·
2V
2V
h2
2πmkB T
3/2
→ 0,
which is consistent with β → 0. Using this result,
χ≈
The result
2µ20 V
N µ20
·
z
=
.
kB T λ3
kB T
χ
N
T →∞
=
µ20
,
kB T
is known as the Curie susceptibility.
χ/Nµο2
χ/Nµο2 ∼ 1/k BT
3/2k BTF
3/2k BTF
1/k BT
(e) Estimate the magnitude of χ/N for a typical metal at room temperature.
• Since TRoom ≪ TF ≈ 104 K, we can take the low T limit for χ (see(d)), and
3µ20
3 × (9.3 × 10−24 )2
χ
=
≈
≈ 9.4 × 10−24 J/T2 ,
N
2kB TF
2 × 1.38 × 10−23
179
where we used
µ0 =
eh
≃ 9.3 × 10−24 J/T.
2mc
********
4. Freezing of He3 :
At low temperatures He3 can be converted from liquid to solid by
application of pressure. A peculiar feature of its phase boundary is that (dP/dT )melting is
negative at temperatures below 0.3 K [(dP/dT )m ≈ −30atm K−1 at T ≈ 0.1 K]. We will
use a simple model of liquid and solid phases of He3 to account for this feature.
(a) In the solid phase, the He3 atoms form a crystal lattice. Each atom has nuclear spin
of 1/2. Ignoring the interaction between spins, what is the entropy per particle ss , due to
the spin degrees of freedom?
• Entropy of solid He3 comes from the nuclear spin degeneracies, and is given by
Ss
kB ln(2N )
ss =
=
= kB ln 2.
N
N
(b) Liquid He3 is modelled as an ideal Fermi gas, with a volume of 46Å3 per atom. What
is its Fermi temperature TF , in degrees Kelvin?
• The Fermi temperature for liquid 3 He may be obtained from its density as
εF
h2
TF =
=
kB
2mkB
3N
8πV
2/3
(6.7 × 10−34 )2
≈
2 · (6.8 × 10−27 )(1.38 × 10−23 )
3
8π × 46 × 10−30
2/3
≈ 9.2 K.
(c) How does the heat capacity of liquid He3 behave at low temperatures? Write down an
expression for CV in terms of N, T, kB , TF , up to a numerical constant, that is valid for
T ≪ TF .
• The heat capacity comes from the excited states at the fermi surface, and is given by
CV = kB
π2 2
3N
π2
T
π2
kB T D(εF ) =
kB T
=
N kB
.
6
6
2kB TF
4
TF
180
(d) Using the result in (c), calculate the entropy per particle sℓ , in the liquid at low
temperatures. For T ≪ TF , which phase (solid or liquid) has the higher entropy?
• The entropy can be obtained from the heat capacity as
T dS
,
CV =
dT
⇒
1
sℓ =
N
Z
T
0
CV dT
π2
T
=
kB
.
T
4
TF
As T → 0, sℓ → 0, while ss remains finite. This is an unusual situation in which the solid
has more entropy than the liquid! (The finite entropy is due to treating the nuclear spins
as independent. There is actually a weak coupling between spins which causes magnetic
ordering at a much lower temperature, removing the finite entropy.)
(e) By equating chemical potentials, or by any other technique, prove the Clausius–
Clapeyron equation (dP/dT )melting = (sℓ − ss )/(vℓ − vs ), where vℓ and vs are the volumes
per particle in the liquid and solid phases respectively.
• The Clausius-Clapeyron equation can be obtained by equating the chemical potentials
at the phase boundary,
µℓ (T, P ) = µs (T, P ),
and µℓ (T + ∆T, P + ∆P ) = µs (T + ∆T, P + ∆P ).
Expanding the second equation, and using the thermodynamic identities
∂µ
S
V
∂µ
= − , and
= ,
∂T P
N
∂P T
N
results in
∂P
∂T
=
melting
sℓ − ss
.
vℓ − vs
(f) It is found experimentally that vℓ − vs = 3Å3 per atom. Using this information, plus
the results obtained in previous parts, estimate (dP/dT )melting at T ≪ TF .
• The negative slope of the phase boundary results from the solid having more entropy
than the liquid, and can be calculated from the Clausius-Clapeyron relation
π2
T
− ln 2
4
TF
∂P
sℓ − ss
=
≈ kB
.
∂T melting
vℓ − vs
vℓ − vs
Using the values, T = 0.1 K, TF = 9.2 J K, and vℓ − vs = 3 Å3 , we estimate
∂P
≈ −2.7 × 106 Pa ◦ K−1 ,
∂T melting
181
in reasonable agreement with the observations.
********
5. Non-interacting fermions:
Consider a grand canonical ensemble of non-interacting
fermions with chemical potential µ. The one–particle states are labelled by a wavevector
~k, and have energies E(~k).
(a) What is the joint probability P ( n~k ), of finding a set of occupation numbers n~k ,
of the one–particle states?
• In the grand canonical ensemble with chemical potential µ, the joint probability of
finding a set of occupation numbers n~k , for one–particle states of energies E(~k) is given
by the Fermi distribution
P ( n~k
i
h
~
exp
β(µ
−
E(
k))n
Y
~
k
h
i,
)=
~
1 + exp β(µ − E(k))
~
k
where
n~k = 0 or 1,
for each ~k.
(b) Express your answer to part (a) in terms of the average occupation numbers
• The average occupation numbers are given by
h
i
exp β(µ − E(~k))
h
i,
n~k − =
~
1 + exp β(µ − E(k))
from which we obtain
h
i
exp β(µ − E(~k)) =
n~k
−
1 − n~k
This enables us to write the joint probability as
n~k Y 1 − n~k
P ( n~k ) =
n~k −
~
k
n
n~k
o
.
−
.
−
−
1−n~k .
(c) A random variable has a set of ℓ discrete outcomes with probabilities pn , where n =
1, 2, · · · , ℓ. What is the entropy of this probability distribution? What is the maximum
possible entropy?
• A random variable has a set of ℓ discrete outcomes with probabilities pn . The entropy
of this probability distribution is calculated from
S = −kB
ℓ
X
pn ln pn
n=1
182
.
The maximum entropy is obtained if all probabilities are equal, pn = 1/ℓ, and given by
Smax = kB ln ℓ.
(d) Calculate the entropy of the probability distribution for fermion occupation numbers
in part (b), and comment on its zero temperature limit.
• Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by
S = −kB
Xh
n~k
~
k
+
1 − n~k
−
ln n~k
−
−
ln 1 − n~k
−
i
.
In the zero temperature limit all occupation numbers are either 0 or 1. In either case the
contribution to entropy is zero, and the fermi system at T = 0 has zero entropy.
(e) Calculate the variance of the total number of particles N 2 c , and comment on its zero
temperature behavior.
• The total number of particles is given by N =
are independent
N
2
c
=
XD
~
k
since
D
n~2k
vanishes.
E
−
= n~k
E
n~2k
c
−
=
X D
~
k
E
n~2k
−
−
P
n~k . Since the occupation numbers
~
k
2
n~k −
=
. Again, since at T = 0, n~k
X
n~k
~
k
−
1 − n~k
−
−
,
= 0 or 1, the variance N 2
c
(f) The number fluctuations of a gas is related to its compressibility κT , and number
density n = N/V , by
N2
c
= N nkB T κT
.
Give a numerical estimate of the compressibility of the fermi gas in a metal at T = 0 in
units of Å3 eV −1 .
• To obtain the compressibility from N 2
c
= N nkB T κT , we need to examine the behavior
at small but finite temperatures. At small but finite T , a small fraction of states around the
fermi energy have occupation numbers around 1/2. The number of such states is roughly
N kB T /εF , and hence we can estimate the variance as
N2
c
= N nkB T κT ≈
183
1 N kB T
×
.
4
εF
The compressibility is then approximates as
κT ≈
1
,
4nεF
where n = N/V is the density. For electrons in a typical metal n ≈ 1029 m−3 ≈ 0.1Å3 , and
εF ≈ 5eV ≈ 5 × 104 ◦ K, resulting in
κT ≈ 0.5Å3 eV −1 .
********
6. Stoner ferromagnetism:
The conduction electrons in a metal can be treated as a
gas of fermions of spin 1/2 (with up/down degeneracy), and density n = N/V . The
Coulomb repulsion favors wave functions which are antisymmetric in position coordinates,
thus keeping the electrons apart. Because of the full (position and spin) antisymmetry of
fermionic wave functions, this interaction may be approximated by an effective spin-spin
coupling which favors states with parallel spins. In this simple approximation, the net
effect is described by an interaction energy
U =α
N+ N−
,
V
where N+ and N− = N − N+ are the numbers of electrons with up and down spins, and V
is the volume. (The parameter α is related to the scattering length a by α = 4πh̄2 a/m.)
(a) The ground state has two fermi seas filled by the spin up and spin down electrons.
Express the corresponding fermi wavevectors kF± in terms of the densities n± = N± /V .
• In the ground state, all available wavevectors are filled up in a sphere. Using the
appropriate density of states, the corresponding radii of kF± are calculated as
N± = V
Z
k<kF±
d3 k
3
(2π)
=V
Z
kF±
0
3
V kF±
2
k
dk
=
,
3
6π 2
(2π)
4π
leading to
kF± = 6π 2 n±
1/3
.
(b) Calculate the kinetic energy density of the ground state as a function of the densities
n± , and fundamental constants.
184
• For each Fermi sea, the kinetic energy is given by
Ekin.± = V
Z
k<kF±
5
h̄2 k 2 d3 k
h̄2 4π kF±
,
=
V
2m (2π)3
2m (2π)3 5
where m is the mass of the electron. The total kinetic energy density is thus equal to
5/3
Ekin
h̄2 1
h̄2 3
5/3
2 2/3
5
5
6π
n+ + n− .
=
k + kF − =
V
2m 10π 2 F +
2m 5
(c) Assuming small deviations n± = n/2 ± δ from the symmetric state, expand the kinetic
energy to fourth order in δ.
• Using the expansion
(x + δ)
5/3
=x
5/3
10 −1/3 δ 2
10 −4/3 δ 3
40 −7/3 δ 4
5 2/3
− x
+ x
+ O(δ 5 ),
+ x δ+ x
3
9
2
27
6
81
24
the kinetic energy calculated in the previous part can be written as
n 5/3 5 n −1/3 2
5 n −7/3 4
h̄2 6
Ekin
2 2/3
6
6π
+
δ +
δ + O(δ ) .
=
V
2m 5
2
9 2
243 2
(d) Express the spin-spin interaction density U/V in terms of n and δ. Find the critical
value of αc , such that for α > αc the electron gas can lower its total energy by spontaneously
developing a magnetization. (This is known as the Stoner instability.)
• The interaction energy density is
n
n
U
n2
= αn+ n− = α
+δ
−δ = α
− αδ 2 .
V
2
2
4
The total energy density has the form
2
E
E0 + αn2 /4
4
2 2/3 h̄
−1/3
3π
=
+
n
− α δ2 + O δ4 .
V
V
3
2m
When the second order term in δ is negative, the electron gas has lower energy for finite
δ, i.e. it acquires a spontaneous magnetization. This occurs for
α > αc =
2/3 h̄2 −1/3
4
3π 2
n
.
3
2m
185
(e) Explain qualitatively, and sketch the behavior of the spontaneous magnetization as a
function of α.
• For α > αc , the optimal value of δ is obtained by minimizing the energy density. Since the
coefficient of the fourth order term is positive, and the optimal δ goes to zero continuously
as α → αc ; the minimum energy is obtained for a value of δ 2 ∝ (α−αc ). The magnetization
√
is proportional to δ, and hence grows in the vicinity of αc as α − αc , as sketched below.
********
7. Boson magnetism: Consider a gas of non-interacting spin 1 bosons, each subject to a
Hamiltonian
H1 (~
p, sz ) =
p~ 2
− µ0 sz B
2m
,
where µ0 = eh̄/mc, and sz takes three possible values of (-1, 0, +1). (The orbital effect,
~ has been ignored.)
p~ → p~ − eA,
(a) In a grand canonical ensemble of chemical potential µ, what are the average occupation
n
o
numbers hn+ (~k)i, hn0 (~k)i, hn−(~k)i , of one-particle states of wavenumber ~k = p~/h̄?
• Average occupation numbers of the one-particle states in the grand canonical ensemble
of chemical potential µ, are given by the Bose-Einstein distribution
ns (~k) =
=
1
eβ [H(s)−µ] − 1
,
(for s = −1, 0, 1)
1
h 2
i
p
exp β 2m
− µ0 sB − βµ − 1
(b) Calculate the average total numbers {N+ , N0 , N− }, of bosons with the three possible
+
values of sz in terms of the functions fm
(z).
186
• The total numbers of particles with spin s are given by
Z
X
1
V
3
h
i
.
d
k
Ns =
ns (~k), =⇒ Ns =
p2
(2π)3
exp
β
−
µ
sB
−
βµ
−
1
0
~
2m
{k}
After a change of variables, k ≡ x1/2
√
Ns =
where
+
fm
(z)
1
≡
Γ(m)
Z
∞
0
2mkB T /h, we get
V +
βµ0 sB
f
ze
,
λ3 3/2
dx xm−1
,
z −1 ex − 1
λ≡ √
h
,
2πmkB T
z ≡ eβµ .
(c) Write down the expression for the magnetization M (T, µ) = µ0 (N+ − N− ), and by
expanding the result for small B find the zero field susceptibility χ(T, µ) = ∂M/∂B|B=0 .
• The magnetization is obtained from
M (T, µ) = µ0 (N+ − N− )
i
V h +
+
−βµ0 sB
βµ0 B
.
− f3/2 ze
= µ0 3 f3/2 ze
λ
Expanding the result for small B gives
∂ +
+
+
+
(z) ± z · βµ0 B f3/2
(z[1 ± βµ0 B]) ≈ f3/2
ze±βµ0 B ≈ f3/2
f3/2
(z).
∂z
+
+
Using zdfm
(z)/dz = fm−1
(z), we obtain
M = µ0
V
2µ20 V
+
+
(z)
=
(z),
(2βµ
B)
·
f
· B · f1/2
0
1/2
λ3
kB T λ3
and
χ≡
∂M
∂B
=
B=0
2µ20 V
· f + (z).
kB T λ3 1/2
To find the behavior of χ(T, n), where n = N/V is the total density, proceed as
follows:
(d) For B = 0, find the high temperature expansion for z(β, n) = eβµ , correct to second
order in n. Hence obtain the first correction from quantum statistics to χ(T, n) at high
temperatures.
187
+
• In the high temperature limit, z is small. Use the Taylor expansion for fm
(z) to write
the total density n(B = 0), as
3 +
N+ + N0 + N−
(z)
= 3 f3/2
V
λ
B=0
z3
3
z2
≈ 3 z + 3/2 + 3/2 + · · · .
λ
2
3
n(B = 0) =
Inverting the above equation gives
z=
nλ3
3
1
−
23/2
nλ3
3
2
+ ···.
The susceptibility is then calculated as
2µ20 V
· f + (z),
kB T λ3 1/2
2µ20 1
z2
χ/N =
z + 1/2 + · · ·
kB T nλ3
2
3
2
1
nλ
2µ0
1
2
.
1 + − 3/2 + 1/2
+O n
=
3kB T
3
2
2
χ=
(e) Find the temperature Tc (n, B = 0), of Bose–Einstein condensation. What happens to
χ(T, n) on approaching Tc (n) from the high temperature side?
• Bose-Einstein condensation occurs when z = 1, at a density
n=
3 +
f (1),
λ3 3/2
or a temperature
h2
Tc (n) =
2πmkB
n
3 ζ 3/2
2/3
,
+
+
(z) = ∞, the susceptibility χ(T, n) diverges
(1) ≈ 2.61. Since limz→1 f1/2
where ζ 3/2 ≡ f3/2
on approaching Tc (n) from the high temperature side.
(f) What is the chemical potential µ for T < Tc (n), at a small but finite value of B? Which
one-particle state has a macroscopic occupation number?
−1
~
• Since ns (~k, B) = z −1 eβ E s (k,B) −1
is a positive number for all ~k and sz , µ is bounded
above by the minimum possible energy, i.e.
for
T < Tc ,
and
B finite,
zeβµ0 B = 1,
188
=⇒
µ = −µ0 B.
Hence the macroscopically occupied one particle state has ~k = 0, and sz = +1.
(g) Using the result in (f), find the spontaneous magnetization,
M (T, n) = lim M (T, n, B).
B→0
• Contribution of the excited states to the magnetization vanishes as B → 0. Therefore the
total magnetization for T < Tc is due to the macroscopic occupation of the (k = 0, sz = +1)
state, and
M (T, n) = µ0 V n+ (k = 0)
= µ0 V n − nexcited
3V
= µ0 N − 3 ζ 3/2 .
λ
********
8. Dirac fermions are non-interacting particles of spin 1/2. The one-particle states come
in pairs of positive and negative energies,
p
E ± (~k) = ± m2 c4 + h̄2 k 2 c2
,
independent of spin.
(a) For any fermionic system of chemical potential µ, show that the probability of finding
an occupied state of energy µ + δ is the same as that of finding an unoccupied state of
energy µ − δ. (δ is any constant energy.)
• According to Fermi statistics, the probability of occupation of a state of of energy E is
p [n(E)] =
eβ(µ−E )n
,
1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ,
p [n(µ + δ)] =
eβδn
,
1 + eβδ
=⇒
p [n(µ + δ) = 1] =
eβδ
1
=
.
1 + eβδ
1 + e−βδ
Similarly, for a state of energy µ − δ,
p [n(µ − δ)] =
e−βδn
,
1 + e−βδ
=⇒
p [n(µ − δ) = 0] =
189
1
= p [n(µ + δ) = 1] ,
1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of
finding an unoccupied state of energy µ − δ.
(b) At zero temperature all negative energy Dirac states are occupied and all positive
energy ones are empty, i.e. µ(T = 0) = 0. Using the result in (a) find the chemical
potential at finite temperatures T .
• The above result implies that for µ = 0, hn(E)i + hn − E)i is unchanged for an tem-
perature; any particle leaving an occupied negative energy state goes to the corresponding
unoccupied positive energy state. Adding up all such energies, we conclude that the total
particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enfrces
µ(T ) = 0.
(c) Show that the mean excitation energy of this system at finite temperature satisfies
E(T ) − E(0) = 4V
Z
d3~k
E + (~k)
(2π)3 exp βE (~k) + 1
.
+
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated
as
E(T ) − E(0) =
X
k,sz
=2
[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]
X
k
2 hn+ (k)i E + (k) = 4V
Z
d3~k
E + (~k)
.
(2π)3 exp βE (~k) + 1
+
(d) Evaluate the integral in part (c) for massless Dirac particles (i.e. for m = 0).
• For m = 0, E + (k) = h̄c|k|, and
∞
4πk 2 dk h̄ck
E(T ) − E(0) = 4V
=
(set βh̄ck = x)
8π 3 eβh̄ck + 1
0
3 Z ∞
2V
x3
kB T
= 2 kB T
dx x
π
h̄c
e +1
0
3
kB T
7π 2
V kB T
.
=
60
h̄c
Z
For the final expression, we have noted that the needed integral is 3!f4− (1), and used the
given value of f4− (1) = 7π 4 /720.
(e) Calculate the heat capacity, CV , of such massless Dirac particles.
190
• The heat capacity can now be evaluated as
∂E
CV =
∂T
V
7π 2
=
V kB
15
kB T
h̄c
3
.
(f) Describe the qualitative dependence of the heat capacity at low temperature if the
particles are massive.
• When m 6= 0, there is an energy gap between occupied and empty states, and we thus
expect an exponentially activated energy, and hence heat capacity. For the low energy
excitations,
E + (k) ≈ mc2 +
and thus
h̄2 k 2
+ ···,
2m
√ Z ∞
2V
2 −βmc2 4π π
dxx2 e−x
E(T ) − E(0) ≈ 2 mc e
π
λ3
0
2
48 V
= √ 3 mc2 e−βmc .
πλ
The corresponding heat capacity, to leading order thus behaves as
C(T ) ∝ kB
V
2 2 −βmc2
βmc
e
.
λ3
********
9. Numerical estimates: The following table provides typical values for the Fermi energy
and Fermi temperature for (i) Electrons in a typical metal; (ii) Nucleons in a heavy nucleus;
and (iii) He3 atoms in liquid He3 (atomic volume = 46.2Å3 per atom).
n(1/m3 )
m(Kg)
εF (eV)
TF (K)
electron
1029
4.4
nucleons
1044
9 × 10−31
5 × 104
liquid He3
2.6 × 1028
1.6 × 10−27
4.6 × 10−27
1.0 × 108
×10−3
1.1 × 1012
101
(a) Estimate the ratio of the electron and phonon heat capacities at room temperature for
a typical metal.
• For an electron gas, TF ≈ 5 × 104 K,
TF ≫ Troom ,
=⇒
Celectron
π2 T
≈
·
≈ 0.025.
N kB
2 TF
191
For the phonon gas in iron, the Debye temperature is TD ≈ 470K, and hence
#
"
2
T
1
Cphonon
+ . . . ≈ 3,
≈3 1−
N kB
20 TD
resulting in
Celectron
≈ 8 × 10−3 .
Cphonon
(b) Compare the thermal wavelength of a neutron at room temperature to the minimum
wavelength of a phonon in a typical crystal.
• Thermal wavelengths are given by
λ≡ √
h
.
2πmkB T
For a neutron at room temperature, using the values
m = 1.67 × 10−27 kg,
T = 300 K,
kB = 1.38 × 10−23 JK−1 ,
h = 6.67 × 10−34 Js,
we obtain λ = 1Å.
The typical wavelength of a phonon in a solid is λ = 0.01 m, which is much longer
than the neutron wavelength. The minimum wavelength is, however, of the order of atomic
spacing (3 − 5 Å), which is comparable to the neutron thermal wavelength.
(c) Estimate the degeneracy discriminant, nλ3 , for hydrogen, helium, and oxygen gases
at room temperature and pressure. At what temperatures do quantum mechanical effects
become important for these gases?
• Quantum mechanical effects become important if nλ3 ≥ 1. In the high temperature
limit the ideal gas law is valid, and the degeneracy criterion can be reexpressed in terms
of pressure P = nkB T , as
nλ3 =
nh3
h3
P
=
≪ 1.
(2πmkB T )3/2
(kB T )5/2 (2πm)3/2
It is convenient to express the answers starting with an imaginary gas of ‘protons’ at room
temperature and pressure, for which
mp = 1.7 × 10−34 Kg,
and
(nλ3 )proton =
P = 1 atm. = 105 Nm−2 ,
(6.7 × 10−34 )3
10−5
= 2 × 10−5 .
−21
5/2
−27
3/2
(4.1 × 10 ) (2π · 1.7 × 10 )
192
The quantum effects appear below T = TQ , at which nλ3 becomes order of unity. Using
3
3
nλ = (nλ )proton
m 3/2
p
m
and TQ = Troom (nλ3 )3/2 ,
,
we obtain the following table:
(d) Experiments on He4 indicate that at temperatures below 1K, the heat capacity is
given by CV = 20.4T 3 JKg −1 K−1 . Find the low energy excitation spectrum, E(k), of He4 .
(Hint: There is only one non-degenerate branch of such excitations.)
• A spectrum of low energy excitations scaling as
E(k) ∝ k s ,
in d-dimensional space, leads to a low temperature heat capacity that vanishes as
C ∝ T d/s .
Therefore, from CV = 20.4 T 3 JKg−1 K−1 in d = 3, we can conclude s = 1, i.e. a spectrum
of the form
E(k) = h̄cs |~k|,
corresponding to sound waves of speed cs . Inserting all the numerical factors, we have
12π 4 N kB
CV =
5
T
Θ
3
,
where
h̄cs
Θ=
kB
6π 2 N
V
1/3
.
Hence, we obtain
E = h̄cs k = kB
2π 2 kB V T 3
5
CV
1/3
k = (2 × 10−32 Jm) k,
corresponding to a sound speed of cs ≈ 2 × 102 ms−1 .
********
10. Solar interior: According to astrophysical data, the plasma at the center of the sun
has the following properties:
Temperature:
Hydrogen density:
Helium density:
T = 1.6 × 107 K
ρH = 6 × 104 kg m−3
ρHe = 1 × 105 kg m−3 .
(a) Obtain the thermal wavelengths for electrons, protons, and α-particles (nuclei of He).
193
• The thermal wavelengths of electrons, protons, and α-particles in the sun are obtained
from
λ= √
h
,
2πmkB T
and T = 1.6 × 107 K, as
λelectron ≈ p
λproton ≈ p
2π × (9.1 ×
10−31
2π × (1.7 ×
10−27
6.7 × 10−34 J/s
Kg) · (1.4 ×
10−23
Kg) · (1.4 ×
10−23
6.7 × 10−34 J/s
and
J/K) · (1.6 ×
107
107
K)
≈ 1.9 × 10−11 m,
≈ 4.3 × 10−13 m,
J/K) · (1.6 ×
K)
1
λα−particle = λproton ≈ 2.2 × 10−13 m.
2
(b) Assuming that the gas is ideal, determine whether the electron, proton, or α-particle
gases are degenerate in the quantum mechanical sense.
• The corresponding number densities are given by
3
nH ≈ 3.5 × 1031 m−3 ,
ρHe
3
≈ 1.5 × 1031 m−3 ,
ρHe ≈ 1.0 × 105 Kg/m
=⇒ nHe =
4mH
ne = 2nHe + nH ≈ 8.5 × 1031 m3 .
ρH ≈ 6 × 104 kg/m
=⇒
The criterion for degeneracy is nλ3 ≥ 1, and
nH · λ3H ≈2.8 × 10−6 ≪ 1,
nHe · λ3He ≈1.6 × 10−7 ≪ 1,
ne · λ3e ≈0.58 ∼ 1.
Thus the electrons are weakly degenerate, and the nuclei are not.
(c) Estimate the total gas pressure due to these gas particles near the center of the sun.
• Since the nuclei are non-degenerate, and even the electrons are only weakly degenerate,
their contributions to the overall pressure can be approximately calculated using the ideal
gas law, as
P ≈ (nH + nhe + ne ) · kB T ≈ 13.5 × 1031 (m−3 ) · 1.38 × 10−23 (J/K) · 1.6 × 107 (K)
2
≈ 3.0 × 1016 N/m .
194
(d) Estimate the total radiation pressure close to the center of the sun. Is it matter, or
radiation pressure, that prevents the gravitational collapse of the sun?
• The Radiation pressure at the center of the sun can be calculated using the black body
formulas,
P =
as
P =
1U
,
3V
1U
π 2 k4 4
c=
T = σT 4 ,
4V
60h̄3 c3
and
4
4 · 5.7 × 10−8 W/(m2 K4 ) · (1.6 × 107 K)4
2
σT 4 =
≈ 1.7 × 1013 N/m .
8
3c
3 · 3.0 × 10 m/s
Thus at the pressure in the solar interior is dominated by the particles.
********
11. Bose condensation in d–dimensions:
Consider a gas of non-interacting (spinless)
bosons with an energy spectrum ǫ = p2 /2m, contained in a box of “volume” V = Ld in d
dimensions.
(a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical
+
potential µ. Express your answers in terms of d and fm
(z), where z = eβµ , and
1
=
Γ (m)
+
fm
(z)
Z
∞
0
xm−1
dx.
z −1 ex − 1
(Hint: Use integration by parts on the expression for ln Q.)
• We have
Q=
=
∞
X
P
eNβµ
N=0
n =N
i
X
i
{ni }
YX
exp −β
eβ(µ−ǫi )ni =
i {ni }
Y
i
X
i
ni ǫi
!
,
1
1 − eβ(µ−ǫi )
P
P
whence ln Q = − i ln 1 − eβ(µ−ǫi ) . Replacing the summation i with a d dimensional
h
i
R
d
d R d−1
integration V dd k/ (2π) = V Sd / (2π)
k
dk, where Sd = 2π d/2 / (d/2 − 1)!, leads
to
ln Q = −
V Sd
d
(2π)
Z
2 2
k d−1 dk ln 1 − ze−βh̄ k /2m .
The change of variable x = βh̄2 k 2 /2m (⇒ k =
results in
V Sd 1
ln Q = −
d
(2π) 2
2m
h̄2 β
d/2 Z
195
p
p
2mx/β/h̄ and dk = dx 2m/βx/2h̄)
xd/2−1 dx ln 1 − ze−x .
Finally, integration by parts yields
V Sd 1
ln Q =
d
(2π) d
2m
h̄2 β
d/2 Z
x
d/2
ze−x
Sd
dx
=
V
1 − ze−x
d
i.e.
Sd
G = −kB T ln Q = −V
d
2m
h2 β
d/2
kB T Γ
2m
h2 β
d/2 Z
dx
d
(z) ,
+ 1 f+
d
2 +1
2
which can be simplified, using the property Γ (x + 1) = xΓ (x), to
G=−
V
(z) .
kB T f +
d
2 +1
λd
The average number of particles is calculated as
d/2 Z
ze−x
Sd 2m
∂
d/2−1
x
dx
ln Q = V
N=
∂ (βµ)
d h2 β
1 − ze−x
,
d/2 Sd 2m
V +
d
+
=V
f d (z) = d f d (z)
Γ
2
2 h2 β
2
λ 2
i.e.
n=
1 +
f d (z) .
λd 2
(b) Calculate the ratio P V /E, and compare it to the classical value.
• We have P V = −G, while
E=−
d ln Q
d
∂
ln Q = +
= − G.
∂β
2 β
2
Thus P V /E = 2/d, identical to the classical value.
(c) Find the critical temperature, Tc (n), for Bose-Einstein condensation.
• The critical temperature Tc (n) is given by
n=
1 +
1
f d (1) = d ζ d ,
d
λ 2
λ 2
for d > 2, i.e.
h2
Tc =
2mkB
n
ζd
2
!2/d
(d) Calculate the heat capacity C (T ) for T < Tc (n).
196
.
xd/2
,
z −1 ex − 1
• At T < Tc , z = 1 and
∂E
C (T ) =
∂T
z=1
d ∂G
=−
2 ∂T
z=1
d
=−
2
d
+1
2
G
d
=
T
2
d
+1
2
V
kB ζ d +1 .
2
λd
(e) Sketch the heat capacity at all temperatures.
•
.
(f) Find the ratio, Cmax /C (T → ∞), of the maximum heat capacity to its classical limit,
and evaluate it in d = 3.
• As the maximum of the heat capacity occurs at the transition,
Cmax
d
= C (Tc ) =
2
Thus
d
+1
2
V
ζ d /n
2
kB f +
d
2 +1
Cmax
=
C (T → ∞)
d
+1
2
d
(1) = N kB
2
ζ d +1
2
ζd
d
+1
2
ζ d +1
2
ζd
.
2
,
2
which evaluates to 1.283 in d = 3.
(g) How does the above calculated ratio behave as d → 2? In what dimensions are your
results valid? Explain.
+
• The maximum heat capacity, as it stands above, vanishes as d → 2! Since fm
(x → 1) →
∞ if m ≤ 2, the fugacuty z is always smaller than 1. Hence, there is no macroscopic
occupation of the ground state, even at the lowest temperatures, i.e. no Bose-Einstein
condensation in d ≤ 2. The above results are thus only valid for d ≥ 2.
197
********
12. Exciton dissociation in a semiconductor:
Shining an intense laser beam on a semi-
conductor can create a metastable collection of electrons (charge −e, and effective mass
me ) and holes (charge +e, and effective mass mh ) in the bulk. The oppositely charged
particles may pair up (as in a hydrogen atom) to form a gas of excitons, or they may
dissociate into a plasma. We shall examine a much simplified model of this process.
(a) Calculate the free energy of a gas composed of Ne electrons and Nh holes, at temperature T , treating them as classical non-interacting particles of masses me and mh .
• The canonical partition function of gas of non-interacting electrons and holes is the
product of contributions from the electron gas, and from the hole gas, as
Ze−h
1
= Ze Zh =
Ne !
V
λ3e
Ne
1
·
Nh !
V
λ3h
Nh
,
√
where λα = h/ 2πmα kB T (α =e, h). Evaluating the factorials in Stirling’s approximation,
we obtain the free energy
Fe−h = −kB T ln Ze−h = Ne kB T ln
Ne 3
λ
eV e
+ Nh kB T ln
Nh 3
λ .
eV h
(b) By pairing into an excition, the electron hole pair lowers its energy by ε. [The binding
energy of a hydrogen-like exciton is ε ≈ me4 /(2h̄2 ǫ2 ), where ǫ is the dielectric constant,
−1
and m−1 = m−1
e + mh .] Calculate the free energy of a gas of Np excitons, treating them
as classical non-interacting particles of mass m = me + mh .
• Similarly, the partition function of the exciton gas is calculated as
1
Zp =
Np !
V
λ3p
Np
e−β(−Np ǫ) ,
leading to the free energy
Fp = Np kB T ln
where λp = h/
p
Np 3
λ
eV p
− Np ǫ,
2π (me + mh ) kB T .
(c) Calculate the chemical potentials µe , µh , and µp of the electron, hole, and exciton
states, respectively.
198
• The chemical potentials are derived from the free energies, through
µe =
µh =
∂Fe−h
∂Ne
∂Fe−h
∂Nh
∂Fp
∂Np
µp =
where nα = Nα /V (α =e, h, p).
T,V
= kB T ln ne λ3e ,
T,V
= kB T ln nh λ3h ,
T,V
= kB T ln np λ3p − ǫ,
(d) Express the equilibrium condition between excitons and electron/holes in terms of their
chemical potentials.
• The equilibrium condition is obtained by equating the chemical potentials of the electron
and hole gas with that of the exciton gas, since the exciton results from the pairing of an
electron and a hole,
electron + hole ⇀
↽ exciton.
Thus, at equilibrium
µe (ne , T ) + µh (nh , T ) = µp (np , T ) ,
which is equivalent, after exponentiation, to
ne λ3e · nh λ3h = np λ3p e−βǫ .
(e) At a high temperature T , find the density np of excitons, as a function of the total
density of excitations n ≈ ne + nh .
• The equilibrium condition yields
np = ne nh
λ3e λ3h βǫ
e .
λ3p
At high temperature, np ≪ ne = nh ≈ n/2, and
n 2
h3
λ3 λ3
np = ne nh e 3 h eβǫ =
3/2
λp
2
(2πkB T )
********
199
me + mh
me mh
3/2
eβǫ .
13. Freezing of He4 :
At low temperatures He4 can be converted from liquid to solid by
application of pressure. An interesting feature of the phase boundary is that the melting
pressure is reduced slightly from its T = 0K value, by approximately 20Nm−2 at its
minimum at T = 0.8K. We will use a simple model of liquid and solid phases of 4 He to
account for this feature.
(a) The important excitations in liquid 4 He at T < 1K are phonons of velocity c. Calculate
the contribution of these modes to the heat capacity per particle CVℓ /N , of the liquid.
• The dominant excitations in liquid 4 He at T < 1K are phonons of velocity c. The
corresponding dispersion relation is ε(k) = h̄ck. From the average number of phonons in
D
E
−1
mode ~k, given by n(~k) = [exp(βh̄ck) − 1] , we obtain the net excitation energy as
h̄ck
exp(βh̄ck) − 1
~
k
Z
4πk 2 dk
h̄ck
=V ×
(change variables to x = βh̄ck)
3
(2π) exp(βh̄ck) − 1
4 Z ∞
4
x3
6
kB T
π2
kB T
V
dx x
,
h̄c
=
V h̄c
=
2π 2
h̄c
3! 0
e −1
30
h̄c
Ephonons =
X
where we have used
1
ζ4 ≡
3!
Z
0
∞
dx
x3
π4
=
.
ex − 1
90
The corresponding heat capacity is now obtained as
dE
2π 2
CV =
=
V kB
dT
15
kB T
h̄c
3
,
resulting in a heat capacity per particle for the liquid of
2π 2
CVℓ
=
kB vℓ
N
15
kB T
h̄c
3
.
(b) Calculate the low temperature heat capacity per particle CVs /N , of solid 4 He in terms
of longitudinal and transverse sound velocities cL , and cT .
• The elementary excitations of the solid are also phonons, but there are now two transverse sound modes of velocity cT , and one longitudinal sound mode of velocity cL . The
200
contributions of these modes are additive, each similar inform to the liquid result calculated
above, resulting in the final expression for solid heat capacity of
2π 2
CVs
=
kB vs
N
15
kB T
h̄
3
×
2
1
+ 3
3
cT
cL
.
(c) Using the above results calculate the entropy difference (sℓ − ss ), assuming a single
sound velocity c ≈ cL ≈ cT , and approximately equal volumes per particle vℓ ≈ vs ≈ v.
Which phase (solid or liquid) has the higher entropy?
• The entropies can be calculated from the heat capacities as
sℓ (T ) =
ss (T ) =
Z
Z
T
0
T
0
CVℓ (T ′ )dT ′
2π 2
=
kB vℓ
T′
45
2π 2
CVs (T ′ )dT ′
=
kB vs
T′
45
kB T
h̄c
kB T
h̄
3
,
3
×
2
1
+ 3
3
cT
cL
.
Assuming approximately equal sound speeds c ≈ cL ≈ cT ≈ 300ms−1 , and specific volumes
vℓ ≈ vs ≈ v = 46Å3 , we obtain the entropy difference
4π 2
kB v
sℓ − ss ≈ −
45
kB T
h̄c
3
.
The solid phase has more entropy than the liquid because it has two more phonon excitation
bands.
(d) Assuming a small (temperature independent) volume difference δv = vℓ − vs , calculate
the form of the melting curve. To explain the anomaly described at the beginning, which
phase (solid or liquid) must have the higher density?
• Using the Clausius-Clapeyron equation, and the above calculation of the entropy difference, we get
∂P
∂T
melting
sℓ − ss
4π 2
v
=
=−
kB
vℓ − vs
45
δv
kB T
h̄c
3
.
Integrating the above equation gives the melting curve
π2
v
Pmelt (T ) = P (0) −
kB
45 δv
kB T
h̄c
3
T.
To explain the reduction in pressure, we need δv = vℓ − vs > 0, i.e. the solid phase has
the higher density, which is normal.
201
********
14. Neutron star core:
Professor Rajagopal’s group at MIT has proposed that a new
phase of QCD matter may exist in the core of neutron stars. This phase can be viewed as
a condensate of quarks in which the low energy excitations are approximately
E(~k)± =± h̄2
2
~
| k | − kF
2M
.
The excitations are fermionic, with a degeneracy of g = 2 from spin.
(a) At zero temperature all negative energy states are occupied and all positive energy
ones are empty, i.e. µ(T = 0) = 0. By relating occupation numbers of states of energies
µ + δ and µ − δ, or otherwise, find the chemical potential at finite temperatures T .
• According to Fermi statistics, the probability of occupation of a state of of energy E is
eβ(µ−E )n
p [n(E)] =
,
1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ,
eβδn
,
p [n(µ + δ)] =
1 + eβδ
=⇒
eβδ
1
p [n(µ + δ) = 1] =
=
.
βδ
1+e
1 + e−βδ
Similarly, for a state of energy µ − δ,
p [n(µ − δ)] =
e−βδn
,
1 + e−βδ
=⇒
p [n(µ − δ) = 0] =
1
= p [n(µ + δ) = 1] ,
1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of
finding an unoccupied state of energy µ − δ. This implies that for µ = 0, hn(E)i + hn(−E)i
is unchanged for an temperature; for every particle leaving an occupied negative energy
state a particle goes to the corresponding unoccupied positive energy state. Adding up all
such energies, we conclude that the total particle number is unchanged if µ stays at zero.
Thus, the particle–hole symmetry enforces µ(T ) = 0.
(b) Assuming a constant density of states near k = kF , i.e. setting d3 k ≈ 4πkF2 dq with
q = |~k | − kF , show that the mean excitation energy of this system at finite temperature is
k2
E(T ) − E(0) ≈ 2gV F2
π
Z
∞
0
202
dq
E + (q)
exp (βE + (q)) + 1
.
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated
as
E(T ) − E(0) =
X
[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]
k,s
=g
X
k
2 hn+ (k)i E + (k) = 2gV
Z
E + (~k)
d3~k
.
(2π)3 exp βE (~k) + 1
+
The largest contribution to the integral comes for |~k | ≈ kF . and setting q = (|~k | − kF )
and using d3 k ≈ 4πkF2 dq, we obtain
4πkF2
E(T ) − E(0) ≈ 2gV
2
8π 3
Z
∞
0
E + (q)
k2
dq
= 2gV F2
exp (βE + (q)) + 1
π
Z
0
∞
dq
E + (q)
exp (βE + (q)) + 1
.
(c) Give a closed form answer for the excitation energy by evaluating the above integral.
• For E + (q) = h̄2 q 2 /(2M ), we have
∞
h̄2 q 2 /2M
=
(set βh̄2 q 2 /2M = x)
βh̄2 q 2 /2M + 1
e
0
1/2 Z ∞
2
gV kF
x1/2
2M kB T
=
dx
k
T
B
π2
ex + 1
h̄2
0
1/2 √ ζ3/2 V kF2
1
gV kF2
π
1
2M kB T
√
√
ζ
=
1
−
= 2 kB T
1
−
kB T.
3/2
π
2
π
λ
h̄2
2
2
k2
E(T ) − E(0) = 2gV F2
π
Z
dq
−
For the final expression, we have used the value of fm
(1), and introduced the thermal
√
wavelength λ = h/ 2πM kB T .
(d) Calculate the heat capacity, CV , of this system, and comment on its behavior at low
temperature.
• Since E ∝ T 3/2 ,
∂E
CV =
∂T
V
3ζ3/2
3E
=
=
2T
2π
√
1
V kF2
1− √
kB ∝ T .
λ
2
This is similar to the behavior of a one dimensional system of bosons (since the density
of states is constant in q as in d = 1). Of course, for any fermionic system the density of
states close to the Fermi surface has this character. The difference with the usual Fermi
systems is the quadratic nature of the excitations above the Fermi surface.
********
203
15.
Non-interacting bosons:
Consider a grand canonical ensemble of non-interacting
bosons with chemical potential µ. The one–particle states are labelled by a wavevector ~
q,
and have energies E(~q).
(a) What is the joint probability P ({nq~ }), of finding a set of occupation numbers {nq~}, of
the one–particle states, in terms of the fugacities zq~ ≡ exp [β(µ − E(~q))]?
• In the grand canonical ensemble with chemical potential µ, the joint probability of
finding a set of occupation numbers {nq~}, for one–particle states of energies E(~q) is given
by the normalized bose distribution
P ({nq~ }) =
=
Y
q
~
Y
q
~
{1 − exp [β(µ − E(~q))]} exp [β(µ − E(~q))nq~]
n
(1 − zq~ ) zq~ q~ ,
with nq~ = 0, 1, 2, · · · ,
for each ~q.
(b) For a particular ~q, calculate the characteristic function hexp [iknq~ ]i.
n
• Summing the geometric series with terms growing as zq~ eik q~ , gives
hexp [iknq~ ]i =
1 − zq~
1 − exp [β(µ − E(~q))]
=
.
1 − exp [β(µ − E(~q)) + ik]
1 − zq~ eik
(c) Using the result of part (b), or otherwise, give expressions for the mean and variance
of nq~ . occupation number hnq~ i.
• Cumulnats can be generated by expanding the logarithm of the characteristic function
in powers of k. Using the expansion formula for ln(1 + x), we obtain
ln hexp [iknq~ ]i = ln (1 − zq~ ) − ln 1 − zq~ 1 + ik − k 2 /2 + · · ·
zq~
k 2 zq~
= − ln 1 − ik
+
+···
1 − zq~
2 1 − zq~
"
2 #
zq~
zq~
zq~
k2
= ik
−
+
+···
1 − zq~
2 1 − zq~
1 − zq~
= ik
zq~
zq~
k2
−
+ ···.
1 − zq~
2 (1 − zq~ )2
From the coefficients in the expansion, we can read off the mean and variance
hnq~ i =
zq~
,
1 − zq~
and
204
nq2~
c
=
zq~
2.
(1 − zq~ )
(d) Express the variance in part (c) in terms of the mean occupation number hnq~ i.
• Inverting the relation relating nq~ to zq~ , we obtain
zq~ =
hnq~ i
.
1 + hnq~i
Substituting this value in the expression for the variance gives
nq2~
=
c
zq~
(1 − zq~ )
2
= hnq~ i (1 + hnq~ i) .
(e) Express your answer to part (a) in terms of the occupation numbers {hnq~i}.
• Using the relation between zq~ and nq~ , the joint probability can be reexpressed as
i
Yh
nq~
−1−nq~
.
P ({nq~}) =
(hnq~ i) (1 + hnq~ i)
q
~
(f) Calculate the entropy of the probability distribution for bosons, in terms of {hnq~ i}, and
comment on its zero temperature limit.
• Quite generally, the entropy of a probability distribution P is given by S = −kB hln P i.
Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by
S = −kB
X
q
~
[hnq~ i ln hnq~ i − (1 + hnq~ i) ln (1 + hnq~i)] .
In the zero temperature limit all occupation numbers are either 0 (for excited states) or
infinity (for the ground states). In either case the contribution to entropy is zero, and the
system at T = 0 has zero entropy.
********
16. Relativistic Bose gas in d dimensions:
Consider a gas of non-interacting (spinless)
bosons with energy ǫ = c |~
p |, contained in a box of “volume” V = Ld in d dimensions.
(a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical
+
(z), where z = eβµ , and
potential µ. Express your answers in terms of d and fm
+
(z)
fm
1
=
(m − 1)!
Z
0
205
∞
xm−1
dx.
z −1 ex − 1
(Hint: Use integration by parts on the expression for ln Q.)
• We have
Q=
=
∞
X
P
eNβµ
N=0
n =N
i
X
i
exp −β
{ni }
YX
eβ(µ−ǫi )ni =
i {ni }
Y
i
X
ni ǫi
i
!
,
1
1−
eβ(µ−ǫi )
P
P
i)
whence ln Q = − i ln 1 − eβ(µ−ǫ
.
Replacing
the
summation
i with a d dimensional
iR
h
R∞
∞
d
d
k d−1 dk, where Sd = 2π d/2 / (d/2 − 1)!,
integration 0 V dd k/ (2π) = V Sd / (2π)
0
leads to
ln Q = −
V Sd
d
(2π)
Z
∞
0
k d−1 dk ln 1 − ze−βh̄ck .
The change of variable x = βh̄ck results in
ln Q = −
V Sd
d
(2π)
kB T
h̄c
d Z
∞
0
xd−1 dx ln 1 − ze−x .
Finally, integration by parts yields
V Sd 1
ln Q =
d
(2π) d
kB T
h̄c
d Z
∞
0
ze−x
Sd
x dx
=
V
1 − ze−x
d
d
kB T
hc
d Z
∞
dx
0
xd
,
z −1 ex − 1
leading to
Sd
G = −kB T ln Q = −V
d
kB T
hc
d
+
kB T d!fd+1
(z) ,
which can be somewhat simplified to
G = −kB T
V π d/2 d! +
f
(z) ,
λdc (d/2)! d+1
where λc ≡ hc/(kB T ). The average number of particles is calculated as
N =−
∂G
∂G
V π d/2 d! +
= −βz
= d
f (z) ,
∂µ
∂z
λc (d/2)! d
where we have used z∂fd+1 (z)/∂z = fd (z). Dividing by volume, the density is obtained as
n=
1 π d/2 d! +
f (z) .
λdc (d/2)! d
206
(b) Calculate the gas pressure P , its energy E, and compare the ratio E/(P V ) to the
classical value.
• We have P V = −G, while
E=−
∂ ln Q
∂β
= +d
z
ln Q
= −dG.
β
Thus E/(P V ) = d, identical to the classical value for a relativistic gas.
(c) Find the critical temperature, Tc (n), for Bose-Einstein condensation, indicating the
dimensions where there is a transition.
• The critical temperature Tc (n) is given by
n=
1 π d/2 d!
1 π d/2 d! +
f
(z
=
1)
=
ζd .
λdc (d/2)! d
λdc (d/2)!
This leads to
hc
Tc =
kB
n(d/2)!
π d/2 d!ζd
1/d
.
However, ζd is finite only for d > 1, and thus a transition exists for all d > 1.
(d) What is the temperature dependence of the heat capacity C (T ) for T < Tc (n)?
• At T < Tc , z = 1 and E = −dG ∝ T d+1 , resulting in
C (T ) =
∂E
∂T
= (d + 1)
z=1
π d/2 d!
G
V
E
= −d(d + 1) = d(d + 1) d kB
ζd+1 ∝ T d .
T
T
λc
(d/2)!
(e) Evaluate the dimensionless heat capacity C(T )/(N kB ) at the critical temperature
T = Tc , and compare its value to the classical (high temperature) limit.
• We can divide the above formula of C(T ≤ T c), and the one obtained earlier for N (T ≥
T c), and evaluate the result at T = Tc (z = 1) to obtain
d(d + 1)ζd+1
C(Tc )
=
.
N kB
ζd
In the absence of quantum effects, the heat capacity of a relativistic gas is C/(N kB ) = d;
this is the limiting value for the quantum gas at infinite temperature.
********
207
17. Graphene is a single sheet of carbon atoms bonded into a two dimensional hexagonal
lattice. It can be obtained by exfoliation (repeated peeling) of graphite. The band structure of graphene is such that the single particles excitations behave as relativistic Dirac
fermions, with a spectrum that at low energies can be approximated by
E ± (~k) = ±h̄v ~k
.
There is spin degeneracy of g = 2, and v ≈ 106 ms−1 . Recent experiments on unusual
transport properties of graphene were reported in Nature 438, 197 (2005). In this problem,
you shall calculate the heat capacity of this material.
(a) If at zero temperature all negative energy states are occupied and all positive energy
ones are empty, find the chemical potential µ(T ).
• According to Fermi statistics, the probability of occupation of a state of of energy E is
eβ(µ−E )n
p [n(E)] =
,
1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ,
eβδn
,
p [n(µ + δ)] =
1 + eβδ
eβδ
1
p [n(µ + δ) = 1] =
=
.
βδ
1+e
1 + e−βδ
=⇒
Similarly, for a state of energy µ − δ,
p [n(µ − δ)] =
e−βδn
,
1 + e−βδ
=⇒
p [n(µ − δ) = 0] =
1
= p [n(µ + δ) = 1] ,
1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of
finding an unoccupied state of energy µ − δ.
At zero temperature all negative energy Dirac states are occupied and all positive
energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0,
hn(E)i + hn − E)i is unchanged for all temperatures; any particle leaving an occupied
negative energy state goes to the corresponding unoccupied positive energy state. Adding
up all such energies, we conclude that the total particle number is unchanged if µ stays at
zero. Thus, the particle–hole symmetry enforces µ(T ) = 0.
(b) Show that the mean excitation energy of this system at finite temperature satisfies
E(T ) − E(0) = 4A
Z
d2~k
E + (~k)
(2π)2 exp βE (~k) + 1
+
208
.
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated
as
E(T ) − E(0) =
X
k,sz
=2
[hn+ (k)i E + (k) − (1 − hn− (k)i) E − (k)]
X
k
2 hn+ (k)i E + (k) = 4A
Z
E + (~k)
d2~k
.
(2π)2 exp βE (~k) + 1
+
(c) Give a closed form answer for the excitation energy by evaluating the above integral.
• For E + (k) = h̄v|k|, and
∞
2πkdk h̄vk
=
(set βh̄ck = x)
4π 2 eβh̄vk + 1
0
2 Z ∞
2A
x2
kB T
=
dx x
kB T
π
h̄v
e +1
0
2
kB T
3ζ3
.
AkB T
=
π
h̄v
E(T ) − E(0) = 4A
Z
For the final expression, we have noted that the needed integral is 2!f3− (1), and used
f3− (1) = 3ζ3 /4.
E(T ) − E(0) = A
Z
E + (~k)
d2~k
(2π)2 exp βE (~k) − 1
.
+
(d) Calculate the heat capacity, CV , of such massless Dirac particles.
• The heat capacity can now be evaluated as
∂E
CV =
∂T
V
9ζ3
=
AkB
π
kB T
h̄v
2
.
(e) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capacity of graphene. The typical sound velocity in graphite is of the order of 2×104 ms−1 . Is the
low temperature heat capacity of graphene controlled by phonon or electron contributions?
• The single particle excitations for phonons also have a linear spectrum, with E p = h̄vp |k|
and correspond to µ = 0. Thus qualitatively they give the same type of contribution to
209
energy and heat capacity. The difference is only in numerical pre-factors. The precise
contribution from a single phonon branch is given by
Z ∞
2πkdk h̄vp k
Ep (T ) − Ep (0) = A
=
(set βh̄ck = x)
4π 2 eβh̄vp k − 1
0
2 Z ∞
A
x2
kB T
=
dx x
kB T
2π
h̄vp
e −1
0
2
2
ζ3
3ζ3
kB T
kB T
= AkB T
, CV,p =
.
AkB
π
h̄vp
π
h̄vp
We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since
the speed of Dirac fermions is considerably larger than that of phonons, their contribution
to heat capacity of graphene is negligible.
********
18. Graphene bilayer:
The layers of graphite can be peeled apart through different
exfoliation processes. Many such processes generate single sheets of carbon atoms, as
well as bilayers in which the two sheets are weakly coupled. The hexagonal lattice of the
single layer graphene, leads to a band structure that at low energies can be approximated
by E 1± layer (~k) = ±tk (ak), as in relativistic Dirac fermions. (Here k = ~k , a is a lattice
spacing, and tk is a typical in-plane hopping energy.) A weak hopping energy t⊥ between
the two sheets of the bilayer modifies the low energy excitations drastically, to
E bilayer
(~k)
±
=±
t2k
2t⊥
(ka)2
,
i.e. resembling massive Dirac fermions. In addition to the spin degeneracy, there are two
branches of such excitations per unit cell, for an overall degeneracy of g = 4.
(a) For the undoped material with one electron per site, at zero temperature all negative
energy states are occupied and all positive energy ones are empty. Find the chemical
potential µ(T ).
• According to Fermi statistics, the probability of occupation of a state of of energy E is
p [n(E)] =
eβ(µ−E )n
,
1 + eβ(µ−E )
for
n = 0, 1.
For a state of energy µ + δ,
p [n(µ + δ)] =
eβδn
,
1 + eβδ
=⇒
p [n(µ + δ) = 1] =
210
eβδ
1
=
.
βδ
1+e
1 + e−βδ
Similarly, for a state of energy µ − δ,
e−βδn
,
p [n(µ − δ)] =
1 + e−βδ
p [n(µ − δ) = 0] =
=⇒
1
= p [n(µ + δ) = 1] ,
1 + e−βδ
i.e. the probability of finding an occupied state of energy µ + δ is the same as that of
finding an unoccupied state of energy µ − δ.
At zero temperature all negative energy Dirac states are occupied and all positive
energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0,
hn(E)i + hn − E)i is unchanged for all temperatures; any particle leaving an occupied
negative energy state goes to the corresponding unoccupied positive energy state. Adding
up all such energies, we conclude that the total particle number is unchanged if µ stays at
zero. Thus, the particle–hole symmetry enforces µ(T ) = 0.
(b) Show that the mean excitation energy of this system at finite temperature satisfies
E(T ) − E(0) = 2gA
Z
d2~k
E + (~k)
(2π)2 exp βE (~k) + 1
.
+
• Using the label +(-) for the positive (energy) states, the excitation energy is calculated
as
E(T ) − E(0) =
X
k,sz ,α
=g
X
k
[hn+ (k)i E + (k) + (1 − hn− (k)i) E − (k)]
2 hn+ (k)i E + (k) = 2gA
Z
d2~k
E + (~k)
.
(2π)2 exp βE (~k) + 1
+
(c) Give a closed form answer for the excitation energy of the bilayer by evaluating the
above integral.
• Let E + (k) = αk 2 , with α = (tk a)2 /(2t⊥ ), to get
∞
2πkdk αk 2
=
(set βαk 2 = x)
2 eβαk2 + 1
4π
0
Z ∞
x
kB T
gA
dx x
kB T
=
2π
α
e +1
0
2
gπ
π A kB T
kB T
=
=
AkB T
t⊥ .
24
α
3 a2
tk
E(T ) − E(0) = 2gA
Z
For the final expression, we have noted that the needed integral is f2− (1), and used f2− (1) =
ζ2 /2 = π 2 /12.
211
(d) Calculate the heat capacity, CA , of such massive Dirac particles.
• The heat capacity can now be evaluated as
∂E
CA =
∂T
A
2π A
=
kB
3 a2
k B T t⊥
t2k
!
.
(e) A sample contains an equal proportion of single and bilayers. Estimate (in terms of the
hopping energies) the temperature below which the electronic heat capacity is dominated
by the bilayers.
• As stated earlier, the monolayer excitations for phonons have a linear spectrum, with
E1
layer
= ±tk (ka). Their contribution to energy and heat capacity can be calculated as
before. Including the various prefactors (which are not required for the solution), we have
Z ∞
2πkdk tk ak
1 layer
E
(T ) = 2gA
=
(set βtk ak = x)
4π 2 eβtk ak + 1
0
2 Z ∞
x2
gA
kB T
dx x
=
kB T
π
tk a
e +1
0
2
2
kB T
kB T
A
A
1 layer
∝ 2 kB
, CA
.
∝ 2 kB T
a
tk
a
tk
The bilayer heat capacity, which is proportional to T is more important at lower temper-
atures. By comparing the two expressions, it is apparent that the electronic heat capacity
per particle is larger in the bilayer for temperatures smaller than T ∗ ≈ t⊥ /kB .
(f) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capacity of graphene. The typical sound velocity in graphite is of the order of 2 × 104 ms−1 .
Is the low temperature heat capacity of (monolayer) graphene controlled by phonon or
electron contributions?
• The single particle excitations for phonons also have a linear spectrum, with E p = h̄vp |k|
and correspond to µ = 0. Thus qualitatively they give the same type of contribution to
energy and heat capacity. The difference is only in numerical pre-factors. The precise
contribution from a single phonon branch is given by
Z ∞
2πkdk h̄vp k
=
(set βh̄ck = x)
Ep (T ) − Ep (0) = A
4π 2 eβh̄vp k − 1
0
2 Z ∞
A
x2
kB T
=
dx x
kB T
2π
h̄vp
e −1
0
2
2
3ζ3
kB T
kB T
ζ3
, CV,p =
.
AkB
= AkB T
π
h̄vp
π
h̄vp
212
We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since
the speed of Dirac fermions is considerably larger than that of phonons, their contribution
to heat capacity of graphene is negligible.
********
213