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UNIVERSITETET I OSLO
Det matematisk-naturvitenskapelige fakultet
Eksamen i: FYS 203 Statistisk mekanikk
Eksamensdag: Fredag 25.mai 2001
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Clark: Physical and Mathematical Tables
Øgrim: Størrelser og enheter i fysikken
Olivber and Boyd: Science Data Book
Tabeller i fysikk for den videregående skole
Rottmann: Matematisk formelsamling
Godkjent numerisk electronisk kalkulator
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Problem 1: Classically Confined Fermi Gas
A Fermi gas containing N noninteractiong particles with spin S is confined in z-direction
by the harmonic-oscillator potential
1
U (z) = mω02 z 2
2
(1)
where m is the mass of the particle, while ω0 is the oscillation frequency.
In this problem, we assume that motion along z-axis obeys classical mechanics.
a: Show that the density of states as a function of the particle energy ε, D(ε), has the form
D(ε) = Ag0 ε/~ω0 ,
(2)
g0 = (2S + 1)m/2π~2
(3)
where
and A is the area of the system.
Hints:
Rβ
(i) By definition, for any γ belonging to the interval β > γ > α, α dx δ(x − γ) = 1. If γ
is outside this interval, then the integral is 0. This definition is often expressed in terms
of the Heaviside step function Θ(ξ) as
½
Z β
1, ξ > 0
dx δ(x − γ) = Θ(β − γ)Θ(γ − α) where Θ(ξ) =
.
(4)
0, ξ < 0
α
1
(ii) An auxiliary intergral:
Z ∞
Z
∞
du
0
0
µ
¶
u2
v2
E
dv Θ E − a − b
=√ .
2
2
ab
(5)
Solution to the problem 1a
The classical Hamiltonian of a particle is
H(p, z) =
p2
mω02 2
+
z ,
2m
2
p2 ≡ p2x + py2 + p2z , .
(6)
Then
Z
¢
d3 p d3 r ¡
δ ε − p2 /2m
3
(2π~)
µ
¶
Z
p2⊥
p2z
mω 2 2
2π(2S + 1)A
p⊥ dp⊥ dpz dz δ ε −
−
−
z
.
=
(2π~)3
2m 2m
2
D(ε) = (2S + 1)
(7)
Here A is the area of the system. Since p⊥ dp⊥ = m d(p⊥ /2m), we integrate over p⊥ in
Eq. (7) using δ-function [see Hint (4)] to get
µ
¶
Z
p2z
mω02 2
(2S + 1)mA
dpz dz Θ ε −
−
z
.
D(ε) =
4π 2 ~3
2m
2
To integrate over pz and z we can use Hint (5). What we actually do is to introduce new
variables
pz = m1/2 ρ cos η , z = (mω02 )−1/2 ρ sin η
and obtain
Z
¡
¢
(2S + 1)mA
D(ε) =
ρdρdη Θ ε − ρ2 /2
2
3
4π ~ ω0
Z
(2S + 1)mA ε
ε
=
,
d (ρ2 /2) = g0 A
3
2πω0 ~
~ω0
0
g0 ≡
(2S + 1)m
.
2π~2
The answer is
D(ε) = Ag0 ε/~ω0 ,
g0 = (2S + 1)m/2π~2 .
which coincides with Eq. (2).
b: Show that the chemical potential µ at zero temperature (let us denote it is as µ0 ) as a
function of particle sheet concentration, ns ≡ N/A, is given by the expression
µ(0) ≡ µ0 = (2ns ~ω0 /g0 )1/2 .
2
(8)
Hint:
Z
∞
0
f (x) dx
≈
ex−z + 1
Z
z
f (x) dx +
0
π2 0
f (z) + . . .
6
at z À 1 .
(9)
Solution to the problem 1b
By definition
Z
∞
N=
£
¤−1
.
f0 (ε) = e(ε−µ)/T + 1
D(ε) f0 (ε) dε ,
(10)
0
Here T is the temperature in energy units while µ is the electrical potential.
According to Eq. (2) we have
Z ∞
Z ∞
D(ε) dε
x dx
Ag0 T 2 ³ µ ´
N=
F1
, F1 (z) =
=
.
(ε−µ)/T
x−z
e
+1
~ω0
T
e
+1
0
0
(11)
The chemical potential µ is given by the equation
F1 (µ/T ) = ns ~ω0 /g0 T 2 .
At zero temperature, z → ∞, using the Hint (9) we obtain F1 (z) =
(12)
Rz
0
x dx = z 2 /2. Thus,
µ(0) ≡ µ0 = (2ns ~ω0 /g0 )1/2 .
which coincides with Eq. (8).
c: Analyze temperature dependence of the chemical potential at high and at low temperatures.
Find proper criterion for crossover between these cases.
Hint: At large negative z, z < 0, |z| À 1,
Z ∞
Z z
f (x) dx
z
≈e
e−x f (x) dx .
x−z + 1
e
0
0
(13)
Solution to the problem 1c
At high temperatures, µ < 0, and |µ| À T . Then, using Eq. (11) and Hint (13), we get
Z ∞
z
xe−x dx = ez
F(z) ≈ e
0
As a result, the Fermi gas behaves as the Boltzmann one with the distribution function
fB (ε) = e(µ−ε)/T .
3
(14)
From Eq. (12) we obtain
µ = −T ln (g0 T 2 /ns ~ω0 ) = −T ln (2T 2 /µ20 ) .
(15)
This equation is valid at T À µ0 .
For low temperatures, we use the Hint (9). In our case, F(z) ≈ z 2 /2 + π 2 /6, and
Eq. (12) acquires the form
π2
z +
= z02 ,
3
2
z0 ≡ µ0 /T À 1 .
Let us search for the solution of this equation as z = z0 + z1 where z1 ≡ (µ − µ0 )T and
assume that |z1 | ¿ z0 . Then we iterate the above equation as
2z0 z1 = −π 2 /3
⇒
z1 = −
π2
6z0
Returning to dimensional variables we get
µ = µ0 −
π2 T 2
.
6 µ0
(16)
This equation is valid at T ¿ µ0 . In this way we get the crossover temperature as T ≈ µ0 .
d: Find internal energy U , Landau free energy Ω, entropy S and specific heat CV of the gas.
Show that Ω = −U/2. Derive equation of state.
Hint: In quasi two-dimensional systems the pressure is defined as a derivative with respect
to the area A rather than with respect to the volume.
Solution to the problem 1d
According to the definition,
Z
Z
Z ∞
ε2 dε
Ag0 T 3 ∞ x2 dx
Ag0 ∞
=
, (17)
U =
dε εD(ε) f0 (ε) =
~ω0 0 e(ε−µ)/T + 1
~ω0 0 ex−µ/T + 1
0
Z
Z ∞
¢
¡
Ag0
Ag0 T ∞
ε2 dε
U
(µ−ε)/T
=−
Ω =
εdε ln 1 + e
=
−
,
(18)
~ω0 0
2~ω0 0 e(ε−µ)/T + 1
2
µ ¶
µ
¶
∂Ω
1 ∂U
S = −
=
,
(19)
∂T A,µ 2 ∂T A,µ
R∞
At T À µ0 we have f0 (ε) = fB (ε), 0 x2 e−x dx = 2. Thus, U = (2Ag0 T 3 /~ω0 )eµ/T and
Ω=−
Ag0 T 3 µ/T
e
~ω0
4
(20)
From Eq. (15) we have eµ/T = ns ~ω0 /g0 T 2 and
U = 2N T ,
Ω = −N T ,
· µ
¶
¸
g0 T 2
S = N ln
+3 .
ns ~ω0
CV = 2N ,
(21)
The latest equality is obtained by differentiantion of Eq. (20) with respect to T and subsequent using of Eq. (15). This result is compatible with the equipartition principle for 4
degrees of freedom.
At low temperatures, T ¿ µ0 , we can use the Hint (9),
Z ∞
x2 dx
π2
z3 π2
z03
2
≈
+
z
≈
+
z
z
+
z0
0 1
ex−z + 1
3
3
3
3
0
µ
¶
1
z3 π2
z03 π 2
+ z0 1 −
= 0 + z0 .
=
3
3
2
3
6
As a result,
Ag0 µ30 π 2 Ag0 µ0 T 2
U=
+
,
3~ω0
6
~ω0
π 2 Ag0 µ0
π2
CV =
T =
3 ~ω0
3
U
Ω=− ,
2
r
2ns g0
AT .
~ω0
(22)
To get the entropy we have to differentiate at constant µ. Thus we re-write Ω as
Ω=−
Ag0 3
(µ + π 2 µT 2 )
6~ω0
to obtain
π 2 Ag0 µ0 T
π2
S=
=
3 ~ω0
3
r
2ns g0
AT .
~ω0
(23)
(24)
The equation of state is given by the expression
Ω
U
P =− =
=
A
2A
½
ns T
at T À µ0
.
g0 µ30 /6~ω0 at T ¿ µ0
(25)
Problem 2: Quantum-Mechanical Confinement
Now let us assume that the the motion along z-direction in the potential (1) obeys quantum
mechanics.
The quantum-mechanical state is specified by the quantum numbers n, p⊥ (p⊥ =
{px , py }) where n is the discrete quantum number describing quantized motion in zdirection. The state with a given n is called the nth transverse mode.
5
The energy eigenvalues are
¶
µ
p2
1
+ ⊥.
εn (p) = ~ω0 n +
2
2m
a: Find density of states as a function of the particle energy ε. Show that its profile is a set of
equidistant steps of the same height,
µ
¶
X
1
D(ε) = Ag0
Θ(ε − εn ) , εn ≡ ~ω0 n +
.
(26)
2
n
Hint:
The motion in the x − y plane remains classical.
Solution to the problem 2a
According to the definition,
D(ε) = A
·
¶
¸
µ
d2 p⊥
p2⊥
1
δ ε − ~ω0 n +
−
.
(2π~)2
2
2m
XZ
n
As a result, using Hint (4) we obtain
D(ε) = Ag0
X
Θ(ε − εn ) ,
n
¶
µ
1
.
εn ≡ ~ω0 n +
2
that coincides with Eq. (26). It shows steps at ε = εn .
b: Derive the equation for the chemical potential µ as a function of particle sheet concentration,
ns . Show that at T = 0 this equation reads as
µ
¶
1
1
nmax (nmax + 1)
µ
ns
= +
−
,
(27)
~ω0
2 nmax + 1 g0 ~ω0
2
where nmax + 1 is the number of occupied transverse modes having εn ≤ µ. Find nmax .
Hint:
Z
∞
0
dx
= ln (ez + 1) .
+1
ex−z
Solution to the problem 2b
6
(28)
The equation for µ is
ns = g0
n
max
X
n
Z
∞
εn
dε
e(ε−µ)/T + 1
Here
Z
∞
F0 (z) =
0
= g0 T
n
max
X
µ
F0
n
µ − εn
T
¶
.
dx
= ln (ez + 1) .
+1
ex−z
nmax + 1 is the number of filled transverse modes having ~ω0 (n + 1/2) ≤ µ.
At T → 0, F0 (x) → z, and we get, see Hint (28),
¶
n
max µ
X
1
ns
µ
−n−
=
~ω0
2
g0 ~ω0
n=0
(29)
(30)
(31)
Denoting the dimensionless quantities µ/~ω0 − 1/2 as ξ and ns /g0 ~ω0 as ν, we can rewrite
Eq. (31) as
µ
¶
1
nmax (nmax + 1)
ξ=
ν−
.
(32)
nmax + 1
2
which coincides with Eq. (27).
The quantity nmax is determined from the condition that the r.h.s. of Eq. (32) is
non-negative. Thus the the transverse mode with number n is occupied at
ν ≥ νn = n(n + 1)/2 .
As a result,
√
nmax = Int ( 8ν + 1 − 1)/2 .
(33)
Here Int (x) means integer part of the quantity x.
c: Consider the case of low concentration for which nmax = 0 at T = 0.
- Show that at T = 0 the chemical potential is given by the equation
µ − ~ω0 /2 = ns /g0 .
(34)
- Find the range of applicability for this expression (i.e. the inequilities limiting temperature T
and concentration ns ).
- Find finite-temperature corrections to the chemical potential at low concentrations.
Solution to the problem 2c
- At small ns only n = 0 we get ns = g0 (µ − ~ω0 /2), and
µ − ~ω0 /2 = ns /g0 .
7
(35)
- This result is valid is the mode with n = 1 is not filled, but µ − ~ω0 /2 À T . We get
T ¿ ns /g0 ≤ ~ω0 .
(36)
This expression coincides with Eq. (34).
- From Eq. (30) we get
¡
¢
F0 (z) = ln (ez + 1) = z + ln 1 + e−z ≈ z + e−z ≈ z0 + z1 + e−z0 .
Thus
µ − ~ω0 /2 = ns /g0 − e−ns /g0 T .
(37)
The correction is exponentially small.
d: Find internal energy U , Landau free energy Ω, entropy S and specific heat CV at low
concentration and temperature.
Solution to the problem 2d
We have
U=
Ag0
2
"µ
ns
g0
¶2
2
+
#
π 2
T .
3
(38)
Here we neglect exponential corrections. Consequently
CV =
Ag0 π 2
T.
3
(39)
In this case,
Ω = −U ,
S = CV .
e: Using grand canonical ensemble derive the expression for mean square fluctuation in the
particle number,
µ
¶
∂N
2
h(∆N ) i = T
.
(40)
∂µ T,A
Apply this expression for calculation of the fluctuation in the particle number for the discussed
above case of low concentration and temperature.
Hint: Calculate ∂hN i/∂µ and make use of the equality hN i = −∂Ω/∂µ.
Solution to the problem 2e
8
For the grand canonical ensemble, the distribution is specified by the by the number of
particles, N , in the subsystem and by a set of quantum numbers, n,
X
wnN = e(Ω+µN −EnN )/T , e−Ω/T =
e(µN −EnN )/T .
(41)
n,N
Consequently,
hN i = eΩ/T
X
N eµN/T
X
e−EnN /T = −∂Ω/∂µ ,
n
N
µ
¶
∂Ω X −EnN /T
1X
∂hN i
2
=
N +N
e
∂µ
T N
∂µ
n
¡
¢
= hN 2 i − hN i2 /T = h(∆N )2 i/T .
(42)
- According to Eq. (34), N = Ag0 (µ − ~ω/2), and ∂N/∂µ = Ag0 . Thus
h(∆N )2 i = Ag0 T = N T /(µ − ~ω0 /2) .
(43)
Problem 3: Stochastic processes
Consider an electron (spin S = 1/2) bounded to an impiruty center and assume that the
temperature T is much less than the energy difference between the levels of orbital motion.
Consequently, only spin degrees of freedom are excited.
Let the system be placed in external magnetic field B, the additional energy in magnetic
field is ±βB where ± corresponds to the values ±1/2 of the spin component along the field.
Here β = |e|~/2mc is the Bohr magneton, c is the velocity of light since the Gaussian system
is used. The magnetic moment of the states with Sz = ±1/2 is ∓β.
a: Find the stationary probabilities for the states with Sz = ±1/2, P± .
- Find average magnetic moment of the system as function of temperature T and magnetic
field B.
Solution to the problem 3a
Denote w± the probabilities of the states Sz = ±1/2. We have
M = −β (w+ − w− ) .
(44)
According to the Gibbs distribution,
w+ /w− = e−2βB/T .
9
(45)
Since only 2 spin levels are involved,
w+ + w− = 1 .
(46)
Combinig Eqs. (45) and (46) we obtain,
w+ =
1
e2βB/T + 1
,
e2βB/T
w− = 2βB/T
,
e
+1
As a result,
µ
M = β tanh
βB
T
µ
w− − w+ = tanh
βB
T
¶
.
(47)
¶
.
(48)
b: Find mean square fluctuation of magnetic moment, h(∆M )2 i and the ratio h(∆M )2 i/hM i2 .
- Show that
β2
.
(49)
langle(∆M 2 )i =
cosh2 (βB/T )
- Discuss limiting cases of large and low temperatures. Explain these results qualitatively.
Solution to the problem 3b
We get:
hM 2 i = β 2 (w+ + w− ) = β 2
hM i2 = β 2 tanh2 (βB/T ) .
(50)
As a result,
h(∆M 2 )i =
β2
,
cosh2 (βB/T )
h(∆M )2 i
1
=
.
2
2
hM i
sinh (βB/T )
(51)
At low temperatures, T ¿ βB, fluctuations are exponentially suppressed because spin is
aligned to magnetic field.
c: Write down the Master Equation for the probability Pi (t) to find the system in the state
i = ± at time t.
- Using the result for the stationary case find the relation between the transition probabilities
W+− and W−+ for the transition from the state “+” to the state “-” and for the reverse
transition, respectively. Express the probabilities W+− and W−+ through the quantity W =
W+− + W+− .
10
- Derive the Master Equation for the population difference, P(t) ≡ w+ (t) − w+ (t).
Solution to the problem 3c
The equation reads,
X
∂wi (t)
[Wi→s wi (t) − Ws→i ws (t)]
=−
∂t
s=±
(52)
- In the stationary situation, using the detailed equilibrium principle,
Wi→s wi (t) = Ws→i ws (t)
we obtain
W+−
w−
=
= e2βB/T , ,
W−+
w+
W−+ =
W
,
1 + e2βB/T
W+− =
W
.
1 + e−2βB/T
(53)
- Coming back to Eq. (52) we get
∂w+ (t)
= −W+− w+ (t) + W−+ w− (t) ,
∂t
∂w− (t)
= −W−+ w− (t) + W+− w+ (t) .
∂t
(54)
Subtracting equations we arrive at the result,
∂P(t)
= −W P(t) .
∂t
(55)
d: Write down the Master Equation for the conditional probability, P (f, t|i, t0 ), to find the
system in the state f at time t under the conditions that at time t0 it was in the state i.
- Specify initial conditions to this equation.
- Show that solution of this equation with proper initial condition is
P (f, t|i, t0 ) = wf + (δif − wf )e−W |t−t0 |
(56)
where wf is the stationary probability to find the system in the state f .
Hint: It is convenient to express the set of Master Equations for the quantities P (f, t|i, t0 )
at different i and f as an equation for the 2 × 2 matrix P̂ with matrix elements Pf i (t|t0 ) ≡
P (f, t|i, t0 ). The solution of the matrix Master
Equation ∂ P̂ /∂T = Ŵ P̂ at t ≥ t0 can
P
then be searched in in the matrix as P̂ ∝ k Ĉ (k) e−λk (t−t0 ) . Here λk are eigenvalues of the
matrix W , while Ĉ (k) are time-independent matrices. They are determined from the intial
11
conditions and from the fact that at t − t0 → ∞ the conditional probabilities tend to the
stationary ones.
Solution to the problem 3d
We consider a stationary random process, and P (f, t|i, t0 ) is a function of t − t0 . Let us
assume t0 = 0, t ≥ 0 and denote Let us for brevity denote P (f, t|i, 0) ≡ Pf i (t). The Master
Equation for Pf i (t) has the form
X
∂Pf i (t)
=−
[Wf →s Pf i (t) − Ws→f Psi (t)] .
∂t
s
(57)
The initial condition to this equation is
Pf i (t) = δif
at t → 0 .
(58)
Let us introduce the matrix P̂ (t) with matrix elements Pif (t) = P (f, t|i, 0). Then Eq. (57)
can be rewritten as
µ
¶
∂ P̂ (t)
−Wf →s
Ws→f
= Ŵ P̂ (t) , with Ŵ =
.
(59)
Wf →s −Ws→f
∂t
P
It is natural to search the solution of Eq, (57) in the form P̂ (t) k Ĉ (k) e−λk t . Then λk
are the eigenvalues of the matrix Ŵ while Ĉk are the numerical matrices to be determined
from intial conditions at t = 0 and from the fact that Pf i (t) → wf at t → ∞. Here wf is
the stationary probability of the state f .
Equating the determinant of Ŵ − λIˆ we get
λ1 = 0 ,
λ2 = −Wf →s + Ws→f ≡ −W .
Here Iˆ is the unit matrix Then, in general,
(1)
(2)
Pf i (t) = Cf i + Cf i e−W t
where Ĉ (k) are time-independent matrices. From the initial conditions (58),
(1)
(2)
Cf i + Cf i = δf i .
On the other hand, at t → ∞ the conditions probability tends to the stationary one, wf .
Consequently,
(1)
Cf i = wf .
Finally, because of the time reversibility, Pf i (t) depends only on |t|. As result,
Pf i (t) = wf + (δf i − wf )e−W |t| .
12
This expression coincides with Eq. (56).
e: Using Eq. (56) show that the correlation function h∆M (t)∆M (0)i has the form
h∆M (t)∆M (0)i = h(∆M )2 i e−W |t|
(60)
where h(∆M )2 i is given by Eq. (49) while W = W+− + W−+ .
- Find the fluctuation spectrum.
Solution to the problem 3e
By definition,
h∆M (t)∆M (0)i = hM (t)M (0)i − hM i2 =
X
Mi Mf [P (f, t|i, 0) − wf ]wi .
if
Using Eq. (56), one can rewrite the above equation as
h∆M (t)∆M (0)i = e−W |t|
X
Mi Mf (δif − wf )wi = h(∆M )2 ie−W |t| .
if
This expresssion coincides with Eq. (60).
The fluctuation spectrum is
Z
Z ∞
−ωt
2
dt e h∆M (t)∆M (0)i = h(∆M ) i
(∆M )ω =
∞
∞
∞
The result reads as,
(∆M )ω = 2h(∆M )2 i .
13
ω2
W
.
+ W2
dt e−ωt−W |t| .