MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions, Problem Set #11
Problem 1: Ripplons
a)
ky
2π/Lx
points/k-volume ≡ D(k) =
2π/Ly
kx
b)
(2π)2
Lx Ly
k-volume/point =
Lx Ly
(2π)2
A
(2π)2
=
ky
#() = πk 2 ()D(k)
k
= π
kx
D() =
4/3
b
= b k 3/2
D(k)
d#
4 A
1/3
A
=
π
m=
1/3
2
4/3
d
3 (2π) b
3πb4/3
D(ε)
ε
c)
∞
Z
U =
(< n >
0
CA =
∂U
∂T
Z
0
(/kB T 2 )e/kB T
(e/kB T − 1)2
∞
x7/3 ex
dx
(ex − 1)2
0
AkB
(kB T )4/3
=
3πb4/3
∞
)D() d =
∞
=
A
Z
+ 12
Z
0
1
1
1
2
+
D() d
e/kB T − 1
A
1
+2
1/3 d
3πb4/3
∝
T 4/3
One could also proceed directly from U .
CA
∞
4/3
d + C
e/kB T − 1
0
Z
A(kB T )7/3 ∞ x4/3
=
dx + C
3πb4/3
ex − 1
0
Z ∞ 4/3
x
∂U
7AkB
4/3
dx
=
=
(k
T
)
B
∂T A 9πb4/3
ex − 1
0
A
U =
3πb4/3
Z
∝
T 4/3
d) There is no energy gap behavior because there is no energy gap. For any kB T there are
always oscillators with ~ω < kB T .
Problem 2: Two-Dimensional Metal
a)
eikx (x+L) = eikx x eikx L = eikx x ⇒ kx = nx
2π
L
nx = ±1, ±2, · · ·
The same holds of ky .
~k = 2π (nx x̂ + ny ŷ)
L
ni = ±1, ±2, · · ·
b)
~ =
D(k)
L
2π
2
for all ~k
c)
ky
2ε/γ
kx
Taking into account the spin degeneracy
2
~ 2
#() = 2D(k)
γ
2
L
8
D() = 2
2π
γ2
2
L
= 4
πγ
2
D(ε)
ε
d)
8L2
N = #(F ) =
(2π)2
π 2 γ 2 (N/A)
1
2
F =
γ
1/2
e)
Z
D() = a N=
0
Z
U =
0
F
F
a
a d = 2F
2
a
a 2 d = 3F = 23 N F
3
f) Thermal agitation only disturbs a fraction kB T /F of the total number of electrons, and
imparts to them an energy of the order of kB T . Thus the total increase in energy from the
T = 0 value is proportional to T 2 and the heat capacity will be linearly proportional to T .
g)
dU =
0
T
dS} +SdA
|{z
at T =0
dU
2 dF
2
1 F
1
S =
= N
= N (− ) = −
dA
3 dA
3
2 A
3
3
N
A
F
Problem 3: Donor Impurity States in a Semiconductor
a)
< n > → e−(−µ)/kB T = e−(∆−µ)/kB T e−(−∆)/kB T
3/2
V
2me
Dstates () =
( − ∆)1/2
2
2
2π
~
Z ∞
< ne > D() d
NC =
∆
=
=
=
V
2π 2
V
2π 2
V
4
2me
~2
3/2
= α(T )e
Z
∞
e
√ −δ/k T
B
δe
dδ
0
2me kB T
~2
2me kB T
π~2
−(∆−µ)/kB T
3/2
3/2
e
−(∆−µ)/kB T
Z
|0
∞
√
x e−x dx
{z
}
√
π/2
e−(∆−µ)/kB T
−∆/kB T µ/kB T
e
V
where α(T ) ≡
4
2me kB T
π~2
3/2
b) < n > is equal to 1/2 where = µ, so if the donors are only half occupied, and they are
located at = 0, then µ = 0. Since half of the donor electrons are now in the conduction
band, the last equation above becomes
ND /2 = α(T )e−∆/kB T
This could be solved numerically for T .
4
c) When the energy is many times kB T below the chemical potential µ,
< n >→ 1 − e(−µ)/kB T
Then remembering that = 0 at the position of the donors
ND = ND (1 − e(−µ)/kB T ) + α(T )e−∆/kB T eµ/kB T
ND e(−µ)/kB T = α(T )e−∆/kB T eµ/kB T
ND /α(T ) = e−∆/kB T e2µ/kB T
kB T ln[ND /α(T )] = −∆ + 2µ
µ = ∆/2 + (kB T /2) ln[ND /α(T )]
Since α(T ) ∝ T 3/2 the temperature dependent term above is proportional to
T [ln(T −3/2 ) + ln(constant)] = −(3/2)T ln T + T ln(constant)
which goes to zero at T = 0. Thus F = ∆/2
5
d)
<n> =
N (e on donor sites) =
1
exp[( − µ)/kB T ] + 1
ND
exp[−µ/kB T ] + 1
NC = α(T ) exp[−∆/kB T ] exp[µ/kB T ]
exp[−µ/kB T ] =
N (e on donor sites) =
N (e on donor sites) + NC =
α(T )
exp[−∆/kB T ]
NC
ND
(α/NC ) exp[−∆/kB T ] + 1
ND
+ NC = N D
(α/NC ) exp[−∆/kB T ] + 1
ND + NC (α/NC ) exp[−∆/kB T ] + NC = ND (α/NC ) exp[−∆/kB T ] + ND
(α/ND ) exp[−∆/kB T ] + (NC /ND ) = (ND /NC )(α/ND ) exp[−∆/kB T ]
(NC /ND )(α/ND ) exp[−∆/kB T ] + (NC /ND )2 = (α/ND ) exp[−∆/kB T ]
NC
ND
2
+
NC
ND
α −∆/kB T
α −∆/kB T
e
−
e
=0
ND
ND
Set NC /ND = 1/2.
1 1 α −∆/kB T
α −∆/kB T
+
e
−
e
=0
4 2 ND
ND
ND = 2α(T )e−∆/kB T
When T → 0, exp[−∆/kB T ] → 0 leaving (NC /ND )2 = 0, so in this limit NC = 0
When T → ∞ α(T ) grows without bound and the first term in the equation can be
neglected leaving
NC
α −∆/kB T
α −∆/kB T
−
e
=0
e
ND ND
ND
with the solution NC = ND .
6
Problem 4: Spin Polarization
a) Recall that in zero field the density of states for spin- 12 Fermions is
V
D0 () =
2π 2
2m
~2
3/2
1/2
= 0
>0
<0
These are equally divided between spin up and spin down. The application of a field shifts
all ms = 12 states down in energy by µ0 H and all ms = − 12 states up by µ0 H (assuming a
~ Thus
positive µ0 , that is, a magnetic moment parallel rather than anti-parallel to S).
D 1 () =
2
D− 1 () =
2
D() =
1
D0 ( + µ0 H)
2
1
D0 ( − µ0 H)
2
1
[D0 ( + µ0 H) + D0 ( − µ0 H)]
2
b) The filling of D() at T = 0 must stop just short of = µ0 H where the ms = − 12 states
would begin to fill.
Z µ0 H
Z 2µ0 H
1
N =
D() d =
D0 () d
2 | {z
}
−µ0 H
0
a1/2
a2
(2µ0 H)3/2
23
"
3/2 #
1 V
2m
=
(2µ0 H)3/2
3 2π 2 ~2
=
2
6π N/V
=
4mµ0 H
~2
7
3/2
4mµ0
H
~2
1 ~2 (6π 2 N/V )2/3
=
µ0
4m
(6π 2 N/V )2/3 =
H0
c) Using cgs units
H0 = 4.22 × 10−54
1
(N/V )2/3
µ0 M
Using M = 9.11 × 10−28 g, µ0 = −9.27 × 10−21 ergs-gauss−1 , and n = 8.45 × 1022 cm−3 gives
H0 = 9.6 × 108 gauss = 9.6 × 104 Tesla.
The negative µ0 means that the electron spins are polarized anti-parallel to the direction of
~
H.
d) For 3 He, M = 5.01 × 10−24 g, µ0 = 1.075 × 10−23 ergs-gauss−1 , and n = 1.64 × 1022 cm−3 .
So in this system
H0 = 5.1 × 107 gauss = 5.1 × 103 Tesla.
Problem 5: T = 0 solubility of 3 He in 4 He
a) Recall that for non-interacting spin 1/2 Fermions in 3 dimensions with an effective mass
m∗
1/3
kF = 3π 2 (N/V )
F
~2
=
2m∗
3π 2 N
V
2/3
When the Fermi energy of 3 He in 4 He equals the potential difference W , any additional 3 He
atoms will begin to fill their own phase where their kinetic energy can begin at zero. Thus
W =
2m∗ W
~2
nmax
3
=
~2
2 max 2/3
3π
n
3
2m∗
3π 2 nmax
3
1
=
3π 2
8
2/3
2m∗ W
~2
3/2
b)
nmax
3
1
=
3π 2
2m∗ W
~2
3/2
3/2
2 × 10−23 × 0.5 × 10−16
10−54
−39 3/2
1
10
1
14 3/2
=
=
10
×
10
3π 2 10−54
3π 2
√
10 10
× 1021 ≈ 1021
=
2
3π
1
=
3π 2
Then
1021
n3max
≈
= 5%
n3 + n4
2 × 1022
This is close to the value indicated in the figure.
Problem 6: Melting Curve of 3 He
a) The entropy of a Fermi gas at T = 0, the model for the liquid 3 He, is zero.
b) The entropy of N distinguishable, stationary spin 1/2 particles in the absence of an
applied magnetic field is N kB ln 2
c) At T = 0 in the presence of a large ”effective magnetic field” each of the spins in the solid
will be oriented with its moment anti-parallel to the effective field. Thus the entropy of the
solid goes to N kB ln 1 = zero.
d) In region II the slope of the melting curve is −kB ln 2/v0 . Yes, it is negative.
e) In region I the slope of the melting curve goes to zero.
f) When heat is added to the system, the entropy must go up. The only way for that to
happen is for some of the liquid to turn into solid, that is, to freeze. In this region on the
melting curve, when one heats the system it solidifies and when one cools it, it melts.
9
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8.044 Statistical Physics I
Spring 2013
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