1

Physics 7440 (Fall 2012), Problem Set # 5 Solutoins

1. (a) We just need to plug in...

"r r #
h i M ωλ (k) i M ωλ′ (k′ ) i
akλ , a†k′ λ′ = ukλ + p pkλ , u−k′ λ′ − p p−k′ λ′ (1)
2~ 2~M ωλ (k) 2~ 2~M ωλ′ (k′ )
s s
−i ωλ (k) i ωλ′ (k′ )
= [u kλ , p −k ′ λ′ ] + [pkλ , u−k′ λ′ ] (2)
2~ ωλ′ (k′ ) 2~ ωλ (k)
s s
−i ωλ (k) i ωλ′ (k′ )
= ′
(i~δλλ′ δkk′ ) + (−i~δλλ′ δkk′ ) (3)
2~ ωλ′ (k ) 2~ ωλ (k)
= δλλ′ δkk′ . (4)

And...
"r r #
h i M ωλ (k) i M ωλ′ (k′ ) i
akλ , ak′ λ′ = ukλ + p pkλ , uk′ λ′ + p pk′ λ′ (5)
2~ 2~M ωλ (k) 2~ 2~M ωλ′ (k′ )
s s
i ωλ (k) i ωλ′ (k′ )
= ′
[ukλ , pk′ λ′ ] + [pkλ , uk′ λ′ ] (6)
2~ ωλ′ (k ) 2~ ωλ (k)
s s
i ωλ (k) i ωλ′ (k′ )
= (i~δ λλ′ δk,−k′ ) + (−i~δλλ′ δk,−k′ ) (7)
2~ ωλ′ (k′ ) 2~ ωλ (k)
= 0. (8)
In the last line we used the fact that ωλ (−k) = ωλ (k).
(b) First we can solve for ukλ and pkλ in terms of the raising and lowering operators:
s
~
ukλ = (a + a†−kλ ) (9)
2M ωλ (k) kλ
r
~M ωλ (k)
pkλ = −i (akλ − a†−kλ ). (10)
2
Now we plug this into the Hamiltonian:
XXh 1 † M [ωλ (k)]2 † i
H= pkλ pkλ + ukλ ukλ (11)
2M 2
k λ
X X h ~ωλ (k) † ~ωλ (k) † i
= (akλ − a−kλ )(akλ − a†−kλ ) + (akλ + a−kλ )(akλ + a†−kλ ) (12)
4 4
k λ
X X ~ωλ (k) † h i
= akλ akλ + a−kλ a†−kλ (13)
2
k λ
X X ~ωλ (k) h † i
= akλ akλ + akλ a†kλ (14)
2
k λ
X X ~ωλ (k) h † i
= akλ akλ + a†kλ akλ + 1 (15)
2
k λ
XX
= ~ωλ (k)(a†kλ akλ + 1/2) (16)
k λ

(c) We can start with the expression for u(0) in terms of raising and lowering operators,
s
1 X ~
µ
aλ (k) + a†λ (−k) .
 
u (0) = √ Sµλ (k) (17)
N 2M ωλ (k)
k,λ
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Plugging this in, we have

hψ0 |u(0)2 |ψ0 i = hψ0 |uµ (0)uµ (0)|ψ0 i (18)
1 ~ X X Sµλ (k)Sµλ′ (k′ ) 
h0| aλ (k) + a†λ (−k) aλ′ (k′ ) + a†λ′ (−k′ ) |0i
 
= p (19)
N 2M ′
ωλ (k)ωλ (k )
′
k,k λ,λ
′ ′

To evaluate the expectation value above, note that hψ0 |aλ (k)aλ′ (k′ )|ψ0 i = hψ0 |a†λ (k)a†λ′ (k′ )|ψ0 i = 0, using
Eq. 7 of the homework. So we have

hψ0 | aλ (k) + a†λ (−k) aλ′ (k′ ) + a†λ′ (−k′ ) |ψ0 i = hψ0 | aλ (k)a†λ′ (−k′ ) + a†λ (−k)aλ′ (k′ )|ψ0 i.
   
(20)

The second term vanishes, because aλ′ (k′ )|ψ0 i = 0. To evaluate the first term, we can use the commutation
relations to rewrite

aλ (k)a†λ′ (−k′ ) = δλλ′ δk,−k′ + a†λ′ (−k′ )aλ (k), (21)

so we have
hψ0 | aλ (k) + a†λ (−k) aλ′ (k′ ) + a†λ′ (−k′ ) |ψ0 i = hψ0 | aλ (k)a†λ′ (−k′ )|ψ0 i
   
(22)
= hψ0 | δλλ′ δk,−k′ + a†λ′ (−k′ )aλ (k) |ψ0 i
 
(23)
= δλλ′ δk,−k′ hψ0 |ψ0 i (24)
= δλλ′ δk,−k′ . (25)
Putting this back in, we have
1 ~ X X Sµλ (k)Sµλ′ (k′ )
hψ0 |u(0)2 |ψ0 i = δλλ′ δk,−k′ (26)
N 2M
p
′ ′ ωλ (k)ωλ′ (k′ )
k,k λ,λ
1 ~ X Sµλ (k)Sµλ (−k)
= (27)
N 2M
p
k,λ
ωλ (k)ωλ (−k)
1 ~ X Sµλ (k)Sµλ (k)
= (28)
N 2M ωλ (k)
k,λ
1 ~ X 1
= (29)
N 2M ωλ (k)
k,λ

a3 ~ X d3 k 1
Z
= . (30)
2M (2π)3 ωλ (k)
λ

Here, the integral is taken over the Brillouin zone.

(d) Note that this is different from what we found in the finite-temperature classical case – in that case, there
was a factor of 1/ω 2 inside the integral, which gives a stonger singularity at k = 0. Here the integral is
certainly finite in 3d – in spherical coordinates, schematically, the integral at small-k looks like
1
Z
hψ0 |u(0) |ψ0 i ∼ dkk 2 ,
2
(31)
k
which is finite. The analogous integral is still finite in 2d, but diverges in 1d, since
1
Z
2
hψ0 |u(0) |ψ0 i1d ∼ dk . (32)
k

Now let’s discuss differences from the classical result. In 3d, the classical result predicts hψ0 |u(0)2 |ψ0 i = 0
at T = 0. However, this can’t really be correct, because we know that there is zero point motion. The fact
that we found hψ0 |u(0)2 |ψ0 i to be nonzero is simply because the quantum calculation takes the zero point
motion into account.
In 2d, the difference is more dramatic. We found that hψ0 |u(0)2 |ψ0 i diverged for T > 0 in the classical
calculation, but here it is finite at hψ0 |u(0)2 |ψ0 i. Physically this makes sense – there are more fluctuations
at finite temperature than T = 0, and in 2d evidently those fluctuations are enough to make hψ0 |u(0)2 |ψ0 i
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diverge (and to destroy crystalline order). What the present result tells us is that, in 2d, quantum fluctu-
ations alone are not enough to destroy crystalline order. While it is not obvious from calculations we have
done, it turns out that hψ0 |u(0)2 |ψ0 i actually diverges in 2d as soon as the temperature becomes nonzero
– this can be checked by repeating the quantum calculation, but at finite temperature. So in 2d there is
really a qualitative difference between zero and nonzero temperature.
In 1d, hψ0 |u(0)2 |ψ0 i diverges both for T = 0 and for T > 0, however the divergence is weaker at T = 0
(a weaker singularity in the integral), which we might expect since at T = 0 we have only quantum
fluctuations. So in 1d, even just quantum fluctuations destroy crystalline order.
All of these results are another manifestation of the fact that order gets weaker as the spatial dimension
gets lower.
2. Ashcroft & Mermin 2.1

(a) Let N be the total number of electrons in the two-dimensional (2d) electron gas. Then
N = total number of filled states, inside the Fermi sea. (33)
In 2d, the fermi sea is a circle of radius kF in k-space. Following the arguments of A+M p.34-35, but for
two dimensions, the number of states per unit k-space area (not including spin) is A/(2π)2 , where A is the
total area of the system. Then we have
N = 2(number states per k-space area)( area of Fermi sea ) (34)
 A 
=2 (πkF2 ). (35)
(2π)2
(The factor of 2 in front of both these expressions accounts for spin.) Therefore,
N k2
n= = F. (36)
A 2π
(b) Recall that in three dimensions, rs is defined as the radius of a sphere, whose volume is equal to the volume
per electron. We generalize this to 2d by saying rs is the radius of a circle, whose area is equal to the area
per electron. Therefore
A 1
πrs2 = ( area per electron ) = = . (37)
N n
Using the result from part (a), we have
√
1 1 2
rs = √ = p = . (38)
πn π(kF2 /2π) kF

(c) We start from the definition of the density of states, and proceed to evaluate the integral. We shall need
to use the following result for k-space sums in 2d, in the limit of large system size (A → ∞):
1 X d2 k
Z
↔ . (39)
A (2π)2
k

This result can be derived in the same way as the corresponding 3d result.
For the density of states we have
2 X
D(ǫ) = δ(ǫ − ǫk ) (40)
A
k
d2 k
Z
=2 δ(ǫ − ǫk ) (41)
(2π)2
d2 k  ~2 k 2 
Z
=2 δ ǫ − (42)
(2π)2 2m
Z ∞
2(2π)  ~2 k 2 
= dk k δ ǫ − (43)
(2π)2 0 2m
Z ∞ 2 2
1  ~ k
= dk k δ ǫ − . (44)
π 0 2m
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We now make the change of variables u = ~2 k 2 /2m, resulting in

Z ∞
m
D(ǫ) = du δ(ǫ − u). (45)
π~2 0
For ǫ > 0, the delta function integrates to unity. However for ǫ < 0, the argument of the delta function is
never zero, and the delta function integrates to zero. So the result is

m/(π~2 ) , ǫ > 0

D(ǫ) = . (46)
0 ,ǫ<0

(d) Recall that

Z ∞
n= dǫ D(ǫ)f (ǫ), (47)
−∞

where f (ǫ) is the Fermi function. In this part of the problem we’re supposed to evaluate this using the
Sommerfeld expansion. The general form of the Sommerfeld expansion is given in A+M Eq. (2.69), and
for the present case we have
Z ∞ Z µ ∞
" #
2n−1
2n d
X
dǫ D(ǫ)f (ǫ) = dǫ D(ǫ) + an (kB T ) D(ǫ) . (48)
−∞ −∞ n=1
dǫ2n−1
ǫ=µ

Now, because D(ǫ) is constant except at ǫ = 0, its derivatives all vanish (except at ǫ = 0). For any
reasonably low temperature, we should expect µ ≈ ǫF 6= 0, so all the above derivatives vanish, and the
Sommerfeld expansion simply gives
Z µ
n= dǫ D(ǫ). (49)
−∞

Therefore,
Z µ Z ǫF
n(T ) = dǫ D(ǫ) = n(T = 0) = dǫ D(ǫ). (50)
−∞ −∞

(This is the statement that density is constant as a function of temperature.) The only way for this to be
consistent is if

µ(T ) = ǫF . (51)

(e) Now we will directly evaluate the integral of Eq. (47) above. We have, plugging in the form of the Fermi
function,
Z ∞
m dǫ
n= 2 (ǫ−µ)/k
. (52)
π~ 0 e BT + 1

We make the change of variables u = (ǫ − µ)/kB T to obtain

mkB T ∞ du
Z
n= . (53)
π~2 −µ/kB T e u+1

While there is probably a more systematic way to do it, I found it easiest to do the integral by simply
guessing the form of a function whose derivative is equal to the integrand, 1/(eu + 1). The result is
du
Z
= − ln(e−u + 1), (54)
eu + 1
which gives the result
mkB T  µ/kB T 
n= ln e + 1 . (55)
π~2
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Now we just need to express this in the form given in A+M. Using the fact that π~2 n/m = ǫF , we have
 
ǫF = kB T ln eµ/kB T + 1 (56)
 
= kB T ln eµ/kB T 1 + e−µ/kB T


(57)
 
= µ + kB T ln 1 + e−µ/kB T , (58)

as desired.
(f) First, we express the result of part (e) in the form
 
ǫF − µ = kB T ln 1 + e−µ/kB T . (59)

We would like to get an estimate of ǫF − µ, the deviation of µ from its zero temperature value. We assume
that kB T ≪ ǫF for any temperature of interest. When this assumption holds we also expect µ ≈ ǫF , so
that e−µ/kB T ≪ 1. Then we can expand the logarithm in powers of e−µ/kB T – we stop at leading order to
obtain

ǫF − µ ≈ kB T e−µ/kB T . (60)

Since we expect µ ≈ ǫF , it is reasonable to replace µ by ǫF in the expression above. This leads to our
estimate:

ǫF − µ ≈ (kB T )e−ǫF /kB T . (61)

This result says that ǫF − µ is exponentially smaller than kB T – this is a tiny correction and is not
significant. The reason the Sommerfeld expansion “failed” here is that the density of states D(ǫ) is not an
analytic function on the entire real axis – it is discontinuous at ǫ = 0. This discontinuity is not captured
in the Sommerfeld expansion, but does appear (albeit as a tiny correction) when the integral is evaluated
exactly.
It is worth noting that a similar issue presumably arises in 3d – there, again, D(ǫ) is not analytic at
ǫ = 0. However, in 3d, any exponentially small temperature dependence coming from this nonanalyticity
is completely dominated by the T 2 temperature dependence given by the (nonvanishing) leading term in
the Sommerfeld expansion.

3. (a) The first thing to understand is the single-particle spectrum of the trapped fermions – we just have to solve
the three-dimensional (3d) Harmonic oscillator, with Hamiltonian H = p2 /2m + (1/2)mωt2 r 2 . The energy
eigenstates are labeled by the vector of integers n = (nx , ny , nz ), where nx = 0, 1, 2, . . . , and similarly for
ny and nz . The energy of the state with quantum numbers n is

ǫn = ~ωt (nx + ny + nz ). (62)

Here I have ignored the zero-point energy – in this problem it will not be important, as it will only give an
overall shift of the chemical potential.
We define Ω(ǫ) to be the total number of states with energies ǫn ≤ ǫ. We have N = Ω(ǫF ), since the
number of particles is obtained by filling all states up to the Fermi energy. Let us calculate Ω(ǫ):
X
Ω(ǫ) = Θ(ǫ − ǫn ). (63)
n

In the limit N ≫ 1, we can replace the sum on n by an integral as follows

X Z ∞ Z ∞ Z ∞ Z
→ dnx dny dnz ≡ d3 n. (64)
n 0 0 0

This replacement is legitimate because the sum in Eq. (63) will be dominated by contributions from large
nx , ny and nz . If nx ≫ 1, then the discrete change nx → nx + 1 can reasonably be thought of as an
infinitesimal change, i.e. nx → nx + dnx . Therefore we approximate n by a continuous variable, and
replace the sum by an integral.
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We have
Z
d3 nΘ ǫ − ~ωt (nx + ny + nz ) .


Ω(ǫ) = (65)

So the integrand is equal to unity in the region nx + ny + nz ≤ ǫ/~ωt , and is zero otherwise – therefore
Ω(ǫ) is just the volume of this region. Defining α = ǫ/~ωt > 0, we then have
Z α Z α−nz Z α−nz −ny
α3 ǫ3
Ω(ǫ) = dnz dny dnx = = 3 3. (66)
0 0 0 6 6~ ωt

Note that Ω(ǫ) = 0 for ǫ < 0, since there are no states at negative energy. Then

ǫ3F
N = Ω(ǫF ) = , (67)
6~3 ωt3
and

ǫF = ~ωt (6N )1/3 . (68)

(b) The density of states D(ǫ) is given by

dΩ(ǫ)
D(ǫ) = . (69)
dǫ
This allows for the usual interpretation that the quantity
dΩ(ǫ)
D(ǫ)dǫ = dǫ (70)
dǫ
is the total number of states in the interval [ǫ, ǫ + dǫ]. Using the result of part (a),

ǫ2
D(ǫ) = , (71)
2~3 ωt3

for ǫ > 0, and D(ǫ) = 0 for ǫ < 0.

(c) At finite temperature, the total number of fermions is given by
Z ∞
N= dǫ D(ǫ)f (ǫ). (72)
−∞

We shall evaluate this integral at low temperature via the Sommerfeld expansion, and use it to determine
the chemical potential µ(T ). Carrying out the Sommerfeld expansion to leading order we have
Z µ
π2
N= dǫD(ǫ) + (kB T )2 D′ (µ). (73)
0 6

As we did in class for the free electron gas, we let µ(T ) = ǫF + c(kB T )2 + O(kB T )4 . Keeping terms up
through order (kB T )2 , we have
Z ǫF
π2
N= dǫ D(ǫ) + c(kB T )2 D(ǫF ) + (kB T )2 D′ (ǫF ). (74)
0 6
Rǫ
Noting that N = 0 F dǫ D(ǫ), we obtain the following expression for the constant c:

π 2 D′ (ǫF ) π2
c=− =− (75)
6 D(ǫF ) 3ǫF
Therefore
π2
µ(T ) = ǫF − (kB T )2 + O(kB T )4 . (76)
3ǫF
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(d) The total energy is given by

Z ∞
E(T, N ) = dǫ ǫD(ǫ)f (ǫ). (77)
−∞

We evaluate this using the Sommerfeld expansion, which, up to terms of order T 2 , gives
Z µ
π2
E(T, N ) = dǫ ǫD(ǫ) + (kB T )2 [D(µ) + µD′ (µ)]. (78)
0 6

Using the result from part (c) above, and keeping only terms up through order (kB T )2 , we have
Z ǫF
π2
E(T, N ) = dǫ ǫD(ǫ) + c(kB T )2 ǫF D(ǫF ) + (kB T )2 [D(ǫF ) + ǫF D′ (ǫF )] (79)
0 6
π2
= E(0, N ) + (kB T )2 D(ǫF ). (80)
6
The ground state energy is
Z ǫF
E(0, N ) = dǫ ǫD(ǫ) (81)
0
ǫ4F 64/3
= 3 3 = ~ωt N 4/3 , (82)
8~ ωt 8
and we have
64/3 π 2 (kB T )2 2/3
E(T, N ) = (~ωt )N 4/3 + N + O(T 4 ). (83)
8 2 · 61/3 ~ωt

(e) The specific heat is given by

 ∂E  π 2 kB2
T 2/3
C(T, N ) = = 1/3
N . (84)
∂T N 6 ~ωt