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

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

[

akλ, a

†k′λ′

]

=

[√

Mωλ(k)

2~ukλ +

i√

2~Mωλ(k)pkλ,

√

Mωλ′(k′)

2~u−k′λ′ − i

√

2~Mωλ′(k′)p−k′λ′

]

(1)

=−i2~

√

ωλ(k)

ωλ′(k′)[ukλ, p−k′λ′ ] +

i

2~

√

ωλ′(k′)

ωλ(k)[pkλ, u−k′λ′ ] (2)

=−i2~

√

ωλ(k)

ωλ′(k′)(i~δλλ′δkk′ ) +

i

2~

√

ωλ′(k′)

ωλ(k)(−i~δλλ′δkk′ ) (3)

= δλλ′δkk′ . (4)

And...

[

akλ, ak′λ′

]

=

[√

Mωλ(k)

2~ukλ +

i√

2~Mωλ(k)pkλ,

√

Mωλ′(k′)

2~uk′λ′ +

i√

2~Mωλ′(k′)pk′λ′

]

(5)

=i

2~

√

ωλ(k)

ωλ′(k′)[ukλ, pk′λ′ ] +

i

2~

√

ωλ′(k′)

ωλ(k)[pkλ, uk′λ′ ] (6)

=i

2~

√

ωλ(k)

ωλ′(k′)(i~δλλ′δk,−k′) +

i

2~

√

ωλ′(k′)

ωλ(k)(−i~δλλ′δk,−k′) (7)

= 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:

ukλ =

√

~

2Mωλ(k)(a

kλ + a†−kλ) (9)

pkλ = −i√

~Mωλ(k)

2(a

kλ − a†−kλ). (10)

Now we plug this into the Hamiltonian:

H =∑

k

∑

λ

[ 1

2Mp†kλpkλ +

M [ωλ(k)]2

2u†kλukλ

]

(11)

=∑

k

∑

λ

[

~ωλ(k)

4(a†

kλ − a−kλ)(akλ − a†−kλ) +~ωλ(k)

4(a†

kλ + a−kλ)(akλ + a†−kλ)]

(12)

=∑

k

∑

λ

~ωλ(k)

2

[

a†kλakλ + a−kλa

†−kλ

]

(13)

=∑

k

∑

λ

~ωλ(k)

2

[

a†kλakλ + a

kλa†kλ

]

(14)

=∑

k

∑

λ

~ωλ(k)

2

[

a†kλakλ + a†

kλakλ + 1]

(15)

=∑

k

∑

λ

~ωλ(k)(a†kλakλ + 1/2) (16)

(c) We can start with the expression for u(0) in terms of raising and lowering operators,

uµ(0) =1√N

∑

k,λ

Sµλ(k)

√

~

2Mωλ(k)

[

aλ(k) + a†λ(−k)]

. (17)

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Plugging this in, we have

〈ψ0|u(0)2|ψ0〉 = 〈ψ0|uµ(0)uµ(0)|ψ0〉 (18)

=1

N

~

2M

∑

k,k′

∑

λ,λ′

Sµλ(k)Sµλ′(k′)√

ωλ(k)ωλ′(k′)〈0|

[

aλ(k) + a†λ(−k)][

aλ′(k′) + a†λ′(−k′)]

|0〉 (19)

To evaluate the expectation value above, note that 〈ψ0|aλ(k)aλ′(k′)|ψ0〉 = 〈ψ0|a†λ(k)a†λ′(k′)|ψ0〉 = 0, using

Eq. 7 of the homework. So we have

〈ψ0|[

aλ(k) + a†λ(−k)][

aλ′(k′) + a†λ′(−k′)]

|ψ0〉 = 〈ψ0|[

aλ(k)a†λ′ (−k′) + a†λ(−k)aλ′(k′)|ψ0〉. (20)

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

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

so we have

〈ψ0|[

aλ(k) + a†λ(−k)][

aλ′(k′) + a†λ′(−k′)]

|ψ0〉 = 〈ψ0|[

aλ(k)a†λ′(−k′)|ψ0〉 (22)

= 〈ψ0|[

δλλ′δk,−k′ + a†λ′(−k′)aλ(k)]

|ψ0〉 (23)

= δλλ′δk,−k′〈ψ0|ψ0〉 (24)

= δλλ′δk,−k′ . (25)

Putting this back in, we have

〈ψ0|u(0)2|ψ0〉 =1

N

~

2M

∑

k,k′

∑

λ,λ′

Sµλ(k)Sµλ′(k′)√

ωλ(k)ωλ′(k′)δλλ′δk,−k′ (26)

=1

N

~

2M

∑

k,λ

Sµλ(k)Sµλ(−k)√

ωλ(k)ωλ(−k)(27)

=1

N

~

2M

∑

k,λ

Sµλ(k)Sµλ(k)

ωλ(k)(28)

=1

N

~

2M

∑

k,λ

1

ωλ(k)(29)

=a3~

2M

∑

λ

∫

d3k

(2π)31

ωλ(k). (30)

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, therewas a factor of 1/ω2 inside the integral, which gives a stonger singularity at k = 0. Here the integral iscertainly finite in 3d – in spherical coordinates, schematically, the integral at small-k looks like

〈ψ0|u(0)2|ψ0〉 ∼∫

dkk21

k, (31)

which is finite. The analogous integral is still finite in 2d, but diverges in 1d, since

〈ψ0|u(0)2|ψ0〉1d ∼∫

dk1

k. (32)

Now let’s discuss differences from the classical result. In 3d, the classical result predicts 〈ψ0|u(0)2|ψ0〉 = 0at T = 0. However, this can’t really be correct, because we know that there is zero point motion. The factthat we found 〈ψ0|u(0)2|ψ0〉 to be nonzero is simply because the quantum calculation takes the zero pointmotion into account.

In 2d, the difference is more dramatic. We found that 〈ψ0|u(0)2|ψ0〉 diverged for T > 0 in the classicalcalculation, but here it is finite at 〈ψ0|u(0)2|ψ0〉. Physically this makes sense – there are more fluctuationsat finite temperature than T = 0, and in 2d evidently those fluctuations are enough to make 〈ψ0|u(0)2|ψ0〉

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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 havedone, it turns out that 〈ψ0|u(0)2|ψ0〉 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 isreally a qualitative difference between zero and nonzero temperature.

In 1d, 〈ψ0|u(0)2|ψ0〉 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 quantumfluctuations. 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 dimensiongets 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 fortwo dimensions, the number of states per unit k-space area (not including spin) is A/(2π)2, where A is thetotal area of the system. Then we have

N = 2(number states per k-space area)( area of Fermi sea ) (34)

= 2( A

(2π)2

)

(πk2F ). (35)

(The factor of 2 in front of both these expressions accounts for spin.) Therefore,

n =N

A=k2F2π

. (36)

(b) Recall that in three dimensions, rs is defined as the radius of a sphere, whose volume is equal to the volumeper 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

πr2s = ( area per electron ) =A

N=

1

n. (37)

Using the result from part (a), we have

rs =1√πn

=1

√

π(k2F /2π)=

√2

kF. (38)

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

1

A

∑

k

↔∫

d2k

(2π)2. (39)

This result can be derived in the same way as the corresponding 3d result.

For the density of states we have

D(ǫ) =2

A

∑

k

δ(ǫ− ǫk) (40)

= 2

∫

d2k

(2π)2δ(ǫ − ǫk) (41)

= 2

∫

d2k

(2π)2δ(

ǫ− ~2k2

2m

)

(42)

=2(2π)

(2π)2

∫ ∞

0

dk k δ(

ǫ− ~2k2

2m

)

(43)

=1

π

∫ ∞

0

dk k δ(

ǫ− ~2k2

2m

)

. (44)

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We now make the change of variables u = ~2k2/2m, resulting in

D(ǫ) =m

π~2

∫ ∞

0

du δ(ǫ− u). (45)

For ǫ > 0, the delta function integrates to unity. However for ǫ < 0, the argument of the delta function isnever zero, and the delta function integrates to zero. So the result is

D(ǫ) =

m/(π~2) , ǫ > 00 , ǫ < 0

. (46)

(d) Recall that

n =

∫ ∞

−∞

dǫD(ǫ)f(ǫ), (47)

where f(ǫ) is the Fermi function. In this part of the problem we’re supposed to evaluate this using theSommerfeld expansion. The general form of the Sommerfeld expansion is given in A+M Eq. (2.69), andfor the present case we have

∫ ∞

−∞

dǫD(ǫ)f(ǫ) =

∫ µ

−∞

dǫD(ǫ) +

∞∑

n=1

an(kBT )2n

[

d2n−1

dǫ2n−1D(ǫ)

]

ǫ=µ

. (48)

Now, because D(ǫ) is constant except at ǫ = 0, its derivatives all vanish (except at ǫ = 0). For anyreasonably low temperature, we should expect µ ≈ ǫF 6= 0, so all the above derivatives vanish, and theSommerfeld expansion simply gives

n =

∫ µ

−∞

dǫD(ǫ). (49)

Therefore,

n(T ) =

∫ µ

−∞

dǫD(ǫ) = n(T = 0) =

∫ ǫF

−∞

dǫD(ǫ). (50)

(This is the statement that density is constant as a function of temperature.) The only way for this to beconsistent 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 Fermifunction,

n =m

π~2

∫ ∞

0

dǫ

e(ǫ−µ)/kBT + 1. (52)

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

n =mkBT

π~2

∫ ∞

−µ/kBT

du

eu + 1. (53)

While there is probably a more systematic way to do it, I found it easiest to do the integral by simplyguessing the form of a function whose derivative is equal to the integrand, 1/(eu + 1). The result is

∫

du

eu + 1= − ln(e−u + 1), (54)

which gives the result

n =mkBT

π~2ln(

eµ/kBT + 1)

. (55)

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Now we just need to express this in the form given in A+M. Using the fact that π~2n/m = ǫF , we have

ǫF = kBT ln(

eµ/kBT + 1)

(56)

= kBT ln(

eµ/kBT(

1 + e−µ/kBT)

)

(57)

= µ+ kBT ln(

1 + e−µ/kBT)

, (58)

as desired.

(f) First, we express the result of part (e) in the form

ǫF − µ = kBT ln(

1 + e−µ/kBT)

. (59)

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

ǫF − µ ≈ kBTe−µ/kBT . (60)

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

ǫF − µ ≈ (kBT )e−ǫF/kBT . (61)

This result says that ǫF − µ is exponentially smaller than kBT – this is a tiny correction and is notsignificant. The reason the Sommerfeld expansion “failed” here is that the density of states D(ǫ) is not ananalytic function on the entire real axis – it is discontinuous at ǫ = 0. This discontinuity is not capturedin the Sommerfeld expansion, but does appear (albeit as a tiny correction) when the integral is evaluatedexactly.

It is worth noting that a similar issue presumably arises in 3d – there, again, D(ǫ) is not analytic atǫ = 0. However, in 3d, any exponentially small temperature dependence coming from this nonanalyticityis completely dominated by the T 2 temperature dependence given by the (nonvanishing) leading term inthe Sommerfeld expansion.

3. (a) The first thing to understand is the single-particle spectrum of the trapped fermions – we just have to solvethe three-dimensional (3d) Harmonic oscillator, with Hamiltonian H = p2/2m+ (1/2)mω2

t r2. The energy

eigenstates are labeled by the vector of integers n = (nx, ny, nz), where nx = 0, 1, 2, . . . , and similarly forny 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 anoverall 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 thenumber of particles is obtained by filling all states up to the Fermi energy. Let us calculate Ω(ǫ):

Ω(ǫ) =∑

n

Θ(ǫ− ǫn). (63)

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

∑

n

→∫ ∞

0

dnx

∫ ∞

0

dny

∫ ∞

0

dnz ≡∫

d3n. (64)

This replacement is legitimate because the sum in Eq. (63) will be dominated by contributions from largenx, ny and nz. If nx ≫ 1, then the discrete change nx → nx + 1 can reasonably be thought of as aninfinitesimal change, i.e. nx → nx + dnx. Therefore we approximate n by a continuous variable, andreplace the sum by an integral.

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We have

Ω(ǫ) =

∫

d3nΘ(

ǫ− ~ωt(nx + ny + nz))

. (65)

So the integrand is equal to unity in the region nx + ny + nz ≤ ǫ/~ωt, and is zero otherwise – thereforeΩ(ǫ) is just the volume of this region. Defining α = ǫ/~ωt > 0, we then have

Ω(ǫ) =

∫ α

0

dnz

∫ α−nz

0

dny

∫ α−nz−ny

0

dnx =α3

6=

ǫ3

6~3ω3t

. (66)

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

N = Ω(ǫF ) =ǫ3F

6~3ω3t

, (67)

and

ǫF = ~ωt(6N)1/3. (68)

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

D(ǫ) =dΩ(ǫ)

dǫ. (69)

This allows for the usual interpretation that the quantity

D(ǫ)dǫ =dΩ(ǫ)

dǫdǫ (70)

is the total number of states in the interval [ǫ, ǫ+ dǫ]. Using the result of part (a),

D(ǫ) =ǫ2

2~3ω3t

, (71)

for ǫ > 0, and D(ǫ) = 0 for ǫ < 0.

(c) At finite temperature, the total number of fermions is given by

N =

∫ ∞

−∞

dǫD(ǫ)f(ǫ). (72)

We shall evaluate this integral at low temperature via the Sommerfeld expansion, and use it to determinethe chemical potential µ(T ). Carrying out the Sommerfeld expansion to leading order we have

N =

∫ µ

0

dǫD(ǫ) +π2

6(kBT )

2D′(µ). (73)

As we did in class for the free electron gas, we let µ(T ) = ǫF + c(kBT )2 + O(kBT )

4. Keeping terms upthrough order (kBT )

2, we have

N =

∫ ǫF

0

dǫD(ǫ) + c(kBT )2D(ǫF ) +

π2

6(kBT )

2D′(ǫF ). (74)

Noting that N =∫ ǫF0

dǫD(ǫ), we obtain the following expression for the constant c:

c = −π2

6

D′(ǫF )

D(ǫF )= − π2

3ǫF(75)

Therefore

µ(T ) = ǫF − π2

3ǫF(kBT )

2 +O(kBT )4. (76)

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(d) The total energy is given by

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

E(T,N) =

∫ µ

0

dǫ ǫD(ǫ) +π2

6(kBT )

2[D(µ) + µD′(µ)]. (78)

Using the result from part (c) above, and keeping only terms up through order (kBT )2, we have

E(T,N) =

∫ ǫF

0

dǫ ǫD(ǫ) + c(kBT )2ǫFD(ǫF ) +

π2

6(kBT )

2[D(ǫF ) + ǫFD′(ǫF )] (79)

= E(0, N) +π2

6(kBT )

2D(ǫF ). (80)

The ground state energy is

E(0, N) =

∫ ǫF

0

dǫ ǫD(ǫ) (81)

=ǫ4F

8~3ω3t

=64/3

8~ωtN

4/3, (82)

and we have

E(T,N) =64/3

8(~ωt)N

4/3 +π2

2 · 61/3(kBT )

2

~ωtN2/3 +O(T 4). (83)

(e) The specific heat is given by

C(T,N) =(∂E

∂T

)

N=

π2

61/3k2BT

~ωtN2/3. (84)