P7654 HW2

Due [Questions 1-4: Wednesday Jan 30, 2013, Question 5: Friday Feb 1

Depending on your quantum mechanics background (in particular, how well you know scattering theory),

this problem set could be easy or “hard”. By hard, I mean therecould be a lot of concepts coming at you. I

will assume that most of the students are in this latter category. I would like you to complete problems 1-4

by Wednesday, and problem 5 by Friday. I will have office hourson both Tuesday and Thursday at 2:30 –

meet at 514A.

Problem 1. Equation of State within the Hartree-Fock approximation In class we derived a perturbationexpansion for the free energy of a weakly interacting gas, described by Hamiltonian

H =∑

k

ǫka†kak +

1

2Ω

∑

kpq

Vqa†ka

†pap+qak−q, (1)

= H0 +H1 (2)

it is easy enough to add spin, but for simplicity we will leaveit out. In free space the dispersion is

ǫk =k2

2m− µ,

but we can just as well takeǫk to correspond to some sort of lattice dispersion. The interactions are parame-

terized byVq which is the Fourier transform of the real space potential, andΩ is the volume of space, which

could be removed or changed to another power depending on your Fourier conventions. I like these Fourier

conventions because they make the thermodynamic limit straightforward

1

Ω

∑

k

→∫

dk

(2π)d.

The free energy is

e−βF = Tre−βH.

Here we will calculate the lowest order correction to the equation of state. On general grounds, if you are

only going to sum a finite number of terms in the perturbation expansion, you might as well stop at this first

correction. If the interactions are very weak, then all further contributions are negligable. If the interactions

are strong, then you need all of the terms.

Note that we do need to use ”diagrams” or ”field theory” to do this calculation – it is just first order pertur-

bation theory – but nonetheless we will frame it in terms of diagrams. If you take a solid state physics class

you will probably see it done using a variational method. It therefore is not really that good of an example

of the utility of many-body field theory – rather it is a warmup. Question 5 is a better example.

1.1. Expand the free energy to first order inH1. Write the resulting expression in terms of an integral of theexpectation value ofH1. (We did this in class)

-1

Solution 1.1. We do this by expanding

Z = Tr e−βĤ = Tr e−β(Ĥ0+Ĥ1) = Tr[

e−βN (Ĥ0+Ĥ1)

]N. (3)

TakingN to be very large, this is

Z ≈ Tr[

1 − βN

(

Ĥ0 + Ĥ1

)

]N

(4)

which we can in turn expand̂H1:

Z ≈ Tr[

1 − βNĤ0

]N

+ Tr

[

N−1∑

s=0

[

1 − βNĤ0

]s(

− βNĤ1

)[

1 − βNĤ0

]N−s]

= Z0 −∫ β

0dt Tr

[

e−Ĥ0tĤ1e−Ĥ0(β−t)

]

= Z0

(

1 −∫ β

0dt〈

Ĥ1 (t)〉

0

)

(5)

whereZ0 = Tr e−βĤ0 ,〈

X̂〉

0= 1Z Tr X̂e

−βĤ0 andĤ1 (t) = e−Ĥ0tĤ1eĤ0t.

From the partition function we can get the free energy

F = − 1β

lnZ ≈ F0 +1

β

∫ β

0dt〈

Ĥ1 (t)〉

0(6)

whereF0 = − 1β lnZ0 is the free energy of the noninteracting system.

1.2. Use Wick’s theorem to write this first order correction to theFree Energy as the sum of two terms,corresponding to the two contractions of this term.

Solution 1.2. Writing Ĥ1 explicitly we have (taking all operators to be thet-dependent version)

∆F =1

β

∫ β

0dt

〈

1

2Ω

∑

k,p,q

Vqâ†kâ

†pâp+qâk−q

〉

0

=1

2Ωβ

∑

k,p,q

Vq

∫ β

0dt〈

â†kâ†pâp+qâk−q

〉

0

=1

2Ωβ

∑

k,p,q

Vq

∫ β

0dt〈

â†kâk−q

〉

0

〈

â†pâp+q

〉

0±〈

â†kâp+q

〉

0

〈

â†pâk−q

〉

0

(7)

where as usual in the case of± the upper sign is for bosons and the lower for fermions.

1.3. These expressions are easy to evaluate, as all of the field operators are at the same time. Write theresulting integrals in terms of the equilibrium occupations nk = 〈a†kak〉.

-2

Solution 1.3. As we have〈

â†kâp

〉

= δk,pnk this becomes

∆F =1

2Ωβ

∑

k,p

∫ β

0dt (V0 ± Vk−p)nknp. (8)

1.4. Specialize to the case of a point interactionVk = const= V0. Calculate the first order interaction shiftto the Free energy. You should find that for fermions it vanishes – this is because two identical fermions can

never be in the same place, so they never see a point interaction. For bosons, however, you will get a finite

result. Express the shift in terms of the densityn = (1/Ω)∑

k nk. Verify that the shift is extensive.

Solution 1.4. Indeed, we see that in the case of a point interaction the shift vanishes for fermions; whilefor bosons we find

∆F =V0Ωβ

∑

k,p

∫ β

0dtnknp = V0Ω

1

β

∫ β

0dtn2

= V0Ωn2

(9)

which is manifestly extensive.

Problem 2. Scattering Phase ShiftsThis should be a reminder of things from your quantum class. If youalready know this, then feel free to skip this question. I am just assigning this because if you don’t know

this stuff, you will have some conceptual issues with the next questions. Our eventual goal (in question

5) is to derive a non-trivial result, namely the superfluid transition temperature of a Fermi gas with strong

local interactions. It is hopeless to do that until we know how to describe scattering off of a local potential.

We also need to do this question so you can understand which ofthe phenomena in question 5 are due to

many-body physics, and which are just two-body physics coming from a 3D point scatterer.

Consider a spherically symmetric potential in three dimensions,V (r), with V (r) = 0 for r > r0. We are

going to solve the single particle Schrodinger equation[−∇2

2m+ V (r)

]

ψ(r) = Eψ(r).

2.1. Assuming that the wavefunctionψ(r) is spherically symmetric,ψ(r) = u(r)/r. Write down the timeindependent Schrodinger equation foru(r).

Solution 2.1.[

−∇2

2m+ V (r)

]

ψ (r) =

[

− 12m

(

∂2

∂r2+

2

r

∂

∂r

)

+ V (r)

]

u (r)

r

=

[

− 12m

1

r

∂2u (r)

∂r2+ V (r)

u (r)

r

]

= Eu (r)

r[

− 12m

∂2

∂r2+ V (r)

]

u (r) = Eu (r) .

(10)

-3

SinceV (r) = 0 for r > r0, this reduces to a free Schrodinger equation forr > r0. All of the details of the

potential can be replaced by a boundary condition atr = r0, namely

u′

u

∣

∣

∣

∣

r=r0

= f(r0, E), (11)

wheref(r0, E) is a well behaved function ofE asE → 0. [I.E. it should have a nice Taylor expansionwhich has a finite radius of convergence.] We will be interested in low energy scattering, so we will replace

f(r0, E) with f(r0, 0).

2.2. Show that if one is only interested in the wavefunction forr > r0, that the solutions to the freeSchrodinger equation with boundary condition (11) can be reproduced by using a free Schrodinger equation

with boundary conditions atr = r0. This boundary condition has the form

u′

u

∣

∣

∣

∣

r=0

= −1a, (12)

wherea, the s-wave scattering length, has dimensions of length. Write a in terms off(r0, 0).

[Hint – since we are looking at the limit of low energy scattering you can linearize the wavefunction about

r0.]

Solution 2.2. At r > r0 we have simply

− 12m

∂2u (r)

∂r2= Eu (r) (13)

and so

u (r) = Aeikr +Be−ikr

= sin(kr + δ)(14)

wherek =√

2mE. As the overall magnitude ofA,B is determined by normalization, we only need one

more constraint, such as

u′

u

∣

∣

∣

∣

r=r0

= k cot(kr0 + δ) = f (r0, E) . (15)

If we assumed Eq (14) worked up tor = 0, we can replace this boundary condition with one atr = 0. In

particular:

−1a

=u′

u

∣

∣

∣

∣

r=0

= k cot(δ) = k cot(cot−1(f/k) − kr0)

= kf cos(kr0) + k sin(kr0)

k cos(kr0) − f sin(kr0)

= f(r0, 0) +

(

k2 +f2

k

)

r0 + · · ·

(16)

To leading order ink, this is justf(r0, 0) = −1/a.

-4

2.3. Finda for a hard sphere potential of radiusr0.

Solution 2.3. A hard sphere solution is given byu (r = 0) = 0, or

u (r) = A sin (k (r − r0)) . (17)

If we extend this tor = 0 we getu′

u

∣

∣

∣

∣

r=0

= − 1r0,

soas = r0.

2.4. Show that the free Schrodinger equation, with boundary condition (12), has a bound state if and only ifa > 0. (Yes, this sign is right, the result is indeed counter-intuitive.) What is the energy of the bound state?

Solution 2.4. For a bound stateu(r) = e−κr. (18)

Direct evaluation givesu′(r)

u(r)= −1

a= −κ, (19)

anda > 0. The energy is

E = − κ2

2m= − 1

2ma2. (20)

2.5. For r > r0 we can writeu(r) = sin(kr + δk). How is the phase shiftδk related tof(0, E)? Linearizeδk aboutk = 0, and relate it toa.

Solution 2.5. Using this alternate formulation, we have

f (r,E) = k cot (kr + δk) (21)

and sof (0, E) = k cot (δk). Solving forf (0, E) = −1/a we find

cot (δk) = −1

ka. (22)

For smallk, then,δk ≈ −ka.

Let’s now calculate the scattering length for a delta-function potential

V (r) = V0δ(r) =V0Ω

∑

k

eik·r

It will be convenient to solve the Schrodinger equation in momentum space.

2.6. Defineψ(r) = 1Ω∑

k eik·rψk. Fourier transform the time independent single particle Schrodinger

equation to write an equation forψk

-5

Solution 2.6. Starting with the Schrodinger equation in real space,[−∇2

2m+ V (r)

]

ψ(r) = Eψ(r)

we substitute in the expressions in terms of the Fourier transforms

1

Ω

∑

k

k2

2mψke

ik·r +1

Ω2

∑

kq

ei(k+q)·rVkψq =1

Ω

∑

k

Eψkeik·r.

We then equate terms in the sums multiplying the same exponentials and specialize to the delta function to

arrive at

(ǫk − E)ψk +V0Ω

∑

q

ψq = 0

where we have dubbedǫk = k2/2m.

With the delta-function interaction, the interaction termV (r)ψ(r) = δ(r)V0ψ(0) =V0ψ(0)

Ω

∑

k eik·r, is a

constant in momentum space. Lets call∆ = V0ψ(0).

2.7. Findψk in terms of∆ andE. [AssumeE = −Eb < 0 – we will look for bound states. That way youdo not have to worry about dividing by zero.]

Solution 2.7.ψk =

∆

E − ǫk.

2.8. We can also write∆ =

V0Ω

∑

k

ψk.

Insert your expression forψk into this formula.∆ should appear on both sides of the equation and can be

cancelled. Write the resulting equation, which involvesE, ǫk = k2/2m, V0 andΩ. Solve for1/V0 in terms

of the other quantities.

Solution 2.8.1

V0=

1

Ω

∑

k

1

E − ǫk

In three dimensions, the right hand side of this equation is formally−∞, telling you that there are no boundstates for delta-function potential in 3D unlessV0 = 0−. This is weird but true. If this was a quantum

mechanics class I would have us explore this a bit more. Here we will accept this, and define

1

V=

1

V0+

1

Ω

∑

k

1

ǫk.

The quantityV will parameterize the deviation ofV0 from zero.

-6

2.9. Using your equation for1/V0, find an equation for1/V . The sum should now be convergent. Convertyour sum into an integral, and find a relationship betweenV andEb = −E.

Solution 2.9.

1

V=

1

Ω

∑

k

(

1

E − ǫk+

1

ǫk

)

= 2m

∫

d3k

(2π)3

(

1

2mE − k2 +1

k2

)

= 4m

∫

k2dk

(2π)2

(

1

2mE − k2 +1

k2

)

=m√

2mEb2π

.

(23)

2.10. Using your knowledge of howEb is related toa, give an expression forV in terms of the scatteringlength.

Solution 2.10.

V =2π

m

√a2 =

2π

ma. (24)

Problem 3. Lipman-Schwinger approach to scatteringHere we are going to rederive some of the keyresults of question 2 in a diagrammatic framework. In order to mirror the language we will use in problem 5,

we will couch it as a two-body problem instead of a one-body problem – question 2 was simply in the center

of mass frame (so in comparing results with those in this section you should replacem with the reduced

massm/2). We will consider a Hamiltonian for two-component fermions with point interactions

H =∑

kσ

ǫka†kσakσ +

V0Ω

∑

kpq

a†k↑a†p↓ap−q↓ak+q↑

whereσ =↑ / ↓ is the spin index, andǫk = k2/2m. Since this is a point interaction, I have only includedinteractions between non-alike fermions. Our one ”field theory” trick is to look at is the resolvent:

R(k, ω) =∑

q

〈ak/2+q↑ak/2−q↓1

ω −Ha†

k/2−q↓a†

k/2+q↑〉, (25)

where the expectation value is taken in the vacuum state containing no particles. This resolvent is nothing

but a trace of a zero temperature two-particle greens function in frequency and momentum space. It is useful

because it is related to the two-particle density of stateρ2(ω, k) via

ρ2(ω, k) =1

πImR(k, ω). (26)

This density of states tells us how many two-particle stateshave energyE and momentumk.

In retrospect, I should have just asked you to calculate the two-particle Greens function

Gk,qq′(ω) = 〈ak/2+q↑ak/2−q↓1

ω −Ha†

k/2−q′↓a†

k/2+q′↑〉,

-7

which is related to the resolvent by

R(k, ω) =∑

q

Gk,qq(ω).

this would save one sum. Not a big deal though.

3.1. Prove Eq. (26). Hint: Use1/(ω − ǫ) = P/(ω − ǫ) ± iπδ(ω − ǫ).

Solution 3.1. Let us first introduce the basis of state|2, k, ǫ〉 denoting a state of two particles of totalmomentumk and

Ĥ |2, k, E〉 = ǫ |2, k, ǫ〉 . (27)

We can diagonalizêH in this subspace because it conserves both total momentum and number of particles.

Then, we can rewrite the trace in this basis,

R (k, ω) =∑

ǫ

〈2, k, ǫ| 1ω − Ĥ

|2, kǫ〉 =∑

ǫ

1

ω − ǫ , (28)

and all that’s left is to isolate the imaginary part

R (k, ω) =∑

ǫ

[

P

ω − ǫ + iπδ (ω − ǫ)]

(29)

to see that

1

πImR (k, ω) =

∑

ǫ

δ (ω − ǫ) (30)

indeed counts the number of states of total momentumk that have energyω.

Following a similar approach to what we did in class for the free energy, we will calculate a perturbation

expansion forR in powers ofV0. We will sum this expansion to all orders. For simplicity we will specialize

to the casek = 0, but thek 6= 0 case is not much harder. The calculation is fairly straightforward. Onetricky bit is that if we want to get a finite scattering length,thenV0 is formally 0−. This is just an quirk of

treating the potential as zero range, and the same problem came up in the last question. Another tricky bit is

that the real part ofR is formally infinite. This is not a big deal, as this real part has no physical significance.

3.2. Write a sum forR0(ω) = R0(0, ω) in the case whereV0 = 0. [The subscript 0 onR0 represents thefact that there are no interactions.] This sum should be familiar. Convert the sum into an integral. Argue

that the real part is infinite.

-8

Solution 3.2. We have

R0 (ω) =∑

k

〈

âk↑â−k↓1

ω − Ĥ0â†−k↓â

†k↑

〉

. (31)

This is easy to calculate sincêa†−k↓â†k↑ |vac〉 is an eigenstate of̂H0 with energyǫk + ǫ−k = 2ǫk. Thus this

is simply

R0 (ω) =∑

k

1

ω − 2ǫk. (32)

This is (up to a volume function) what we found for1/V before. We can rewrite it as an integral

R0 (ω) = Ω

∫

d3k

(2π)31

ω − k2/m =4πΩ

(2π)3

∫

dkk2

ω − k2/m. (33)

Clearly the integrand goes to unity ask → ∞ and so the integral is infinite.

3.3. DefineΠ0 =∑

k−12ǫk

, andΠ(ω) = R0(ω)−Π0(ω). ClearlyΠ0 is infinite, on the other handΠ(ω) is anice analytic function in the complexω plane, except on the positive real axis, where it has a branchcut.

Takeω to be real and negative. Perform the integral to calculateΠ(ω).

Solution 3.3.

Π(ω) = Ω

∫

d3k

(2π)3

[

1

ω − k2/m −1

k2/m

]

=Ω

4πm√−mω. (34)

3.4. Calculateρ2(ω, 0) for the non-interacting case. Does this match what you know from elementaryarguments? [Remember in the center of mass frame this is a 1-body problem, so (up to a factor of

√2) this

should agree with the single particle density of states fromelementary statistical mechanics.

Solution 3.4. We’ve shown

ρ2 (ω) =1

πImR (ω) = Ω

1

4π2m3/2

√ω. (35)

This is, indeed, the number of states (i.e. density of statestimesΩ) for a three-dimensional free fermion

gas, up to a factor of√

2.

3.5. We can expand the resolvant in a power series in

H1 =V0Ω

∑

kpq

a†k↑a†p↓ap−q↓ak+q↑,

formally writing

1

ω −H =1

ω −H0+

1

ω −H0H1

1

ω −H0+

1

ω −H0H1

1

ω −H0H1

1

ω −H0+ · · · (36)

-9

By constructionH1 is always acting on a two-particle state, with total momentum 0. Thus the only terms in

H1 which matter are those withp = −k. Moreover, since we are acting on two-particle states, we can inserta vacuum state in the middle, and use

H1 →V0Ω

(

∑

k

a†k↑a†−k↓|vac〉

)(

∑

q

〈vac|ak↓a−k↑

)

The state|vac〉 is the vacuum containing no particles. A potential that can be decomposed into a creationterm times an annihilation term is known as ”separable.”

Substitute this form forH1 into Eq. (36), and write the resolventR(ω) as a power series inR0(ω).

Solution 3.5. We start by restricting ourselves to the the two-particle, zero-momentum subspace. Explic-itly this is

Ĥ11̂2p = Ĥ1∑k′

â†k′↑â†−k′↓ |vac〉 〈vac| â−k′↓âk′↑

=V0Ω

∑

k′kpq

â†k↑â†p↓âp−q↓âk+q↑â

†k′↑â

†−k′↓ |vac〉 〈vac| â−k′↓âk′↑

=V0Ω

(

∑

k

â†k↑â†−k↓ |vac〉

)(

∑

k

〈vac| â−k↓âk↑)

.

(37)

Within this space then-th term (n ≥ 1) of the expansion is[

1

ω − Ĥ0Ĥ1

]n 1

ω − Ĥ0=

(

V0Ω

)n(

∑

k

1

ω − 2ǫk

)n−1

×(

∑

k

1

ω − 2ǫkâ†k↑â

†−k↓ |vac〉

)(

∑

k

1

ω − 2ǫk〈vac| â−k↓âk↑

)

.

(38)

This is immediately obvious forn = 1 and can be proven for anyn by induction. It then follows

∑

k

〈

âk↑â−k↓

[

1

ω − Ĥ0Ĥ1

]n 1

ω − Ĥ0â†−k↓â

†k↑

〉

=∑

k

(

V0Ω

)n(

∑

k

1

ω − 2ǫk

)n−1(

1

ω − 2ǫk

)2

.

(39)

The sum here is finite

∑

k

(

1

ω − 2ǫk

)2

= Ω

∫

d3k

(2π)31

(ω − k2/m)2=

Ω

4πm

√

m

−ω (40)

and so

R (ω) = R0 (ω) +V04πm

√

m

−ω

(

∞∑

n=0

V0ΩR0 (Ω)

)

(41)

-10

3.6. Sum the power series.

Solution 3.6.

R (ω) = R0 (ω) +

[

1 − V0ΩR0 (ω)

]−1 V04πm

√

m

−ω(42)

3.7. SinceΠ(ω) is real for negativeω,R(ω) will also be real. Since the imaginary part ofR corresponds tothe density of states, this would imply that there are no bound states. The loophole is thatR(ω) could have

a pole. From the conditionR(ω)−1 = 0 find the energy of the bound state. Write the result in terms ofV0andΠ0.

Solution 3.7. R (ω) clearly has a pole atV0Ω R0 (ω) = 1, or

1

V0=

1

ΩR0(ω)

=Π0Ω

+Π(ω)

Ω

(43)

Following the notation of the last question we write

1

V=

1

V0− Π0

Ω. (44)

Using our explicit expression forΠ, we have

ω = −(4π)2

mV 2. (45)

3.8. Using the result from problem 2,

Eb =~

2

2mra2s,

wheremr = m/2, relateV0 to the scattering length and the infinite constantR0. [You know which branch

of the square root to use, since a positive scattering lengthcorresponds having a bound state.]

Solution 3.8. We find finally, given that~ = 1,

(

V

4πm

)−2

=1

as(46)

or

V =4π

mas. (47)

Note 1: In your quantum mechanics class you may have done essentially this same calculation in terms of

solving the Lipmann-Schwinger equation. This treatment interms of the resolvent is equivalent. You might

want to look at your notes and compare.

-11

Problem 4. Diagramatic derivation of Lipman-Schwinger equation

Here we just connect the calculation in the previous sectionto the Feynman Rules introduced in lecture. We

first note that our Resolvent is nothing but a frequency spaceGreens function. We first use the fact that we

are taking a vacuum expectation value to write:

R(ω) =∑

q

〈ak/2+q↑ak/2−q↓1

ω −Ha†

k/2−q↓a†

k/2+q↑〉

=∑

q

〈ak/2+q↑ak/2−q↓1

ω −Ha†

k/2−q↓a†

k/2+q↑ − a†

k/2−q↓a†

k/2+q↑

1

ω+Hak/2+q↑ak/2−q↓〉.

The second term we added is just a fancy name for zero. For the first term we take the branch where

1

ω −H−iη =1

i

∫

dtei(ω−H)tθ(t)e−ηt.

For the second term we instead use

1

ω +H+iη= −1

i

∫

dtei(ω+H)tθ(−t)eηt

to yieldR(ω) =∫

dteiωtR(t) with

R(t > 0) = −i∑

q

〈ak/2+q↑ak/2−q↓e−iHta†k/2−q↓a†

k/2+q↑〉

R(t < 0) = −i∑

q

〈a†k/2−q↓a†

k/2+q↑eiHtak/2+q↑ak/2−q↓〉

But since the vacuum state has zero energy we can just as well write this as

R(t > 0) = −i∑

q

〈eiHtak/2+q↑ak/2−q↓e−iHta†k/2−q↓a†

k/2+q↑〉

R(t < 0) = −i∑

q

〈a†k/2−q↓a†

k/2+q↑eiHtak/2+q↑ak/2−q↓e

−iHt〉,

which we can recognize as a time ordered response function. We can either work with these, or if we really

want to shoe-horn it into our old approach, we can write the vacuum expectation value as a finite temperature

expectation value forµ < 0 andT → 0. We would probably also taket → iτ . We will see this in the nextsection though.

Expanding theH ’s in powers ofH1 and using Wick’s theorem gives us a diagrammatic expansion similar

to the one we had for the Free energy. The same Feynman Rules apply here as applied there. As before, we

will focus on the casek = 0, the more general case is a straightforward extension.

The zeroth order term has only one contraction:

R0 =

-12

This is nothing but the statement that

R0(t > 0) = −i∑

q

〈eiH0tak/2+q↑e−iH0ta†k/2+q↑〉〈eiH0tak/2−q↓e

−iH0ta†k/2−q↓〉 (48)

R0(t < 0) = −i∑

q

〈a†k/2+q↑eiH0tak/2+q↑e

−iH0t〉〈a†k/2−q↓eiH0tak/2−q↓e

−iH0t〉

4.1. WriteR0(t) in terms ofG0σ(k, t) = −i〈Takσ(t)a†kσ(0)〉 = −iθ(t)〈akσ(t)a†kσ(0)〉+iθ(−t)〈a

†kσ(0)akσ(t)〉.

Solution 4.1.

R0 (k, t) = i∑

q

G0↑ (k/2 + q, t)G0↓ (k/2 − q, t) . (49)

4.2. Use this result to writeR0(ω) as a convolution ofG0(ω). Someone experienced with diagrams wouldimmediately write down this expression without going through the exercise of going to the time domain.

[Of course in this particular case, the calculation is easier in the time domain.]When we come to the next

question, we will see some subtleties with this argument, but here the naive approach works.

Solution 4.2.

R0 (k, ω) =

∫

dt eiωtR0 (k, t)

= i∑

q

∫

dt e−iωt∫

dη

2π

dη′

2πe−i(η+η

′)tG0↑ (k/2 + q, η)G0↓(

k/2 − q, η′)

= i∑

q

∫

dηdη′

2πδ(

η + η′ − ω)

G0↑ (k/2 + q, η)G0↓(

k/2 − q, η′)

= i∑

q

∫

dη

2πG0↑ (k/2 + q, ω/2 + η)G0↓ (k/2 − q, ω/2 − η) .

(50)

4.3. SubstituteG0(ω) = 1/(ω − ǫk). Use a partial fraction expansion, and the residue theorem to recoverthe result in question 3.2,

R0(ω) =∑

k

1

ω − 2ǫk(51)

-13

Solution 4.3. Specifying now toR0 (ω) = R0 (k = 0, ω) we write

R0 (ω) = i∑

k

∫

dη

2πG0↑ (k, ω/2 + η)G0↓ (−k, ω/2 − η)

= i∑

k

∫

dη

2π

1

(ω/2 + η − ǫk)1

(ω/2 − η − ǫk)

=∑

k

∑

Res

[

1

(ω/2 + η − ǫk)1

(ω/2 − η − ǫk)

]

(52)

where we’ve completed the integral over, say, the arcη = Reiχ with R → ∞ andχ = 0 . . . π. Takingωto have a positive imaginary component, we get

R0 (ω) =∑

k

1

ω − 2ǫk. (53)

4.4. As an alternative way to calculateR0(ω), write the explicit time domain expression forR0(t). Fouriertransform this result. Often I find that this approach is quicker than doing a contour integral.

Solution 4.4. Using the fact that̂a†k |vac〉 are eigenstates of̂H0, we can write

R0 (k, t > 0) = −i∑

q

〈

eiĤ0tâk/2+q↑e−iĤ0tâ†k/2+q↑

〉〈

eiĤ0tâk/2−q↓e−iĤ0tâ†k/2−q↓

〉

= −i∑

q

〈

âk/2+q↑e−iǫk/2+qtâ†k/2+q↑

〉〈

âk/2−q↓e−iǫk/2−qtâ†k/2−q↓

〉

= −i∑

q

e−i(ǫk/2+q+ǫk/2−q)t

R0 (k,< 0) = 0

(54)

and so takingω again to have a positive imaginary component,

R (ω) =

∫

dt eiωtR (t) = −i∑

k

∫ ∞

0dt ei(ω−ǫk−ǫ−k)t

= −i∑

k

1

i (ω − 2ǫk)[

ei(ω−ǫk−ǫ−k)t]∞

0=∑

k

1

ω − 2ǫk.

(55)

Note: I won’t ask you to do it (since it is just the calculationyou already did in the last problem) but now

the full calculation of R can be done diagrammatically, you just sum the series

R = + + + · · ·

=

1 −

-14

where the dot represents the point interaction, and in the diagrammatic language the dot is assigned the value

V0/Ω. Sometimes this series is instead drawn

R = + + + · · ·

In the next question we will see how to do this calculation in the ”Matsubara” formalism at finite temperature

and density.

Problem 5. Superfluid transition temperature within the Nozieres-Schmidt Rink approximation

You are aware that a Fermi gas with attractive interactions is unstable towards superfluidity at low tem-

peratures. Here we will calculate the equation of state in the normal state, and find the phase transi-

tion as a thermodynamic singularity. You may find it useful tolook at the original paper where this

result was first given: P. Nozières and S. Schmitt-Rink, J. Low Temp. Phys. 59, 195 (1985).http:

//link.springer.com/article/10.1007/BF00683774 I caution you however that there is a

lot in that paper, and we are just doing a subset of it here. That paper has a great description of the ”physics.”

Here we will predominantly worry about the mathematics.

We are going to start by thinking of the ”BEC” limit. By this, we will think about what happens when the

interactions become so strong that there is a two-body boundstate. Physically, when the temperature is of

order the binding energy, pair form. At a lower temperature they Bose condense. The Bose condensation

condition is the superfluid transition, and will show up as a singularity in the Free energy.

How do we capture the physics of these pairs? Well, we know howto get bound states in the 2-body

problems. The idea of Nozieres and Schmidt-Rink is to include the same set of diagrams in the many-body

problem. As before we get a series we can sum.

As I have emphasized before, the field theory is a language, and it doesn’t tell you what approximation to use.

The NSR technique captures the fact that two-body bound states can form. It misses out on other physics

(such as charge density waves). As with many calculations inmany-body physics, it is an uncontrolled

approximation. Its principle value is in the qualitative insight it gives.

We again consider a Hamiltonian of the form

H =∑

kσ

ǫka†kσakσ +

V0Ω

∑

kpq

a†k↑a†p↓ap−q↓ak+q↑

denoting these two terms asH0 andH1. We are going to take terms in the perturbation expansion forthe

free energy which involve all ”multiple scatterings of two particles”. These ”pair ring diagrams” look like:

δF = + + + + · · ·

-15

We will implicitly assume that the spins on the inner ring are”down” and on the outer ring ”up”. These

are exactly the ladder graphs we used in the 2-body problem, just ”wrapped up.” By construction we will

be able to exactly reproduce the physics of two-particles infree space. This approximation is the simplest

many-body extension of those results. Note, if we stopped atthe first term we would have ”Hartree-Fock.”

There is no ”Fock” term because the spins on the two propagators are distinct.

We now have 5 tasks:

1. Express each ring diagram in terms of the interactionV0 and the elementary particle-particle bubble

(which looks likeR0)

2. Calculate the multiplicity of the ring diagrams

3. Sum the series

4. CalculateR0

5. Find the thermodynamic instability

To start with, lets consider a typical ring diagram appearing in the perturbative expansion of the partition

function,

R4 =

For now lets not worry about the multiplicity (which we discuss shortly). This diagram has 4 powers ofV0,

so its expression as an intergral must begin with

(−βV0Ω

)4 ∫ β

0dτ1dτ2dτ3dτ4

Next we have a sum over a bunch ofk’s. We could write down the most general form, then start taking

contractions. A little thought however makes it easier. At each vertex, the total momentum going in equals

the total momentum going out. Thus the center of mass momentum of the ”two particles” going around the

ring is fixed. We need to sum over this global momentum,K, as well as the four relative momentak1, . . . k4.

With the point interaction all of these are weighted equally, so we get

1Ω

∑

K1

Ω4∑

k1···k4〈TψK/2+k1(τ1)ψ

†

K/2+k1(τ2)〉〈TψK/2−k1(τ1)ψ

†

K/2−k1(τ2)〉

× 〈TψK/2+k2(τ2)ψ†

K/2+k2(τ3)〉〈TψK/2−k2(τ2)ψ

†

K/2−k2(τ3)〉

× 〈TψK/2+k2(τ3)ψ†

K/2+k2(τ4)〉〈TψK/2−k2(τ3)ψ

†

K/2−k2(τ4)〉

× 〈TψK/2+k2(τ4)ψ†

K/2+k2(τ1)〉〈TψK/2−k2(τ4)ψ

†

K/2−k2(τ1)〉

5.1. ExpressR4 in terms ofR0(τ), which is defined by Eq. 48, witht→ −iτ .

-16

Solution 5.1. We have

R0 (k, t) = −i∑

q

〈

T âk/2+q↑ (t) â†

k/2+q ↑〉〈

T âk/2−q↓ (t) â†

k/2−q ↓〉

. (56)

Going back to the definition in terms of

1

ω −H =1

i

∫

dt ei(ω−H)tΘ (t) (57)

we see that

R0 (k, τ ) = −∑

q

〈

T âk/2+q↑ (τ) â†

k/2+q↑〉〈

T âk/2−q↓ (τ) â†

k/2−q↓〉

(58)

and so

R4 = (−V0)4∫ β

0dτ1dτ2dτ3dτ4

1

Ω

∑

K

1

Ω4

(−1)4 [R0 (K, τ1 − τ2)R0 (K, τ2 − τ3)R0 (K, τ3 − τ4)R0 (K, τ4 − τ1)] .(59)

This expression is simpler in frequency space, writing

R0(τ) =1

β

∑

ωn

eiωnτR0(iωn) (60)

whereωn = 2πn/β.

5.2. By using your knowledge of the antiperiodicity of the Fermion Greens functions, explain whyωn =2πn/β.

Solution 5.2. We have seen thatR0 (t) ∼ G0↑ (t)G0↓ (t); hence

R0 (t+ β) ∼ G0↑ (t+ β)G0↓ (t+ β) = (−1)2G0↑ (t)G0↓ (t) ∼ R0 (t) (61)

is periodic inβ and so we must haveωn = 2πn/β.

5.3. What isR4 in terms ofR0(iωn)?

-17

Solution 5.3.

R4 = (V0)4∫ β

0dτ1dτ2dτ3dτ4

1

Ω

∑

K

1

Ω4

(

1

β

)4∑

ω1,ω2,ω3,ω4[

eiω1(τ1−τ2)R0 (K, iω1) eiω2(τ2−τ3)R0 (K, iω2)×

eiω3(τ3−τ4)R0 (K, iω3) eiω4(τ4−τ1)R0 (K, iω4)

]

= (V0)4 1

Ω

∑

K

1

Ω4

∑

ω1,ω2,ω3,ω4

δω1,ω2δω2,ω3δω3,ω4δω4,ω1

[R0 (K, iω1)R0 (K, iω2)R0 (K, iω3)R0 (K, iω4)]

= (V0)4 1

Ω

∑

K,ω

[

1

ΩR0 (K, iω)

]4

.

(62)

5.4. What is the genericm’th order ring diagramRm in terms ofR0(iωn)? Having worked outR4, youshould be able to instantly write it down.

Solution 5.4. It’s clear

Rm =1

Ω

∑

K,ω

[

V0ΩR0 (K, iω)

]m

. (63)

Next we have to calculate the multiplicity. First of all, them’th order diagram appears with a factor of

(1/m!). Next, we note that there are vertices atm times. We arbitrarily take one to bet1. There are then

(m− 1)! ways to order the other ones so you form a ring. ThusRm appears with a factor of1/m in front ofit, leading to the expression:

βδF =∑

ωn

∑

K

∑

m

1

m

(

V0R0(iωn)

Ω

)m

.

5.5. Sum the series to simplify the expression forβδF .

Solution 5.5. Using the expression

log(1 − x) = −∑

j

xj

j

we get

βδF = −∑

ωn

∑

K

log

(

1 − V0R0(K, iωn)Ω

)

.

[In comparing with Nozieres and Schmidt-Rink, they callχ = V0R0/Ω.] Now we just have to calculate

R0(iωn). Our result should reduce to the expression in Eq. 51 in the limit of zero density (ie. the two particle

limit). Just as in the zero temperature case there are two approaches.

5.6. Write the explicit expression forR0(τ) in the imaginarytime domain (don’t do the sum overq). Theonly τ dependence comes fromθ-functions and exponentials.Use the fact that

G0(q, τ) = −〈Tψq(τ)ψ†q(0)〉 = −θ(τ)(1 − f(ǫq))e−ǫqτ + θ(−τ)f(ǫq)e−ǫqτ ,

-18

wheref(x) = 1/(eβx + 1.

Solution 5.6.

R0 (K, τ ) = −∑

q

Θ (τ)〈

eH0τ âK/2+q↑e−H0τ â†K/2+q↑

〉〈

eH0τ âK/2−q↓e−H0τ â†K/2−q↓

〉

+ Θ (−τ)〈

â†K/2+q↑eH0τ âK/2+q↑e

−H0τ〉〈

â†K/2−q↓eH0τ âK/2−q↓e

−H0τ〉

= −∑

q

e−(ǫK/2+q+ǫK/2−q)τ [θ(τ)f+(1 − f−) − θ(−τ)f−(1 − f+)]

(64)

wheref± = 1eβǫk/2±q+1

.

5.7. Prove that for−β < τ < 0, the bubble obeysR0(K, τ + β) = R0(K, τ), and hence can be expandedas a Fourier series

R0(K, τ) =1

β

∑

n

e−iωnτR0(K, iωn)

whereωn = 2πn/β.

5.8. Use your explicit expression forR0(K, τ) to calculate

R0(K, iωn) =

∫ β

0R0(K, τ)e

iωnτ

5.9. We should be able to reproduce our two-particle result, by going to the limitf(ǫ) → 0. Verify this istrue.

Here is the harder way to do the calculation. Strangely, thisis the way that appears in all the textbooks.

Given thatR0(K, τ) = 1Ω∑

q G0(K/2 + q, τ)G0(K/2 − q, τ), the Fourier transform is a convolution:

R0(K, iωn) =1

β

∑

ν

1

Ω

∑

q

1

iωn − iν − ǫK/2+q1

iν − ǫK/2−q. (65)

5.10. What are the values thatν are summed over?

Solution 5.7. The sum is over all Fermi Matsubara frequencies:ν = (2m+ 1)π/β.

The functionf(z) = 1/(eβz + 1) has poles with residue1/β whenz = (2m+ 1)π/β. Thus by the residue

theorem, we can rewrite the sum in Eq. (65) as an integral in the complex plane:

R0(K, iωn) =β

2πi

∮

dzf(z)1

Ω

∑

q

1

iωn − z − ǫK/2+q1

z − ǫK/2−q,

where the contour is circumnavigated in the counterclockwise direction, and includes all of the poles off .

5.11. Calculate this contour integral by closing the contour at infinity.

-19

Solution 5.8. We can calculate this integral using the residue theorem. Taking a contour clockwise at|z| → ∞, we have

β

2πi

∮

dzf (z)1

Ω

∑

q

1

iωn − z − ǫK/2+q1

z − ǫK/2−q

= β∑

zF

Res [f, zf ]1

Ω

∑

q

1

iωn − zF − ǫK/2+q1

zF − ǫK/2−q

+β

Ω

(

∑

q

f(

z = ǫK/2−q)

iωn − ǫK/2−q − ǫK/2+q+∑

q

f(

z = −iωn + ǫK/2+q)

−iωn − ǫK/2+q − ǫK/2−q

)

=∑

zF

1

Ω

∑

q

1

iωn − zF − ǫK/2+q1

zF − ǫK/2−q

+β

Ω

(

∑

q

nK/2−q

iωn − ǫK/2−q − ǫK/2+q+∑

q

nK/2+q

−iωn − ǫK/2+q − ǫK/2−q

)

(66)

wherezF = (2m+ 1) π/β are the poles off (z), and having identifiedf (z = ǫk) as the Fermi-Dirac

distribution we rewrite it asnk.

All that’s left is to identify that the integrand goes as1/z2 and so vanishes asz → ∞, so that the integrandvanishes along the contour and the integral is identically zero. Thus we find the sum over frequenciesz is

equal to (minus) the term in the last line,

R0 (K, iωn) =1

Ω

∑

q

nK/2+q

iωn + ǫK/2+q + ǫK/2−q−

nK/2−q

iωn − ǫK/2−q − ǫK/2+q. (67)

At this point we are almost done. We have an explicit expression forR0 in terms of a sum over momenta

q. We have a further expression for the free energy in terms of the sum overk andω of a function ofR0.

These sums are readily done numerically. Even without doingthem, however, we can see some interesting

physics. In particular, the free energy has a singularity when1− βV0R0(K,iωn)Ω = 0 for some term in the sum.A singularity in the free energy corresponds to a phase transition. It turns out that this is the pairing phase

transition, and the term which first diverges is the one withK = 0 andiωn = 0. This gives the ”Thouless”

criterion for superconductivity:1

V0=

1

ΩR0(0, 0).

5.12. Rewrite this expression in terms of the scattering length.

Solution 5.9. Having foundV0 = 4πm as we have

m

4πas=

1

ΩR0 (0, 0) . (68)

Unlike the two particle problem, we actually find that this equation can be solved for both positive and

negative scattering length. For positive scattering length, this just corresponds to BEC of pairs. For negative

scattering length this instability is a non-trivial many-body effect. The remarkable result of Nozieres and

-20

Schmidt-Rink is that these two forms of condensation are continuously connected. In a future homework

we might explore this further: taking the derivative of the free energy with respect toµ in order to get an

expression for the density. Regardless, at this point you probably have enough knowledge to make a stab at

reading the original paper.

-21