Questions and Answers in Quantum Theory of Solids · Questions and Answers in Quantum Theory of Solids Jimmy Qin and Yunchao Zhang Spring 2019 Yunchao and I wrote this document based

Questions and Answers in Quantum Theory of Solids

Jimmy Qin and Yunchao Zhang

Spring 2019

Yunchao and I wrote this document based on the class Physics 295b: Quantum Theory of Solids(Harvard), lectured by Eugene Demler.

Contents

1 Useful links 2

2 Useful Formulae 2

2.1 Response functions and Fermi-Liquid theory . . . . . . . . . . . . . . . . . . . . . 2

2.2 Second Quantization and Gaussian Approximations . . . . . . . . . . . . . . . . . 3

2.3 Quantum Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.4 Superconductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Response and correlations 7

4 Landau Fermi-liquid theory 12

5 Second quantization: Equations of motion and Hartree-Fock approximation 16

6 Response and correlations, reprise: RPA and screening 18

7 Magnetism: Hubbard and Anderson models 23

8 Magnetism: Spin waves 27

9 Magnetism: Theories of indirect coupling 28

10 Superconductivity: BCS and Landau theory 32

1

Physics 295b Questions and Answers

11 Superconductivity: Unconventional SC and Josephson effect 39

1 Useful links

• http://www.phys.lsu.edu/~jarrell/COURSES/SOLID_STATE/Chap6/chap6.pdf

• http://eduardo.physics.illinois.edu/phys561/FL.pdf

• https://www.quantumoptics.ethz.ch/fileadmin/content/lectures/quantumgases/FrontiersOfQuantumGasResearch/

lecturePaivi8.pdf

• https://courses.physics.ucsd.edu/2014/Spring/physics239/LECTURES/CH03.pdf

• http://large.stanford.edu/courses/2008/ph373/

• https://www.physics.uu.se/digitalAssets/433/c_433300-l_1-k_rkky.pdf

• https://en.wikipedia.org/wiki/Peierls_transition

2 Useful Formulae

2.1 Response functions and Fermi-Liquid theory

1. Probe coupling

Hint =∑

V (ri −R) =∑

Vqρ†qe−iqR

2. Excitation Probability and Dynamic Form Factor

Pqω = 2π|Vq|2∑n

|(ρ†q)no|2δ(ω − ωno)︸︷︷︸S(q,ω)

3. Density-density response function

χqω =∑k

|(ρq)k0|22ωko

ω − ωko + iη

4. Causality and other relations

χ′(q, ω) =

∫ ∞0

dω′ S(q, ω′)P(

2ω′

ω2 − ω′2

)= − 1

π

∫dω′ χ′′(q, ω′)P

(1

ω − ω′

)χ′′(q, ω) = −π[S(q, ω)− S(q,−ω)] =

1

π

∫dω′ χ′(q, ω′)P

(1

ω − ω′

)χ(q, t) =

∫dt′ χ(q, ω)e−iωt

′= 0 ∀t < 0

2

http://www.phys.lsu.edu/~jarrell/COURSES/SOLID_STATE/Chap6/chap6.pdf

http://eduardo.physics.illinois.edu/phys561/FL.pdf

https://www.quantumoptics.ethz.ch/fileadmin/content/lectures/quantumgases/FrontiersOfQuantumGasResearch/lecturePaivi8.pdf

https://www.quantumoptics.ethz.ch/fileadmin/content/lectures/quantumgases/FrontiersOfQuantumGasResearch/lecturePaivi8.pdf

https://courses.physics.ucsd.edu/2014/Spring/physics239/LECTURES/CH03.pdf

http://large.stanford.edu/courses/2008/ph373/

https://www.physics.uu.se/digitalAssets/433/c_433300-l_1-k_rkky.pdf

https://en.wikipedia.org/wiki/Peierls_transition


5. Free electron properties

n =1

3π2p3F

C =1

3π2ν(0)T

ν(0) =m∗pFπ2

S(q, ω) =∑pσ

npσ(1− np+q,σ)δ(ω − ωpq)

6. FL free energy

F − F0 =∑

(εp − µ)δnp +1

2V

∑fpσp′σ′δnpσδnp′σ′

f ↑↑ = fS + fA, f ↑↓ = fS − fA

Stability (Pomeranchuk relations): Fs(a)l > −(2l + 1)

7. Quasiparticle energy

εpσ = εpσ +∑

fpσp′σ′δnp′σ′

8. Transport equation

∂tnp +∇rnp · ∇pεp −∇pnp · ∇r εp = I[np]

=⇒ (~q · ~vp − ω)δnpσ − ~q · ~vp∂no

∂εp

∑σ′

∫d3p′

2π3fpσp′σ′δnp′σ′ = I[np]

9. Collective modes: Take δnpσ = δ(εp − µ)vFuσp , so

(cos θ − λ)u(θ, ϕ) +cos θ

4π

∫dΩ′ F (ξ)u(θ′, ϕ′) = 0

10. First soundλ

2log

λ+ 1

λ− 1− 1 =

1

F0

F0 > 0, real solution−1 < F0 < 0, complex λF0 < −1, imaginary λ

2.2 Second Quantization and Gaussian Approximations

11. Basic definitions

Electron field operator: ψ(r) =1√V

∑eiprc†p, with equation of motion

dψ

dt= i[H,ψ]

T = − 1

2m

∫d3rψ(r)†∇2ψ(r)

V =1

2

∑σσ′

∫~r,~r′

V (~r − ~r′)ψ†σ(~r)ψ†σ′(~r′)ψσ′(~r

′)ψσ(~r) =1

2

∑σσ′;~p~p′~q

V (~q)c†~p+~q,σc†~p′−~q,σ′c~p′,σ′c~p,σ.

3


12. HF equations

εjφjσ(r) =

[−∇

2

2m+

∫d3r′ V (r′ − r)

∑iσ′

niσ′φ∗iσ′(r

′)φiσ′(r′)

]φjσ(r)

−∫d3r′V (r′ − r)φjσ(r′)

∑i

niσφ∗iσ(r′)φiσ(r)

13. Jellium model

ε(~k) =~k2

2m− e2

2π

[~k2F − ~k2

klog

(∣∣∣∣kF + k

kF − k

∣∣∣∣)+ 2kF

]=

[2.21

r2s

− 0.916

rs

]Ry

14. HF Correlation

S(r) =1

N

∫d3r′ 〈ψ0|ρ(r + r′)ρ(r)|ψ0〉

g(r) =1

N(N − 1)〈ψ0|

∑δ(r + ri − rj)|ψ0〉

15. Landau Parameters

fpσp′σ′ =∂2E0

∂npσ∂np′σ′= V0 −

1

2Vk−pδσσ′

16. ScreeningStatic:

ϕ(r) =z

rexp(−qFT r), qFT =

(6πρ0e

2

EoF

)1/2

Dynamic:

ε ~E = ~D,1

εqω= 1 +

4πe2

q2χqω

εqω = 1 +4πi

ωσqω

17. RPA

χoqω =2

V

∑ nop − nop+qω − ωopq + iη

χRPA =χo

1− Vqχo, χZZRPA = − χo

1 + V0︸︷︷︸contact int

χo

CDW: χ0 ∝ −ν(0) log1

|2kf − q|

4


2.3 Quantum Magnetism

18. Heisenberg ModelFerromagnetic, J > 0

H = −1

2

∑Jrr′Sr · Sr′ − gµBH

∑Szr

Spin wave: |~k〉 =1√N

∑R e

i~k·~R[

1√2SS−r |0〉

]

Ek − E0 = 2SJ∑R

sin2

(1

2~k · ~R

)+ gµBH

Antiferro, J < 0

∂t~Si = −J∑j

~Si × ~Sj

19. Hubbard Model

H = −t∑

c†iσcjσ + U∑(

ni↑ −1

2

)(ni↓ −

1

2

)+ µ

∑niσ

χ0 = −ν(0) log1

|q −Q|

Dual to Heisenberg for U t with J =4t2

UMF Hamiltonian, assume 〈Szi 〉 = 〈ni↑ − ni↓〉 = S(−)i, εk = −2t(cos kx + cos ky)

H =∑

εkc†kσckσ − UN0

∑k

σzαβc†k+Q,αck,β

Bogoliubov transform so that

ck = ukγck + vkγ

vk , ck+Q = vkγ

ck − ukγvk

=⇒ uk =

[1

2

(1 +

εkEk

)]1/2

, vk = −[

1

2

(1− εk

Ek

)]1/2

Ek =√ε2k + ∆2, ∆ = US

20. Hubbard Self-consistency2

N

∑k∈BZ′

1

Ek=

1

U

∆ ≈ U for U large, ≈ te−t/U for U small

21. Anderson model (U |Vk|, leads to local interaction)

H =∑

εdndσ + Und↑nd↓ +∑

εkc†kσckσ +

∑kσ

(Vkc†dσckσ + V ∗k c

†kσcdσ)

∼ J ~S ·∑kk′

c†kα~σαβck′β

5


22. RKKY

H =∑

εkc†kσckσ +

1

2

∑Jn~Inψ†α(rn)~σαβψβ(rn)

∼∑n>m

JnJm~In · ~ImΦ(~rn − ~rm) =4mk4

f

(2π)3

∑n>m

JnJm~In · ~ImFRKKY (2kfrnm)

FRKKY (x) =x cosx− sinx

x4

23. Kondo Problem

H = H0 +J

N

∑kk′

ei(k−k′)RnSiσiαβc

†k′αckβ

From 3rd order perturbation theory, Tk ∼ µ expn

Jν(0), or more accurately (|J |µ)1/2 exp

n

Jν(0)Flow:

dJ

d logD= −2

ν(0)

nJ2, De−n/(2ν(0)J) = fixed

2.4 Superconductivity

24. Phenomenology

F = Fn + α|ψ|2 +β

2|ψ|4 +

1

2m

∣∣∣∣∣(−i~∇− 2e ~A

c

)ψ

∣∣∣∣∣2

+h2

8π

ξT =~2

2m|α|,

1

λ2L

=16πe2ψ2

0

mc2=

32πe2

~2c2Cδ2

0, κ =λ

ξ

25. Cooper’s problem

Vq =2ω(q)|g(q)|2

[ε(k + q)− ε(k)]2 − ω(q)2

−λ = 2ωDe−2/(ν(0)V )

26. BCS Theory

H =∑

ξkc†kσckσ −

1

2

∑Vk,k′,qc

†k+q,σc

†k′σ′ck′+q,σ′ckσ

|ψG〉 =∏k

(uk + vkc†k↑c−k↓)|0〉, 〈c−k↓ck↑〉 = ukvk

1 = V∑ 1

2Ek, ∆ = 2ωDe

−1/(ν(0)V )

We set ∆k =∑

k′ Vkk′uk′vk′ , uk =

[1

2

(1 +

ξkEk

)]1/2

, vk =

[1

2

(1− ξk

Ek

)]1/2

Scales: ∆ ∼ Tc ωD EFCondensation energy: −ν(0)∆2/2Coulomb Renormalization:

Veff = Vph −Vc

1 + ν(0)Vc log(EF/ωD)

6


27. Bogoliubov Theory, define ∆k =∑Vklbl =

∑Vkl〈c−l↓cl↑〉

H =∑

ξkc†kσckσ −

∑(∆kc

†k↑c†−k↓ + ∆∗kc−k↓ck↑ −∆kb

∗k)

=∑

(ξk − Ek + ∆kb∗k) +

∑Ekλγ

†kλγkλ

Self Consistency:

∆k =∑k′

Vkk′〈c−k′↓ck′↑〉 =⇒ 1 =V

2

∑k∈|ξk|<ωD

tanh(βEk)

2Ek

Tc = 1.13ωDe−1/(ν(0)V ) =⇒ ∆(0) = 1.764Tc

Density of states:

ρ(E) = ν(0)E√

E2 −∆2

3 Response and correlations

Question 1. What is a linear response? Explain why the pertubative part of the energy is

Hext =

∫~r

ρ(~r, t)φ(~r, t).

Why is it useful to work in the linear regime?

Answer 1. In linear response, we assume our system responds linearly to the measuring apparatus.Therefore, we assume the system only responds to the probe and not to itself responding. In thisregime, we can obtain exact, theoretical results. We often consider the perturbative part of theenergy that looks like

Hext =

∫~r

ρ(~r, t)φ(~r, t)

because it resembles how an electric potential couples to a charge density. Under such linearresponse, each Fourier component of ρ and φ also acts independently. Therefore, φ(~q, ω) induces

a density fluctuation only at (~q, ω), and we can use χ(~q, ω) =〈ρ(~q, ω)〉〈φ(~q, ω)〉

to characterize the linear

response.

Question 2. Show that for a scattering probe, we have

d2σ

dΩdω∼ S(~q, ω),

where S(~q, ω) is the dynamic form factor. Interpret this experimentally.

Answer 2. We have a probe at R with momentum P , our system in ground state state |0〉. Takingour

Hint =∑q

vqρ†qe−i~q·~R, 〈n, P − q|Hint|0, P 〉 = vq(ρ

†q)no

7


Fermi’s golden rule then gives the scattering rate

P = 2π|vq|2∑n

|(ρ†q)no|2δ(ω − ωno) = 2π|vq|2S(q, ω)

We see that the scattering rate for momentum transfer P and energy loss ω is related to thedynamic form factor by the square of the q Fourier component of the interaction potential.

Question 3. Prove the fluctuation-dissipation theorem,

S(ω) = 2~(1 + nB(ω))χ′′(ω) = coth(ω

2T)χ′′(ω),

where nB(ω) is the value of the Bose-Einstein distribution at energy ω.

Answer 3. At finite temperature, the correlation function must be modified by the appropriateBoltzmann factors so that

S(~q, ω) = Z−1∑mn

e−βEm|(ρ†~q)nm|2δ(ω − ωnm)

Interchanging indices gives

S(~q, ω) = Z−1∑mn

e−βm|(ρ†~q)nm|2δ(ω − ωnm)

= Z−1∑mn

e−βEn|(ρ†~q)mn|2δ(ω + ωnm)

= Z−1∑mn

e−β(En−Em)−βEm|(ρ~q)nm|2δ(ω + ωnm)

= eβωS(−~q,−ω)

By time reversal, S(−~q,−ω) = S(~q,−ω). Then, using that by definition, χ′′(~q, ω) = −π[S(~q, ω)−S(~q,−ω)], we get what we desired, that

S = − 1

2πcoth

( ω2T

)χ′′

Question 4. Explain why the definition of density-density response,

χ(~q, ω) =〈ρ(~q, ω)〉φ(~q, ω)

is reasonable. Show that it can be written in Lehmann representation as

χ(~q, ω) =∑k

|〈k|ρq|0〉|2(1

ω − ωk0 + iη− 1

ω + ωk0 + iη),

where ωk0 := Ek − E0. Here, |0〉 is the ground state and |k〉 is any excited state.

Answer 4. The density is supposed to couple to an external field. For example, in classicalelectromagnetism, the charge density couples to the electric potential, and the energy is

Eext =

∫r

ρ(r, t)φ(r, t). (1)

8


The external field, φ, deforms the density, ρ. The same idea holds in solid state physics, for whichthe external field can be more general than an electric potential.

In fact, the deformation of ρ due to φ is generally nonlinear in φ. However, for small-enough φ,the deformation is linear. If this is the case,

δρ ∝ φ, (2)

then it makes sense to characterize the deformation by the ratio δρφ

. This is the density-density

response. There are response functions even in undergraduate physics, such as χ(q, w) = P(q,ω)E(q,ω)

.

The derivation of Lehmann-form of the response is done later in these notes.

Question 5. Explain why the above density-density response function satisfies causality.

Answer 5. This is true because we took the adiabatic boundary condition that the interactionis turned on very slowly by making He ∝ eηt, causing the system to evolve over an infinitely longperiod of time. This boundary condition ensures that the response follows the probe (AKA theresponse is causal). Mathematically, it is because χ is analytic in the upper half plane (UHP),which leads to the Kramers-Kronig relations. Essentially taking ω → ω+ iη pushes the poles intothe LHP and allows us to include the real axis in the UHP.

Question 6. Show the above density-density response function can be written

χ(t) = 〈[ρ(t), ρ(0)]〉θ(t)

where θ is the Heaviside function. Explain why this real-time form of the correlation also makessense.

Answer 6. We will first rewrite

χ(q, ω) =∑k

(〈0|ρq|k〉〈k|ρ−q|0〉ω − ωk0 + iη

− 〈0|ρ−q|k〉〈k|ρq|0〉ω + ωk0 + iη

)The Fourier transform to r them sets

χ(r − r′, ω) =∑k

(〈0|ρr|k〉〈k|ρr′|0〉ω − ωk0 + iη

− 〈0|ρr′|k〉〈k|ρr|0〉

ω + ωk0 + iη

)Then the Fourier transform to real time (residue theorem) leads to

χ(r − r′, t) = −ieηt∑n

[〈φ0|ρ(r)|k〉〈k|ρ(r′)|φ0〉eiωk0t − 〈φ0|ρ(r′)|k〉〈k|ρ(r)|φ0〉e−iωk0t

]θ(t)

= −ieηt〈φ0|[ρ(r, t), ρ(r′, 0)]|φ0〉θ(t)

The step function ensures causality.

Question 7. How do commutators pop up in the linear response regime?

9


Answer 7. We will work in the interaction picture so that i∂t|φ(t)〉 = eiHtV e−iHt|φ(t)〉 ≡ VI |φ(t)〉.integrating this gives

|φ(t)〉 = |φ0〉 − i∫ t

∞dt′ VI(t

′)|φ(t′)〉 ≈ |φ0〉 − i∫ t

∞dt′ VI(t

′)|φ0〉

by the Jimmy method. Then, we can calculate that to linear order in the interaction (note alloperators are in the interaction picture here)

〈ρ(r, t)〉 = 〈φ0|ρ|φ0〉 − i∫ t

∞dt′ eηt

′〈φ0|[ρ(r, t), VI(t′)]|φ0〉

Then we can find the change in ρ to be

〈δρ〉 = −ieηt∫ t

∞dt′ eη(t′−t)〈φ0|[ρ(r, t), VI(t

′)]|φ0〉

= −ieηt∫d~r′

∫ t

∞dt′ eη(t′−t)〈φ0|[ρ(r, t), ρ(r′, t′)]|φ0〉U(r′, t′)

Note that this is valid for VI =∫dr ρU . Then, by definition, the correlation function is the kernel

of this integralχ(r, r′, t− t′) = −ieη(t′−t)〈φ0|[ρ(r, t), ρ(r′, t′)]|φ0〉θ(t− t′)

The step function comes in and allows us to extend the limits of the integral over dt to ∞.

Another way to say it is

ρH(t) = (T e−i∫Hint(t)dt)ρI(t)(T e

i∫Hint(t)dt) ≈ ρI(t) + i

∫ t

−∞[ρI(t), ρI(t

′)]φ(t′)dt′.

Here, H means Heisenberg picture and I means interaction picture.

Question 8. Explain why different correlation functions are, in fact, intimately related. Consider:optical conductivity, scattering, density-density, etc.

Answer 8. One way to think of this is that there are only ”so many things to measure.” In fact,the correlation function measured by scattering is the same as the density-density correlation.What comes up from a scattering experiment is the absolute value of the square of the Fouriertransform of the density-density correlation.

The detector detects the intensity at a given angle θ. If the scattering is elastic and the projectilehas momentum k, then θ and the wavevector q of the transferred momentum is easy to get,

q = 2k sin θ. (3)

Why does this give us something like the density or density-density correlation? Consider thephase shift, like from elementary optics. Suppose the positions of two scatterers are r1, r2. Themomentum in is p0 and the momentum out, what hits the detector, is pf . The phase shift betweenthese paths (from the source to the scatterer to the detector) is(

p0 ·(r1−rsource)+pf ·(rdetector−r1)

)−(

p0 ·(r2−rsource)+pf ·(rdetector−r2)

)= q ·(r2−r1). (4)

10


Here, q = pf − p0. Therefore, the incoming intensity is the time-averaged quantity

dσ

dΩ=

⟨∣∣∣∣∑j

bjeiq·rj

∣∣∣∣2⟩. (5)

bj is the amplitude associated with the jth scatterer. If bj = b for all j (the scatterers are allidentical), then we have

dσ

dΩ= |b|2NS(q) (6)

where S(q) is the structure factor

S(q) :=1

N

⟨∣∣∣∣∑j

eiq·rj∣∣∣∣2⟩ =

1

N

⟨∣∣∣∣ ∫r

eiq·rρ(r)

∣∣∣∣2⟩ (7)

As promised, the scattering intensity I(q) is the square of the magnitude of the Fourier transformof the density ρ(r). What about the other interpretation, the one with correlation function?

∣∣∣∣ ∫r

eiq·rρ(r)

∣∣∣∣2 =

(∫r

e−iq·rρ∗(r)

)(∫r′eiq·r

′ρ(r′)

)=

∫r,r′

eiq·(r′−r)ρ(r)ρ(r′). (8)

Wow! The two-point correlation magically popped out. If the system is translationally invariant,we can perform one of the integrals (and add an expectation) and rewrite∣∣∣∣ ∫

r

eiq·rρ(r)

∣∣∣∣2 = V

∫r

eiq·r〈ρ(0)ρ(r)〉 = VF

(〈ρ(0)ρ(r)〉

). (9)

There are various definitions of the two point correlation to get rid of delta functions which popup when the scatterer sees itself, etc.

How about optical conductivity? Recall that the response function χ(q, ω) measures the responseρ(q, ω) of the electron fluid to an external potential φ(q, ω). In fact, we can think of the electricalpolarization P as a response to the displacement field D. (To make this more exact, we refer tothe later answer (18).) Thus, χ(q, ω) gives the dielectric constant ε(q, ω).

Why does this give the optical conductivity? Suppose we know

J(ω) = σ(ω)E(ω),D(ω) = ε(ω)E(ω). (10)

The Maxwell equation (Ampere’s law) can be written

∇×H =1

c

∂D

∂t+

4π

cJ =

ε′

c

∂E

∂t(11)

where

ε′ = ε+4πiσ

ω. (12)

The dielectric constant,n =√ε′ = nR + ik (13)

is related to the optical properties.

In conclusion, an optical experiment can measure the reflectivity R or absorption coefficient α.This enables one to reconstruct the index of refraction, n, and hence the permittivity ε′. Oncethis is done, you can combine ε′ with the response ε(q, ω) to solve for the optical conductivity,σ(q, ω).

11


4 Landau Fermi-liquid theory

Question 9. What is a quasiparticle? Explain why the quasiparticle and real-electron distributionfunctions have different shape. Explain why the change in quasiparticle number equals the changein real-electron number.

Answer 9. In Landau FL theory, a quasiparticle is a dressed electron or an excitation of thesystem that is adiabatically connected to a free electron. The change in quasiparticle numberequals the change in real-electron number by construction-it is a fundamental assumption of thetheory and the defintion of a normal Fermi liquid. The quasiparticle distribution function willbe different from the real electron one when there is any excitation in the system δnpσ, whichwill occur when interactions are turned on. However, note that bare particles in a Fermi gas andquasiparticles in a Fermi liquid have the same distribution in momentum space (p and σ remaingood quantum numbers), and both follow Fermi statistics.

Question 10. When is Landau Fermi-liquid theory likely to be valid? Argue for a relationshipamong the following quantities: T, εF , µ. What happens outside the region of validity?

Answer 10. Landau FL theory is an effective expansion in low T/εF , which is the same as anexpansion in low number of quasiparticle excitations. Outside this regime, we will need to takeinto account higher order effects, and phenomena such as phase transitions can occur.

Question 11. Explain the density expansion of the free energy,

F − F0 =∑~p

(ε~p − µ)δn~p +1

2V

∑~p~p′;σσ′

f~pσ,~p′σ′δn~pσδn~p′σ′ .

Show that both the linear and quadratic corrections are O(T 2). Show that the correspondingquasiparticle energy is

ε~pσ = ε~p +1

V

∑~p′σ′

f~pσ,~p′σ′δn~p′σ′

Answer 11. The first term in the free energy expansion just comes from considering the con-servation of quasiparticle number. We note that δn will only be nonzero in a window of δ ∼ Taround the Fermi surface. Therefore, when δn 6= 0, we see that (εp − µ)δnp is actually of orderδ2. Therefore, we need to push our expansion one term further and include a quadratic term thatcaptures interactions between quasiparticles. Then, it is evident that the energy of adding a singlequasiparticle is ε~pσ = ε~p + 1

V

∑~p′σ′ f~pσ,~p′σ′δn~p′σ′ .

Question 12. Explain why the definition of symmetric and antisymmetric parameters,

f s~p~p′ and fa~p~p′ ,

makes sense. Define the Landau parameters for each angular momentum channel l as

F a,sl = ν(0)fa,sl

and express F a,sl in terms of fa,s~p~p′ .

12


Answer 12. There are two kinds of electrons: ↑ and ↓. There are also two kinds of forces orinteractions – those which treat ↑ and ↓ the same, and those which treat ↑ and ↓ differently. So,f s tells us the effect of those forces and interactions which are spin-symmetric (treat ↑ and ↓ thesame), and fa tells us how the responses of ↑ and ↓ are different.

The interactions are generally not isotropic, so we classify them according to angular momentum.The intuition is like the quasiclassical picture of scattering theory in atomic processes, wheresemiclassically, a projectile in the l-channel orbits the scattering center l times before exiting the”bound state.” Since Landau Fermi Liquid theory is a theory about interactions, it isn’t surprisingthat there are also angular momentum channels in this case.

We decompose

F s,a(cos θ) =∞∑l=0

F s,al Pl(cos θ). (14)

Actually, some people use a different convention with an extra (2l + 1).

The way to think about this is that the Landau correction in the free energy looks kind of like

Fint ∼ fσσ′(cos θ)npσnRpσ′ (15)

where R rotates by an angle θ. So, the energy of interaction has some angular dependence.

Question 13. Derive the magnetic susceptibility of a Fermi liquid. Without any computation,argue that F a

0 is the only possibly relevant Landau parameter. (For reference, χ = 1V∂M∂B

, whereM is the total magnetization of the system.)

Answer 13. We argue that F a0 is the only relevant parameter because l = 0 couples to the density

and the magnetic susceptibility arises from an interaction antisymmetric on spin. In other words,the energy is

Hmag ∼ −∫r

(n↑(r, t)− n↓(r, t))Bz. (16)

Obviously, this couples to the antisymmetric component of the density. The density is isotropicin k-space, so the Landau interactions will (for example, for Bz > 0) make the Fermi sphere for ↑get bigger and that for ↓ get smaller, but the spheres will stay spherical.

After turning on a field, we will have a new local equilibrium

npσ = no(εpσ − µ− gµBσH) =⇒ δnpσ = ∂εno(εpσ − µ− gµBσH − εpσ)

Our condition then becomes

δnpσ − ∂εno1

V

∑fpσp′σ′δnp′σ′ = −gµBHσ∂εno

Putting in an ansatzδnpσ = Csign(σ)∂εn

o

gives

χ =(gµB

2

)2 ν(0)

1 + F a0

13


Question 14. Show the effective mass of a quasiparticle is

m∗ = m(1 +F s

1

3).

Why is it the l = 1 channel that contributes?

Answer 14. The l = 1 channel contributes because the energy (as shown below) changes bysomething like p · v = |p||v| cos θ. Since there is a P1(cos θ) = cos θ, we are in l = 1. Going to amoving frame at velocity ~v = ~q/m, the Hamiltonian transforms as

H ′ = H − ~P · ~v.

That’s because εf − ε0 = (P + v)2−P2 = 2P · v. Then, for a translationally invariant system, wehave

~j = −∂E∂~q

∣∣∣∣~q=0

The particle current for a single particle of momentum ~k is then

~jk = −∂ε~k−~q∂~q

∣∣∣∣~q=0

= ~vk +1

V

∑fkσk′σ′~vk′∂εn

o = ~vk

(1 +

F s1

3

)≡~k

m

Question 15. In a FL, why the specific heat is given by the free electron value,

C =π2

6ν(0)T 2

at low T?

Answer 15. The reason this is true is because the integral∫dp δnpp

2

is of order T 2 in any direction of momentum space. Then, the interaction term will be of orderT 4, so we just have to consider the free electron term in the free energy.

Question 16. Describe the transport equation for quasiparticles, i.e. identify the different termsand what they represent. What is the “collisionless approximation” and why is it useful?

Answer 16. In order to describe the transport properties, Landau considered the quasiparticles asindependent, described by a classical Hamiltonian ε. Then, the Boltzmann equation for transportgives

∂tnp +∇rnp · ∇pεp −∇pnp · ∇r εp = I[np]

The Boltzmann equation arises from considering the flow of particles through small volumes inphase space. The second term takes the velocity of the quasiparticle into account and the thirdterm is like a diffusion force term. The RHS is the collision integral. Keeping first order terms inδn leads to

(~q · ~vp − ω)δnpσ − ~q · ~vp∂no

∂εp

∑σ′

∫d3p′

2π3fpσp′σ′δnp′σ′ = I[np]

The collisionless regime allows us to take the collision integral on the RHS to be zero, whichis important because the collision integral it makes any computation notoriously difficult. Thisregime is physically valid at low temperatures. You can make an argument based on comparingthe de Broglie wavelength (which is temperature-dependent) to the effective cross-section for ascattering process.

14


Question 17. What is zero sound? What makes it different from first sound? Describe, inpictures, the effects of both zero and first sound on the shape of the Fermi surface. Explain whythe formula for first sound

ω2 =q2v2

F

3(1 + F s

0 )(1 +F s

1

3)

makes sense (i.e. explain what each term means).

Answer 17. Zero sound ignores the particle collisions, so holds for a different regime of ω. Infirst sound, the fermi surface maintains its spherical shape. In first sound, the Fermi surfaceends up looking like a guitar pick. For first sound, formula makes sense because it is literallyidentical to the equation for macroscopic sound velocity obtained from the compressibility by theusual hydrodynamic arguments. Physically, taking u = a+ b cos θ in first sound corresponds to adensity oscillation superimposed on a uniform translation of the fluid: this is just what we expectto find for an ordinary acoustic wave.

So, zero sound oscillations have l = 0 and first sound oscillations have l = 1; the formula givenin the question has the l = 1 symmetry, so it makes sense. For the original paper, see http:

//www.jetp.ac.ru/cgi-bin/dn/e_006_01_0084.pdf. Actually, both zero and first sound arealways present, but zero sound dominates at low temperatures; see https://en.wikipedia.org/

wiki/Zero_sound. So, first sound is the usual acoustic wave. Zero sound is not a wave inthe particles’ positions, but rather a wave in the distribution function. It’s true that the wavepropagates through space, but the particles don’t really ”oscillate” around their positions like infirst sound. For example, the zero sound propagation is

δf(p,x, t) = δ(εp − EF )ei(k·x−ωt)ν(p). (17)

(Actually, the δ-function is only approximate, and it should really be a smeared-out δ-function ofwidth ∼ kT about the Fermi surface.) Basically what this means is that at each point x, there isa Fermi sphere of radius kF . In zero sound, the sizes of the Fermi spheres at each x oscillate; if Idefine a local radius of the Fermi sphere kF (x), then kF (x, t) ≈ kF + δkF e

i(k·x−ωt).

So, there is no classical analogue to zero sound. It is a quantum-mechanical phenomenon – aspatiotemporal wave of spherically-symmetric deformations of the Fermi sphere, or equivalentlyof the distribution function f .

Question 18. What is the zero sound instability? When does it occur?

Answer 18. The zero sound instability comes from when the transport equation has two imag-inary roots, one of which would correspond to an exponentially growing wave. In such a case,the state which we assumed to be the ground state spontaneously evolves toward another state,containing permanent fluctuations of the quantity associated with the unstable collective mode.For zero sound, we can look for an imaginary root λ− iα which becomes equivalent to

α−1 arctan1

α=

1

F0

=⇒ F0 < −1

Therefore, we see if the interaction (from the Landau parameters) is too attractive, damped zerosound is replaced by an unstable mode. Physically, the stability conditions for l = 0 insure thatthe compressibility and spin susceptibility are positive. The stability condition for F s

1 ensures theeffective mass is positive.

15

http://www.jetp.ac.ru/cgi-bin/dn/e_006_01_0084.pdf

http://www.jetp.ac.ru/cgi-bin/dn/e_006_01_0084.pdf

https://en.wikipedia.org/wiki/Zero_sound

https://en.wikipedia.org/wiki/Zero_sound


Question 19. Suggest experiments to determine the following Landau parameters: F s1 , F a

0 , F s0 .

Answer 19. From above, we see that we can probe F s1 by probing m∗ (measuring specific heat).

F a0 corresponds to spin susceptibility, which can be measured by putting our system in a magnetic

field. F s0 , or compressibility, can be measured through sound velocity. All of these can be measured

through the dynamic form factor and probing collective modes.

5 Second quantization: Equations of motion and Hartree-

Fock approximation

Question 20. What is a Slater-determinant state? What is Wick’s theorem and why does itapply to Slater-determinant states?

Answer 20. A Slater determinant state is a multi-fermionic wave function that is a properlyantisymmetrized product of single-electron wave functions. In second-quantized form, it can bewritten as |n1, n2, · · · 〉. Wick’s theorem says that we can factor operators through contractionsthat conserve number such that

〈c†1c†2c3c4〉 = 〈c†1c4〉〈c†2c3〉 − 〈c†1c3〉〈c†2c4〉

This equivalence only holds for Slater determinant states.

Question 21. Explain why the interaction part of the Hamiltonian looks like

V =1

2

∑σσ′

∫~r,~r′

V (~r − ~r′)ψ†σ(~r)ψ†σ′(~r′)ψσ′(~r

′)ψσ(~r) =1

2

∑σσ′;~p~p′~q

V (~q)c†~p+~q,σc†~p′−~q,σ′c~p′,σ′c~p,σ.

Explain why we chose this ordering of the creation and annihilation operators. What is the physicalinterpretation of this interaction, and why is it the simplest interaction to write down?

Answer 21. This interaction is the simplest form of the two body interaction such that

Vint =∑i≤j

V (ri − rj) =1

2

∫d3r d3r′ V (r − r′)[ρ(r)ρ(r′)− ρ(r)δ(r − r′)], ρ(r) =

∑δ(r − ri)

Using anticommutation relations, this can be put into the form of creation and annihilation oper-ators and Fourier decomposed as desired. A convenient form for the interaction uses the densityoperator in momentum space ρq =

∑c†pcp+q such that

Vint =1

2V

∑Vq : ρ(q)ρ(−q) :

Question 22. What is the goal of the equations of motion method? Explain why the Hartree-Fock approximation is necessary to close the equations of motion if the Hamiltonian containsan interaction term. Explain why this is a “generalization” of Wick’s theorem. Interpret theHartree-Fock approximation in terms of a “pseudopotential” or “mean-field theory.”

16


Answer 22. The goal of this method is to show the electron field (or quasiparticle) operatorssatisfy the normal Schrodinger equation,

d

dtψ(r) = i[H,ψ]

For the noninteracting case, ψ =∑clφl(r) =⇒ d

dtψ = −i

∑εlclφl(r). However, turning on the

interaction leads to

[Hint, ψ] = −∑

c†kσ′clσ′cmσ

∫d3r′ V (r′ − r)φ∗kσ′(r′)φlσ′(r′)φmσ(r),

which is no longer closed on φj(r). Therefore, we take∫d3r′ V (r′ − r)φ∗kσ′(r′)φlσ′(r′)φmσ(r) =

∫d3r′ V (r′ − r)〈φ∗kσ′(r′)φlσ′(r′)〉φmσ(r)

−∫d3r′ V (r′ − r)〈φ∗kσ′(r′)φmσ(r)〉φlσ′(r′)

This is a generalized Wick’s theorem, where even without expectation values, we take

c†1c2c3 = 〈c†1c2〉c3 − 〈c†1c3〉c2,

which is an operator equivalence on the manifold of Gaussian states.

Question 23. Consider an interacting gas of N electrons confined in volume V , in a neutralizingbackground. Show that the quasiparticle energy in Hartree-Fock approximation is

ε(~k) =~k2

2m− e2

2π[~k2F − ~k2

klog(|kF + k

kF − k|) + 2kF ].

Answer 23. In the Jellium model, the uniform background of positive charge cancels the Hartree

term. The exchange term, for a single plane wave state φj(r) = 1√Vei~kj ·~r, is

−∫d3r′ φjσ(r)V (r′ − r)

∑i 6=j

φ∗iσ(r′)φiσ(r) = − 1

V)∑i 6=j

∫d3r′ V (r′ − r)niσei(kj−ki)(r

′−r)φj(r)

= − 1

V

∑i 6=j

niσVki−kjφjσ(r)

= −4πe2

∫|k|<kF

d3k

(2π)3

1

|kj − k|

= − e2

2π

[~k2F − ~k2

j

kjlog

(∣∣∣∣kF + kjkF − kj

∣∣∣∣)+ 2kF

]Adding the kinetic energy then gives the desired energy for quasiparticle j.

Question 24. Derive the Landau Fermi-liquid parameters from the Hartree-Fock approximationon charged liquid. Show that it is

f~pσ,~p′σ′ = V (~0)− V (~p− ~p′)δσσ′ , or

f s~p,~p′ = V (~0)− 1

2V (~p− ~p′), fa~p,~p′ = −1

2V (~p− ~p′).

If V is taken to be the Coulomb interaction, what problems arise and why?

17


Answer 24. The ground state in HF is simply the Fermi Sea, so

E0 =∑

εpnpσ +1

2V

∑V0〈c†pσc

†kσ′ckσ′cpσ〉+

1

2V

∑Vq〈c†(p+q)σc

†(k−q)σ′ckσ′cpσ〉

=∑

εpnpσ +1

2V

∑V0npσ(nkσ′ − δkpδσσ′) +

1

2V

∑Vqnp+q,σnpσ

The Landau parameters then emerge as

fpσp′σ′ =∂2E0

∂npσ∂np′σ′= V0 −

1

2Vk−pδσσ′

The unscreened Coulomb interaction is long-range and can lead to unphysical results such as avanishing density of states at the Fermi energy.

Question 25. Describe the dependence on σ, σ′, r of the pair correlation function gσσ′(r),where σ, σ′ are the spins of the two electrons in question and r is the (scalar) distance betweenthem. Explain how this follows from the Hartree-Fock approximation.

Answer 25. The pair correlation function

g(r) =1

(N)(N − 1)〈φ0|

∑i 6=j

δ(r + ri − rj)|φ0〉 =1

2(g↑↑ + g↑↓)

In HF, g(0) = 1/2 and g(∞) = 1. We see g↑↑(0) = 0 due to the Pauli exclusion principle, enforcedby our Slater determinant wave functions, but g↑↓(0) = 1 because particles of different spin areuncorrelated in HF.

6 Response and correlations, reprise: RPA and screening

Question 26. Describe the different limits in which the following are valid: Thomas-Fermi screen-ing, plasma oscillations, zero sound, first sound. Provide a physical description of each phe-nomenon.

Answer 26. Thomas Fermi screening is a static limit ω = 0, in a high electron density limit (sothat electron interactions are weak), and when inhomogeneities correspond to distances which arelong compared to the interparticle spacing (assumed local relationship between εF and density ρ).Plasma oscillations can exist in all regimes, but the familiar formula we derived for ω2

p is only truein the q → 0 limit.Zero sound is an FL collective excitation that represents the high frequency component of normalsound. It holds in the collisionless regime, in which ω ν. Collisions disrupt the self-consistentfields responsible for collective modes and zero sound is damped. Ordinarily, collective modescorrespond to a single m, but a mixture of l channels. Zero sound is the longitudinal (m=0)symmetric mode that couples to the total density (l=1).First sound is a hydrodynamic mode, in the regime ω ν. Normally, a full calculation of firstsound requires a suitable collision intgral, but in the super small ω limit, we can guess that thedisplacement of the Fermi surface u will only contain l = 0 and l = 1 components, for those coupleto density and current, which are unaffected by collisions. Then, the longitudinal mode (m = 0)gives what we discussed in class.

18


Question 27. Derive the form of Thomas-Fermi screening,

φ(r) =Z

re−qTFr, where q2

TF =6πρ0e

2

E0F

.

Explain heuristically the dependence of qTF on ρ0, e, E0F .

Answer 27. Thomas Fermi is at its core a theory of static screening. Putting an external chargeof z at the origin also leads to an induced polarization charge e〈ρ〉 so that

∇2ϕ = −4πzδ(r) + e〈ρ(r)〉

The approximation comes in when we assume that the chemical potential of the electrons is dueto the potential energy (EM) and the noninteracting kinetic energy, such that εoF (r) = εoF − eϕ(r).This leads to a fluctuation in the local particle density as

n =(2mεoF )3/2

3π2

in a free electron gas. To first order,

〈ρ(r)〉 =3

2

neϕ(r)

εoF=⇒ (q2 + q2

TF )ϕq = 4πz

from which the results follow. We see that charge is effectively screened within a distance of orderq−1TF . The interaction between the electrons acts to bring about screening; the kinetic energy of

the electrons, which represents their essentially random motion, opposes it.

Question 28. Derive the form of plasma oscillation,

x = −ω2px, where ω2

p =4πρe2

m.

What does the plasma frequency represent?

Answer 28. We displace a slab of charge of thickness d by a distance x. This will lead to aconstant electric field in between the two ends of the slab

E = −4πρex

An electron inside obeysmx = eE = −mω2

px

and the result follows. Such a simple derivation neglects altogether the random motion of theelectrons, so this is only justified for q → 0. The plasma frequency is a collective oscillation of theelectrons.

Question 29. Let ρe(~r, t) be the particle-number density of free (external) charged particles andρ(~r, t) be the particle-number density of electrons. Starting from the definitions

∇ · ~D = 4πZρe(~r, t),∇ · ~P = 4πe〈ρ(~r, t)〉, ~E = ~D + ~P ,

show that for suitable choice of χ(~q, ω), we can write

~E(~q, ω) =1

ε(~q, ω)~D(~q, ω), where

1

ε(~q, ω)= 1 +

4πe2

q2χ(~q, ω). (18)

Why does this choice of χ(~q, ω) make sense? What does it represent?

19


Answer 29. From the definitions, we have that ∇ · ~E = 4πZρe(~r, t) + e〈ρ(~r, t)〉, so that

i~q · ~D(~q, ω) = 4πZρe(~q, ω), i~q · ~E(~q, ω) = 4πZρe(~q, ω) + e〈ρ(~q, ω)〉

Now we make our key assumption: linear response such that ~E(~q, ω) = ~D(~q, ω)/ε(~q, ω). It isevident that

1

ε(~q, ω)= 1− 4πie〈ρ(~q, ω)〉

~q · ~D(~q, ω)= 1 +

e〈ρ(~q, ω)〉zρe(~q, ω)

= 1 +4π2

q2

e〈ρ(~q, ω)〉Vqρe(~q, ω)

Identifying Vqρe(~q, ω) = ϕ, we get

1

ε(~q, ω)= 1 +

4π2

q2χqω

Question 30. Describe logically the calculation of χ(~q, ω) or ε(~q, ω) in the RPA regime. Thischoice of χ makes sense as it would be the response of a neutral system with the same properties.

Answer 30. The Hamiltonian has three parts:

H = Hkin +Hint +Hext.

Since both Hkin,Hext ∼ c†c, they give linear equations of motion and we will not need to makeapproximations when we compute the equation of motion for ρ. Thus, we compute the equationof motion for the non-interacting Hamiltonian: i∂tρpq,σ = [ρpq,σ,Hkin +Hext]. This gives the exactresponse function χ0, which is called the Lindhardt function.

We now add the interactions, which do not give linear equations of motion, and use the “time-dependent Hartree approximation,” which takes only the Hartree terms, because those are themost badly divergent. The Hartree approximation is equivalent to introducing another part to theexternal potential, such that the equation of motion is again linear in ρ.

The result is

χ(~q, ω) =χ0(~q, ω)

1− V (~q)χ0(~q, ω)=⇒ εRPA = 1− V (~q)χ0(~q, ω).

It is the behavior of the denominator in χ(~q, ω) that leads to possible divergences and hence plasmaoscillations - when the magnitude of the response blows up.

Question 31. Explain why 〈ρpq,σ〉 = 〈ρq〉 = 0 in a system with complete translational symmetry.Argue that the Hext term breaks the translational symmetry and makes 〈ρq〉 6= 0 possible.

Answer 31. Here are two ways to think about it. Let the ground state of a symmetric system be|Ω〉. Because ρq moves particles to non-symmetric positions in ~k-space, clearly

|Ω〉 and ρq|Ω〉

must be orthogonal, so 〈ρq〉 = 0.

Another way to think about it is to note that ρq is an operator which subtracts momentum ~q from

the system, so ρq ∼ e−i~q·~r. Shifting the entire system ~r → ~r + ~R sends

ρq → ρq × e−i~q·~R.

This doesn’t make sense, because 〈ρq〉 is supposed to be translationally-invariant. So, 〈ρq〉 = 0.

20


Question 32. Heuristically, explain how εRPA returns the Thomas-Fermi screening and plasmaoscillations in different limits. What are the respective conditions on εRPA in each case?

Answer 32. We first note that with zero external charge,

ε(~q, ω)~q · ~E(~q, ω) = 0, i~q · ~E(~q, ω) = 4π〈ρ(~q, ω)〉

For plasma oscillations, ε(~q, ω) = 0, and we can have a nonzero charge density and electric field. Inthe q → 0 limit, ε(0, ω) = 1−ω2

p/ω2, which yields the plasmon as a zero of the dielectric constant.

In the Thomas Fermi approximation, with ω = 0 and qr0 1, we get

ε(~q, 0) = 1 +q2FT

q2

which is equivalent to

ϕq =4πz

q2 + q2FT

Question 33. Give an interpretation of the RPA approximation in terms of Feynman diagrams.Explain why it is a non-perturbative approximation, when it is useful, and when it fails to capturethe important divergences.

Answer 33. Recall that correlation functions (such as the RPA) are like those kinds of ”eye”diagrams. It was shown by Gell-Mann and Brueckner that in fact, RPA is the summation of thefollowing chain of Feynman diagrams:

Obviously, this is non-perturbative since it is an infinite summation. What’s more, the series itselfis divergent but its sum is convergent, or something like that. (See the paper by Gell-Mann andBrueckner.)

RPA is the first-order correction in 1/N , where N is the number of types of fermions in the system.In the electron gas, N = 2. RPA is exact for N → ∞ and holds in the limit of large particledensity. For more, see Altland and Simons, page 216.

Question 34. Give a physical explanation of the following instabilities: Stoner instability andcharge density waves (CDW). Explain why the former is considered macroscopic while thelatter is considered microscopic.

21


Answer 34. Stoner instability is the development of ferromagnetic order (instability) in a solid.For a system with contact interactions, the criterion for such an instability in the q → 0 limit isν(0)V0/2 > 1. This instability corresponds to a divergence in the RPA correlation function

χspin = − χ0

1 + V0χ0

The charge density wave arises due to a gap opening up in the band structure, which generatesthe potential that leads to a periodic density modulation. In 1 dimension, this occurs no matterthe interaction, so Fermi Liquid theory fails in 1 dimension.

Question 35. Derive the Stoner criterion in one dimension from the Hubbard model.

Answer 35. At half-filling and large U , the Hubbard model leads to antiferromagnetism. Weshow here that the Hubbard model can also lead to ferromagnetism, via the Stoner criterion.Let 〈n↑〉 = 1

2n + m, 〈n↓〉 = 1

2n −m. Hartree-Fock approximation on the interaction term of the

Hubbard model gives ∑i

ni↑ni↓ ≈∑i

〈ni↑〉ni↓ + ni↑〈ni↓〉 − 〈ni↑〉〈ni↓〉.

The diagonalization of the energies for ↑ and ↓ are

H↑ =∑k

(εk + U(n

2−m))c†k↑ck↑ and H↓ =

∑k

(εk + U(n

2−m))c†k↓ck↓.

(Originally, H =∑

kσ(εk + U n2)c†kσckσ.) The new maximum kinetic energies are

(εmaxk )↑ = µ+ Um and (εmin

k )↑ = µ− Um.

Thus, the change in kinetic energy is

∆Ekin = (∆Eavg)ν(0)Um =ν(0)

2U2m2.

The change in interaction energy, remembering the scalar term 〈ni↑〉〈ni↓〉, is

∆Eint = (−Um)m+ (−Um)m+ Um2 = −Um2.

Thus, the total chnge in energy is

∆Etot = (ν(0)

2U2 − U)m2 =⇒ ν(0)U

2> 1 ,

the Stoner criterion. This is for small interactions, U .

Question 36. What is the generalized RPA?

Answer 36. While both RPA and GRPA generalize Wick’s theorem, RPA contains only Hartreecontributions (which are the most divergent contributions) and GRPA also contains Fock contri-butions. The Fock contributions are not expressible in terms of the original operators, such as ρq,so GRPA gives equations of motion that are not closed in the operators ρq.

22


Question 37. Show that GRPA implies the possibility of Stoner instabilities. Interpret thisphysically.

Answer 37. Recall from the last question that RPA contains only Hartree (i.e. spin-symmetric)contributions. GRPA contains also the Fock (spin-nonsymmetric) contributions. Since the Stonerinstability is an instability to a magnetized state, obviously we have to include the Fock contribu-tions if we have any hope of getting magnetism.

Indeed, Eugene’s derivation of the response function was for the antisymmetric channel of thedensity response

〈(ρ†qω)a〉 = χzzqωeφaqω, where χzzqω = −

χ0qω

1 + V0χ0qω

(19)

where V0 is the strength of the (assumed point) interaction. Why he called this response func-tion χzz is a true mystery to me. (Or, maybe he just thinks the spins are aligned along the zaxis?) Clearly, this response function can blow up. Because it is a response of the antisymmetriccombination of the densities, near its divergence the system is unstable to the formation of aferromagnetic state.

So, it seems that generally, for spin density waves (SDW) you must include the Fock contributions(i.e. use GRPA). For the charge density waves (CDW) (see the next question), it’s okay to dropthe Fock contributions (i.e. use RPA).

Question 38. Show that the RPA density-density correlation function, χ(~q, ω), implies the pos-sibility of charge density waves (CDW).

Answer 38. We know thatχRPA =

χ0

1− Vqχ0

so that CDW can emerge when 1−Vqχ0 = 0. In 1-D at T = 0, χ0 ∝ log 1|2kF−q|

. With any V , there

will be some temperature at which χRPA diverges (denominator goes to 0), leading to a finite ρqwithout any φext. This will be the state with CDW order.

7 Magnetism: Hubbard and Anderson models

Question 39. In the Heisenberg model H ∝∑〈ij〉 J ~Si · ~Sj, the ferromagnetic state, with all spins

aligned, is an eigenstate of the Hamiltonian. Why is the antiferromagnetic state not an eigenstateof the Hamiltonian?

Answer 39. From SU(2) symmetry of the Hamiltonian, we have [~Stot,H] = 0. This means anyeigenstate should have a well-defined eigenvalue S2

tot. However, the antiferromagnetic state withtwo sub-lattices is a linear combination of |00〉 and |10〉. Hence, it cannot be an eigenstate.

However, in the N →∞ limit, the classical picture of the antiferromagnetic ground state is valid,due to the Anderson tower.

Question 40. The Anderson tower is the following relation:

∂E

∂S2tot

∼ 1

N.

23


Show this is true from the Antiferromagnetic Heisenberg Hamiltonian

H =1

2

∑〈ij〉

~Si · ~Sj.

Argue that the Anderson tower allows us to use the classical picture of the AF ground state, inthe N →∞ limit.

Answer 40. I argue that E(S2tot) ∼ J S2

tot/N. We have S2tot = (

∑r~Sr)

2 ∼ N2s2. However,∑〈rr′〉 ~sr · ~sr′ has approximately N terms in the sum, so H ∼ JNs2. This means H ∼ J S2

tot/N.

In the N →∞ limit, the classical ground state (which is made of a bunch of states with differenteigenvalues of S2

tot) is roughly at the same energy as the quantum ground state. So, we can justuse the classical ground state.

Question 41. Explain why a tight-binding model is not appropriate for sp-band electrons butmay be valid for d-band electrons. Interpret the terms in the Hubbard model energy,

H = −∑〈ij〉

tc†iσcjσ + U∑i

ni↑ni↓,

and argue physically that this should lead to a magnetic ground state.

Answer 41. sp-band electrons are typically thought of as an ”electron sea,” which means theyare highly delocalized. In other words, the free-electron approximation is better for them. d-band electrons have smaller orbital overlap, so they are more localized on individual atoms. Itis therefore more reasonable to use the tight-binding model to describe them, which becomes theHubbard model in the second-quantized regime.

Let’s interpret the terms in the above Hamiltonian. The first is the kinetic or ”hopping” termand the hopping parameter is basically the overlap integral t =

∫~rψi(~r)V(~r)ψi+1(~r) from the tight-

binding model. The second is the Coulomb repulsion of electrons localized on the same ion. Weknow the spins are different because electrons of the same spin can’t occupy the same ionic site,due to Pauli exclusion principle.

Why should this lead to a magnetic ground state? Consider the U → ∞ limit at half-filling, wherethere is exactly one electron per site and its spin can be oriented in either direction. Consider twoneighboring spins. If they are aligned, hopping is impossible. If they are antialigned, there is avirtual process described by second-order perturbation theory (H = H0 + λV):

∆E = −λ2∑γ 6=α

VαγVγαEγ − Eα

.

Here, γ is the intermediate state where a pair of antialigned spins are on the same site. BecauseEγ > Eα =⇒ ∆E < 0, the antiferromagnetic ordering is favored.

Question 42. For the half-filled Hubbard model in d = 2, show that

χ0(~q, ω = 0) ∼ ν(0) log2 1

|~q − ~Q|.

Interpret the singularity physically. (Hint: χ0(~q, ω) =∑

~pn~p−n~p+~q

ω+(ε~p+~q−ε~p)+iε.) Why does this imply the

existence of spin-density waves (SDW)? How to do the calculation? probably too hard...

24


Answer 42. The correlation function blows up at ~q = ~Q. This makes sense, because in the AFground state of the Hubbard model, there is perfect correlation when jumping from a spin tosecond spin diagonally next to the first one. Therefore, the correlation function should blow upfor some diagonal momentum, which is just ~Q.

In fact, the nesting vector ~Q makes the system look quasi-one dimensional. Note that ~Q = (π, π)corresponds to a real-space vector (2a, 2a). This periodicity is not the fundamental periodicity inthe system, which would be (a, a). It turns out that it is energetically favorable to introduce a gapin the spectrum by dimerizing the two bonds in the real-space vector (2a, 2a): making one of thebonds longer and one shorter, but preserving their total length. This is why the nesting vector is(π, π) but not (2π, 2π) as we would expect just by looking at the symmetry of the lattice. For more,see https://www.quora.com/How-can-charge-density-wave-be-explained-using-the-Fermi-nesting-wave-vector.

Because the Lindhardt function blows up, this implies the original Fermi surface was unstable.The true ground state is the one with a static spin-density wave, which looks simply like the AFground state.

Question 43. Show that the Heisenberg model is a low-energy limit of the Hubbard model. Inother words, show that for large U , the Hubbard model

H = −∑〈ij〉

tc†iσcjσ + U∑i

ni↑ni↓ for U > 0

can be mapped onto the Heisenberg model

H = J ~Si · ~Sj,

where J = 4t2

U .

Answer 43. Consider the U → ∞ limit at half-filling, where there is exactly one electron persite and its spin can be oriented in either direction. Consider two neighboring spins. If they arealigned, hopping is impossible. If they are antialigned, there is a virtual process described bysecond-order perturbation theory (H = H0 + λV):

∆E = −λ2∑γ 6=α

VαγVγαEγ − Eα

.

Here, γ is the intermediate state where a pair of antialigned spins are on the same site. BecauseEγ > Eα =⇒ ∆E < 0, the antiferromagnetic ordering is favored.

Here, we have Eγ − Eα = −U , λ = t. Now, actually things get tricky. Let us call the twointeracting sites 1 and 2. They find it increasingly hard to talk to each other in the U → ∞ limit,so the ground state is

|α〉 =1

2(↑1 + ↓1)⊗ (↑2 + ↓2). (20)

Since the numerator has both a 〈α| and a |α〉, we have a factor of 1/4. Now, let’s look at whichparts of |α〉 contribute. Note that there are two possible starting states: ↑1 ⊗ ↓2 and ↓1 ⊗ ↑2.There are also two possible ending states. This gives a factor of 4.

Finally, when we hop into the intermediate state |γ〉, there are two ways to do so since we canmove either of the electrons to the other’s spot. Similarly when we hop out of the intermediate

25

https://www.quora.com/How-can-charge-density-wave-be-explained-using-the-Fermi-nesting-wave-vector


state, there are also two ways to choose the mover. This gives a factor of 4. The overall factor istherefore 1

4× 4× 4 = 4.

That gives −4t2/U .

Compare this to Heff = J ~S1 · ~S2. The singlet has energy −3J /4 and the triplet has energy J /4.So, 0− (−4t2

U ) = J4− −3J

4.

Question 44. Explain how the Anderson model modifies the Hubbard model.

Answer 44. The Hubbard model considers only d-band electrons and treats them as interactingonly with other d-band electrons (on the same atomic site). The Anderson model introduces s-d interactions. The s electrons are interpreted as the conduction band and the d electrons areinterpreted as stationary impurities.

Question 45. Starting from the Anderson model

H =∑

εdndσ + Und↑nd↓ +∑

εkc†kσckσ +

∑kσ

(Vkc†dσckσ + V ∗k c

†kσcdσ),

explain heuristically why this leads to an effective Hamiltonian of

Hsd = J ~S ·∑kk′

c†kα~σαβck′β.

Explain why this makes sense only when both εd < EF and εd + U > EF are true.

Answer 45. Given a single empty site, adding a single electron costs energy εd and adding asecond electron costs energy εd + U . If εd > EF , there are zero d-band electrons anywhere. Ifεd + U < EF , the d-band is completely filled with both spin ↑ and ↓, so there can be no preferredorientation of the d-band electrons. For εd < EF < εd + U , we expect each impurity to have oned-band electron, with unknown direction of spin. If the direction of spin has some preference, thenthere can be magnetism.

So for there to be magnetism, the highest-energy s-band electrons, at EF , must have higher energythan the lowest-energy d-band electrons, at εd, but lower energy than the higher-energy d-bandelectrons, at εd + U . We claim that the interaction terms

H ⊃∑kσ

Vkc†dσckσ + V ∗k c

†kσcdσ

yield an effective interaction Hsd between the conduction and impurity electrons. We use second-order perturbation theory on the following processes:

1. Conduction electron at k′ hops to high band εd + U and returns to conduction band at k.Here, ε′k ≈ εk ≈ EF .

2. Impurity electron at εd hops to top of conduction band at k and then returns to impurity atεd. Here, εk ≈ EF .

The multiplication of matrix elements together in the perturbation theory gives the ~S ·~σ structure.For example, there might be a term which looks like c†k↑cd↑c

†d↓′ck′↓′ ⊂ S−σ+.

26


Due to the denominators of the perturbation theory, the effective spin interaction parameter endsup looking like

Jkk′ = V ∗k Vk′(1

U + εd − εk′+

1

εk − εd).

Though the fractions are always positive, the parameter V ∗k Vk′ doesn’t have to be positive... orreal (?). So, the interaction can be either ferro- or antiferro-magnetic.

8 Magnetism: Spin waves

Question 46. Derive the spin wave dispersion relation in both the ferromagnetic and AF Heisen-berg Hamiltonians,

H =J2

∑〈ij〉

~Si · ~Sj −H∑i

Szi .

Here, the sign of J can change. Argue that in both cases, the spin wave is a Goldstone mode.

Answer 46. In the ferromagnetic case, the ground state will be a state of broken SU(2) symmetry,while the AF ground state breaks the symmetry in Stot. In the ferromagnet, we can construct alow lying excitation

|~k〉 =1√N

∑R

ei~k·~R[

1√2SS−r |0〉

]which turns out to have energy

Ek − E0 = 2SJ∑R

sin2

(1

2~k · ~R

)+ gµBH

This is quadratic dispersion in the long wavelength limit. For AF states, the appropriate spinwave dispersion is ω ∼ vsk, from the Landau-Lifschitz equation.

Question 47. What is a Mott insulator?

Answer 47. A Mott insulator is a material that is predicted to conduct under normal band theory,but cannot due to strong electron-electron interactions. In the case of the half-filled Hubbardmodel, although the energy levels do not have a band gap, there is an on site repulsion term Uthat can prevent electrons from hopping and conducting. In essence, the Coulomb repulsion Umakes the excitation gapped, which causes the material to become a Mott insulator at low enoughtemperature and large enough U t.

Question 48. Explain how to derive Landau-Lifshitz equation,

∂t~Si = −J∑j

~Si × ~Sj,

from the Heisenberg model H = J∑〈ij〉

~Si · ~Sj. Argue that this leads to spin waves. Explain why

the Goldstone modes (ω(~k)→ 0) are located at ~k = ~0, ~Q.

27


Answer 48. We start with H = J∑ ~Si~Sj and assume Neel order such that 〈Szi 〉 = eiQrN0.

Equations of motion givedSaidt

= i[H,Sai ] = −JεabcSbi∑j

Scj

Linearize around the Neel state so that

d

dtδSxi = −Jεxzy〈Szi 〉

∑(δSyj )− Jεxyz(δSyi )

∑〈Szj 〉

= JeiQriN0

(∑δSyj − δS

yi

∑1)

Taking δSyi ∝ eikri−iωt and δSyi ∝ eiQrieikri−iωt leads to ω = vsk in the limit of k → 0 (gaplessGoldstone mode).

Question 49. Explain heuristically why an interaction term U in the Hubbard-model Hamilto-nian,

H = −t∑〈ij〉

c†iσcjσ + U∑i

ni↑ni↓,

leads to a nonzero gap ∆ in the spectrum. What does it mean, physically, and why does it leadto spin-density waves (SDW)?

Answer 49. It’s easy to explain why nonzero U leads to spin-density waves of the ground state.This is just the magnetic ordering of the ground state that was discussed earlier - neighboringelectrons tend to be antialigned.

Generally, if neighboring electrons tend to be antialigned, then U has generated a kind of ”extraperiodicity” on the lattice. Generally in physics, if we have periodicity along with an interaction(which here, is functionally given by the quadratic term), then we can open up a gap.

Again, for the half-filled Hubbard model, spin-density waves are a result of the nesting vector ~Qwhich leaves open the possibility of dimerization, opening up energetically favorable gaps in thespectrum near the edges of the Fermi surface. Although the energy levels of the free electrons donot have a band gap, there is an on site repulsion term U that can prevent electrons from hoppingand conducting. In essence, the Coulomb repulsion U makes the excitation gapped, which causesthe material to become a Mott insulator at low enough temperature and large enough U t.

9 Magnetism: Theories of indirect coupling

Question 50. What are the four names in RKKY?

Answer 50. Ruderman, Kittel, Kasuya, Yosida.

Question 51. What is RKKY? Explain heuristically the path from Anderson model to RKKY.

Answer 51. RKKY is also called ”indirect exchange interaction via conduction electrons.” Sup-pose there is a low density of spin impurities in a conducting material (they could be nuclear spinsor d-band electron impurities). The s-band electrons can hop between the impurities. The energyof the electrons doing so changes with the relative orientation of the impurities, so this indirectexchange may favor a certain relative orientation of the impurities.

28


This follows from Anderson model because it led to the contact-point interaction

HAnderson → HHeisenberg ∼ ~S · ~σ,

where ~S is the spin of the impurity and ~σ is the spin of the conduction electron sitting on theimpurity. In fact, that Heisenberg model is the starting Hamiltonian for deriving the RKKYinteraction,

H =1

2

∑n

Jn~In · ψ†α(~rn)~σαβψβ(~rn) =1

2V

∑kk′n

Jn~Inei(~k−~k′)·rnc†kα~σαβck′β.

Above,∑

n is a sum over all the impurities, n. Second-order perturbation theory on the above

Hamiltonian will couple the spins ~I1 and ~I2 together.

Question 52. Show that the RKKY effective Hamiltonian is

Heff = JnJm~In · ~ImΦ(~rn − ~rm).

Don’t try to compute the function Φ, but explain why it behaves like Friedel oscillations,

Φ(x) ∼ cos(2kFx)

x3.

Answer 52. The above effective Hamiltonian is really the perturbation in the full Hamiltonian

H = H0 + V ,

where H0 =∑

kσ c†kσckσ refers to conduction electrons and V ∼ Jn~In · c†kα~σαβck′β is the impurity-

conduction electron interaction.

Let the state of a conduction electron (i.e. according to H0 be described by a momentum andspin, |kσ〉). Second-order perturbation theory gives

〈kα|Heff|k′β〉 =∑k′′γ

〈kα|V|k′′γ〉〈k′′γ|V|k′β〉Ek′′ − Ek

so〈~rα|Heff|~r′β〉 ∼ JnJm~In · ~ImΦ(~rn − ~rm),

where Φ(~x) ∼∫~kei(

~k−~k′)·~x fk(1−fk′′ )Ek′′−Ek

. This means that the electron |k′′γ〉 cannot be in the Fermi sea,

which makes sense because it is an intermediate state.

Question 53. Why are there Friedel-like oscillations in the RKKY effective potential?

Answer 53. Recall that RKKY is a strong-coupling effect, and consider the spin electronic densityaround a spin ~I1. In the AF regime, (1) the electrons very close to ~I1 will tend to be spin-polarizedin the opposite direction. (2) Those electrons a little farther away will tend to be spin-polarized in

the original direction of ~I1 because it interacts most strongly with the electrons in (1). Electrons

farther away than (2) will tend to flip again, and so on. Another impurity ~I2 which feels the

potential from ~I1 is actually probing the spin electronic density created by ~I1, which oscillates.

Question 54. Why are the oscillations of wavevector kF ?

29


Answer 54. You can think of the RKKY interaction as what happens when an electron scatterselastically off an impurity. The electron has momentum on the order of kF , and if the impu-rity is treated as a point potential, the scattered wave is spherically symmetric with wavevectorapproximately kF .

Question 55. What is the Kondo effect?

Answer 55. The Kondo effect explains why the resistance of pure metals, such as Au and Ag,increases with decreasing temperature, for very low temperatures. The resistivity passes througha minimum and then diverges logarithmically as T → 0, as

ρ(T ) = ρ0 + aT 2 + cm log(µ

T),

Where log( µT

) is the Kondo contribution. This is derived from the third-order perturbation theoryon the Anderson model of magnetic impurities. The third-order term becomes logarithmicallydivergent because there turns out to be a singular dependence on the energy of the initial state.

Physically, at very low temperatures, the conduction electron is moving slower and slower. So, itcan bind for a significant lifetime to the magnetic impurity due to magnetic interaction. If thisis true, then the conduction electrons are harder to push through the system as T → 0, and theresistivity shoots to infinity. The magnetic coupling becomes non-perturbatively strong as T → 0,and the value at which it becomes important is called the Kondo temperature,

TK ∼ µe−n/J ν(0).

Question 56. Explain where the ρ(T ) ⊃ log( µT

) dependence of the Kondo effect comes from, viathird-order perturbation theory.

Answer 56. Recall that in the Drude model, resistivity is derived by arguments about the meanfree path, etc. So, resistivity is a scattering phenomenon. Resistivity ρ has a term proportional tothe square of the scattering amplitude, which in the regular Fermi golden rule is

W (1)(a→ b) ∝ δ(Ea − Eb)〈a|V|b〉〈b|V|a〉.

However, the new term in the second Born approximation is

W (2)(a→ b) ∝ δ(Ea − Eb)∑c 6=a

〈a|V|b〉〈b|V|c〉〈c|V|a〉Ea − Ec

,

and the denominator causes a logarithmic behavior in W (2).

Question 57. What is Anderson’s ”Poor Man’s Scaling?”

Answer 57. Poor man’s scaling is a renormalization-group approach to the same problem thatKondo solved using third-order perturbation theory. Instead of doing perturbation theory, An-derson’s idea was to integrate out the ”fast modes” or ”high-energy modes” of the conductionband. What remains are the slower modes, which are the more important ones for divergence ofresistivity, anyway.

30


Schematically, Anderson took the conduction band to have width D and integrated out all high-energy electrons in the band [D−δD,D] via second-order perturbation theory. These high-energymodes are absorbed into the coupling constant J ,, which for SU(2)-symmetric theories flows as

∂J∂ logD

= −2ν(0)

nJ 2.

This turns out to return the Kondo temperature, which is interpreted as the scaling invariant,

kBTK ∼ De−n

2J ν(0) = const.

This gives us more insight into the Kondo resistivity. If J → ∞ as T → 0, it makes sense thatlow-energy conduction electrons get ”stuck” on spin impurities at low temperatures - the couplingconstant is very large!

In fact, only the AF isotropic (SU(2)-symmetric) model runs to strong coupling. For the ferro-magnetic isotropic model, the starting point is J < 0 and so J runs to zero.

Question 58. Outline the calculation of the RG flow equations in Anderson’s ”Poor Man’s Scal-ing.”

Answer 58. When we say we ”integrate out the fast modes,” what we mean is we integrate outthe intermediate states of high energy. An electron in the core of the Fermi sea must jump overa great energy barrier to access the shore, so in fact we also integrate out electrons of low energy(because they correspond to intermediate states of high energy).

Take the band diagram and divide it into two halves, both of size D. The top half is typicallyunfilled and represents the conduction band. The bottom half is typically filled and represents thevalence band. Poor man’s scaling chops off the (a) top part of the conduction band and (b) thebottom part of the valence band.

There are two processes. Because the Hamiltonian looks like ~S · ~σ, you can either flip the spin ornot flip the spin. So generally, the spins of the initial, intermediate, and final states can be either↑ or ↓, and choosing the particular spin is like choosing which of J+,J−,Jz to renormalize:

1. A conduction electron jumps up to (a), then hops back down to another spot in the conduc-tion band. Schematically,

kσ → qσ′ → k′σ′′.

2. A valence electron in (b) jumps up to the conduction band. Then, a conduction electrontakes the spot that used to be filled by the valence electron. Schematically,

qσ → k′σ′ and kσ′′ → qσ.

In the end, the functional form of Heff will look like the functional form of the original H, so theparameters have been renormalized. For example, consider the first kind of process. If σ = σ′′, wehave renormalized Jz. If σ =↑ and σ′ =↓, we have renormalized J−.

Another good thing is to know where the log comes from. Suppose the Hamiltonian is SU(2)-symmetric. Then, the renormalization of J will look like

δJ ∝ ν(0)J 2 δD

D.

The ν(0)δD comes from the sum over states and the 1D

comes from 1Eq−Ek

≈ 1D.

31


10 Superconductivity: BCS and Landau theory

Question 59. Describe the Meissner effect. Explain how the equation ∇2 ~B − λ−2L~B = 0 arises,

either from the London equations or from gauging a U(1) theory.

Answer 59. Superconductors will expel magnetic fields from their interior. The penetration depthof the magnetic field will be λL, which diverges at Tc. For a current ~j, we have from Maxwell’sequations that

∂t

(∇× m

ne2~j +

~B

c

)= 0

In the bulk of a SC, j and B are 0. Therefore, ∇× m

ne2~j +

~B

c= 0. Along with ∇× ~B =

4π

c~j, we

get

∇2 ~B − λ−2L~B = 0, λ2

L =mc2

4πne2

Question 60. The canonical two-body interaction is Hint =∑

kk′q,σσ′ V (q)c†k+q,σc†k′−q,σ′ck′,σ′ck,σ.

In the BCS Hamiltonian, why does this get approximated as

Hint =∑k

(∆kc†k↑c†−k↓ + h.c.) ?

Answer 60. There are different ways to think about it. Here is one way: recall that the BCSwavefunction has structure |ΨG〉 ∼ c†k↑c

†−k↓|0〉. BCS theory is a variational problem, which involves

computing〈ΨG|Hint|ΨG〉.

There are generally 3 kinds of contributions: Hartree, Fock, and pairing, which is a new contri-bution. The Hartree and Fock contributions look like 〈c†c〉, but the pairing contribution lookslike 〈c†c†〉 or 〈cc〉 and is nonzero only if particle number is not conserved (such as in the BCSwavefunction). The pairing contribution is the one that contributes to superconductivity and hask = −k′, q = −k − k′. Taking the mean-field approximation

Hint ∼ 〈c†−k′,σc†−k,σ′〉ck′,σ′ck,σ + c†−k′,σc

†−k,σ′〈ck′,σ′ck,σ〉

gives the pairing contributions.

Another way to think about it is that the Hamiltonian either creates or annihilates a Cooper pair.

Question 61. Describe physically the phonon-mediated formation of Cooper pairs. Why do theelectrons in a Cooper pair create a singlet?

Answer 61. Consider two electrons. The first electron impinges on the lattice, disturbing theionic positions and creating a local region of positive charge. The second electron is attracted tothis local region of positive charge, which tends to have a long lifetime because ions are heavy.Thus, the two electrons are attracted to each other.

In fact, there can be Cooper pairs with L = 1 as well, but for s-wave superconductor, the singletconfiguration is more energetically favorable. Heuristically, the two electrons are almost at thesame place in real space, so we expect them to be oppositely aligned.

32


Question 62. Describe mathematically the phonon attraction process, by first writing down themicroscopic electron-phonon interaction. In what regime can we confidently say the interaction isattractive, and which phonons are most important?

Answer 62. We expect the electron-phonon interaction to involve a momentum transfer q betweenthe electron and phonon:

Hint =∑kqσ

c†kσck+q,σg(q)(b†q + b−q).

We want to use Hint to construct an effective two-body interaction between electrons. To do so,consider the following two-body process:

|k + q, σ〉|k′, σ′〉 → |k, σ〉|k′ + q, σ′〉.

It can happen in two ways:

1. |k + q, σ〉 emits a photon of momentum q, which is eaten by |k′, σ′〉

2. |k′, σ′〉 emits a photon of momentum −q, which is eaten by |k + q, σ〉

Second-order perturbation theory gives

(〈k, σ|〈k′ + q, σ′|)Hint(|k + q, σ〉|k′, σ′〉) =|g(q)|2

ε(k + q)− ε(k)− ω(q)+

|g(q)|2

ε(k′)− ε(k′ + q)− ω(q)

=2ω(q)|g(q)|2

(ε(k + q)− ε(k))2 − ω(q)2

which follows by energy conservation.

• The interaction is attractive for |ε(k) − ε(k + q)| ω(q), and also becomes independent of~k. Note that such a situation is easiest to achieve when k and k + q are on opposite sidesof the Fermi sea, since they will have similar energy but the phonon connecting them willhave high energy. This pairs k with −k – a Cooper pair!

• Because d3~q ∼ q2dq, and

Hint ∼1

q,

the most important photons are those with the largest q. This is the cutoff qD, where Dstands for Debye frequency. Hence, the interaction is retarded on the scale ∆t ∼ ω−1

D .

Question 63. What is a Cooper instability? How is it related to the BCS theory?

Answer 63. Cooper found that for attractive interactions, regardless of the strength of interaction,the Fermi sea is unstable and therefore not the true ground state. Namely, it is unstable tothe formation of Cooper pairs which interact from opposite shores of the Fermi sea. BecauseCooper instability shows the Fermi sea is unstable, the BCS wavefunction allows for arbitraryparticle number to account for this instability; BCS wavefunction is a variational wavefunctionwith parameters uk, vk chosen to minimize the energy.

33


Question 64. Outline the calculation of Cooper instability.

Answer 64. We start with the interacting Hamiltonian and use the variational wavefunction ona single Cooper pair,

|Ψ〉 =∑k

α(k)c†k↑c†−k↓|F 〉, where α(k) = α(−k) and 1 =

∑k

|α(k)|2.

The α(k) = α(−k) guarantees the Cooper pair will be a singlet, and 1 =∑

k |α(k)|2 guaranteesthere is only one Cooper pair.

The interaction must take a Cooper pair to another Cooper pair, so only the Cooper pair entersin the scattering process. The relevant interactions for

(k ↑,−k ↓)→ (k + q ↑,−k − q ↓)

arec†k+q,↑c

†−d−q↓c−k↓ck↑ and c†−d−q↓c

†k+q,↑ck↑c−k↓.

Because there are two contributions to this one process, the factor of 12

goes away. So, the energyof the state is

Etot = E|FS〉 + 2∑k>kF

ξk|α(k)|2 −∑k>kF ,q

V (q)α∗(k + q)α(k)− λ∑k

|α(k)|2,

where we inserted the Lagrange multiplier. Variation on this function returns a negative energy

λ = Etot − E|FS〉 = −2ωDe−2/ν(0)V < 0.

Question 65. Describe the competition between attractive phonon interaction and repulsiveCoulomb interaction. Why is superconductivity still possible?

Answer 65. The two interactions oppose each other; therefore, the repulsive Coulomb interactionwill work to weaken superconductivity. However, because the phonon interaction is retarded onscales of ω−1

d E−1f , it is much more relevant at low temperatures compared to the Coulomb

interaction. The Coulomb interaction is actually strongly renormalized such that

Veff = Vph −Vc

1 + ν(0)Vc log(EF/ωD),

which can be attractive even if Vc Vph. However, what will happen is that the superconductinggap ∆ will change sign at ωD in order to minimize energy.

Question 66. Describe the ”criterion for superconductivity.” Comment on the dependence of Veff

on EF and ωD.

Answer 66. The criterion for superconductivity is that Veff > 0 . This means the relevant phononcoupling overpowers the irrelevant Coulomb interaction. The phonon coupling will always over-power the Coulomb interaction at low enough energy; the question is whether the energy is lowenough for the Coulomb interaction to scale away enough. The dimensionless quantity EF/ωDmeasures ”how long” the Coulomb interaction has had to scale away. The longer it has to scaleaway, the weaker the renormalized Coulomb interaction, so Veff will be stronger.

34


Question 67. Describe the connection between BCS variational theory and the superconductivitysolution by Bogoliubov transform.

Answer 67. The BCS wavefunction is

|ΨG〉 =∏k

(uk + vkc†k↑c†−k↓)|0〉

and the parameters uk, vk are varied to obtain the minimum energy. Some useful formulas toremember are

ukvk =∆k

2Ek, Ek =

√ξ2k + ∆2

k.

In fact, the Bogoliubov transform on the mean-field Hamiltonian

H =∑kσ

ξkc†kσckσ −

∑k

(∆kc†k↑c†−k↓ + ∆∗kc−k↓ck↑)

with

ck↑ = ukγk↑ + vkγ†−k↓, c

†−k↓ = −vkγk↑ + ukγ

†−k↓

yields the same values of uk, vk and hence the same energy Ek, gap equation, etc. For example,the BCS wavefunction is normalized for every wavevector k as u2

k + v2k = 1, and we must also have

u2k + v2

k = 1 for the Bogoliubov transformation to preserve anticommutations.

Then what is the difference?

〈c−k↓ck↑〉 = ukvk in BCS wavefunction

and〈c−k↓ck↑〉 = ukvk(1− 2f(Ek)) in BCS mean-field Hamiltonian,

but it seems that if you take T > 0 into account, BCS wavefunction might return the same result?No-the meaning is that BCS wave function is zero temperature limit. With Bogoliubov, we cantake into account finite temperature.

Question 68. What is the gap equation? Describe its importance and how it can yield a valuefor TC .

Answer 68. The gap equation is a self-consistency condition in superconductivity. Its impor-tance is that it establishes superconductivity as a robust phenomenon; the larger the gap, themore stable the superconductor. Here is is:

1 =V

2

∑k

1

Ek=⇒ ∆ = 2ωde

− 1ν(0)V .

The solution of∆(T = TC) = 0

gives a value for the critical temperature TC . In other words, if the gap disappears, there is noenergetic reason for the system to have Cooper pairs. You can think about this in the mean-fieldtheory,

H =∑kσ

ξkc†kσckσ −

∑k

(∆kc†k↑c†−k↓ + ∆∗kc−k↓ck↑).

35


If ∆k > 0, then Hint can be negative if we have a Cooper pair, because of the negative sign infront of the sum. Equivalently, you can think about this in the BCS theory, where the energyexpectation turns out to be

Etot = 2∑k

ξkv2k −

∑kk′

Vkk′ukvkuk′vk′ = 2∑k

ξkv2k −

∑k

∆kukvk.

Equivalently, the condensation energy is

Es − En ≈ −1

2ν(0)∆2.

Question 69. Derive the condensation energy expression,

Es − En ≈ −1

2ν(0)∆2.

You can use u2k = 1

2(1 + ξk

Ek), v2

k = 12(1− ξk

Ek).

Answer 69. First, let’s think about why this is plausible. Opening a gap in the spectrum meansmaking the occupied states have lower energy and the unoccupied states have higher energy; thedifference is approximately ∆. The range of states affected is roughly proportional to ∆, whichmakes sense if you just draw a dispersion relation. So the energy difference is

En − Es ≈ ∆

∫ EF

EF−∆

ν(E)dE ∼ ν(0)∆2.

Now let’s derive it rigorously. We will first use that

Es =∑

(ξk − Ek) +∑

Vkk′〈c†k↑c†−k↓〉〈c−k′↓ck′↑〉+

∑Ekγ

†kσγkσ

We use that

〈c†k↑c†−k↓〉 =

∆k

2Ek, ∆k =

∑k′

Vkk′〈c−k′↓ck′↑〉

and En = 2∑

k<kfξk so that

Es − Ek =∑(

ξk − Ek +∆2k

2Ek− 2ξkθ(kf − k)

)Linearize about the fermi energy and write

∑(2Ek)

−1 = V −1 so that

Es − En =∆2

V− 2ν(0)

∫ ωd

0

dξ (Ek − ξk) ≡∆2

V− 2ν(0)Γ

We calculate Γ =∆2

4+

∆2

2log

2ωd∆

=∆2

4+

∆2

2ν(0)V, leading to

Es − En =∆2

V− 2ν(0)∆2

4− ∆2

V= −ν(0)∆2

2

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Question 70. Derive the gap equation at T = 0 and for T > 0.

Answer 70. At T = 0, we can use the BCS wavefunction, |ΨG〉 =∏

(uk+vkc†k↑c†−k↓)|0〉. We would

like to focus only on the pairing contribution of the canonical two-body interaction,

Hint = −1

2

∑kk′qσσ′

V (q)c†k+q,σc†k′σ′ck′+q,σ′ckσ →

−1

2

∑kk′qσσ′

V (q)〈c†k+q,σc†k′σ′〉〈ck′+q,σ′ckσ〉 = −

∑kk′

Vkk′ukvkuk′vk′ .

Here, q = −k − k′. Also, the number operator is 〈c†k↑ck↑〉 = v2k, so the total energy is

E = 2∑k

ξkv2k −

∑kk′

Vkk′ukvkuk′vk′ = 2∑k

ξkv2k −

∑kk′

∆kukvk.

Remembering uk =√

1− v2k and differentiating w.r.t. vk gives

2ξkukvk + (v2k − u2

k)∆k = 0 =⇒ ukvk =∆k

2Ek.

The integral equation for ∆ is then

∆k =∑k′

Vkk′uk′vk′ =1

2

∑k′

Vkk′∆k′

Ek′.

In the BCS approximation where Vkk′ = V θ(ωD − ξk)θ(ωD − ξk′), ∆k = ∆ and we can divide outthe ∆’s from both sides. This gives the gap equation at T = 0.

The gap equation for T > 0 has to do, of course, with the Fermi-Dirac distribution. Specifically,it has to do with the Fermi distribution of quasiparticles at a certain temperature. Bogoliubovdiagonalization brings the BCS mean-field Hamiltonian to

H =∑kσ

Ek(γ†k↑γk↑ + γ†−k↓γ−k↓)

and then use〈c−k↓ck↑〉 = ukvk(1− γ†k↑γk↓ − γ

†k↓γk↓) = (1− 2f(Ek))

in the self-consistency condition, ∆k =∑

k′ Vkk′〈c−k′↓ck′↑〉. The BCS approximation ∆k = ∆ gives

1 =V

2

∑k

tanh(βEk/2)

Ek,

which is similar to before. (Recall that tanh(x) approaches 1 asymptotically as x→∞.) Setting∆ = 0 gives TC .

Question 71. What is the BCS-BEC crossover?

37


Answer 71. See https://www.physics.rutgers.edu/grad/601/CM601/talks/bec.pdf. Thisis called a crossover rather than a phase transition because there is no phase transition: it is aperfectly smooth process, because both limits can be captured by the same wavefunction. In BEC,the interaction is so strong that we can have a bound state in vacuum. Not only is the Fermi seaunstable but the vacuum is unstable as well.

BEC stands for Bose-Einstein Condensate. Basically, if the interactions are weak enough, we canthink of the system as a collection of bosons b, all in the ground state q = 0. Namely, let theoperator

b† =∑k

vkukc†k↑c

†−k↓

create a bosonic quasiparticle of momentum q = 0. Then the BCS wavefunction can be rephrasedas

|ΨG〉 = e∑kvkukc†k↑c

†−k↓|0〉 = eb

†|0〉 = |ψBEC〉.

Heuristically, the BEC boson is when the coherence lengthscale ξ (i.e. size of the Cooper pairs)gets much smaller than the characteristic interparticle spacing, so we are justified in talking about”spatially-bound electron pairs,” which are just the BEC bosons.

Question 72. The Landau free energy functional for superconductivity is

F = Fn + α|ψ|2 +β

2|ψ|4 +

1

2m

∣∣∣∣∣(−i~∇− 2e ~A

c

)ψ

∣∣∣∣∣2

+h2

8π

Consider the situation with no currents or magnetic fields to extract the coherence length ξ.

Answer 72. With no currents or fields, the minimization condition becomes

− ~2m

d2ψ

dx2+ αψ + βψ3 = 0

Defining ψ = ψ0f , ψ20 = −α/β, we get

−ξ2T

d2f

dx2− f + f 3 = 0, ξT =

~2

2m|α|

The coherence length ξT characterizes the length over which the order parameter varies.

Question 73. How can we use Landau theory to ge the London penetration depth λL?

Answer 73. In weak fields, the current is

~j =e

im(ψ∗∇ψ − ψ∇ψ∗)− 4e2

mc~A|ψ|2 =⇒ ∇×~j = −4e2

mc|ψ0|2 ~B

Using ∇× ~B =4π

c~j, we get

1

λ2L

=16πe2ψ2

0

mc2=

32πe2

~2c2Cδ2

0

Question 74. How does Landau theory explain type I and II superconductors?

38

https://www.physics.rutgers.edu/grad/601/CM601/talks/bec.pdf


Answer 74. The ratio

κ =λ

ξ=

~c4Ce

(B

2π

)1/2

determines the type. Note that B here is from the non-normalized Landau Ginzburg free energy(with δ instead of ψ as the order parameter). κ < 1/

√2 is type I, meaning that beyond a critical

field, superconductivity is destroyed. κ > 1/√

2 is type 2, in which above a first critical field Hc1,vortices form in the bulk and increase in density until a higher critical field Hc2, above whichsuperconductivity is destroyed.

11 Superconductivity: Unconventional SC and Josephson

effect

Question 75. Describe the magnon mediation of unconventional superconductivity.

Answer 75. Conventional superconductivity hinges on phonon attraction via the lattice ions.Unconventional superconductivity can occur in fermionic systems even without a lattice. Recallthat we found the spin correlation function

gσσ′(r)

for the interaction electron gas in the Hartree-Fock approximation, and found that opposite spinsare uncorrelated while like spins are anticorrelated; namely, Pauli exclusion says that an ↑ electrontends to push away other ↑ electrons. An electron at (~r, t) produces a polarization cloud inthe neighboring electron liquid. Really, it produces Friedel oscillations of polarization, so forcertain distances from the electron, other electrons can be attracted to it. In field theory, this isinterpreted as exchange of magnons, which are quantums of spin imbalance.

Question 76. Explain why the order parameter

∆dx2−y2(~k) = cos(kx)− cos(ky)

makes sense for the d-wave superconductors.

Answer 76. The d-pairing in the cuprate superconductors is described by the Hubbard model,which as we know implies an antiferromagnetic ground state. The AF ground state suffers a (−1)phase shift upon rotation by π, so it’s reasonable that the order parameter should obey the samesymmetry.

To derive this more rigorously, recall that the interaction basically depends on the product ofspins,

Question 77. What is a Josephson junction?

Answer 77. A Josephson junction is a ”weak link” between two superconductors. Generally, asuperconductor has a well-defined phase, but two superconductors next to each other may have anonzero phase difference, ∆θ. Josephson junction takes advantage of this nonzero ∆θ by using itto generate a current. The charge carriers are Cooper pairs instead of electrons.

39


The way to think about this is that ∇xθ 6= 0 implies a nonzero superfluid velocity, and hence anonzero supercurrent.

The order parameter ∆ decreases in the weak-link region. It does not usually go to zero; in fact,in the L ξ approximation (where L is the thickness of the junction), it goes to zero exactlyonce every 2π cycle of the relative phase, ∆θ.

So, one way to think about Josephson junction is that the order parameter from the superconduct-ing electrodes ”bleeds into” the junction, and you have to tie them together (and continuouslymatch the phase!). Another way to think about it is that the Cooper pairs quantum tunnelthrough the barrier.

Question 78. Describe mathematically the current characteristics of a Josephson junction. Inwhat sense is it like an inductor?

Answer 78. The Josephson junction is most easily described in the framework of Landau theory,where there is a U(1) gauge symmetry of the superconducting order parameter. The junctionoperates by ”connecting” the phase on either side of the junction.

The DC Josephson effect isIsupercurrent = Ic sin(∆θ) (21)

where Ic is the Josephson critical current. Ic decreases when you make the weak link havegreater thickness, for example. DC Josephson effect says that you can have a current without avoltage drop across the junction.

The AC Josephson effect is

∂t(∆θ) =2e

~V (22)

where V is the voltage across the junction. Combining this with the DC effect suggests that thecurrent across a Josephson junction is of amplitude Ic and oscillates with frequency ν = 2eV

h. Note

that δE = hν = 2eV is the energy of a Cooper pair transferred across the junction.

If you put the DC and AC effects together, you can write

V = LJdI

dt, where LJ =

Φ0

2πIc cos(∆θ). (23)

Thus, the Josephson junction is like an inductor, with nonlinear inductance which depends on therelative phase.

Question 79. What is a SQUID?

Answer 79. A SQUID is the best word you can throw at people when playing hangman. Inphysics, it is when you connect two Josephson junctions in parallel:

40


The phase θ(x) has to be continuous around the loop. If the loop gets penetrated by a nonzero Bfield, then the Josephson periodicity of the current through the junctions allows for precise mea-surement of the magnetic flux through the loop. How precise? The smallest unit of measurementis the fluxoid, Φ0.

In particular, such measurements have suggested that the order parameter in YBCO (i.e. high-Tc superconductor) is d-wave. If the YBCO is shaped like a square, van Harlingen’s SQUIDexperiment detected a π-phase shift in the phase θ between the right side and bottom sides of thesquare.

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Questions and Answers in Quantum Theory of Solids · Questions and Answers in Quantum Theory of Solids Jimmy Qin and Yunchao Zhang Spring 2019 Yunchao and I wrote this document based