2 Models for Electrons in Solids 6 2.1 Jellium Model ...... 6 2.2 Blöch Functions ...... 6 2.3 Hubbard Model ...... 7 2.3.1 Two Particle Hamiltonian ...... 7 2.3.2 High-Temperature Superconductors ...... 8 2.4 Perturbation Theory in U ...... 9 2.4.1 Limits of Perturbation Theory ...... 9 2.4.2 Luttinger Theorem ...... 9 2.5 Hartree Approximation ...... 10 2.5.1 Steps ...... 10 2.6 Fock (exchange) Approximation ...... 10 2.6.1 Hartree Term ...... 10 2.6.2 Hartree Term Cont...... 11 2.7 Dynamical Correlations ...... 13 2.8 Response of Electrons to External Perturbation ...... 14 2.8.1 Linear Response ...... 15 2.9 Models of Dielectric Functions ...... 17 2.9.1 Thomas-Fermi Approximation ...... 17 2.9.2 Special Case: Point Charge ...... 18 2.9.3 Self-Consistent Theory ...... 18 2.9.4 Limiting Cases ...... 21 2.9.5 Eﬀect of ﬁnite relaxation time ...... 25 2.9.6 Connection in RPA ...... 25 2.10 Response to Magnetic Perturbation ...... 27

3 Scattering Experiments 28 3.1 Electron Scattering ...... 28 3.2 Neutron Scattering ...... 30 3.2.1 Interaction with the nuclei via short range forces ...... 30 3.2.2 Interaction with the nucleon spin ...... 31 3.3 Light Scattering ...... 31 3.4 Magnetic Scattering of Neutrons ...... 31 3.5 Neutron Scattering Experiment ...... 33

4 Total Energy of the Electron Gas 33

5 Fermi Liquid Theory 36 5.1 Transport in Fermi Liquid ...... 38 5.2 Spin-Dependent Case ...... 41

1 6 Theory of Superconductivity 42 6.1 Insights ...... 42 6.2 Microscopic Theory ...... 43 6.3 Variational Approach ...... 45 6.4 Quasi-particle Excitations ...... 48 6.4.1 Particle number expectation value ...... 49 6.4.2 One particle (quasi-particle) states ...... 49 6.4.3 Bogoliubov-Valatin Transformation ...... 49 6.4.4 Calculation of Measurable Quantities ...... 50

Lectures

Lecture 1 3

Lecture 2 5

Lecture 3 8

Lecture 4 11

Lecture 5 14

Lecture 6 17

Lecture 7 19

Lecture 8 21

Lecture 9 25

Lecture 10 27

Lecture 11 30

Lecture 12 33

Lecture 13 36

Lecture 14 38

Lecture 15 40

Lecture 16 42

Lecture 17 45

Lecture 18 48

Lecture 19 51

2 30 March 2006 PHY240C Lecture 1

Three Topics • Consequences of electron-electron interactions. • Magnetism • Superconductivity

Schrieﬀer - Theory of Superconductivity Oﬃce Hour: Tues. 2pm. 10 minute quiz the end of Thursday lecture.

Grade Breakdown 1. Quizzes - 30% 2. Midterm - 30% 3. Final Exam - 40%

1 Interactions Between Electrons 1.1 Models

~2 H = 2 + U (r ) + U (r r ) (1.1) −2m ∇i ion i e-e i − j i e i X X Xi6=j U (r ) = Ze2 r R −1 (1.2) ion i − | i − m| m X e2 U (r r ) = (1.3) e-e i − j r r | i − j | Now we apply a new notation.

n(r) = δ(r r ) (1.4) − i i X U = dr U (r)n(r) (1.5) ion ˆ ion 1 U = dr dr′ n(r)n(r′)U (r r′) (1.6) e-e 2 ¨ e-e − 4πe2 U = Fourier Transform (Speciﬁc to 3D) (1.7) e-e q2 2πe U = (2D); ln q (1D) (1.8) e-e q ∼ | | | | (NOTE: The motivated student should show the 3D result.) Diagonal terms vanish due to phase space argument?

1 ′ −iq·r ′ −iq′·r′ ′′ −q′′·(r−r′) Ue-e = dr dr n(q)e n(q )e Ue-e(q )e (1.9) 2 ¨ ′ Xq,q ′′ δ(q + q′′) = dr e−i(q+q )·r (1.10) ˆ U = n(q)n( q)U(q) (1.11) e-e − q X

3 30 March 2006 PHY240C Lecture 1

1.2 Second Quantization † • Creation operator: akσ (Metals, heavily doped semiconductors). † r ∗ r † • Field operator: ψσ(b) = φk,σ( )ak,σ (insulator, metal oxides, lightly doped semiconductors). k X • Principle of transcription:b matrix elementsb of operators should be the same.

1.2.1 One Particle Operators

1 ′ † H = ℓ H ℓ aℓ′ aℓ (1.12) ′ h i Xℓ,ℓ b b b (H1 is a one particle second quantized operator. H is a ﬁrst quantized operator)

b 1 ′ † † ′ ℓ1 H ℓ2 = ℓ H ℓ 0 aℓ1 aℓ aℓaℓ2 0 (1.13) ′ h i D E Xℓ,ℓ D E b † aℓa 0 = δℓ,ℓ b b b b (1.14) ℓ2 | i 2 Xℓ ℓ bHb1 ℓ = ℓ H ℓ (1.15) 1 2 h 1 2i D E b

4 04 April 2006 PHY240C Lecture 2

Review

• We are most interested in the interactions between electrons and the ions and the electron-electron interaction.

• It is most physically reasonable to discuss the physics in terms of creation and annihilation operators.

• How do we go about doing this? The goal of Heisenberg is to produce the same observables as the ﬁrst quantized operators (get the same matrix elements).

1.2.2 Two Particle Operators

(2) (2) † † H = Hℓℓ′ℓ′′ℓ′′′ aℓaℓ′ aℓ′′ aℓ′′′ (1.16) ′ ′′ ′′′ ℓℓXℓ ℓ (2)b ′ ∗ b∗ b ′b (2)b ′ ′ H ′ ′′ ′′′ = dr dr φ (r)φ ′ (r )H (r r )φ ′′ (r)φ ′′′ (r ) (1.17) ℓℓ ℓ ℓ ¨ ℓ ℓ − ℓ ℓ

(ℓ refers to a generic quantum number)

1.3 Equivalent formulation using ﬁeld operators

ψ(r) = φk(r)ak (1.18) k X b H(1) = dr ψ†(br)H(1)(r)ψ(r) (1.19) ˆ

(Motivated student can prove this.) b b b

1.3.1 For the electron system

1 H = H a† a + U a† a† a a (1.20) k,l,σ k,σ l,σ 2 klmn k,σ lσ mσ nσ ~2 X ∗ 2 X Hklb dr φ (r) + U (r) φl(r) (1.21) ≡ ˆ k −2m∇ ion ′ ∗ ∗ ′ ′ ′ Uklmn = dr dr φ (r)φ (r )U (r r )φm(r)φn(r ) (1.22) ¨ k l e-e − • Intuition tells us that propagation does not cause spin to flip is reason why spins are the same in kinetic term. Counterexample: spin in magnetic field. In heavier elements, spin-orbit coupling will flip the spin w/o electron-electron interaction.

• Feynman diagram can be used to describe an interaction.

m,σ k,σ

Uklmn

l,σ′

n,σ′

5 04 April 2006 PHY240C Lecture 2

• Density interaction is a†a, not a†a†aa.

• Normal Ordering: all destruction operators should be moved to the right.

• If we calculate the effective mass of electrons coming from a photon, the effective mass is infinite, so we do Renormalization.

• Imagine that the mass of the electron is infinite as well. When you absorb the infinity from the vacuum fluctuations, the resulting mass is the one you observe.

• Vacuum fluctuations: Viewpoint is that as a photon is travelling there is a particle-antiparticle pair and if we calculate the contributions to the mass, it is infinite. Electron’s effective energy and mass are infinity. When combined with vacuum fluctuations then result is what we observe. Putting annihilation operators on the right gets rid of these “ghost” particles.

2 Models for Electrons in Solids 2.1 Jellium Model • Smear out the positive ions into a homogeneous background charge.

• Eigenstates are plane waves

1 1 0 φ (r) = eik·rη η = η = (2.1) k,σ √ σ ↑ 0 ↓ 1 V (1) H φk = εk φk (2.2) | ,σi ,σ | ,σi (1) φl H φk = δklεk (2.3)

D E † 1 H = εkc ck + U (q)n(q)n( q) (2.4) kσ σ 2V e-e − k q Xσ X b b† b 1 b † b † = ε c c + U (q)c c ′ ′ c ′ c (2.5) k kσ kσ 2V e-e k−q,σ k +q,σ k ,σ k,σ kσ q,k,k′ X Xσ,σ′ b b b b b b Second term is Fourier transform of the interaction Hamiltonian (interaction between density waves). n(q) is a density ﬂuctuation of the gas with wavenumber q. (Claim High Tc superconductors have charge density waves.) KNOW THIS HAMILTONIAN. Claim that four out of ﬁfteen will need to know this for our orals!

† † n(q) = ck+qck n(r) = ψ (r)ψ(r) (2.6) k X b b (Motivated student canb take ﬁeld formulationb to get second quantizedb formulation of the above equations). First equation: we annihilate and create an electron with momentum diﬀerence q.

2.2 Blöch Functions • Restore ionic structure.

• We introduce bands (denoted by index n) to the system.

ψn,k,σ(r) (2.7)

• Momentum is conserved up to a reciprocal latticeb vector G (crystal momentum is conserved).

6 04 April 2006 PHY240C Lecture 2

k, n k q, n′′ −

k′ + q + G, n′′′

k′, n′

Two more summation indices: 1. band index n 2. reciprocal lattice vector G

G = 0 direct G = 0 Umklapp (2.8) 6 2.3 Hubbard Model • Anderson claimed it holds key for High Tc Superconductivity (they are metal oxides) • Uses ﬁeld operators. • Suitable for localized orbitals: insulators, metal oxides (High Tc S.C.), lightly doped semiconductors • Use Wannier states as our basis instead of Blöch functions. • Most studied Hamiltonian of the last 20 years.

~2 H(1) = dr φ∗(r R ) 2 + U (r) φ (r R ) (2.9) ˆ i − i −2m∇ ion j − j Biggest is when i = j, the diagonal term.

(1) † H = ε ci ci (2.10) i X Assume energy is the same on every site since the atoms are identical. Next biggest: i = j + δ, nearest neighbor.

(1) † H = t ci cj (2.11) hXi,ji Physical meaning is that this term is the hopping term.

2.3.1 Two Particle Hamiltonian The biggest term is the diagonal term.

(2) H = U ni,σni,−σ (2.12) H = t X c† c + U n n (2.13) b i,σbj,σ i,σ i,−σ i,σ hi,jXi,σ X b Can also consider b b b b • Next nearest neighbor hopping t′ • Neighbor repulsions V .

7 13 April 2006 PHY240C Lecture 3

Last Time

• Derived Hubbard Model, simplest model capturing interactions between electrons.

Additional terms

First(second) neighbor Coulomb interaction V ninj (long-range portion of Coulomb dropped because of screening) (2.14) ij X ′ † t ciσcjσ Next nearest neighbor hopping (2.15) ij n.n.n.X

2.3.2 High-Temperature Superconductors

Typical values for a high-Tc superconductor

t 0.5 eV U 5 eV (2.16) ∼ ∼ t′ 0.1 eV V 1 2 eV (2.17) ∼ ∼ −

What happens when we add disorder? (Example: La2−xSrxCuO4)

CuO Layer

La, Sr

Figure 1: Blue: Cu, Grey: O

Zhang-Rice claim that we view the chemically correct (more complicated) structure as a series of a single type of site.

8 13 April 2006 PHY240C Lecture 3

2.4 Perturbation Theory in U

U 1 † † φ = FS + + ck+qc ′ ck′ ck FS (2.18) ′ ′ k −q | i | i V ′ εk + εk εk+q + εk −q | i kX,k ,q −

φ nk φ = h i z

kf k

z = 1 1 ln 2 [U ρ(ε )]2 “Fermi step” (2.19) − 4 · F 1. Fermi distribution nk gets rounded 2. Fermi step is diminished (z < 1), but non-vanishing (for small U). 3. Finite T the step still exists? 4. Step in Fermi function translates to the strengths of the pole of the Greens function which in turn represents the strength or amplitude of the quasi-particle interaction.

2.4.1 Limits of Perturbation Theory 1. U ρ(ε ) 1 · F ∼ • Fermi step goes to zero. 2. 1D (previous results given for 3D) n 2 [U ρ(ε )] ln k k (2.20) h ki ∼ · F | − F | • Zero radius of convergence Non-perturbative analysis is required.

2.4.2 Luttinger Theorem Within radius of convergence of perturbation theory, volume of Fermi surface is unchanged.

⇒

Figure 2: An example of a deformation of a Fermi surface where the volume is unchanged

• Gives rise to zero sound (Oscillation of the Fermi surface).

9 13 April 2006 PHY240C Lecture 3

2.5 Hartree Approximation Represent the interactions as if the electrons propagate in the average ﬁeld of the other electrons. ~2 H = dr ψ† (r) 2 + U ψ (r) ˆ σ −2m∇ ion σ σ X (2.21) b b ′ † † ′ b ′ ′ + dr dr ψσ(r) ψσ′ (r )ψσ′ (r ) ψσ(r)Ue-e(r r ) ′ ˆ − Xσσ D E b b b b We replaced ψ†(r′)ψ(r′) with its average (or mean-ﬁeld). This makes the Hamiltonian quadratic (which can be diagonalized). Is our solution self-consistent?

2.5.1 Steps 1. Assume a form for expectation value ψ†ψ . h i 2. Diagonalize H.

3. Determine modiﬁedb ψ. 4. Recalculate ψ†ψ h i 5. Repeat from step 2.

This will make the theory self-consistent. Theories that are self-consistent or are mean-ﬁeld are non- perturbative.

2.6 Fock (exchange) Approximation

† ′ • Fock: ψσ(r)ψσ′ (r ) D E • H is quadraticb b (good), but non-local (bad)

b † † ′ U(q)ck+q,σck′−q,σ′ ck ,σ′ ck,σ (2.22) ′ kX,k ,q b b b b † † ′ ′ ′ ck+q,σck ,σ′ = δk+q,k δσ,σ′ ck′,σ′ ck ,σ′ (2.23) D E D E Remaining term b b b b † † ck′−q,σ′ ck,σ = ck,σck,σ (2.24)

b b b b ′ † † H = εk U(k k ) c ′ ck′ c ck (2.25) ,σ − − k ,σ ,σ k,σ ,σ k ( ′ ) X,σ Xk D E b b b b b 2.6.1 Hartree Term

† U(q = 0) ck′ ck′ = U(q = 0)Ne (2.26) k X D E Shifts energy of every electron by the same amount.b b

• Turns out this number is canceled out by ionic contribution (Only true for simplest interaction problems).

10 18 April 2006 PHY240C Lecture 4

Last Time

• Eﬀects of the interaction • Perturbation Theory – For Strong interaction, Perturbation Theory is not valid. – 1D: Breaks down for small interaction • Approximate quartic interaction term of the Hamiltonian as a quadratic term times the expectation value of the remaining quadratic terms (Hartree Approximation and Fock Approximation). • Self-Consistent theories are good ways to go beyond perturbative approaches.

2.6.2 Hartree Term Cont.

′ † † H = εk U(k k ) c ′ ′ ck′ ′ c ck (2.27)  ,σ − − k ,σ ,σ  k,σ ,σ k ′ ′ X,σ kX,σ D E Fb  ′ †  ′ b b b b ∆εk,σ = U(k k ) ck′,σ′ ck ,σ′ (2.28) ′ ′ − kX,σ D E Calculation of Fock energy shift for spherical FS. b b

′ |k |

1 x

~2k2 2k e2 k εF = F F (2.31) k 2m − π k F Can ﬁnd the eﬀective mass by taking the derivative of the Fock energy with respect to k and evaluate at the FS since all physical processes take place at the FS.

1 ∂εF 1 1 c 2 = 1 + ln (2.32) kF ∂k ∼ meff m  2 k  kF 1 − kF   Effective mass diverges logarithmically! m 0 eff → • Okay to use Fock approximation away from Fermi surface • Effective mass diverges logarithmically near the Fermi surface Failure of Fock approximation ⇒

11 18 April 2006 PHY240C Lecture 4

Correlations Beyond Hartree-Fock Density-Density Correlator (Correlation Function)

′ V ′ Sσ,σ′ (r, r ) = nσ(r)nσ′ (r ) (2.33) Ne h i Reminder: b b

† nσ(r,t) = ψσ(r,t)ψσ(r,t) (2.34) n (q) = c† c (2.35) σ b k,σ bk+q,σ b k X b b b We wish to work in Fourier space.

1 Sσ(q) = nσ(q)nσ( q) (2.36) Ne h − i 1 † † ′ = b ckb′,σ′ ck +q,σ′ ck,σck−q,σ (2.37) Ne ′ kX,k D E 1 b† b b† b ′ = ck′,σ′ ck +q,σ′ ck+q,σck,σ (2.38) Ne ′ kX,k D E b b b b Consider case when q = 0. Only non-vanishing contribution comes from when k = k′. 6

2 † † S(q) = ( 1) ck q c c ck (2.39) − + ,σ k+q,σ k,σ ,σ † D E D E ckck = f(εk) (2.40)

D † E ck qc = 1 f(εk q) (2.41) + bk+bq − + D E 2 S(q) = f(εk) [1 f(εk q)] (2.42) b b N · − + e k X

k + q k q

Figure 3: Non-zero contribution is denoted by the shaded area

k + q must be outside the FS and k must be inside the FS. • q 0: S(q) 0. → → • q > 2kF : No more overlap, so we have S(q) = 1

2 3 |q| 1 |q| • 0q < 2kF : S(q) = 5 kF − 16 kF 12 18 April 2006 PHY240C Lecture 4

HF g↑↓ (r) = 1 S(r) = g(r) = sin k r k r cos k r 2 (2.43) gHF(r) = 1 9 F − F F  ↑↑ − (k r)3 F Within H-F(Hartree-Fock) approximation, the plane wave basis remains appropriate. Only the parameters F (εk, meﬀ) change. (For the H-F approximation within the Jellium model, the radius of the Fermi sphere remains constant due to Luttinger Theorem.)

HF HF g↑↓ g↑↑ Exchange hole

r r

Figure 4: Correlation function in the Hartree-Fock approximation

From the above, we see that correlations for unlike electrons are ignored in the HF approximation (i.e. there no energy cost to put two electrons on the same site). Also, the correlations for like electrons show that HF two electrons of the same spin cannot be near each other (Pauli). The oscillations in g↑↑ were noticed by Friedel. (RKKY)

true true g↑↓ g↑↑ Bigger than HF

r r

Figure 5: “True” Correlation function

true true For g↑↓ , the results for small r are not agreed upon. With regard to g↑↑ , the size of the exchange hole is due to two things:

• Pauli principle

• Coulomb repulsion

Spin glasses are due to the fact that the interactions between spins are long range (greater than exponential).

2.7 Dynamical Correlations

1 † † ′ S(q,t) = ck′,σ′ (t)ck +q,σ′ (t)ck+q,σ(0)ck,σ(0) (2.44) Ne ′ ′ kkXσσ D E † † iεkt/~ ck(t) = cke b b b b (2.45)

b b

13 20 April 2006 PHY240C Lecture 5

Last Time

• Finished discussion of eﬀects of Fock term • Discussed eﬀects of correlations • Described density-density correlation function (static). – Hartree-Fock: opposite spin electrons do not see each other – HF: like-spins cannot be close to each other (Pauli exclusion principle)

Dynamical Correlations Cont.

† † iεkt/~ ck(t) = cke (2.46)

1 i(εk−εk+q)t/~ S(q,t) = f(εk)[1 f(εk q)]e (2.47) b bN − − e k X Only diﬀerence between static and dynamical correlations is a complex phase factor. Taking the Fourier transform yields

∞ eiωt = 2πδ(ω) (2.48) ˆ−∞ 2π~ S(q, ω) = f(εk)[1 f(εk q)] δ (~ω (εk q εk)) (2.49) N − + · − + − e k X S = 0 only if ω = εk q εk, “Fermi’s Golden Rule”. 6 + − ω S(q, ω) = 0 “two particle continuum”

S(q, ω) = 0

1 2 3 q/kF

Figure 6: In the shaded region, S(q, ω) = 0. 6

2.8 Response of Electrons to External Perturbation

Insert extra charges into your system, i.e. place ρext(r,t), and calculate response.

ρ(r,t) = ρext(r,t) + ρind(r,t) (2.50) 1 ϕ(r,t) = ϕext(r,t) (2.51) ǫr

2ϕ = 4πρ q2ϕ(q) = 4πρ(q) (2.52) ∇ − 2ϕ = 4πρ q2ϕ (q) = 4πρ (q) (2.53) ∇ ext − ext ext ext (2.54)

14 20 April 2006 PHY240C Lecture 5

1 ρ(q) = ρext(q) (2.55) ǫr 1 ρ + ρ ρ = ext ind = 1 + ind (2.56) ǫr ρext ρext 4π ρind = 1 + 2 (2.57) q ϕext ρ = en (2.58) ind − ind U = eϕ (2.59) ρ − n ind = e2 ind (2.60) ϕext U 2 1 4πe nind = 1 + 2 (2.61) ǫr q U

4πe2 is the Fourier transform of the Coulomb interaction. q2

n(r,t) = ψ ei(H+Hext)tn(r)e−i(H+Hext)t ψ (2.62) D b b b b E If Hext = Hext(t), then b

t t b b ′ ′ iHt −iHt ′ ′ n(r,t) = ψ exp i Hext(t ) dt e n(r)e exp i Hext(t ) dt ψ (2.63) ˆ−∞ − ˆ−∞ b b The above formula is the “interaction representation”.b b b

2.8.1 Linear Response Expand the exponential to ﬁrst order

n(r,t) = eiHtn(r)e−iHt (2.64) b b t t n(r,t) = ψ 1 + i H (t′) dt′ n(r,t) 1 i H (t′) dt′ ψ (2.65) b b ˆ ext − ˆ ext −∞ −∞ = n0 + nind b b b (2.66) t ′ ′ nind(r,t) = i dt ψ Hext(t ), n(r,t) ψ (2.67) ˆ−∞ D h i E H (t′) = dr′ U (r′,t′b)n(r′,t′)b (2.68) ext ˆ ext t b ′ ′ ′ ′ ′ nind(r,t) = i dt dr ψb [n(r ,t ), n(r,t)] ψ U(r ,t ) (2.69) ˆ−∞ ˆ h i Π(r r′,t t′) = ψ [n(r′,t′), n(r,t)] ψ (2.70) − − h b i b Next, since we note that we have a convolution, we immediately go to Fourier space since the convolution is b b a product.

nind(q, ω) = Π(q, ω) Uext(q, ω) (2.71) n · ind = Π(q, ω) (2.72) U 1 4πe2 q = 1 + 2 Π( , ω) (2.73) ǫr q · Kubo-Greenwood discovered that the commutator-type correlation function gives the linear response to external perturbations.

15 20 April 2006 PHY240C Lecture 5

Π is a retarded correlator. There are no restrictions in space, but t′

t Π(q, ω) = i ψ [n(q,t′), n( q,t)] ψ dt′ (2.74) ˆ−∞ h − i 0 = i dt eibωt ψ [nb(q,t′), n( q, 0)] ψ dt′ (2.75) ˆ−∞ h − i “ ”: Adiabatic switching. b b −∞

δt lim Uexte (2.76) δ→0 t∈[−∞,0]

i(ω−∆ε)t 1 Re e U(t) (2.77) ˆ → ω ∆ε + iδ − Turning on the interaction slowly pushes the pole oﬀ the real axis.

16 25 April 2006 PHY240C Lecture 6

Last Time

• Studying collective behavior of interaction electrons • Determined generic formula of the linear response of the system. • Correlation functions of the system will capture the response of the system more appropriately. • Commutator correlation function • Derived for speciﬁc case but is generally true.

2.9 Models of Dielectric Functions 2.9.1 Thomas-Fermi Approximation

2U(r) = 4πe [ρ (r) + ρ (r)] = 4πeρ(r) (2.78) ∇ ext scr ρ(r) = en(r) n(r) Number density (2.79) − ≡ (Note: U is potential energy.) Assume that locally the system is a free electron system.

k3 (r) 4 k 3 n = F 2 F π = n (2.80) 3π2 · 3 2π 0 " # Introduce a space-dependent Fermi momentum as a result of the test charge. Since the test charge is a spatially varying quantity, it is reasonable that its eﬀect on the Fermi momentum is also spatially variable. k2 (r) E = F + U(r) (2.81) F 2m k3 (r) = [2m(E U(R))]3/2 (2.82) F F − U(r) 3/2 n(r) = n 1 (2.83) 0 · − E F k3 n = F,0 (2.84) 0 3π2 U(r) 3/2 2U(r = 4πe ρ + en en 1 (2.85) ∇ ext 0 − 0 − E " F # U(r) 3/2 ρ (r) = en en 1 (2.86) scr 0 − 0 − E F When U = 0, we have ρscr = 0. When U = 0, we have a non-trivial screening charge. Since we are performing a small perturbation, we can expand the6 term containing U(r) in the parentheses, yielding

2 2 6πn0e U(r) = 4πeρext + U(r) (2.87) ∇ EF Transforming to Fourier space yields

q2U(q) = 4πe[ρ (q) + ρ (q)] (2.88) − ext scr 2 2 2 2 6πn0e q qTF U(q) = 4πeρext(q) qTF = (2.89) − − EF 4πeρ (q) q ext U( ) = 2 2 (2.90) − q + qTF ρ (q2 + q2 )U(q)/( 4πe) q2 + q2 ǫ (q) = ext = TF − = TF (2.91) r ρ + ρ q2U(q)/( 4πe) q2 ext scr − Note: Kondo problem is that of magnetic impurities.

17 25 April 2006 PHY240C Lecture 6

2.9.2 Special Case: Point Charge

ρ (r) Q δ(r) (2.92) ext ≡ · ρ (q) = Q const (2.93) ext · ddq 4πe Q r iq·r U( ) = d e −2 ·2 (2.94) ˆ (2π) q + qTF ∞ 1 4πeQ 2 iqrx 1 = 3 2π dq q dxe 2 2 (2.95) − (2π) · ˆ0 ˆ−1 q + qTF eQ ∞ q2 2 sin qr = dq 2 2 (2.96) − π ˆ0 q + qTF qr 2eQ 1 ∞ q = dq 2 2 sin qr (2.97) − π r ˆ0 q + qTF eQ 1 ∞ q = dq 2 2 sin qr (2.98) − π r ˆ−∞ q + qTF

We wish to do a contour integral since we have poles at iq . ± TF eQ 1 E 1/2 U(r) = e−qTFr = F = r (2.99) − r q 6πe2n TF TF The motivated student will calculate the induced charge density and calculate the integral of the charge density and ﬁnd that it equals

Total screening charge ddr ρ (r) = Q (2.100) ≡ ˆ scr −

(Derivation assumed that our electrons were free fermions. Thus, our work is not valid for a dielectric or an insulator.) Next level: Fock approximation ln(k kF ) in ε(k). This non-analyticity DESTROYS exponential decay, gives rise to − cos 2k r U(r) F Friedel Oscillations (2.101) ∼ r3

2.9.3 Self-Consistent Theory

2 1 4πe nind(q, ω) = 1 + 2 (2.102) ǫr q Uext(q, ω) 1 U(q, ω) = Uext(q, ω) (2.103) ǫr 2 1 4πe nind(q, ω) 1 = 1 + 2 (2.104) ǫr q U(q, ω) ǫr 4πe2 n (q, ω) ǫ = 1 ind (2.105) r − q2 U(q, ω)

Both ǫr and 1/ǫr represent a response to the system. 1/ǫr is the response to an external perturbation. ǫr is the response to any kind of potential.

18 27 April 2006 PHY240C Lecture 7

Last Time

• System’s response will be related to its correlation

• Referred to as retarded since they require times from negative inﬁnity

• Thomas-Fermi approximation: Assume perturbation is not too strong. Can view the electron locally as being a Fermi system with the parameters (Fermi momentum) being spatially dependent. Fermi energy must remain constant since the system is in equilibrium.

k2 (r) E = F + U(r) F 2m

• For case of a single point charge: Get induced screening cloud with exponential decay.

• 1/ǫr is the response to external perturbations.

• ǫr is the response to all of the perturbations.

Question Why do we work in second quantization? It is very natural to work in this formulation. Many of the physical processes result in changes in particle number (absorption of a photon by a solid). It is first quantized functions times bookkeeping. Will later learn that superconducting state will be very hard to write in first quantized form. It is very hard to mix wavefuctions that describe different particle numbers.

Self-Consistency Screening cont.

2 1 nin 4πe = 1 + U0 U0 = 2 (2.106) ǫr Uext q

U0 is the bare Coulomb interaction.

1 = 1 + U0Π(q, ω) (2.107) ǫr n ǫ = 1 U ind = 1 U Π(q, ω) (2.108) r − 0 U − 0 1 1 + U0Π = e (2.109) 1 U Π − 0 Π Π = e (2.110) 1 U0Π 1 −e = 1 + U0Π = 1 + UeﬀΠ (2.111) ǫr e U0 Ueﬀ = = U0 + U0eΠU0 + (2.112) 1 U Π · · · − 0 Looks like a geometric series. Indeed, one can visualize this as e e

+ + + · · · U 0 U0 Π U0 U0 Π U0 Π U0

Viewing as an eﬀective interaction ande viewing as an eﬀective correlatione are physicallye equivalent.

19 27 April 2006 PHY240C Lecture 7

Eﬀective Interaction as a geometric series:

• Originally called the Random Phase Approximation(RPA)

• David Bohm and David Pines

• First attempt to take a series and sum it up to inﬁnite order in the interaction.

• Contains high order ( ) powers of U . ∞ 0 • Non-perturbative result.

• Need U0Π

• It is a subset.e Need to convince audience that this is the most important subset of perturbative corrections. It is conceivable that another subset will be more important. For example, in superconductivity we have a solution which is similar to RPA but yields a diﬀerent set of diagrams.

Now, let us ﬁnd the correlation function Π.

0 Π = i dt eiωt [n(q,t), n( q, 0)] e (2.113) ˆ−∞ h − i e Assume free electron gas b b

0 iωt 1 i(εk−εk+q)t i(εk−εk+q)t Π = i dt e f(εk)[1 f(εk q)]e f(εk q)[1 f(εk)]e (2.114) ˆ V − + − + − −∞ k X,σ n o e 0 1 i(ω−(εk+q−εk))t = i dt e [f(εk) f(εk q)] (2.115) ˆ V − + −∞ k X,σ 1 f(εk) f(εk q) = − + (2.116) V ω (εk q εk) + iδ k + X,σ − − Plugging this back into our formula for the dielectric constant yields

4πe2 2 f(ε ) f(ε ) RPA q k k+q ǫr ( , ω) = 1 2 − (2.117) − q · V ω (εk q εk) + iδ k + X − − Memorize the above formula!

′ ′′ ǫr = ǫr + iǫr (2.118) 2 ′ 4πe 2 f(εk) f(εk+q) ǫr = 1 2 − (2.119) − q V P ω (εk q εk) + iδ k − + − 2 2 X ′′ 4π e 2 ǫ = [f(εk) f(εk q)]δ(ω (εk q εk)) (2.120) r q2 · V − + − + − k X Principle Part( ): (Look this up in Boas and Arfken to refresh) P 1 f(x) = = P ( 1 ) + iπδ(x) (2.121) x x ∞ 1 −ǫ ∞ 1 dx g(x) = + dx g(x) + iπg(0) (2.122) ˆ−∞ x ˆ−∞ ˆǫ x

20 02 May 2006 PHY240C Lecture 8

Last Time

• Linear response

• Studying model dielectric functions

• Thomas-Fermi approach

• Diﬀerence between 1/ǫ and ǫ. 1/ǫ is response to external perturbation. ǫ is response to total potential.

• Determined ǫ for a point charge.

2 ′ 4πe 2 f(εk) f(εk+q) ǫr = 1 2 − (2.123) − q V P ω (εk q εk) + iδ k + X − − 2.9.4 Limiting Cases The order of the limits is important. Many of these response functions are not analytic, thus the order is important and physical circumstances can aid in determining the proper order. Example: a photon is wiggling electrons at very high frequency but low momentum. In contrast, the phonons are wiggling at low frequency but high momentum.

1. ω 0, q 0 → → As q 0, we Taylor expand f(εk) f(εk q) → − + 2 ∂f ′ 4πe 2 ∂ε (εk+q εk) ǫr = 1 2 − − (2.124) − q V (εk q εk) k + X − − 8πe2 d3k ∂f = 1 (2.125) − q2 ˆ (2π)3 ∂ε

Convert to energy integral.

k2 ε = k = √2mε (2.126) 2m → 1 dk = √2m dε (2.127) 2√ε ∂f = δ(ε ε ) (2.128) ∂ε − − F 8πe2 4π ∞ 1 1 ′ √ ǫr = 1 + 2 3 dε 2m 2mε δ(ε εF ) (2.129) q 8π ˆ0 2 √ε · − e2 4 ∞ √ 3/2 √ = 1 + 2 2 m dε ε δ(ε εF ) (2.130) q · π · · ˆ0 · − e2 4√2 4me2k = 1 + m3/2 √ε = 1 + F (2.131) q2 · π · F πq2 3 2 2 2 kF 2 q 6πe n 6πe 2 4me k ǫTF = 1 + TF q2 = 0 = 3π = F (2.132) 2 TF k2 q εF F π 2m

2. q 0, ω 0 → → 2 ′ 4πe 2 1 1 ǫr = 1 2 f(εk) (2.133) − q · V · ω (εk q εk) − ω (εk q + εk) k + + X − − −

21 02 May 2006 PHY240C Lecture 8

Shifted k k q in second term. → − 2 ′ 4πe 2 2(εk+q εk) ǫr = 1 2 f(εk) 2 − 2 (2.134) − q · V · ω (εk q εk) k + X − − k·q q2 4πe2 2 2 2 2m + 2m = 1 f(εk) (2.135) − q2 · V · h ω2 i k X In the denominator, we let q 0, resulting in no k or q dependence. Next, in the numerator k q = kq cos θ and dkk q = 0. → · ´ · 2 2 ′ 4πe q 2 ǫ = 1 f(εk) (2.136) r − q2 · mω2 · V k X Thus, we have

ω2 4πe2n ǫ′ = 1 pl ω2 = (2.137) r − ω2 pl m

where ωpl is the plasma frequency! Next, recall

2 1 4πe nind = 1 + 2 (2.138) ǫr q Uext

If ω of Uext is ωpl, the system will produce a huge response (plasma oscillations). This is a homogeneous excitation with the overall center of charge for the positive charges and negative charges oscillate out of phase.

3. ω 0, q finite → q2 1 1 x + 1 q ǫ′ = 1 + TF + (1 x2) ln | | x (2.139) r q2 2 4x − x 1 ≡ 2k | − | F As q 0 (x 0) the above reduces to Thomas-Fermi results. First calculated by Lindhard. Note: Non-analytic→ at→x 1. If we FT back to real space, we find the screening cloud n to be ∼ ind cos 2k r n (r) = Q F (2.140) ind · r3 This differs from Thomas-Fermi as the screening cloud oscillates in density. Because of the non-analyticity, the large q behavior influences the large r behavior. A consequence of the Fermi surface is that response functions are oscillatory with a power law envelope. Decays exponentially?

22 02 May 2006 PHY240C Lecture 8

4. Low dimensions q2 q ǫ (ω = 0, q) = 1 + TF F (2.141) r q2 2k F

F (x) 1D

2D 3D

1 x

Figure 7: Fermi function in 1(red), 2(black) and 3(blue) dimensions

1/ǫ diverging is a good way of capturing excitations of the system. ǫ diverging is a good way of ﬁnding instabilities of the system. The natural excitations of the system will be bosons (electron-hole pairs). The instability manifests itself by the nature of the excitations from electronic to bosonic. Example: Superconductivity.

5. “Nesting” Everything we have done previously assumed that the Fermi surface is spherical, i.e. that everything is isotropic or that there is no dependence on the angle. What happens if the Fermi surface is anisotropic?

Figure 8: Example of nesting for a rectangular Fermi surface

The phase space for a given q where the energy diﬀerence is small ( 0) is much larger when we move away from a spherical Fermi surface. ∼ 1 ǫr large (2.142) ∼ εk+q εk X − ǫr can diverge.

23 02 May 2006 PHY240C Lecture 8

n > 1

n < 1 ky

n = 1

Figure 9: Fermi surface for the 2D Hubbard model at less than (red), equal to (black), and greater than (blue) half-ﬁlling

Nesting in 2D Hubbard model represents instability of anti-ferromagnetism. Reference for Nesting by Ruralds-Virosztek “Nested Fermi-Liquid Theory”.

24 04 May 2006 PHY240C Lecture 9

Last Time / Announcement

• No class next Tuesday. Replacement class 5/16 from 12:30-2:00pm.

• Looked at general formula given by RPA theory.

• Studied small q behavior and found Thomas-Fermi result

• Studied small ω behavior and recovered behavior of plasmas.

2.9.5 Eﬀect of ﬁnite relaxation time Previously assumed in RPA that wave functions were plane waves. What happens if there is disorder? The plane waves will cease to be eigenstates. Let us add an exponential decay to the plane waves.

p p p p −i ε − i t c (t) = c e−iεkt c e−iεkt e−t/2τ = c e ( k 2τ ) (2.143) k k ⇒ k · k Shift frequency to complex plane. For holes, we ﬁnd b b b b p p −i ε + i t c (t) = = c e ( k 2τ ) (2.144) k · · · k Holes can be viewed as particles moving backwards in time. Applying to RPA formula b b 2 4πe 2 f(εk) f(εk q) ǫ = 1 − + (2.145) r − q2 V i k ω (εk+q εk) + τ X − −

εk+q is the energy of an electron, and εk is the energy of the corresponding hole. RPA is describing the creation of a particle-hole pair. i/τ is referred to as the “self-energy” Σ of the Green’s function.

2.9.6 Connection in RPA Maxwell’s equation

E = ϕ (2.146) −∇ E = iqϕ (2.147) − eE = iq( eϕ) = iqU (2.148) − ieq E = q2U (2.149) − · The dielectric constant becomes 4πe2 n 1 ρ 1 ǫ = 1 ind =1+i ind (ρ = en) extra factor (2.150) r − q2 U ǫ q E − 4πǫ 0 · 0 Conductivity is related to currents. Continuity equation relates density to current.

∂ρ(r,t) j(r,t) + = 0 (2.151) ∇ · ∂t iq j(q, ω) iωρ(q, ω) = 0 (2.152) · − Solving for the density q j ρ = · (2.153) ω Recall Ohm’s Law

j(q, ω) = σ(q, ω)E(q, ω) (2.154)

25 04 May 2006 PHY240C Lecture 9

Inserting into our equation for ǫr yields

1 σ(q, ω) σ(q, ω) ǫr(q, ω)=1+i ǫ(q, ω) = ǫ0 + i (2.155) ǫ0 ω ω

Now recall that 4πe2 1 e2 q q q ǫr( , ω) = 1 2 Π( , ω) = 1 2 Π( , ω) (2.156) − q − ǫ0 q iωe2 σ(q, ω) = Π(q,e ω) e (2.157) q2

In the ﬁnite ω, q 0 limit we have calculatedeΠ for the plasmons → n q2 Π(q, ω) =e e (2.158) mω in e2 e σ(ω) = lim σ(q, ω) = e (2.159) q→0 mω

This is the conductivity for a free electron. If the electron lifetime is ﬁnite, then ω ω + i → τ 2 nee 1 σ(ω) = i (2.160) m ω + τ In the limit where ω 0, we ﬁnd → n e2τ σ(ω = 0) = e Recovered Drüde form (2.161) m

For general q, ω 1 1 ρ = q j n = q j (2.162) ω · ω · n The conductivity becomes

iωe2 i ∞ σ = dt eiωt [n(q,t), n( q, 0)] (2.163) q2 −V ˆ h − i 0 e2 1 ∞ = dt eiωt [j(q,t), j( q, 0)] Kubo-Greenwood (2.164) ω V ˆ0 h − i Full formula is e2 1 ∞ ine2 iωt j q j q σ = dt; e [ ( ,t), ( , 0)] + 2 (2.165) ω V ˆ0 h − i mω Where does the second term come from?

p2 (p e A)2 p2 e H = − c = + p A + A p (2.166) 2m → 2m 2m 2mc · · bp be h i bv = i[bH, r] = + A b b b (2.167) m mc b b b b b b

26 11 May 2006 PHY240C Lecture 10

Last Time

• Studied eﬀect of ﬁnite relaxation time

• Determined relationship between ǫ and σ in general.

σ(q, ω) ǫ(q, ω) = ǫ + i 0 ω

• What happens when we add this eﬀect to RPA.

• In the q 0, ω 0 limit, we get the Drüde form for the conductivity. → → • Related conductivity to current-current correlation function (Kubo-Greenwood).

2.10 Response to Magnetic Perturbation In a magnetic ﬁeld 1 ε(H) = ε(H = 0) gµ σ H (2.168) − 2 B ·

σ is the spin of the electron. µB is the Bohr magneton. g is the gyro-magnetic ratio. Response of the system (susceptibility) is

∞ M i iωt χ = ~µ0 dt e [m(q,t),m( q, 0)] (2.169) ≡ Hext ˆ0 h − i Correlation function of the quantity the external ﬁeld couples to. 1 χ(0, 0) = χ = g2µ2 ρ(ε ) Pauli magnetic susceptibility (2.170) 0 2 B F mk ρ(ε ) is the DOS at the Fermi Surface, F (3D). The susceptibility in the RPA approximation is F 2π2 χ χRPA = 0 (2.171) 1 U ρ(ε ) − · F A magnetic response is enhanced because of an electric interaction. Stoner instability is when χ as U increases. Stoner criterion indicates the onset of a ferromagnetic instability. M = 0 → ∞ h i 6

M = 0 M = 0 h i h i 6 Uρ(εF) 0 1

Hartree approximation: Allow diﬀerent densities for up and down electrons.

ψ†ψ = ψ†ψ (2.172) ↑ ↑ 6 ↓ ↓ D E D E Energies of up and down spin electrons will shift diﬀerently. Systems that want to be antiferromagnetic have the same Stoner relation but not at q = 0, instead at q = 0. χ(q ) 6 AF → ∞

27 11 May 2006 PHY240C Lecture 10

n↑,n↓

n↑ 1 2 n↓

Uρ(εF) 1

Figure 10: Electron density versus Uρ(εF) N(ε)

↑

εF

↓

Figure 11: Density of States for up and down spin electrons with (solid black) and without (dashed red) applied magnetic ﬁeld

3 Scattering Experiments

Previously looked at what happens when you expose a system to a large external perturbation.

3.1 Electron Scattering Interaction: Coulomb

V = e eαV (R rα) = e ddrV (R r)ρ(r) (3.1) · − ˆ − R rα X, Fermi’s Golden Rule 2π P = V 2δ(E E ~ω) (3.2) i→f ~ | f i| f − i − t e V t e = V (3.3) h f f i ii f i 1. electron matrix element

k V (R r) k = V (k)e−ik·r k = k k (3.4) h F − ii f − i 2. target/ion matrix element

V = eV (k) ddr e−ik·r i ρ(r) f (3.5) f i ˆ h i b 28 11 May 2006 PHY240C Lecture 10

Insert eiHte−iHt before and after the density operator, yielding

b b ∞ iωt i ρ(r) f δ(Ef Ei ~ω) = dt e i ρ(r,t) f (3.6) h i · − − ˆ−∞ h i 2 2 ∞ e V (k) ′ P = | | ddr ddr′ dt eiωte−ik·(r−r ) i ρ(r,t) f f ρ(r, 0) i (3.7) i→f ~2 ˆ ˆ ˆ−∞ · h i h i Summing over all ﬁnal states yields

2 2 e V (k) ′ P total = | | ddr ddr′ dt eiωte−ik·(r−r ) i ρ(r,t)ρ(r, 0) i (3.8) ~2 ˆ ˆ ˆ · h i

Response to scattering experiments is similar to linear response. We always measure correlation functions.

29 16 May 2006 PHY240C Lecture 11

Last Time

• Considered response to a magnetic perturbation. χ χRPA = 0 (3.9) 1 U ρ(ε ) − · F • Started discussing scattering experiments

• Calculated the total probability of going from the initial state to the ﬁnal state. Learned it was related to a correlation function.

• Correlation function is not retarded and does not have a commutator.

• Fluctuation Dissipation Theorem (FDT): There is a numerical factor coth kBT/~ω relating the retarded commutator correlation function with the simple correlation function.

• If you will study disordered systems (glasses), there is a violation of the Fluctuation Dissipation Theorem.

Electron Scattering Cont. What you measure is not the transition probability but the scattering cross-section σ.

incoming flux dσ = P (3.10) · i→f Xf ~k incoming flux = i (3.11) m 1 1 1 m Final state phase space ddk = k2 dk dΩ = e k dε dΩ (3.12) (2π)3 f (2π)3 f f f (2π)3 ~2 f f f d2σ k m2 e2 f e k = 3 5 4 Sρρ( , ω) Differential cross section (3.13) dεf dΩf ki (2π) ~ k

Diﬀerential cross section is a geometric factor times the density-density correlation function.

′ ′ S (k, ω) = dt dt′ dr dr′ eiω(t−t )−ik·(r−r ) ρ(r,t)ρ(r′,t′) (3.14) ρρ ˆ h i

3.2 Neutron Scattering If you bombard a charged particle with an electron, it will interact strongly with the particle. The electron will not penetrate deep into the sample and are better suited for surface probes. Neutrons do not interact strongly with the particle and will probe deeply into the sample.

3.2.1 Interaction with the nuclei via short range forces

V = V (R rα) = a ddr δ(R r)n(r) (3.15) − ˆ − α X The δ function describes a contact interaction. n(r) is the density of nucleons.

d2σ k m2 a2 f n k = 3 5 Snn( , ω) (3.16) dεf dΩf ki (2π) ~

2 e/k a, the scattering length and ρ(r,t) n(r,t). The ﬁrst term in Snn(k, ω): the Bragg peaks at ω = 0, i.e. elastic→ scattering. →

30 16 May 2006 PHY240C Lecture 11

3.2.2 Interaction with the nucleon spin The neutron spin interacts with the nucleon spin via a dipole interaction.

V = 2µ S (R)V (R r)M (r) (3.17) i ij − γ α X 1 1 V (r) = kˆ kˆ (3.18) ij −4π ∇i∇j r → i · j d2σ k (µ m )2 f n n kˆ kˆ k = 3 5 i j SMiMj ( , ω) (3.19) dεf dΩf ki (2π) ~ · i and j denote the polarization directions.

3.3 Light Scattering

d2σ 1 √ǫ ω 4 f k = 3 Sǫǫ( , ω) (3.20) dεf dΩf 4 (2π) c S is the correlation function of the ﬂuctuations of the dielectric constant. In an isotropic liquid, S S . ǫǫ ǫǫ ∼ nn In a liquid crystal, Sǫǫ consists of many contributions.

3.4 Magnetic Scattering of Neutrons 1. No thermal motion T = 0 ∞ ′ S(k, ω) = A dt eiωt e−ik·R(0)eik·R (t) = δ(ω) δ(k G) (3.21) ˆ − −∞ ′ G RX,R D E X If your system is at T = 0, then when neutrons are scattered their energy and momentum will be conserved up to a reciprocal lattice vector. We will have peaks in the directions of the reciprocal lattice vectors. (NOTE: Fermi’s Golden rule only applies when the interaction strength is small. If the interaction is strong, then we can have multiple scattering events.)

2. With thermal motion T = 0. The ions are moving at ﬁnite temperatures. R(t) = Ri + ui(t). The ui(t) form phonons, characterizing6 ionic displacements from equilibrium position.

′ S(k,t) = eik·(R−R ) e−ik·u(R,0)eik·u(R,t) (3.22) ′ RX,R D E Theorem:

1 2 −iA iB − 2 ( (A−B) −[A,B]) e e = e h i (3.23) D b bE b b b b This is true only in systems where the expectation value is taken for Hamiltonians which are quadratic in A and B and if A, A, B . If and only if these two criteria are satisﬁed, then the above relation is true. Both criteriah areh satisﬁedii for phonons. b b b b b 1 i(q·R−ωqt) † −i(q·R−ωqt) u(R,t) = aqe + aqe (3.24) q 2NMωq X We have to produce: p

′ ′ 1 i(q·R −ωqt) iq·R † i(q·R−ωqt) −iq·R u(R,t) u(R, 0) = aq e e + aq e e (3.25) − q 2NMωq − − X h i 2 1p 1 b † † b [u(R,t) u(R, 0)] = (2 2 cos θRR′ )(aqaq + aqaq) + non-diag. (3.26) − 2NM q ωq − X ′ θRR′ = ωqt + q (R R ) b b b b (3.27) · −

31 16 May 2006 PHY240C Lecture 11

Can ignore non-diagonal terms since we are taking an expectation value and

† H = ωqaqaq (3.28) q X b Thus, the expectation value of all non-diagonal terms isb zero.b

′ 1 1 † iθ ′ 2i 1 [u(R ,t), u(R, 0)] = [a , a ]e RR + h.c. = sin θ ′ (3.29) 2NM ω q q 2NM ω RR q q q q X X Total exponent:b b b b

k2 1 = (2 nq + 1)(1 cos θRR′ ) i sin θRR′ (3.30) 2NM q ωq { h i − − } X

The θRR′ terms depend on time. We will expand the time dependent piece from the exponential. We will keep the 1 in the exponential and expand the cos and sin.

2 ′ ′ ik·(R−R ) k 1 iωqt i(q−k)·(R−R ) S(k,t) = e + nq + 1 e e 2NM ωq h i RR′ ( q X X h (3.31) ′ −iωqt −i(q+k)·(R−R ) −2W + nq e e e h i ) i where

k2 1 e−2W = exp (2 n + 1) (3.32) 2NM ω q (− q q h i ) X The Fourier Transform of the density autocorrelator factor is called the structure factor and it is given by

S(k, ω) = e−2W δ(ω) δ(k G) − ( G X k2 1 + δ(ω ωq) δ(k q G) nq + 1 (3.33) 2NM ωq − − − h i " q G X X 1 + δ(ω + ωq) δ(k + q G) nq + two phonon processes ωq − h i G #) X The first term yields Bragg peaks. The second term describes one phonon emission and absorption. e−2W is the Debye-Waller factor. The zero-point fluctuations scatter the neutrons. The sum is logarithmically divergent in 2D. This suppresses the scattering correlation function to zero. Mermin-Wagner Theorem: In d = 2 the zero point fluctuations destroy the correlation function S(k). NO LATTICE in d = 2. There are no elastic lattices in two dimensions.

32 16 May 2006 PHY240C Lecture 12

Question What happens if the magnetic susceptibility from the Stoner criterion becomes negative? We can only believe the Stoner criterion up to divergence. The system becomes unstable and develops a ﬁnite magnetization.

3.5 Neutron Scattering Experiment Spallation Neutron Sources produce neutrons without radiation. target

kf ﬁlter

ki ﬁlter

two choppers

Figure 12: Schematic of a neutron scattering experiment where the energy is measured via the “time of ﬂight” technique.

Scan for ω, k( k k ). ≡ out − in k

spin waves S = S in 6 out

acoustic phonons Sin = Sout

optical phonons ω

4 Total Energy of the Electron Gas

E = H (4.1) h i Natural length scales: b ~2 r¯s 4 3 V rs = a0 = 2 πr¯s = (4.2) a0 me 3 N

33 16 May 2006 PHY240C Lecture 12

a0 is the radius of s orbit in Hydrogen. r¯s is the space per electron. When rs is large, the system is dilute. When rs is small, the system is dense. When rs is large, the dominant contribution to come from the potential/interaction term. When rs is small, the dominant term is from the kinetic term. 1. If there is no interaction, we have k2 V k2 V ~2 k5 3 E = = 2 d3k = F = N ε (4.3) kin 2m (2π)3 ˆ 2m π2 2m 5 5 F k X 2. Hartree: cancels the electron-ion interaction 2.21 EH = 2 Ry (4.4) rs 3. Fock:

εk εk U(q)εk−q (4.5) → − q X 2.21 0.91 EHF = 2 (4.6) rs − rs 4. RPA approximation singles out diagrams which form rings

(a) Dense case, rs small 2.21 0.91 E = 2 0.096 + 0.62 ln rs + (rs) (4.7) rs − rs − O 2.21 2 2 = 2 1 + a1rs + a2rs + a3rs ln rs (4.8) rs · · · This is a power series in positive powers of rs and is expected to be valid for small rs

(b) Dilute case, rs large Start with interaction (put ~ = 0). To optimize the repulsive energy, electrons will form a lattice. (Wigner crystal (WX))

Figure 13: Optimal Wigner crystal, a triangular lattice.

1 1.8 EWX = c = Ry (4.9) r − r ij ij s X | | WX 1.8 3.0 1.8 −1/2 −1 E = + 3/2 = 1 + a1r + a2rs + (4.10) ⇒ − rs r − rs · · · s h i

34 16 May 2006 PHY240C Lecture 12

Expansion in negative increasing powers of rs.

There is a phase transition when moving from small rs to large rs. The Wigner crystal is an insulator.

E HF ρ

n small (insulator)

rs ?

WX n large (metal)

Phase transition T

Figure 14: Left: Energy versus rs. Right: Resistivity versus temperature. HF WX rs electron gas 38 0 ?

Figure 15: Phase diagram in 2D. Possibilities for the unknown phase in the middle include stripes or electronic microemulsions

35 18 May 2006 PHY240C Lecture 13

Last Time

• Neutrons pick up the structure of the system and phonon peaks as well. • Neutrons have a weight comparable to the ions. Electrons are so much lighter that you cannot simulta- neously satisfy momentum and energy conservation with phonons. • Acoustic phonons have no gap. • Phonons in metals also have acoustic behavior. • Energy of electron gas in small and high density limit.

5 Fermi Liquid Theory

• Developed by Landau in the early forties. • Started out as a phenomenology but became very powerful. • Pursuit of a new type of small parameter in the problem of strong interactions.

Basic question: Why can the dense, strongly interacting fermion gas (like the electrons in metals) be described as more or less the free electrons only with changed mass, etc. ? Phenomonology: We accept this fact as a basis. 1. The states are still characterized by a momentum quantum number p. 2. The interaction only changes parameters: m m∗. → p2 ε = (5.1) 2m∗

3. n(p) still has a jump at the Fermi surface ε = εF .

n(p)

pF p

where z is the “pole strength of the Green’s function.” Also, the Fermi volume remains unchanged (Luttinger Theorem). 4. 1-4 deﬁne the Landau quasi-particles. The eﬀect of strong interaction is captured through the parameters m∗ and z. The remaining interaction is weak. 5. Lifetime of the quasi-particles is long. 1 ε2 (ε = E E ) (5.2) τ(ε) ∼ − F

36 18 May 2006 PHY240C Lecture 13

1 3 F = E µN = F + (ε µ)δnp + fp p′ ′ δnp δnp′ ′ + ((δn) ) (5.3) − 0 p − 2V σ, σ σ σ O p pp′ X X

F0 results when the interaction is zero. δnp is the small ﬂuctuation or variations in occupation number of the quasi-particle caused by the interactions. The interaction is weak because the δn’s are hypothesized to be small. Parametrization:

s a fp p′ ′ = δ ′ f + σ ′′ σ ′′ ′ f ′ (5.4) σ, σ σσ pp σσ · σ σ pp s stands for symmetric. a stands for anti-symmetric.

• No spin-orbit interaction, would be relevant for heavy elements. • No crystalline anisotropies

2. All p, p′ have a magnitude p . ≈ F ∞ s,a 1 s,a f ′ = F P (cos θ) (5.5) pp N(0) ℓ ℓ Xℓ=0 m∗p N(0) = ρ(ε ) = F (3D) (5.6) F π2~3

We want to calculate the particle current in two ways. Equating them will give a constraint on the F ’s. Continuity equation:

d ∂n (r,t) n = p + n r˙ + n p˙ = 0r ˙ = v = ε p˙ = ε (5.7) dt p ∂t ∇r p · ∇p p p ∇p p −∇r p

37 23 May 2006 PHY240C Lecture 14

Last Time

• Knowing that the interactions between electrons are strong, we capture these strong interactions by forming quasi-particles (characterized by m∗ and z). • The quasi-particles interact weakly.

Fermi Liquid Theory Cont.

Recall dn ∂n p = p + n r˙ + n p˙ = 0 r˙ = ε p˙ = ε dt ∂t ∇r p · ∇p p · ∇p p −∇r p Also

0 np(r,t) = np + δnp(r,t) (5.8) 0 ∂ δnp(r,t) + vp δnp(r,t) pn fpp′ δnp′ (r,t) = 0 (5.9) t · ∇r − ∇ p · ∇r p′ X ∂ n0 Recall np = vp . Thus, the last two terms can be written ∇ − ∂ε

∂n0 r vp δnp fpp′ δnp′ (5.10) ∇ ·   − ∂εp  p′  X    The terms in the braces must be the current j since by the continuity equation ∂ δn + j = 0  p  t p ∇r · 0 ˜ ∂n J = jp = δnpvp vpfpp′ δnp′ (5.11) − ∂εp p p pp′ X X X ∂n0 = δnpvp vp′ fpp′ δnp (5.12) − ∂ε ′ p pp′ p X X ∂n0 = δnp vp fpp′ vp′ = δnpjp (5.13)  − ∂ε ′  p p′ p p X X X   • Eqn. (5.10): δnp induces other ﬂuctuations with amplitude δnp′ .

• Eqn. (5.13) vp current experiences/induces other current vp′ . The second term in (5.13) is called drag current or back ﬂow.

5.1 Transport in Fermi Liquid Insert an extra electron into the Fermi Liquid with momentum p. We calculate the total current in two diﬀerent ways. 1. If eﬀects of the crystal are disregarded (i.e. the system is Galilean invariant), then p J = (5.14) m

2. J after a while will be given by our jp p p ∂n0 p′ J ′ = = ∗ fpp ∗ (5.15) m m − ∂εp′ m p′ p Xp = + fpp′ z δ(εp µ) cos ϑ (5.16) m∗ m∗ · − p′ X

38 23 May 2006 PHY240C Lecture 15

ϑ is the angle between p and p′

p p s = ∗ 1 dΩ cos ϑ Fℓ Pℓ(cos ϑ) F = zN(0)f (5.17) m m − ˆ ! Xℓ p 1 = 1 + F s (5.18) m∗ 3 1 ′ The δ(εp′ µ) forces the sum over p to be an angular integral over the Fermi Surface. Thus, the eﬀective mass ratio− is

m∗ 1 = 1 + F s (5.19) m 3 1

∗ s s,a Thus, a measurement of m /m gives F1 . This is a non-perturbative results. Fℓ can be > 1, as long as the system is a Fermi-liquid. Formula will be true until we hit a phase transition!

39 23 May 2006 PHY240C Lecture 15

• Because of the interactions, the eﬀective mass of the particles increases. m∗ 1 = 1 + F s m 3 1

∗ s • If the interaction is completely isotropic, m = m (F1 = 0). • In terms of momentum exchange, the Coulomb interaction is anisotropic. • Eﬀective mass can decrease (superconductivity)

Other similar calculations yield

s 1 + F0 Compressibility ? = 1 s (5.20) 1 + 3 F1 c 2 1 + F s Normal sound velocity = 0 (5.21) c 1 + 1 F s 0 3 1 s s We have two equations and two unknowns (F0 and F1 ). We have no predictive power yet. Landau predicted s zero sound, which is related to F0 . Landau theory now has predictive power and is falsiﬁable. Let us go back to the continuity equation (summing over all p).

∂δn ∂n0 + r vp δnp fpp′ δnp′ = 0 (5.22) ∂t ∇   − ∂εp  p p′ X X    In Fourier space  

i(k·r−ωt) δnp(r,t) = δnpe (5.23) ∂n0 (v k ω)δnp v k fpp′ δnp′ = 0 (5.24) · − − · ∂εp p′ X ∂n0 Recall = zδ(εp µ). Let us consider the case when ∂εp − −

δnp = δ(εp µ)ν(ϑ, ϕ) (5.25) − − ϑ and ϕ describe the angle between p and k. It follows that

(v k ω)ν + v k dΩ F s(ϑ′)ν(ϑ′) = 0 (5.26) · − · ˆ ℓ Xℓ The normal sound of the system is a density ﬂuctuation. ω u˜ = velocity of non-isotropic sound (5.27) k | | v = v is the Fermi velocity. | | F u˜ s = (5.28) vF Let us rewrite our previous result in terms of s.

(s cos ϑ)ν(ϑ) = cos ϑ dΩ F s(ϑ′)ν(ϑ′) (5.29) − ˆ ℓ Xℓ cos ϑ v(ϑ) ∝ (5.30) s cos ϑ −

40 23 May 2006 PHY240C Lecture 15

s Question: Can s be chosen such that the LHS = RHS? We assume that only F0 is large, which yields s s + 1 1 ln 1 = (5.31) 2 s 1 − F s − 0

3 1 In He this anisotropic sound “zero sound” was indeed observed. Normal sound vn.s. = vF . The ex- √3 perimental conﬁrmation is that people in fermionic systems observed sound velocities that were profoundly diﬀerent from the expected and Landau’s prediction was the only “game in town” to account for the change in normal sound.

5.2 Spin-Dependent Case

1 s χ 1 + 3 F1 = a (5.32) χ0 1 + F0

a This formula is of analogous form as the Stoner criterion (RPA). F0 1 is the equivalent of the Stoner criterion. This yields a magnetic instability (onset of ferromagnetism).→ − K. Bedell (foremost expert in ferromagnetism) worked out the relationship between DMFT and Fermi-liquid theory (see Phys. Rev. Lett. ∗ 2 from past few years). The numerator comes from the density of states. N(0) = m pF /2π Values for 3He: F s 10, F s 6 and F a 0.67. 0 ∼ 1 ∼ 0 ∼ − Question: How do we measure m∗? 1 1. R = to get value for n H nec ne2τ 1 2. σ = m · 1 + ω2τ 2

41 25 May 2006 PHY240C Lecture 16

6 Theory of Superconductivity

• Observed in 1911 by Kammerlimph Onnes (Leiden) ρ Hg

T 4

• 1957: Bardeen, Cooper and Schrieﬀer

6.1 Insights 1 1. Isotope effect Tc (1952-3) Serin et. al. ∼ √M 2. Total absence of scattering: electrons formed single quantum state. Every classical liquid has a finite viscosity (resistivity). The effect that the electrons are showing no viscosity (resistivity) is an inherently quantum phenomena. Only a single quantum state is unable to scatter off itself. 3. Macroscopic (even double) occupation of a quantum state is prohibited by Pauli. Cooper: PAIR the electrons form bosons. → 4. Pairing explains • Gap in specific heat C

e−∆/T

T Tc

∆: pair binding energy • Resistivity = 0 explained To make resistivity ﬁnite one has to BREAK the pair. • Tunneling I

V ∆/e

42 25 May 2006 PHY240C Lecture 16

Question: How can phonons bind repulsive electrons? The answer is retardation.

positively polarized channel optimal e- path e- e-

− 1. Electron 1 (e1 ) positively polarizes the channel 2. Polarization remains long after electron 1 is gone. − − 3. Electron 2 (e2 ) can lower its energy by retracing e1 ’s channel.

6.2 Microscopic Theory 1. Two electrons (Cooper 1956) in the presence of a Fermi surface.

k −

Only ǫk > 0 states are available. iq·R ψ(r1, r2) = ϕq(ρ)e (6.1) q CM momentum. Let us consider only q = 0. ≡ ik·r1 −ik·r2 ψ(r1, r2) = ake e (6.2) k X The Hamiltonian is † † ψ H ψ = 2ǫka ak + Vkk′ a ak′ (6.3) h i k k k ′ X Xkk The ﬁrst term is the kinetic energy. The second describes an attractive interaction.

H ψ = E ψ (E 2ǫk) ak = Vkk′ ak′ (6.4) | i | i ⇒ − ′ Xk The interaction is b

V 0 < ǫk, ǫk′ < ωD Vkk′ = −| | (6.5) (0 otherwise The coeﬃcients are

V ak′ k′ ak = X (6.6) E 2ǫk − 1 ak = V ak′ (6.7) E 2ǫk k ′ k X Xk X −

43 25 May 2006 PHY240C Lecture 16

Since the sums over the ak’s are the same, we ﬁnd 1 1 = V V φ (6.8) E 2ǫk ≡ · k X −

1 repulsive V

1 attractive V

1 Figure 16: The red line indicates a positive or repulsive V . The green line indicates a negative or attractive 1 V . The diﬀerence between dotted lines is 2ǫk.

For attractive V , a single eigenvalue becomes negative. Thus, a BOUND state is present for arbitrary weak attraction.

1 ωD 1 N(0) 2ωD 1 N(0) 2ω + w N(0) 2ω = N(0) dǫ = dǫ = ln D | | ln D (6.9) V − ˆ w 2ǫ 2 ˆ ǫ + w 2 w ≈ 2 w | | 0 − 0 | | | | | | 2 w = 2ω exp − (6.10) | | D N(0)V This is inherently non-perturbative since there is no Taylor expansion for e−1/x

44 30 May 2006 PHY240C Lecture 17

Last Time

• From the isotope eﬀect Bardin concluded that phonons play an important role in superconductivity. • Since the resistivity is so low the electrons do not experience scattering and the electrons stay in a particular quantum state. • How can many electrons stay in the same quantum state when this is forbidden by the Pauli principle? We pair the electrons and they become bosons! • Arbitrarily weak attraction can form a bound state.

Microscopic Theory Cont. We neglected exchange! 2. Full many body calculation (following Schrieﬀer)

f ˆ† ˆ † † † H = ǫk nˆ + vkk′ b ′ bk b ′ = c c ǫk εk µ (6.11) red s ks k k k↑ −k↓ ≡ − k ′ Xs Xkk b † † ′ • Hred keeps only q = 0 in U ck′−qck+qck ck. This is down by a factor of 1/N (bad). ′ kX,k ,q • Theb ground state will be occupied by N bosons, ˆb†ˆb will bring out a factor of N.

ˆ† ˆ ˆ† ˆ Hred = 2ǫkbkbk + vkk′ bk′ bk (6.12) k ′ X Xkk b H seems to be quadratic and hence diagonalizable. However, our bosons are hard core bosons.

ˆ ˆ† b bk, b ′ = δkk′ [1 (nk↑ + n−k↓)] (6.13) k − h i2 ˆ† bk = 0 (6.14) The k is a quantum number, but it is not the momentum of the boson. The k reminds us that the boson is composed of two electrons with momentum k and k. − 6.3 Variational Approach Coherent state representation for bose-ﬁelds

† g (ˆa +ˆa- ) ψ = e k k k 0 (6.15) | i | i k Y b is a hard core boson, thus we have

† ψ = N 1 + gkˆb 0 Variational wave function (6.16) | i k | i k Y The normalization is given by

ψ ψ = N 2 (1 + g2 ) (6.17) h i k k Y Thus, the wave function becomes ˆ† 1 + gkbk † ψ = 0 = (uk + vkˆb ) 0 (6.18) | BCSi (1 + g2 )1/2 | i k | i k k k Y Y

45 30 May 2006 PHY240C Lecture 17

where

1 gk 2 2 uk = 2 1/2 vk = 2 1/2 uk + vk = 1 (6.19) (1 + gk) (1 + gk) The ground state energy is

2 E = ψ H ψ = 2ǫkv + vkk′ ukvkuk′ vk′ (6.20) GS h BCS| red| BCSi k k ′ X Xkk The second term is given by ˆ ˆ ˆ† ˆ ˆ† ˆ† 0 (uk′ + vk′ bk′ )(uk + vkbk)vkk′ bk′ bk(uk′ + vk′ bk′ )(uk + vkbk) 0 (6.21) D E We have to assure that there are no electrons with momentum k′. We wish to minimize the energy with 2 2 respect to gk. Using uk + vk = 1 1 u2 = (1 + a ) (6.22) k 2 k 2 1 v = (1 ak) (6.23) k 2 − ∂E 1 2 1/2 Minimizing with respect to a = 0 and noting ukvk = (1 a ) , yields ∂a 2 − k 1 2 1/2 1 2 1/2 E = ǫk(1 ak) + vkk′ (1 a ) (1 a ′ ) (6.24) GS − 2 − k 2 − k k ′ X Xkk ∂EGS 1 1 2 1/2 1 ak ′ ′ = 0 = ǫk + 2 vkk (1 ak ) 2 − 2 1/2 (6.25) ∂ak − · 2 2 − · · 2 (1 a ) k ′ k X Xk − 1 2 1/2 ∆k vkk′ (1 ak′ ) (6.26) ≡ ′ 2 − Xk ak ǫk = ∆k (6.27) (1 a2 )1/2 − k 2 2 ∆k 2 (1 ak) = 2 ak (6.28) − ǫk ∆2 −1 a2 = 1 + k (6.29) k ǫ2 k From this, we can get a self-consistent equation for ∆k.

1 2 1/2 ∆k = vkk′ (1 ak′ ) (6.30) −2 ′ − Xk 1/2 1 1 ′ = vkk 1 2 (6.31)  ∆ ′  −2 ′ − k k 1 + ǫ2 X  k′   2 2 1/2 1 ∆k′ /ǫk′ ′ = vkk 2 2 (6.32) −2 ′ 1 + ∆k′ /ǫk′ Xk ′ 1 ∆k 2 2 = vkk′ Ek = ǫk + ∆k (6.33) ′ −2 ′ Ek Xk q

V 0 < ǫk , ǫk′ < ωD vkk′ = | | | (6.34) (0 otherwise

∆ 0 < ǫk , ǫk′ < ωD ∆kk′ = | | | (6.35) (0 otherwise

46 30 May 2006 PHY240C Lecture 17

∆ ∆ = V 2 2 1/2 (6.36) − 2(ǫ ′ + ∆ ) k′ k X ωD dǫ 2ωD 1 = V N(0) 2 2 1/2 = V N(0) ln (6.37) | | · ˆ−ωD 2(ǫ + ∆ ) | | ∆ 1 ∆ = 2ω exp − (6.38) D N(0) V | | This is similar to Cooper’s formula, but the 2 is not there. This is a many body eﬀect.

47 30 May 2006 PHY240C Lecture 18

We found 1 ǫ 1 ǫ u2 = 1 + k v2 = 1 k (6.39) k 2 2 k 2 2 2 ǫk + ∆ ! 2 − ǫk + ∆ ! Thus the energy diﬀerence is p p 1 1 2 ∆E = N(0)∆2 = N(0)ω2 exp (6.40) GS 2 −2 D −N(0) V | | 1 1 −3 −7 ∆ 1 K, N(0) 4 . Thus, ∆E 10 K. ∆E 10 E . First principles calculations ǫF 10 K GS GS F cannot∼ pick up such∼ minute∼ diﬀerences. ∼ ∼ 1 ∆E = H2 H = 2[πN(0)]1/2∆ (6.41) GS 8π c c 6.4 Quasi-particle Excitations How do we create an excitation? 1. Take away a Cooper pair with (p, p). −

1 2 2 ∆E = 2ǫpv 2 vpkukvk upvp = 2ǫpv + 2∆upvp (6.42) − p − − p " k # X 2. Add electron with p. 2 ∆E = ǫp (6.43)

The total energy change is 1 2 2 ∆E + ∆E = ǫp(1 2v ) + 2∆upvp (6.44) − p 1 ǫ2 1 ∆2 u2v2 = 1 = (6.45) 4 − ǫ2 + ∆2 4 E2 1 ∆k ukvk = (6.46) 2 Ek 1 2 ǫp ∆ ∆E + ∆E = ǫp + ∆ (6.47) · Ep · Ep

1 2 2 ∆E + ∆E2 = Ep = ǫp + ∆ (6.48) q

∆

p pF

When we break a Cooper pair, we create a particle with p and a second one with p and the creation energy for each is ∆. Thus, the energy for breaking a pair is 2∆. From this we see that− the binding energy of a Cooper pair is 2∆.

48 30 May 2006 PHY240C Lecture 18

6.4.1 Particle number expectation value

2 nk v (6.49) h i ≡ k † † = ψ nk ψ = 0 (uk + vkˆbk)ˆb ˆbk(uk + vkˆb ) 0 (6.50) h BCS BCSi k k 1 ǫ D E v2 = 1 k (6.51) k 2 − E k nk is smooth. There is no jump at the Fermi surface (z = 0). This a non-Fermi liquid system.

6.4.2 One particle (quasi-particle) states

Two diﬀerent ways of creating a quasi-particle with momentum p that are spin-up. Both require energy Ep. The Hilbert space of BCS Hamiltonian consists of 2D degenerate subspaces. Small perturbations will couple to a linear combination of these eigenstates. (= “diagonalize H”)

6.4.3 Bogoliubov-Valatin Transformation

† † † † γˆ = upcˆ vpcˆp γˆ ψ = ψ (6.55) p↑ p↑ − ↓ p↑ | BCSi | p↑i The reverse transformation is

† † cˆp↑ = upγˆp↑ + vpγˆ−p↓ (6.56)

49 30 May 2006 PHY240C Lecture 18

6.4.4 Calculation of Measurable Quantities

1. Hc(T ) Critical Field

H2(T ) c = F (T ) F (T ) (6.57) 8π N − s.c.

Hc(T ) Hc(0)

1.0 T 2 1 ∼ − T c

T 2

Tc 1.0

2. Speciﬁc heat C(T ) ln γTc

10−1

10−2

Tc/T 1

The speciﬁc heat is gapped, as expected. For T > 0, we assumed 1 fk = ∆(T ) (6.58) eβEk + 1 ⇒

50 01 June 2006 PHY240C Lecture 19

3. NMR relaxation rate

∗ † † † † HI·S = A ak′ ak Iz ck′↑ck↑ ck′↓ck↓ + I+ck′↓ck↑ + I−ck′↑ck↓ (6.59) ′ − Xkk n o The term in the braces is = I S. For the speciﬁc case of superconductivity {· · · } · † † † † c ′ c + c c ′ = (u u ′ + vbv b′ ) γ ′ γ + γ γ ′ + (u u ′ v v ′ ) γ ′ γ + γ γ ′ k ↑ k↓ −k↑ −k ↓ k k k k k ↑ k↓ −k↑ −k ↓ k k − k k k ↑ −k↑ k↓ −k ↓ h i h (6.60)i

The second term is called a pair breaking term. It does not contribute below energies < ∆ (ωNMR 2∆). The transition rate is from Fermi’s Golden rule ≪

1 2 4 = 2π A a [fk(1 fk′ ) fk′ (1 fk)] δ(Ek′ Ek ω) T1(s) | | | | ′ − − − − − · Xkk 1 ǫkǫk′ + ∆k∆k′ 1 + (6.61) 2 E E ′ k k ∂f ∞ ∆2 E(E + ω)ω ∂E = 2π A 2 a 4N 2(0) dE 1 + · − | | | | ˆ E(E + ω) (E2 ∆2)1/2[(E +ω)2 ∆2]1/2 ∆ − − For the normal state, we have

1 1 ∞ ∂f 1 = = # dE ω = ω #f(0) = ω # (6.62) T (n) T (s, ∆ = 0) ˆ −∂E · · 2 1 1 0 The ratio is

∂f ∞ (E(E + ω) + ∆)2 T1(n) − ∂E = 2 dE 2 2 1/2 2 2 1/2 (6.63) T1(s) ˆ (E ∆ ) ((E + ω) ∆) ∆ − − In the limit as ω 0, we have → T (n) ∞ E2 + ∆2 ∂f 1 = 2 dE (6.64) T (s) ˆ E2 ∆2 −∂E 1 ∆ − Factors:

1 (a) DOS 2 2 → E −∆ (b) Coherence terms 2 → DOS ln divergence of integral. ω cuts it oﬀ. → NMR 1 ∆ ln (6.65) T1 ∼ max(ωNMR, ωZeeman)

51 01 June 2006 PHY240C Lecture 19

TN (T ) TS(T ) Hebel-Schlicter peak 3.0

1.0

T 0.2 1.0 Tc

4. Ultrasonic attenuation Basic process: the phonons are absorbed by the electrons of the matrial.

α(s) ∞ E2 ∆2 ∂f = dE − (6.66) α(n) ˆ E2 ∆2 −∂E ∆ −

αS(T ) αN (T )

1.0

0.0 T/Tc 0.2 1.0

The coherence factors eliminate the DOS log singularity.

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