SHANGHAI JIAO TONG UNIVERSITY LECTURE

Transcription

1 Lecture 7 SHANGHAI JIAO TONG UNIVERSITY LECTURE Anthony J. Leggett Department of Physics University of Illinois at Urbana-Champaign, USA and Director, Center for Complex Physics Shanghai Jiao Tong University

2 BCS theory T 0 Recap: at T = 0 the structure of the MBWF is SJTU 7.1 Ψ = Φ, Φ = u 00 > + υ 11 and the specific values of u and υ were found by minimizing Η μν. For T 0 we expect intuitively that the description of the many-body system can still be factored into a product of descriptions of the occupation of the individual pair states, : technically ρ = ρ density matrix but now (a) all 4 occupation states will be realized with some probability (b) quantities lie Δ will be T dependent (c) at some T c ~ T = 0 Τ B the collective bound state will cease to exist.

3 Recall: for given, 4 occupational states 00, 11, 01, 10 SJTU 7. GP EP BP 1 BP and ground state has ψ = u 00 + υ 11 corresponding to σ H = ε z + Δ x with an energy E H ε + Δ 1Τ. The limit Δ 0 corresponds to the normal GS, and then E ε. So the energy of the ground pair state relative to the normal ground state is E GP = ε E. The EP ( excited pair ) state is formed by simply reversing the pseudospin, so that ψ,ep = υ 00 u 11 (orthogonal to ψ,gf ) This evidently costs an energy E, so E EP = ε + E What about the BP ( broen pair ) states BP 1,? These each correspond (relative to the N ground state) to inetic energy KE ε and zero PE (no partner to scatter!), hence E BP1, = ε Thus the relative energies of the various states are E BP E GP = E, E EP E GP = E

4 Conventional language: SJTU 7.3 State BP 1 BP has Bogoliubov quasiparticle in state ; state ΕΡ has quasiparticles in both and (hence E EP = Ε BP ). ( : but ΕΡ is really an excitation of the condensate whereas BP 1, are not). Population of states: since all 4 states distinguishable, simple MB-Gibbs statistics applies, i.e. P n exp βe n. Thus (taing E GP as zero of E) P GP = Z 1, P BP 1 = P BP = Z 1 exp βe, P EP = Z 1 exp βe E E T Z = 1 + exp βe + exp βε A quantity of special interest is F T 1 σ x T = Τ Δ T E P GP P EP = Δ T /E T tanhβe T / Putting this into the equation Δ T = V 0 F T we find Δ T = V 0 Δ T /E T tanhβe Τ T or in the more general case V 0 V 1 Δ T = V Δ T ΤE T tanhβe T Τ Finite-temperature BCS gap equation

5 As T increases from 0, Δ T decreases from Δ 0 to zero at a temperature T c given by the linearized equation SJTU 7.4 Δ T c = V Δ T c Τ E tanhβ c E Τ β c Τ 1 B T c For the BCS contact potential V V 0 this yields ε c tanhβ ετ Ν 0 V 1 o = න dε = ln 1 14β ε c ε c 0 so comparing this with zero-t gap equation we have Ν 0 V o 1 = ln(ε c /Δ T = 0) Δ T = 0 = 1.76 B T c reasonably well satisfied for most classical superconductors Examination of the gap equation at arbitrary T < T c shows that it is a function only of Τ T T c Δ T = 1.76 B T c f T/T c with f z 1 z 4 Τ 1 (so for T T c, T 1 Τ T T c Τ 1 )

6 Properties of a BCS superconductor at non zero T. SJTU 7.5 A. Condensate: As we saw, the (F.T. of the) condensate wave function has the form at T 0 F T = T /E T )tanhβe Τ T so, in the wave function N 0 න dε sin r r F r = F expi r Δ T ε + Δ T 1Τ tanh β ε + Δ 1Τ the low energy cutoff (which determines the long distance behavior) gradually changes from ~Δ T = 0 to ~ B T. Since for T T c these are of same order of magnitude, we have approximately F r: T Δ T N 0 sin Fr F r exp rτξ T where ξ T ~ξ 0. i. e., Cooper-pair radius is not sharply T-dependent (in particular, does not diverge for T T c from below). The number of Cooper pairs, N c T න F r: T dr is proportional to T, hence for T T c N c T 1 Τ T T c

7 B. The Normal Component SJTU 7.6 Condensate is very inert, e.g. cannot be spin-polarized or (usually) flow in a way determined by walls. This applies both to GP and EP states (both have S = 0, COM momentum = 0). Hence such responses determined entirely by BP states. However, response is not simply proportional to the probability of occupation of BP states: Ex: Pauli spin susceptibility In field H, ΔΕ = μ B H. Hence, does not affect 00 or 11, but i real spin not pseudospin! shifts energies of BP states, S i z ΔE BP1 = μ B H, ΔE BP = +μ B H Hence: P BP1 = exp β Ε μ B H, P BP = exp β Ε K + μ B H and M z μ B S z = μ B Z 1 (exp β Ε μ B H exp β Ε + μ B H. with Z H = Z H

8 For μ 8 H B T, T this gives M z = μ d B H de and so exp βe SJTU 7.7 Z = μ B Hβ sech β E Τ χ Μ z Τ H = μ dn B β dε 0 sech Τ βe dε In the normal state E ε this correctly gives χ = μ B Τ dn dε, so χ T χ n = β න sech βe T Τ dε 0 Yosida function Τ χ χ 1 Note: Reason argument is relatively simple is that energy eigenstates and carry a spin + 1Τ ( 1Τ) respectively T T c

9 Ƹ SJTU 7.8 The normal density The normal density is defined as the fraction of the electrons which can respond to a (transverse) static vector potential, in following sense: In presence of vector potential Α r p p eα r So KE becomes i p i eα r i m i and the current density j r is pƹ i m e m pƹ i Α + Α r i m (ignore the order of operators) j r = 1 i δ r r i pƹ i eα r i Τm + H. C. We already saw that the explicit term in A(r i ) gives rise in the S phase, to the Meissner effect. But in the normal phase it is cancelled by the response of pƹ i to the perturbation p i Α r i. Τ δj δα pert = + Ne m

10 So: in S phase at 0 < T < T c what is perturbative response of p to Α? SJTU 7.9 (almost) exact analogy to calculation of spin susceptibility: 00 and 11 have total P = 0, so cannot respond 10 has momentum p = ħ, 01 has p = ħ. Hence ΔE BP1 = eħ Total induced momentum is AΤm ΔE BP = +eħ AΤm 1 P = ħ Z exp β E eħ A m exp β E + eħ A m and for ħ A B T, T this reduces to J e P m e ħ F 3m A Z 1 d exp βe de e p F 3m β sech βe Τ A directional averaging In N state E ε this correctly reduces to Ne Τm, so ratio ( ρ n Τρ )of response in S state at temperature T to N state value is ρ n Τρ = β න 0 sech Τ βe dε Yosida function χ and ρ n Τρ are untypically simple, because energy eigenstates are also eigenstates of σ and p.

11 Summary of lecture 7 SJTU 7.10 At T 0 the BCS description is still a product over the different pair states,, but now all four states GP u 00 + v 11 BP1 10 BP 01 EP v 00 u 11 are populated, and u and v are functions of T. The relative energies of the 4 states are E BP T E GP T = E T E EP T E GP T = E T The self-consistent equation for the gap is Δ T = E T ε + Δ T 1/ V Δ T /E T tanh βe T / and has a nontrivial Δ 0 solution only for T < T c, where B T c = Δ T = 0 /1.76 Condensate wave function F r: T not strongly T-dependent: no. of Cooper pairs N c T ~Δ T, near T c ~ 1 T/T c Normal component is essentially BP states: contributes to simple quantities χ, P n an amount Y T, e.g. β χ T / χ n = Y T β න sech βe T / dε 0