Lecture 10 Spontaneous Symmetry Breaking II: Superconductivity We have seen in a second quantization exercise that the interactions of electrons and phonons can lead to an effective attractive interaction between electrons in a material, as long as the energies of the participating electrons are close to the Fermi surface. By close we mean within ±!D, with !D the Debye energy which is the maximum frequency for phonons. We can express this as an effective attractive electron potential given by g Veff:(q) ' 2 2 ; (10.1) ! − !q where g is a positive constant and !q < !D. Under the right circumstances, this attractive interactions will lead to the formation of Cooper pairs, which in a superconductor will dominate the ground state. In what follows we will consider a phenomenological model of this interaction that was first proposed by Bardeen, Cooper and Schrieffer (BCS). Just as for the case of the superfluid, we first consider the second quantization formulation and then we go on to the functional integral treatment. 10.1 Superconductivity in Second Quantization The approximation proposed in the BCS formulation involves assuming that the inter- action takes place (predominantly) among electrons of the same energy and opposite momentum. This is a good approximation since the energy difference needs to be smaller than !D, which is much smaller than the typical electron energy. The BCS Hamiltonian is then given by X y X y y H = p cpσcpσ − g ck"c−k#c−p#cp" ; (10.2) p,σ p;k 1 2LECTURE 10. SPONTANEOUS SYMMETRY BREAKING II: SUPERCONDUCTIVITY y where cpσ and cpσ are fermion annihilation and creation operators, and g is a positive constant. The opposite spins in the interaction term are a reflection of the fact that it is supposed to come from integrating out a scalar field (phonons) interacting with electrons. To solve the Hamiltonian in (10.2) we consider a coherent state jΨBCSi where the zero- momentum mode is macroscopically occupied when Cooper pairs (pairs of electrons with opposite spins) condense to form a superconducting ground state. We define an operator that creates Cooper pairs out of the non-interacting ground state j0i as y y y Pp ≡ cp"c−p# ; (10.3) which we then use to build the BCS coherent state y Y αpPp jΨBCSi = Ape j0i ; (10.4) p y where Ap and αp are complex coefficients. First, we notice that Pp and Pp obey the commutation rules y y [Pp;Pp] = 0; [Pp;Pp] = 0;; (10.5) which might tempt us to claim that Cooper pairs are just bosons. However, y [Pp;Pq] = δpq (1 − Np" − N−p#) ; (10.6) y with Npσ = cpσcpσ, shows that their properties as not the usual ones for bosons. In particular, it is straightforward to show that (exercise) y y PpPp = 0 ; (10.7) which betrays the fermionic character of the Cooper pair constituents. Using this property the coherent BCS state (10.4) can be rewritten as Y y jΨBCSi = Ap 1 + αpPp j0i : (10.8) p Furthermore, if we impose the normalization to be unity, we can determine the Ap's. This is 10.1. SUPERCONDUCTIVITY IN SECOND QUANTIZATION 3 hΨBCSjΨBCSi = 1 2 ∗ y = jApj h0j 1 + αpPp 1 + αpPp j0i 2 2 y = jApj 1 + jαpj h0jPpPpj0i : (10.9) However, using (10.6), it is easy to see that y h0jPpP pj0i = h0j1 − Np" − N−p#j0i = 1 ; (10.10) which results in 1 Ap = ; (10.11) p 2 1 + jαpj where we ignored the irrelevant overall phase of the state jΨBCSi. Then, defining 1 αp up ≡ ; vp ≡ ; (10.12) p 2 p 2 1 + jαpj 1 + jαpj the BCS ground state can be written as Y y jΨBCSi = up + vpPp j0i : (10.13) p where the coefficients defined in (10.12) satisfy 2 2 jupj + jvpj = 1 : (10.14) We can interpret the coefficients up and vp by noticing that Y ∗ ∗ y y hΨBCSjNp"jΨBCSi = h0j (uk + vkPk) cp"cp" uq + vqPq j0i k;q Y ∗ y 2 = vkvph0jPkPpj0i = jvpj ; (10.15) k and similarly for Np#. Then, noticing that 4LECTURE 10. SPONTANEOUS SYMMETRY BREAKING II: SUPERCONDUCTIVITY X y X N = cpσcpσ = (Np" + Np#) ; (10.16) p,σ p we obtain X 2 hΨBCSjNjΨBCSi = 2 jvpj : (10.17) p 2 Then, 2jvpj gives the number of occupied states with momentum p. Through (10.14) we 2 can see that jupj gives us the number of unoccupied states. Next, we are interested in obtaining the ground state energy, i.e. using (10.2) we want to compute E = hΨBCSjHjΨBCSi X 2 X ∗ ∗ = 2 pjvpj − g vpvkukup ; (10.18) p p;k where in the second line we used the properties of up and vp (check!), and the 2 in the first term there comes from summing over spins. The expression above can be used to obtain the ground state energy. The solution must minimize the energy subject to the constraints X 2 N = 2 jvpj p 2 2 jupj + jvpj = 1 : (10.19) Then, the function to be extremized is X 2 2 f ≡ E − µN + Ep jupj + jvpj − 1 ; (10.20) p where µ and Ep are the Lagrange multipliers. We need to solve @f @E @N ∗ = − µ + Epup = 0 @up @up @up (10.21) @f @E @N ∗ = − µ + Epvp = 0 : @vp @vp @vp 10.1. SUPERCONDUCTIVITY IN SECOND QUANTIZATION 5 Defining X ∗ ∆ ≡ g upvp ; (10.22) p the minimization equations (10.21) can be written in matrix form as 0 1 0 ∗ 1 0 ∗ 1 (p − µ) ∆ up up @ A @ A = Ep @ A ; (10.23) ¯ ∗ ∗ ∆ −(p − µ) vp vp which has eigenvalues q 2 2 Ep = ± (p − µ) + j∆j : (10.24) ∗ ∗ Obtaining the eigenvectors (i.e. the solution for up and vp ) can be made easier by rewriting (10.23) as 0 1 0 ∗ 1 0 ∗ 1 cos 2θp sin 2θp up up @ A @ A = @ A ; (10.25) ∗ ∗ sin 2θp − cos 2θp vp vp where we defined p − µ ∆ cos 2θp ≡ ; sin 2θp ≡ : (10.26) Ep Ep The solution for the eigenvectors of (10.25) results in ∗ ∗ up = cos θp; vp = sin θp : (10.27) Also worth noticing, we now have ∗ 1 ∆ upvp = cos θp sin θp = sin 2θp = : (10.28) 2 2Ep This results in the transcendental equation 6LECTURE 10. SPONTANEOUS SYMMETRY BREAKING II: SUPERCONDUCTIVITY X X ∆ ∆ = g u∗ v = g : (10.29) p p 2E p p p Then, the energy gap ∆ must satisfy X 1 1 = g : (10.30) p 2 2 p 2 (p − µ) + ∆ This is the so-called the gap equation. Before we go on exploring it, we can now write the BCS ground state as Y y y jΨBCSi = cos θp + sin θpcp"c−p# j0i : (10.31) p Now examining (10.30), we must remember that in order for the Cooper pairs to form the electron-electron interaction must be attractive, which means that jp − µj < !D : (10.32) Then, in replacing the sum in (10.30) by an integral we will have X Z !D ! ν()d ; (10.33) p −!D where ν() is the density of states. But since the integration will be in a small band around the Fermi surface, we will make the approximation ν() ' νF = constant : (10.34) Then the gap equation (10.30) can be written as Z !D 1 1 = gνF d p ; (10.35) 2 2 −!D 2 + ∆ where we defined = p − µ. Since the integrand in (10.35) is even we arrive at 10.1. SUPERCONDUCTIVITY IN SECOND QUANTIZATION 7 1 Z !D 1 = dp 2 2 gνF 0 + ∆ (10.36) h! i = arcsinh D : ∆ Then we arrive at 2! ∆ = D : (10.37) e1=gνF − e−1=gνF For instance, for gνF 1 we have that the gap obeys −1=g νF ∆ ' 2!De ; (10.38) which results in ∆ !D as expected. We will later see that there is a crucial temperature dependence of the energy gap. But for now we see that essentially at zero temperature, or rather for T Tc, the ground state has an energy gap ∆ given by (10.37). The excitation spectrum has energies Ep. The next question is: what are the exited states, or quasi-particles. One way to see this is the consider the Hamiltonian written as X y X y y y y H − µN = (p − µ)cpσcpσ − g hcp"c−p#ic−k#ck" + cp"c−p#hc−k#ck"i ; (10.39) σ p;k where we averaged over pairs of operators in the interaction term. The expectation values in (10.39) are taken with respect to the BCS ground state, jΨBCSi. Explicitly, we have ghΨBCSjcp"c−p#jΨBCSi = ∆ ; (10.40) where we used that X ∗ ∆ = g upvp : (10.41) p Then, we can write (10.39) as X y X ¯ y y H − µN = (p − µ)cpσcpσ − ∆ c−p#cp" + ∆ cp"c−p# : (10.42) σ p 8LECTURE 10. SPONTANEOUS SYMMETRY BREAKING II: SUPERCONDUCTIVITY The expression above can be written in matrix form as 0 1 0 1 p − µ −∆ cp" X y H − µN = cp" c−p# @ A @ A : (10.43) y p ¯ −∆ −(p − µ) c−p# The diagonalization of this Hamiltonian now can proceed through Bogoliuvov's transfor- y mations defining the annihilation and creation operators of the eigenstates, bpσ and bpσ, and satisfying the usual anticommutation relations n y o bpσ; bp0σ0 = δpp0 δσσ0 ;::: (10.44) In terms of them, we can rewrite (10.43) as 0 1 0 1 Ep 0 bp" X y H − µN = bp" b−p# @ A @ A : (10.45) y p 0 −Ep b−p# where Ep is given by (10.24). Then, we arrive at X y H − µN = Ep bpσbpσ : (10.46) p,σ y The states created and annihilated by bpσ and bpσ are excitations, as can be seen by the fact that bpσjΨBCSi = 0 ; (10.47) meaning that the BCS ground state contains none of them.