Physics 223b Lecture 5 Caltech, 04/16/18
2.2 Phonon induced interaction
Historically, the theoretical understanding of superconductivity started from macroscopic and phe- nomenological descriptions like the London equation. The successful microscopic description – the BCS theory – didn’t appear until 1957. In this lecture, we are going to start directly from the microscopic theory and see how the macroscopic properties emerge.
Given the similarity between superconductors and superfluids, people knew that superconductors are related to Bose Einstein condensation somehow. But the moving charges in a superconductor are electrons, which are fermions, and fermion cannot condense by themselves. One possibility is for fermions to form pairs so that each pair is bosonic and then condense, but that would require attractive interaction between the electrons, which is contradictory to effect of the Coulomb repulsion.
In 1950s, it was realized that phonons – the vibration of the crystal lattice of the material – can mediate attractive interaction between electrons. The idea is that as an electron travels through the lattices, it exerts attractive force on the positive ions around it. The ions are attracted towards the trajectory of the electron. And if a second electron passes by this region before the ions have got the chance to relax back to their original position, then the electron sees an increase in positive charge density and is attracted to this region. Therefore, effectively, moving ions can mediate attractive interaction between electrons. Then in 1957, Cooper made the essential observation that an electron fermi sea is always unstable in the presence of attractive interaction, no matter how small it is.
To see this instability, first recall the notion of a Fermi sea. A Fermi sea is a region in momentum space where at zero temperature all the single particle states are occupied while states outside of the Fermi sea are not occupied. In the simplest case of free space, the Fermi sea takes the shape of a solid ball around k = 0. We are going to discuss using this solid ball shaped Fermi sea, although the conclusion does not really depend on the exact shape of the Fermi sea as long as it occupies a region of small momentum and therefore if more electrons are added, they have a lower bound on their momentum.
The instability is referring to the fact that in the presence of a Fermi sea, if two more electrons are added and if there is attractive interaction between them, then the total energy of the two electrons is smaller than if there is no interaction between the electrons, so that they form a bound state and do not fly away from each other.
Suppose that two electrons are added in the presence of a Fermi sea and occupy states just outside of it. By interacting with phonons, the electrons can change their energy, but only by a small amount (the phonon energy) as compared to the energy at the Fermi level. So we have two electrons moving in a narrow shell above the Fermi sea and they have attractive interaction between them. Due to this interaction, the two electrons can exchange momentum and energy.
1 It turns out that such an exchange is strongest if the total momentum is zero. To see this, note that the total energy and momentum of the two electrons are conserved. If the two electrons have a total momentum of K, then the region where they can scatter to is the shaded area in the following figure, which is small unless the total momentum of the two electrons is zero. Therefore, in Coppers discussion, he focused only on the case where the total momentum is zero.
Now consider the Schrodinger equation of the two electrons
2 − ~ (∇2 + ∇2)ψ(r , r ) + V (r − r )ψ(r , r ) = ( + 2E )ψ(r , r ) (1) 2m 1 2 1 2 1 2 1 2 F 1 2 When the interaction V is zero, is zero and the total energy is just twice the Fermi energy 2 ~ 2 EF = 2m kF . And the wave function is the product of two plane waves with k and −k.
ik(r1−r2) ψ(r1, r2) ∝ e (2) where k can take value over the whole shell.
With small V , we assume that the wave function is a superposition of such products
X ik(r1−r2) ψ(r1, r2) = g(k)e (3) k and the sum is over momentum in the shell. Plugging this into the Schrodinger equation and applying a Fourier transform to V we get
2 ~ 2 X 0 g(k)k + g(k )V 0 = ( + 2E )g(k) (4) m kk F k0 where Z 1 3 −i(k−k0)(r −r ) V 0 = d (r − r )V (r − r )e 1 2 (5) kk V ol 1 2 1 2
For simplicity, assume Vkk0 is a constant, a negative constant −V0, so that the equation becomes 2 X ~ g(k)k2 − V g(k0) = ( + 2E )g(k) (6) m 0 F k0 That is 2k2 X ~ − − 2E g(k) = V g(k0) (7) m F 0 k0 The right hand side does not depend on k and we are going to assume for now that it is some value A, then A g(k) = (8) ~2k2 m − − 2EF
2 Of course, g(k) has to satisfy the consistency condition
X X A A = V0 g(k) = V0 (9) ~2k2 k k m − − 2EF That is X V0 1 = (10) ~2k2 k m − − 2EF The sum over k can be converted to an integral over the energy of the shell
Z EF +∆E V0ZE 1 2∆E − 1 = dE = V0ZEF ln (11) EF 2E − − 2EF 2 − where ZE is the density of states, the number of single particle states per energy internal, and we have used the approximation that ZE is roughly constant over the energy shell. From this relation, we see that 2∆E = < 0 (12) 2/V Z 1 − e 0 EF which gives rise to a bound state for the two electron system.
Note that for the existence of a bound state it is important that there is a Fermi Sea. Suppose that we are doing this calculation for two electrons in free space. We can go through the derivation as above and reach the step
Z ∆E X V0 1 1 = = V0 dEZE (13) ~2k2 E − k m − 0 √ where ZE, the density of state, is proportional to E close to k = 0. Therefore, for ≤ 0, the R ∆E 1 integration 0 dEZE E− is some finite value and cannot be as large as needed. Therefore, for some small enough V0, it might become impossible to satisfy this equation and a bound state no longer exists.
On the other hand, if we are in 2D, a bound state always exists even if there is no Fermi surface. And this is exactly because the density of states in 2D goes to a constant even in the limit of E → 0.
In the above calculation, we did not worry about the fermionic statistics of the electrons. The wave function we wrote down is actually invariant under exchanging r1 and r2. This would not be a problem if the electrons carry spins, and they do, because the spin part of the wave function can be anti-symmetric | ↑i| ↓i − | ↓i| ↑i (14) so that when put together with the spatial part, the total wave function is anti-symmetric. The two electrons are said to be in a singlet pairing state – a Cooper pair.
2.3 BCS Wave Function
The last section deals with the case of two electrons with attractive interaction. How to deal with the case of all the electrons in a metal with attractive interaction? This problem is solved by Bardeen, Cooper and Schrieffer. First, a wave function is written down for the many-body state using the second quantization language.
3 In the second quantization language, a Cooper pair with the wave function given above can be written as X † † gkck↑c−k↓|0i (15) k Now if we want to write the wave function of N/2 such Cooper pairs, we have
!N/2 X † † gkck↑c−k↓ |0i (16) k But this turns out to be not adequate. The situation is similar to a superfluid. In a superfluid, we have a macroscopic number of bosons occupying the lowest energy single particle state with k = 0. But instead of writing the total wave function as
N † |ψi = N bk=0 |0i (17) we found that we need to write the wave function as a coherent state √ Nb† |ψi = N e k=0 |0i (18) which is a superposition of all boson number states and breaks particle number conservation.
Here, each Cooper pair is like a boson. Instead of writing the wave function with a fixed number of Cooper pairs, BCS realized that we need to write it as a coherent state
P † † αkc c |ΨBCSi = N e k k↑ −k↓ |0i (19)
Therefore, |ΨBCSi also breaks particle number conservation symmetry and is a superposition of all EVEN fermion number states.
† † † † The operator Pk = ck↑c−k↓ looks very similar to a boson creation operator b in that
† † † † † † [Pk,Pk0 ] = [ck↑c−k↓, ck0↑c−k0↓] = 0 (20)
0 † when k 6= k and similarly [Pk,Pk0 ] = 0.
† † On the other hand, Pk is also very different from b in that
2 2 † † † Pk = ck↑c−k↓ = 0 (21)
That is, each Cooper pair state with fixed k can actually only be occupied once. Of course, if we have a big Fermi surface with many single particle modes in the thin energy shell, we can still accommodate many Cooper pairs. Using this property, we can simplify the expression for the coherent state of Cooper pairs. In particular, notice that
† † αkPk e = 1 + αkPk (22) Therefore, the total BCS wave function can be expressed as
P † Y † Y † k αkPk αkPk |ΨBCSi = N e = N e |0i = N 1 + αkPk |0i (23) k k
4 The wave function is used so much that we actually want to find its normalization and write it out more carefully. 1 Y = (1 + |α |2) (24) N 2 k k and the normalized wave function is
Y † |ΨBCSi = uk + vkPk |0i (25) k where 1 αk uk = , vk = (26) p 2 p 2 1 + |αk| 1 + |αk| The average particle number in the system is
2 X 2|αk| N = 2 (27) 1 + |αk| k
Deep down in the Fermi Sea with k << kF , |αk| >> 1 so that the two electron modes with k and −k are both occupied. Only close to the Fermi surface where k ∼ kF is |αk| on the order of 1 so that there is significant fluctuation of the occupation number of the two modes. In fact, for modes below the Fermi surface, we should interpret ck as the creation operator of a hole – the removal of an electron from the Fermi sea. So taking into account Cooper pairs both above and below the Fermi surface, we should write the BCS wave function as
Y † Y |ΨBCSi = uk + vkPk (ukPk + vk) |FSi (28) k>kF k 5