Importance of Fermi Surface in Cooper Pair Formation

[WalkThePlanck]

Hello,

This problem is about cooper pair formation and what happens with the calculations if there is an attractive potential between electrons but it is not in the presence of a filled fermi surface.

1. Homework Statement

Two electrons just above the filled Fermi Surface of a material can form a bound state as long as there is some attractive interaction between these two electrons. The magnitude of the interaction is not important. In the following problem, we will assess the importance of the filled Fermi Sea in the context of such bound state. Recall, the energy of the Cooper pair i.e., the bound state of two electrons just above the Fermi surface, is determined from 1 = Σ_k>kf|V|/(2ε_k-E)

Now, remove the fermi surface by setting k_f = 0. Show that in 3-D a bound state solution no longer exists for all |V|>0. What is the critical |V_c| such that a bound state exists for |V|>|V_c| but not |V|<|V_c|.

Part 2: Consider the problem in 1-D and 2-D.

Homework Equations

Can convert the sum into an integral over k from 0 to k_c, where k_c=Sqrt(2mω_d/ħ).

The Attempt at a Solution

I tried to solve it the same way as I solved for a filled fermi surface, in which we can see there is a bound state because after solving and isolating E we see that E<2ε_f. I did the integrals, which were somewhat complicated functions of logs in the 3-D and 2-D case, and an inverse hyperbolic tangent in the 1-D case, but I'm not sure how to tell if it is a bound state. Would a bound state mean that E<0 because adding two electrons to an empty fermi surface cost 0 energy? If that is the case then I found a critical |V| in 3-D but in 1-D and 2-D I found that bound states of electrons can form for arbitrarily small |V| even without a filled fermi surface and I'm not sure if that is right.

 

[DrDu]

Sounds good. In 1 and 3 D your conclusions about the critical V are correct. I suppose in 2-D also, but I didn't check.