# A Critical Ingredient for Cooper Pairing in Superconductors within the BCS Picture

I’m sure that readers that are experts in superconductivity are aware of this fact already, but there is a point that I feel is not stressed enough by textbooks on superconductivity. This is the issue of reduced dimensionality in BCS theory. In a previous post, I’ve shown the usefulness of the thinking about the Cooper problem instead of the full-blown BCS solution, so I’ll take this approach here as well. In the Cooper problem, one assumes a 3-dimensional spherical Fermi surface like so: 3D Fermi Surface

What subtly happens when one solves the Cooper problem, however, is the reduction from three dimensions to two dimensions. Because only the electrons near the Fermi surface condense, one is really working in a shell around the Fermi surface like so, where the black portion does not participate in the formation of Cooper pairs: Effective 2D Topology Associated with the Cooper Problem

Therefore, when solving the Cooper problem, one goes from working in a 3D solid sphere (the entire Fermi sea), to working on the surface of the sphere, effectively a 2D manifold. Because one is now confined to just the surface, it enables one of the most crucial steps in the Cooper problem: assuming that the density of states ( $N(E)$) at the Fermi energy is a constant so that one can pull it out of the integral (see, for example, equation 9 in this set of lecture notes by P. Hirschfeld).

The more important role of dimensionality, though, is in the bound state solution. If one solves the Schrodinger equation for the delta-function potential (i.e. $V(x,y)= -\alpha\delta(x)\delta(y)$) in 2D one sees a quite stunning (but expected) resemblance to the Cooper problem. It tuns out that the solution to obtain a bound state takes the following form: $E \sim \exp{(-\textrm{const.}/\alpha)}$.

Note that this is exactly the same function that appears in the solution to the Cooper problem, and this is of course not a coincidence. This function is not expandable in terms of a Taylor series, as is so often stressed when solving the Cooper problem and is therefore not amenable to perturbation methods. Note, also, that there is a bound state solution to this problem whenever $\alpha$ is finite, again similar to the case of the Cooper problem. That there exists a bound state solution for any $\alpha >0$ no matter how small, is only true in dimensions two or less. This is why reduced dimensionality is so critical to the Cooper problem.

Furthermore, it is well-known to solid-state physicists that for a Fermi gas/liquid, in 3D $N(E) \sim \sqrt{E}$, in 2D $N(E) \sim$const., while in 1D $N(E) \sim 1/\sqrt{E}$. Hence, if one is confined to two-dimensions in the Cooper problem, one is able to treat the density of states as a constant, and pull this term out of the integral (see equation 9 here again) even if the states are not confined to the Fermi energy.

This of course raises the question of what happens in an actual 2D or quasi-2D solid. Naively, it seems like in 2D, a solid should be more susceptible to the formation of Cooper pairs including all the electrons in the Fermi sea, as opposed to the ones constrained to be close to the Fermi surface.

If any readers have any more insight to share with respect to the role of dimensionality in superconductivity, please feel free to comment below.

### 5 responses to “A Critical Ingredient for Cooper Pairing in Superconductors within the BCS Picture”

1. Brian

In 2D any shallow potential well has an exponentially shallow bound state. i.e., if the depth $U$ and width $R$ of a potential well are such that $U << \hbar^2/mR^2$, then the bound state in the potential well is $\sim \exp(-\textrm{const.}\times mR^2 U/\hbar^2)$. As it was explained to me, the great insight of the Cooper problem is that this same result appears even when it is the momentum that is 2D and not the physical coordinate.

I’m not sure I understand your comment about constant density of states, though. If I’m only looking at a narrow window of energy around the Fermi surface, wouldn’t the density of states always look constant?

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2. Brian

Oops, I messed that equation up. It should be $\sim \exp(-const. \times \hbar^2/mR^2U)$

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• Anshul Kogar

Thanks for your comment — I have changed the post to (hopefully) make what I was trying to say about the density of states more transparent.

Also, you are right about the 2D shallow potential — I had chosen the delta function potential because that is, in some sense, the potential that Cooper chose. What I mean is that the Fourier transform of a constant, what Cooper chose in momentum space, is a delta function in real space. I think that Cooper’s great insight is therefore realizing how to turn an effectively 3D problem into an effectively 2D problem — this is at the end of the day what enables a bound state.

The video lecture from Week 3 Lecture IV from a coursera series expresses well what I am trying to get across (perhaps better than me!): https://class.coursera.org/eqp-003/lecture

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3. Brian

Thanks for the clarification. And that link is great!

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4. Alfred

Regarding a constant N(E) at Ef, I was thinking along the same line as Brian, i.e., assuming a smooth N(E), if we zoom in close enough to Ef, N(E) will roughly be constant. In this case, N(E) still follows the expression of a 3D N(E) ~ sqrt(E), even though the Fermi surface itself is 2D. Similarly, in 2D, the Fermi “surface” is 1D; and in 1D, Fermi “surface” is 2 points.

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