The BCS energy gap at low density

We show that the energy gap for the BCS gap equation is Ξ=μ8e-2+o(1)expπ2μa\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \varXi = \mu \left( 8 {\mathrm{e}}^{-2} + o(1)\right) \exp \left( \frac{\pi }{2\sqrt{\mu } a}\right) \end{aligned}$$\end{document}in the low density limit μ→0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu \rightarrow 0$$\end{document}. Together with the similar result for the critical temperature by Hainzl and Seiringer (Lett Math Phys 84: 99–107, 2008), this shows that, in the low density limit, the ratio of the energy gap and critical temperature is a universal constant independent of the interaction potential V. The results hold for a class of potentials with negative scattering length a and no bound states.


Introduction and Main Result
The BCS gap equation at zero temperature where , is an important part of the BCS theory of superfluidity and -conductivity [1].The function ∆ has the interpretation of the order parameter describing pairs of Fermions (Cooper pairs).The potential V models an effective local interaction.(In the case of superconductivity it is between electrons.)We will assume that V ∈ L 1 (R 3 ), in which case V has a Fourier transform given by V (p) = (2π) −3/2 R 3 V (x)e −ipx dx.The chemical potential µ > 0 controls the particle density.We study here the limit of low density, i.e. µ → 0. In the low density limit, it is known that superfluid/-conducting behaviour is well described by BCS theory [10,12].
The low density limit for BCS theory has previously been studied in [6], where the critical temperate has been calculated.The critical temperature satisfies that, for temperatures below the critical temperature the system is in a superfluid/-conducting state.For temperatures above, it is not.We will here study the energy gap (at zero temperature) in this limit.The function E has the interpretation as the dispersion relation for the corresponding BCS Hamiltonian, and so Ξ has the interpretation of an energy gap, see [5,Appendix A].
The potential non-uniqueness of such functions ∆ gives rise to difficulty in evaluating this Ξ.However, under the assumption we impose on V (namely that its Fourier transform is non-positive), it is proved in [7] that ∆ is unique (up to a constant global phase).
An analysis of the energy gap in the low coupling limit is given in [7].There one considers a potential λV , for V fixed and λ → 0. In this limit it is shown that the energy gap satisfies Ξ ∼ A exp(−B/λ) for explicit constants A and B. We are here instead interested in the limit µ → 0 for V fixed.Similarly as for the critical temperature in the low density limit [6] this turns out to be related to the scattering length of 2V , which we now define.
Here, operators that are functions of p are to be interpreted as multiplication operators in Fourier space.
In [6, Appendix A] it is explained, why it is sensible to call this a scattering length.
With this, we may now state our main theorem.
Theorem 2. Let V be radial and assume that , and that the scattering length a < 0 is negative.Then, That is, in the limit of low density, the energy gap satisfies This is known in the physics literature [10].Here S 3 = 3 4 2 2/3 π 4/3 ≈ 5.4779 is the best constant in Sobolev's inequality [11,Theorem 8.3].The assumption that V L 3/2 < S 3 gives that p 2 + λV > 0 for any 0 < λ ≤ 1 by Sobolev's inequality, see [11, section 11.3].Thus, by the Birman-Schwinger principle, the operator λV 1/2 1 p 2 |V | 1/2 does not have −1 as an eigenvalue.Varying λ we thus get that the spectrum of In particular, the scattering length is finite.Also, for a V satisfying the assumption it also satisfies the assumptions of [6,Theorem 1].This states that the critical temperature satisfies We thus immediately get following.
Corollary 3. Let V be radial and assume that V , and that the scattering length a < 0 is negative.Then, where γ ≈ 0.577 is the Euler-Mascheroni constant.
That is, in the low density limit, the ratio of the energy gap and critical temperature tends to some universal limit.This is known in the physics literature [4].Also, this property has been observed before in the low coupling limit [1,7,12].Moreover, the universal constant is the same in both the low density and weak coupling limits.The assumptions we impose on the potential V is more or less the assumptions of [6,7].The only difference is the assumption that V L 3/2 < S 3 instead of the assumption that V 1/2 1 p 2 |V | 1/2 has spectrum contained in (−1, ∞).As discussed above our assumption here is stronger.We need such a stronger assumption, since we need to control different scalings of the potential.As discussed in [6] our assumption captures that the operator p 2 + V does not have any bound states.
We will follow the description of BCS theory made in [2,3,[5][6][7][8][9].There the BCS gap equation at zero temperature arises as the Euler-Lagrange equations for minimisers of the BCS functional at zero temperature For a minimiser α one then defines This ∆ then satisfies the BCS gap equation, see [9].We now give the proof of our theorem.
Part of the proof is inspired by and based on the methods of [6].

Proof
One of the key ideas in the proof is to study the asymptotics of This is similar to what is done in [6,7] for the study of the critical temperature and energy gap in the low coupling limit and for the critical temperature in the low density limit.The structure of the proof is as follows.First we find bounds on the minimiser α of the BCS functional.These then translate to bounds on the function ∆, which gives some asymptotic behaviour of m µ (∆).Armed with this, we employ the methods of [6] to prove our theorem.
In [7,Lemma 2] it is proven, that there exists a unique minimiser α of the BCS functional at zero temperature with (strictly) positive Fourier transform.This we will denote by α µ,V .By scaling it follows that With this, we thus see that the minimisers with positive Fourier transform satisfy We now bound this.Proposition 4. In the limit µ → 0 the minimiser satisfies α µ,V H 1 ≤ Cµ 3/4 .
Proof.With the scaling argument above we compute .
We now show, that this latter norm is bounded uniformly in µ.
Sobolev's inequality, see [11, section. 11.3].Thus we may bound for any α, we get for the minimiser that α 1, √ µV √ µ H 1 is bounded uniformly in µ.We conclude that Proposition 5.For small enough µ, the minimiser satisfies for a small ε > 0 and a constant C, both independent of µ.
Proof.By the continuity of V we may find ε > 0 such that 2 V (0 Then p 2 λ + V ≥ 0 and so for the minimiser α = α µ,V we compute We now bound the two remaining integrals.For the first integral we do the following by the bound ĝ L 3/2 ≤ C g H 1 , valid for any function g.To see this, simply write For the double-integral we use the Young and the Hausdorff-Young inequalities [11,Theorems 4.2 and 5.7].This gives Combining all this we get the bound where we absorbed the factors of V into the constants C 1 , C 2 > 0. The right-hand-side above is a second degree polynomial in α1 {|p|>ε} L 3/2 .Moreover, the minimiser α = α µ,V satisfies F µ,V (α) ≤ 0. We conclude that α1 {|p|>ε} L 3/2 is between the two roots of the second degree polynomial.In particular We now bound ∆ µ,V = −2 V α µ,V = −2(2π) −3/2 V * αµ,V .Now, V ≤ 0, and so ∆ µ,V ≥ 0. By the BCS gap equation it follows that even ∆ µ,V > 0, see [7].
Proof.We compute by the Hausdorff-Young inequality [11, Theorem 5.7] and the fact that ĝ The bound for the difference is similar, using that where we used that V (p For the sake of simplifying notation, we will just write ∆ for the function ∆ µ,V from now on.With this bound on ∆ we get some control over m µ (∆).By computing the spherical part of the integral, splitting the integral according to p 2 < 2µ and p 2 > 2µ, and using the substitutions s = µ−p 2 µ and s = p 2 −µ µ we may rewrite m µ (∆) as Here by ∆( √ µ √ 1 ± s) we mean the value of ∆ on a sphere with the given radius.Since ∆ is radial, this is well-defined.We now claim that Proposition 7. In the limit µ → 0 the value m µ (∆) satisfies Proof.For the first and last integrals this follows by a dominated convergence argument.One considers the difference between the claimed value and the known value and uses a dominated convergence argument to shows that this vanishes.For the middle integral we use propositions 5 and 6.The argument is as follows.
Define the function(s) . We must then show that First, the function ∆ satisfies (with ε > 0 chosen from proposition 5)

Also, for any |p|
by the Hausdorff-Young inequality [11, Theorem 5.7] and proposition 5 above.Thus, the function(s) x(s) satisfies the desired |x(s)| ≤ C|x(0)|.With this, we may now prove the desired convergence of integrals.
. Now, one may compute that This shows that vanishes as desired.We conclude the desired.
The remainder of this paper uses the methods of [6].We decompose where A ∆,µ is defined such that this holds.That is, its kernel is In order to see this, note that S 2 e ipx dp = 4π sin |x| |x| .The operator B ∆ is the Birman-Schwinger operator associated to E + V .One easily checks that E + V has its lowest eigenvalue 0, see [7].(This follows from the fact that V ≤ 0 is negative and so the ground state of E + V can be chosen to have non-negative Fourier transform.Hence it is not orthogonal to α µ,V , which is an eigenfunction with eigenvalue 0.) Thus B ∆ has −1 as its lowest eigenvalue.Proposition 8.In the limit µ → 0 the function ∆ satisfies ∆( √ µ) = o(µ).
Proof.Suppose for contradiction that does not vanish.That is, suppose that there is some subsequence with ∆( √ µ) > Bµ for µ → 0 for some constant B > 0. We use the decomposition By the assumptions on V , the spectrum of We show that the remaining two terms in the decomposition above vanish in the limit µ → 0, and so that the spectrum of B ∆ approaches that of V 1/2 1 p 2 |V | 1/2 .Since the latter has its lowest eigenvalue strictly larger than −1, we get a contradiction.
For m µ (∆) we use proposition 7 above.The only term that does not immediately vanish in the limit µ → 0 is the term By splitting this integral according to s < we see that this term may be bounded by Cµ −1/2 ∆( √ µ) ≤ Cµ 1/4 by proposition 6. Hence this term indeed also vanishes.
For the kernel of A ∆,µ we use that For the first integral we bound E(p) ≥ ∆(p) ≥ Bµ and so We bound the second integral as follows.First, with the substitution s = p 2 −µ µ we get where we used that |∆(p)| ≤ Cµ 3/4 .The integral |V (x)||V (y)||x − y| dx dy < ∞ is finite by the assumptions on V .Thus A ∆,µ 2 ≤ Cµ 1/2 vanishes as desired.
Using this refined bound, ∆( √ µ) = o(µ), we may use a dominated convergence argument to show that These integrals can be computed (somewhat easily by hand).This is done in [7].We conclude that in the limit µ → 0. In particular m µ (∆) ≫ √ µ.Now, we are interested in bounding A ∆,µ .
Proposition 9.The operator A ∆,µ vanishes in the following sense.
Proof.The proof is similar as above, only we give a more refined bound on the kernel.We bound the sin b b term by Where Z > 0 is arbitrary, and the constant C does not depend on Z. Then Now, the first and second integral may be bounded by m µ (∆)µ and m µ (∆)µ 1/4 as follows.For any γ we may bound Similarly as before, the last integral may be bounded by µ 1/2 ≪ m µ (∆).Again, by the assumptions on V it follows that |V (x)||V (y)||x − y| dx dy < ∞ is finite.Thus we get We may decompose Since −1 is an eigenvalue of B ∆ we get that −1 is an eigenvalue of which is perhaps most easily seen by writing the left-hand-side as a power series in A ∆,µ .Plugging this into equation (3) above we get 4πa for the first term.The second term gives This function f is the same function f , which was studied in [6].There it was shown that f satisfies f (x)|V (x)| 1/2 (1 + |x|) ∈ L 1 .The third term in the expansion above is o(µ 1/2 ) by equation ( 2) above.We show that the second term is o(µ 1/2 ) as well.Proposition 10.In the limit µ → 0 we have f |sgn V A ∆,µ |f = o(µ 1/2 ).
Proof.This is similar to the bound on A ∆,µ above.We bound the kernel of A ∆,µ by This concludes the proof of theorem 2.