The critical behavior of the BCS order parameter: a straightforward derivation

Textbooks on Solid State Physics, such as[¹, ², ³], include a mandatory chapter on Superconductivity. Usually the basic item to start with is the famous Bardeen-Cooper-Schrieffer model(BCS)[⁴]. One of the main results concerns the way in which the superconducting order parameter Δ(T) vanishes at the critical temperature T_c, namely as

(1) $$\mathbf{\Delta }(T)\sim B{(T_c-T)}^{1/2}$$

where the prefactor B is a non-universal coefficient and the exponent has the classical value α = 1/2¹.

Then one may read that this is a standard result for any mean-field theory, although the student may wonder, why it does not follow straightforwardly from the model?

Yet in the literature this outstandingly simple statement is obtained in a rather roundabout manner. Furthermore α and B are computed only in the weak-coupling limit ℏω_D ≫ k_BT_c, where ω_D is the Debye frequency. This is certainly an aesthetically not very pleasing situation and I doubt the student really wants to grind through the approximations just to get this simple result.

The following lines show a little trick straightening out this situation. It will hopefully find its way to the textbooks.

In the BCS theory the order-parameter Δ(T) satisfies the non-linear integral equation²

(2) $$1=g{\int }_0^{\mathrm{\hslash }{\omega }_D}\mathit{d}ϵ\frac{\tanh \left(\frac{\beta E}{2}\right)}{2E},$$

with $E=\sqrt{{ϵ}^2+{\mathbf{\Delta }}^2},\beta =1/k_{B}T$ and g is some coupling constant.

We extract the critical behavior of the order parameter straightforwardly and without approximations. For this purpose we choose Δ to be real and parametrize it as

(3) $$\mathbf{\Delta }(\beta )=a{\left(\frac{\beta -{\beta }_c}{{\beta }_c}\right)}^{\alpha };\beta \sim {\beta }_c.$$

This yields for the derivative ${\partial }_{\beta }{\mathbf{\Delta }}^2\equiv \frac{\partial {\mathbf{\Delta }}^2}{\partial \beta }$:

(4) $$\underset{T\to T_c}{lim}{\partial }_{\beta }{\mathbf{\Delta }}^2=\{\begin{array}{cc}0\hfill & \alpha >1/2\hfill \\ a^2/{\beta }_c\hfill & \alpha =1/2\hfill \\ \mathrm{\infty }\hfill & \alpha <1/2\hfill \end{array}$$

The non-linear integral equation (2) for the order parameter has the solution Δ(β, ω_D, g), depending on three parameters. Substituting this solution into equation (2) yields an identity. Differentiating this identity with respect to β easily yields the following relation

(5) $${\partial }_{\beta }{\mathbf{\Delta }}^2(\beta ,{\omega }_D,g)=\frac{{\int }_0^{\mathrm{\hslash }{\omega }_D}\frac{dϵ}{{\cosh }^2\frac{\beta E}{2}}}{{\int }_0^{\mathrm{\hslash }{\omega }_D}\frac{dϵ}{E^3}\left(\tanh \frac{\beta E}{2}-\frac{\beta E}{2{\cosh }^2\frac{\beta E}{2}}\right)}.$$

Taking the limit T → T_c, Δ → 0, we obtain

(6) $$0<a^2=\frac{2{(k_B{T}_c)}^2\tanh \frac{\mathrm{\hslash }{\omega }_D{\beta }_c}{2}}{{\int }_0^{\mathrm{\hslash }{\omega }_D{\beta }_c}\frac{dx}{x^3}\left(\tanh \frac{x}{2}-\frac{x}{2{\cosh }^2\frac{x}{2}}\right)}<\mathrm{\infty }$$

implying α = 1/2. Notice that the above integrand is finite at x = 0.

As illustration we evaluate the integral for ℏω_Dβ_c = 10 to get

(7) $$\mathbf{\Delta }(T)=3.10\cdot k_B{T}_c{\left(1-\frac{T}{T_c}\right)}^{\frac{1}{2}},T\lesssim T_c.$$

References

• ^1.
  A.L. Fetter and J.D. Walecka, Quantum Theory of Many-Particle Systems (McGraw-Hill Company, New York, 1971).
• ^2.
  A. Atland and B. Simon, Condensed Matter Field Theory (Cambridge University Press, Cambridge 2010).
• ^3.
  E.C. Marino, Quantum Field Theory Approach to Condensed Matter Physics (Cambridge University Press, Cambridge, 2017).
• ^4.
  J. Bardeen, L.N. Cooper and J.R. Schrieffer, Phys. Rev. 106, 162 (1957).

• 1
  Although traditionally the exponent is referred to as β = 1/2, we use α to avoid confusions with β = 1/k_BT
• 2
  See e.g [³] equation(23.20) or [²] equation (6.28).

Publication Dates

• Publication in this collection
  28 May 2021
• Date of issue
  2021

History

• Received
  06 Apr 2020
• Reviewed
  03 May 2021
• Accepted
  04 May 2020