How does one go about evaluating the behavior of the BCS gap $ \Delta = \Delta(T) $ for $ T \to 0^+ $ under the weak coupling approximation $ \Delta/\hbar\omega_D \ll 1 $?

In Fetter & Walecka, Quantum theory of Many-Particle Systems, Prob. 13.9 it is said that the starting point is $$\tag{1} \ln\frac{\Delta_0}{\Delta} = 2\int_0^{\hbar\omega_D}{\frac{\mathrm d\xi}{\sqrt{\xi^2+\Delta^2}}\frac{1}{e^{\beta\sqrt{\xi^2+\Delta^2}}+1}}, $$ which I have no problem deriving from the theory, but I can't find a way to actually evaluate this integral even under the approximations $ \hbar\omega_D \to \infty $, $ \Delta \approx \Delta_0 $ (in the RHS) and $ \beta\Delta \to \infty $. [Of course $ \beta = (k_BT)^{-1} $ and $ \Delta_0 = \Delta(T = 0)$.] I have tried several approaches, used different Taylor-expansions and changes of variables, but I am simply stuck.

For the record, the expected behavior is supposed to be $$\tag{2} \Delta(T) \sim \Delta_0\left(1 - \sqrt{\frac{2\pi}{\beta\Delta_0}}e^{-\beta\Delta_0}\right) .$$

EDIT: Just leaving it here for the posterity. I found a more complete way to tackle this integral; specifically, under the WC approximation one has $$ \int_0^{+\infty}{\frac{\mathrm d x}{\sqrt{x^2 + 1}}\frac{e^{-\beta\Delta\sqrt{x^2 + 1}}}{1 + e^{-\beta\Delta\sqrt{x^2 + 1}}}} = \int_1^{+\infty}{\frac{\mathrm d y}{\sqrt{y^2 - 1}}\frac{e^{-\beta\Delta y}}{1 + e^{-\beta\Delta y}}} = $$ $$ = \sum_{k=1}^{+\infty}\int_1^{+\infty}{\frac{\mathrm d y}{\sqrt{y^2 - 1}} (-1)^{k+1}e^{-k\beta\Delta y}} = \sum_{k=1}^{+\infty}(-1)^{k+1}\int_0^{+\infty}{\mathrm d t\; e^{-k\beta\Delta \cosh{t}}} = \sum_{k=1}^{+\infty}(-1)^{k+1} K_0(k\beta\Delta), $$

$ K_0 $ being the 0-order modified Bessel function of the second kind, whose asymptotic behavior is known and may be used to solve the problem in a relatively clean way (and even find the corrections at higher orders, which are $ \in O(e^{-\beta\Delta}(\beta\Delta)^{-k - 1/2}) $. Cf. Abrikosov, Gorkov, Dzyaloshinski, Methods of Quantum Field Theory in Statistical Phyisics, 1963. Pagg. 303-304.

  • 1
    $\begingroup$ I guess the answer of this question is contained in an old answer of mine: physics.stackexchange.com/a/65444/16689 please tell me if you need more details. $\endgroup$
    – FraSchelle
    Jul 3, 2014 at 16:34
  • $\begingroup$ @FraSchelle: I did read that post, but I couldn't find this specific problem addressed. I might have missed it, though. $\endgroup$
    – derpy
    Jul 4, 2014 at 12:45

1 Answer 1



  1. Define difference $\delta:=\Delta-\Delta_0$. Deduce from $|\delta|\ll |\Delta_0|$ that the lhs. of eq. (1) is $$\tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.$$

  2. Substitute $\xi=x\Delta $ in the integral on the rhs. of eq. (1). Deduce using $\hbar \omega_D \gg \Delta$ that the rhs. is $$\tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! \frac{dx}{\sqrt{1+x^2}} \frac{1}{e^{\beta\Delta \sqrt{1+x^2}}+1}. $$

  3. Deduce from $\beta\Delta\gg 1$ that we can simplify the rhs. further to a Gaussian integral $$\tag{C} \text{rhs}~\approx~ \int_{\mathbb{R}} \! dx~ e^{-\beta\Delta (1+\frac{1}{2}x^2)}~=~\sqrt{\frac{2\pi}{\beta\Delta}}e^{-\beta\Delta} . $$ Such arguments are closely related to the method of steepest descent.

  4. Deduce eq. (2).

  • $\begingroup$ I see, so $ x $ should be treated as a small quantity in the integrand because $ e^{-\beta\Delta} $ warrants doing so. Thank you very much, this had been bugging me for a while. $\endgroup$
    – derpy
    Jul 3, 2014 at 8:39

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