3
$\begingroup$

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.

$\endgroup$
  • 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 '14 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 '14 at 12:45
3
$\begingroup$

Hints:

  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).

$\endgroup$
  • $\begingroup$ 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 '14 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.