Put $\xi=v_F(k)\ell$ and use the same $\nu_F$. With a symmetric cutoff $|\xi|<\Lambda$, the <BCS gap equation> becomes
$$
\frac1{|V|}
=\nu_F\int_{-\Lambda}^{\Lambda}d\xi
\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}
\frac1{\omega^2+\xi^2+\Delta^2}.
$$
The frequency integral is $1/[2\sqrt{\xi^2+\Delta^2}]$, so
$$
\frac1{\nu_F|V|}
=\operatorname{arsinh}\frac\Lambda\Delta
\sim\log\frac{2\Lambda}{\Delta}
\qquad(\Delta\ll\Lambda).
$$
Thus
$$
\boxed{
\Delta\sim2\Lambda
\exp\left[-\frac1{\nu_F|V|}\right].}
$$
The gap and the one-loop strong-coupling scale have the same nonperturbative exponential dependence; their order-one prefactors depend on the cutoff convention and microscopic completion. The renormalization-group divergence is therefore the normal-state signal of the paired, gapped phase found from the self-consistent gap equation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 337 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-337/2/c/solution