BCS dimensionless coupling ($\lambda =N_{0}V$)

Use $N_0={\nu }_s$ for the normal-state single-spin density of states per unit energy and volume at the Fermi surface. Write the attraction as $-V/\mathcal{V}$, with $V>0$, and define $\lambda =N_{0}V$.^[ref]

Since $V/\mathcal{V}$ has units of energy, $V$ has units of energy times volume, making $\lambda$ dimensionless. Set ${\Omega }_D=\hslash {\omega }_D$, an energy cutoff using the Debye frequency ${\omega }_D$, rather than a frequency itself. These conventions also fix the meaning of the earlier constant-shell BCS gap equation.

Here is the reduction of the BCS gap equation in these conventions. Substituting the zero-temperature BCS anomalous average givesThe $k$ sum counts one spin species. For $\Delta >0$, cancelling $\Delta$ and replacing this sum by an integral with constant $N_0$ givesThe factor two from the symmetric integral cancels the $1/2$ in the BCS anomalous average; it is not an extra spin degeneracy. Substituting $\xi =\Delta \sinh t$ evaluates the integral as $\mathrm{arsinh}({\Omega }_D/\Delta )$. Solving givesThe first equality is exact within the constant-density of states, sharp-cutoff mean-field approximation; the last is the weak coupling asymptote. Extending the first equality to large $\lambda$ does not make this simplified model an accurate description of a strongly coupled material.

For an original numerical example, choose $\hslash {\omega }_D=20\mathrm{meV}$ and $\lambda =0.25$. The exact constant-shell BCS gap equation gives $\Delta (0)=0.732871\mathrm{meV}$; the weak coupling formula gives $0.732626\mathrm{meV}$. Its fractional underestimate, measured against the exact result, is exactly $e^{-2/\lambda }=e^{-8}\simeq 0.03355\%$. At $\lambda =1$ that error is $e^{-2}\simeq 13.53\%$, so the two formulas are no longer interchangeable. No material parameters have been fitted here.

At nonzero temperature, substitute the finite-temperature BCS anomalous average instead. The same cancellation yieldsAt the superconducting transition temperature take the continuous $\Delta \to 0$ limit. The integrand at $\xi =0$ then has the finite limiting value $1/(2k_B{T}_c)$, so no divergent endpoint needs to be inserted into a numerical quadrature. For $k_B{T}_c\ll {\Omega }_D$, the resulting integral gives $k_B{T}_c\simeq 1.134{\Omega }_D{e}^{-1/\lambda }$ and $2\Delta (0)/(k_B{T}_c)\simeq 3.528$. These ratios concern the isotropic weak coupling model.^[ref]