= BCS dimensionless coupling
{c}
{title2=$\lambda=N_0V$}
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 $\lambda=N_0V$.<Theory of Superconductivity by John Bardeen and others>[p.1182, equations (2.34)–(2.40), one-spin density of states; volume normalization stated here]{ref}
Since $V/\mathcal V$ has units of <energy>, $V$ has units of <energy> times <volume>, making $\lambda$ dimensionless. Set $\Omega_D=\hbar\omega_D$, 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> gives
$$
\Delta=\frac{V}{\mathcal V}\sum_{|\xi_k|<\Omega_D}\frac{\Delta}{2\sqrt{\xi_k^2+\Delta^2}}.
$$
The $k$ sum counts one <spin> species. For $\Delta>0$, cancelling $\Delta$ and replacing this sum by an <integral> with constant $N_0$ gives
$$
1=\frac{N_0V}{2}\int_{-\Omega_D}^{\Omega_D}\frac{d\xi}{\sqrt{\xi^2+\Delta^2}}
=\lambda\int_0^{\Omega_D}\frac{d\xi}{\sqrt{\xi^2+\Delta^2}}.
$$
The 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 $\operatorname{arsinh}(\Omega_D/\Delta)$. Solving gives
$$
\Delta(0)=\frac{\hbar\omega_D}{\sinh(1/\lambda)}
=\frac{2\hbar\omega_De^{-1/\lambda}}{1-e^{-2/\lambda}}
\simeq2\hbar\omega_D\exp\left[-\frac{1}{N_0V}\right],\qquad \lambda\ll1.
$$
The 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 $\hbar\omega_D=20\ \mathrm{meV}$ 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 $e^{-2/\lambda}=e^{-8}\simeq0.03355\%$. At $\lambda=1$ that error is $e^{-2}\simeq13.53\%$, so the two formulas are no longer interchangeable. No material parameters have been fitted here.
\Image[/bcs-cutoff-gap-and-weak-coupling-error.png]
{title=The exact cutoff gap and the range of the weak-coupling approximation}
{description=Original zero-<temperature> <constant-shell BCS gap equation> calculation. The upper panel compares $1/\sinh(1/\lambda)$ with $2e^{-1/\lambda}$; the lower panel gives the exact fractional underestimate $e^{-2/\lambda}$. The shaded interval is a chosen small-coupling illustration, not an experimental boundary. The large-coupling part tests the asymptote within the same model.}
At nonzero <temperature>, substitute the finite-<temperature> <BCS anomalous average> instead. The same cancellation yields
$$
1=\lambda\int_0^{\Omega_D}\frac{\tanh\!\left(\sqrt{\xi^2+\Delta(T)^2}/(2k_BT)\right)}{\sqrt{\xi^2+\Delta(T)^2}}\,d\xi.
$$
At the <superconducting transition temperature> take the continuous $\Delta\to0$ limit. The integrand at $\xi=0$ then has the finite limiting value $1/(2k_BT_c)$, so no divergent endpoint needs to be inserted into a numerical quadrature. For $k_BT_c\ll\Omega_D$, the resulting <integral> gives $k_BT_c\simeq1.134\Omega_De^{-1/\lambda}$ and $2\Delta(0)/(k_BT_c)\simeq3.528$. These ratios concern the isotropic <weak coupling> model.<Microscopic Quantum Interference Effects in the Theory of Superconductivity by Leon Neil Cooper>[1972 lecture, thermodynamic properties of the ideal superconductor, pp.80–82]{ref}
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