BCS gap equation (${\Delta }_k=-{\sum }_{k^{\prime }}{V}_{k{k}^{\prime }}{\Delta }_{k^{\prime }}\tanh (E_{k^{\prime }}/(2k_{B}T))/(2E_{k^{\prime }})$)

The BCS gap equation makes the superconducting gap function self-consistent with the BCS anomalous average that creates it. For a translation-invariant spin-singlet mean-field approximation, define ${\xi }_k={ϵ}_k-\mu$ and $E_k=\sqrt{{\xi }_k^2+|{\Delta }_k{|}^2}$. ThenThe interaction $V_{k{k}^{\prime }}$ and its normalization/cutoff are part of the model. A nonzero solution describes pairing; ${\Delta }_k=0$ is always a solution, so finding any solution alone does not establish which state minimizes the free energy.

For constant attraction $-V$ in a symmetric shell $|\xi |<\Lambda$ with constant single-spin density of states ${\nu }_s$, the zero-temperature equation reduces toAt weak coupling, this gives $\Delta \simeq 2\Lambda e^{-1/({\nu }_{s}V)}$. The exponential explains how a small attraction produces an extremely small pairing scale, matching the Cooper instability. The finite-temperature equation can be linearized as the gap tends to zero to determine the mean-field transition. ^[ref]

Differentiate the subtracted grand potential of the reduced BCS model with respect to the squared pairing amplitude. The displayed fermionic Matsubara sum converts the derivative of the trace logarithm into ${\sum }_k\tanh (\beta E_k/2)/(2E_k)$. Setting it equal to $V/\lambda$ gives the uniform nonzero-gap saddle. The limit E→0 is finite at positive temperature, $\beta /4$. The sum relies on antiperiodic temporal boundary conditions from the fermionic coherent-state trace.

For a fixed-density balanced two-spin Fermi gas with masses $(m,m,\gamma m)$, the Fermi energy scales as ${\gamma }^{-1/3}$ while the single-spin Fermi-level density of states scales as ${\gamma }^{1/3}$. A constant attractive interaction $g$ within a narrow symmetric energy window ${\epsilon }_c\ll {\epsilon }_F$ therefore gives $1={\lambda }_0{\gamma }^{1/3}\mathrm{arsinh}({\epsilon }_c/\Delta )$. Solving gives $\Delta ={\epsilon }_c/\sinh [1/({\lambda }_0{\gamma }^{1/3})]$, whose weak-coupling limit is displayed. Extrapolating to arbitrarily large anisotropy violates the assumptions that the Fermi energy dominates the window and that the pairing coupling is small.

An attractive interaction $V_{k,k^{\prime }}=-V/\mathcal{V}$ within $|{\xi }_k|,|{\xi }_{k^{\prime }}|<{\Omega }_D$ gives a constant gap inside that energy shell and zero outside. With approximately constant single-spin density of states ${\nu }_s$ per volume,The weak coupling limit is $\Delta \simeq 2{\Omega }_D{e}^{-1/({\nu }_{s}V)}$. A density counting both spin species is twice ${\nu }_s$ and must be divided by two. The shell cutoff is a relative-energy condition, even when every Cooper pair has zero total momentum.

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]