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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.

In each factor of the BCS ground state, the BCS anomalous average is $⟨b_k⟩=u_k{v}_k={\Delta }_k/(2E_k)$. Self-consistency therefore gives the zero-temperature BCS gap equationPut ${\Omega }_D=\hslash {\omega }_D$. The specified constant attractive interaction makes ${\Delta }_k$ independent of $k$ inside the energy shell and zero outside it:The printed ${\delta }_{k,0}$ is incompatible with this interaction: $k$ labels relative pair momentum, not the total momentum of a Cooper pair. Every pair here has zero total momentum, while many relative momenta contribute.

For the nonzero solution, the constant-shell BCS gap equation becomes $1=(V/L^3){\sum }_{|{\xi }_k|<{\Omega }_D}(2\sqrt{{\xi }_k^2+{\Delta }^2}{)}^{-1}$. Let ${\nu }_s$ denote the approximately constant single-spin density of states per unit volume at the Fermi level. ThenIn weak coupling, $\Delta \simeq 2\hslash {\omega }_D{e}^{-1/({\nu }_{s}V)}$. The displayed answer in the question uses $\nu ={\nu }_s$. If “total electronic density of states” includes both spin species, ${\nu }_{\mathrm{tot}}=2{\nu }_s$, the argument instead reads $2/({\nu }_{\mathrm{tot}}V)$. If the density counts the whole box rather than unit volume, divide it by $L^3$ before using this formula.

Weak coupling means a suitably dimensionless interaction strength is small. In BCS theory, ${\nu }_{s}V\ll 1$ gives a gap exponentially small compared with the pairing-shell energy cutoff; the expansion parameter is not the dimensional quantity $V$ alone.