I [J] = σ^{2} = \frac{1}{V} \sum_{q} \frac{1}{J ( q )} = \frac{2}{V} \sum_{q}^{+} \frac{1}{J ( q )} .

The factor two in the last expression restores the omitted negative wavevectors. For one independent mode

J_{k} = J (k)

\frac{\partial I}{\partial J _{k}} = - \frac{2}{V J _{k}^{2}} .

Differentiating the bound from parts (d) and (e) gives

\frac{\partial F}{\partial J _{k}} = \frac{J _{k} - G ( k )}{J _{k}^{2}} - \frac{12 g}{2 π} \frac{I}{J _{k}^{2}} .

J (q) = G (q) + Δ, Δ = \frac{12 g}{2 π} I [J] = 6 g \frac{2}{π} [\int \frac{d ^{d} k}{( 2 π ) ^{d}} \frac{1}{J ( k )}]^{1/2} .

The interaction has generated a positive, wavevector-independent mass shift.

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