Set
$$
I[J]=\sigma^2
=\frac1V\sum_{\mathbf q}\frac1{J(q)}
=\frac2V\sum_{\mathbf q}^{+}\frac1{J(q)}.
$$
The factor two in the last expression restores the omitted negative wavevectors. For one independent mode $J_k=J(k)$,
$$
\frac{\partial I}{\partial J_k}
=-\frac{2}{VJ_k^2}.
$$
Differentiating the bound from parts (d) and (e) gives
$$
\frac{\partial\widetilde F}{\partial J_k}
=\frac{J_k-G(k)}{J_k^2}
-\frac{12g}{\sqrt{2\pi}}\frac{\sqrt I}{J_k^2}.
$$
Its stationary point therefore obeys the <self-consistency equation>
$$
J(q)=G(q)+\Delta,
\qquad
\Delta=\frac{12g}{\sqrt{2\pi}}\sqrt{I[J]}
=6g\sqrt{\frac2\pi}
\left[
\int\frac{d^d\mathbf k}{(2\pi)^d}\frac1{J(k)}
\right]^{1/2}.
$$
The interaction has generated a positive, wavevector-independent mass shift.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 1 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-344/1/f/solution