Third order GL term

Here we derive the form of the 3rd order GL term due to the interaction between ions mediated by itinerant electrons.

$$\Delta \Omega^{(3)} = -\frac{v_{q_0}^3 \rho_{q_1} \rho_{q_2}\rho_{q_3}}{3 \beta} \sum_{\omega_n, k} G_1(\omega_n) G_2(\omega_n)G_3(\omega_n),$$

where $\epsilon_1 = \epsilon_k$, $\epsilon_2 = \epsilon_{k - q_1}$, and $\epsilon_3 = \epsilon_{k - q_1 - q_2}$, and $q_1 + q_2 + q_3 = 0$. By contour integration,

$$\Delta \Omega^{(3)} = -\frac{v_{q_0}^3 \rho_{q_1} \rho_{q_2}\rho_{q_3}}{3} \sum_{ k} \frac{n_F(\epsilon_1)}{(\epsilon_1 - \epsilon_2) (\epsilon_1 - \epsilon_3)}+ \frac{n_F(\epsilon_2)}{(\epsilon_2 - \epsilon_1) (\epsilon_2 - \epsilon_3)} + \frac{n_F(\epsilon_3)}{(\epsilon_3 - \epsilon_1) (\epsilon_3 - \epsilon_2)}\\ \equiv \lambda(q_0) \rho_{q_1} \rho_{q_2}\rho_{q_3}.$$

The last equality follows from the assumption that densities condense only with a preferred wavevector magnitude, $|q_{1,2,3} | = q_0$, and hence $q_i$ form an equilateral triangle. Naturally, one expects that $\lambda(q_0)$ is singular when $q_0 = 2k_F \cos(\pi/6) = \sqrt{3} k_F$, i.e. when the triangle is inscribed into the Fermi sphere/circle.