We are told in equation 5.25:

$$F^{(2), 1} = - \frac{T^3}{L^6} \sum_{p_1, \, p_2, \, q} \; G_{p_1} \; G_{p_1 + q} \; G_{p_2} \; G_{p_2 + q} \; V(q)^2$$

$$F^{(2), 2} = \frac{1}{2} \frac{T^3}{L^6} \sum_{p, \, q_1, \, q_2} G_p \; G_{p - q_1} \; G_{p - q_1 - q_2} \;\; G_{p - q_2} \; V(q_1) \; V(q_2)$$

The text tells us:

... only configurations where all momentum arguments carried by the Green function are close to the Fermi surface contributes significantly to the sums (5.25). Considering the first sum, we see that, for small $|q|$ and $|p_i| \simeq p_F$, this condition is met, i.e. there are two unbound summations over momentum shells around the Fermi surface. However, with the second sum, the situation is less favourable. For fixed $|p_1| \simeq p_F$ [I think $p_1$ should be $p$], fine-tuning of both $q_1$ and $q_2$ is necessary to bring all momenta close to $p_F$, i.e. effectively one momentum summation is frozen out.

The text argues that the second equation in equation 5.25 has more fine-tuning necessary than required for the first equation in equation 5.25 in order to argue that at high densities, we can neglect the second equation.

What confused me is that it seems that the same amount of fine-tuning is required for both the first and second equations. In particular, for the first equation we need small $|q|$, $|p_1| \simeq p_F$, and $|p_2| \simeq p_F$. Don't we just need for the second equation small $|q_1|$, small $|q_2|$, and $|p| \simeq p_F$?


1 Answer 1


In particular, for the first equation we need small $|q|$, $|p_1|≃p_F$, and $|p_2|≃p_F$. Don't we just need for the second equation small $|q_1|$, small $|q_2|$, and $|p|≃p_F$?

This is correct. But notice, while small $|q|$ requires fine-tuning in all 3 directions in k-space, the condition $|p_1|\simeq p_f$ requires fine-tuning only in one direction - perpendicular to the Fermi surface. In other words, if the relevant scale of $|q|$ beyond which $V(q)$ can be neglected is $\Delta q$, and the thickness of the relevant states around $p_F$ is $\Delta p\approx \Delta q$, then the first sum, $F^{(2),1}$, goes over the volume $\Delta V_1=(4 \pi p_F^2\Delta p)^2 \Delta q^3$, while the second sum goes over the volume $\Delta V_2=(4 \pi p_F^2\Delta p) \Delta q^6$. The ratio of the two $$ \frac{\Delta V_1}{\Delta V_2}=\frac{4 \pi p_F^2\Delta p}{\Delta q^3}\gg1. $$ Hope, this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.