2D: numerics for 4th order term

Here we present the results of for the box diagram of the type $\mu(q_0, \alpha)|\rho_{q_1}|^2 |\rho_{q_2}|^2$ in 2D. This diagram only depends on the value of $q_0$ and the angle $2\alpha$ between $q_1$ and $q_2$. From the geometry, it is clear that there is a "resonant" condition satisfied when $2k_F \cos(\alpha) = q_0$. It is also clear that $\mu(q_0,\alpha) = \mu(q_0,\pi/2-\alpha)$.

The main numerical observations are:

1. for $q>2k_F$, the mutual and the self interactions are attractive, signifying instability

2. there is an anomaly "on resonance", which has a Fano shape, i.e. switches between attraction and repulsion over narrow range of momenta.

3. The limit of $\alpha \to 0$ is non-singular, i.e. mutual interaction connects smoothly with self interaction. However, the combinatorial multiplicty of mutual interaction is higher than for self interaction, and hence the interaction between two nearly identical momenta is 4 times (need to check) stronger than self-interaction.

To do the numerics, we removed singularities:

$$\frac{1}{\epsilon_1 - \epsilon_2} \to Re \frac{1}{\epsilon_1 - \epsilon_2 + i\Gamma},$$

and chose $\Gamma = 1e-14$.

Attached figures illustrate behavior of full 4th order vertex as a function of $Q$ and $\alpha$, as well as Matlab code used to calculate mutual and self-interaction. (.fig files can be downloaded and opened with Matlab).

Consequences for quasicrystals

The fact that interaction has only a good large minimum at angle between $q$'s equal to $\pi/2$, makes the formation of quasicrystals in 2D problematic.The energy can be more effectively saved by forming a 4-fold symmetric square lattice instead of more complex crystals with $2\alpha = 2\pi n/N$, where $N > 0$. A graph of Free Energy denominators (see ) is shown in the attached figure. For system to be stable, all the denominators have to be positive, and the lowest energy state corresponds to the smallest of them. If we take the calculated vertex as all there is (no other terms), then beyond $Q\sim 1.6 k_F$ at $T =0.1$, the square lattice wins the stripe, and for $Q> 1.8 k_F$ it denominator becomes negative, signifying the break down on 4th order expansion.

One can add another angle-independent term, which may be generated by central interactions perhaps, to stabilize the vertex. However, even that is not sufficient to stabilize non-trivial quasicrystals, at least at the mean field level, even though it allows hexagonal crystal to win square, even without cubic term.