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Variational analysis of GL functional

need to correct this discussion to allow for non-coplanar energy terms . current discussion only applies in the absence of non-coplanar terms

Using the general form or GL functional,

$$F = r(q-Q)\sum_i|\rho_{q_i}|^2 + \sum_{i,j,k}\lambda (\{q_i\}) \rho_{q_i}\rho_{q_j}\rho_{q_k}\delta(\sum{q_i}) +\sum_{i\ne j} u (\alpha_{ij}) |\rho_{q_i}|^2|\rho_{q_j}|^2+ \frac{u_0}{2} \sum_i|\rho_{q_i}|^4$$

we can determine the energies of different variational states.

For simplicity, we will assume that it is sufficient to keep only harmonics of density with $|q_i| = Q$. Then the energy only depends on all the relative angles between $q_i$'s. We will consider now only the fully symmetric states that contain $N$ pairs of $\pm q_i$, where all $q_i$'s are symmetry related, and hence has exactly the same set of neighbors. It is then natural to assume that all the Fourier amplitudes are identical, $|\rho_{q_1}| = \rho_0$. The free energy in this case is

$$F = N r|\rho_0|^2 + 2 M_{tri}\lambda (q_0) |\rho_0|^3 \cos \phi +N \left[\sum_{j\ne 0} u (\alpha_{0j}) + \frac{u_0}{2}\right]|\rho_0|^4,$$,

where $M_{tri}$ is the number of equilateral triangles that vectors $q_i$ form, and $\phi$ is the sum of the phases angles of the density components forming the triangles. For now we will consider only states where no triangles are present, $M_{tri} = 0$. Then, it only remains to minimize the energy to obtain,

$$ |\rho_0|^2 = -\frac{r}{2\left[\sum_{j\ne 0} u (\alpha_{0j}) + \frac{u_0}{2}\right]}$$


$$F = -\frac{Nr^2}{4\left[\sum_{j\ne 0} u (\alpha_{0j}) + \frac{u_0}{2}\right]} = -\frac{r^2}{4\left[\frac 1 N\sum_{j\ne 0} u (\alpha_{0j}) + \frac{u_0}{2N}\right]}$$.

As an example, if we take local in space interaction, $\mu |\rho(r)|^4$, then $u_0 = u(\alpha) = 12 \mu $, and hence denominator is proportional to $1 - 1/(2N)$, which makes the stripe state favored. This is the minimal form of the 4th order vertex, commonly used in GL analysis of crystallization.

More generally, if $u(\alpha)$ is smooth as $\alpha \to 0$, as is the case also for electron-mediated interaction, then $u_0 = u(\alpha \to 0)$. An immediate consequence of that is that splitting of one Bragg peak into a pair is unfavorable. Indeed, assume that there is an energetically favorable situation with a spot at $q_0$ with amplitude $\rho_{q_0}$. Now, suppose we split it into two at $q_0'$ and $q_0''$, both approximately equal to $q_0$. To keep the interaction with the other momentum components the same, which we assumed to be optimal, we have to have $ |\rho_{q_0'}|^2 + |\rho_{q_0''}|^2 = |\rho_{q_0}|^2$. That keeps the second order (r) and the interaction terms intact. However, instead of old self interaction we now have $u_0|\rho_{q_0}|^4 \to u_0|\rho_{q_0'}|^4 +u_0|\rho_{q_0''}|^4 + 4 u(\alpha) |\rho_{q_0'}|^2 |\rho_{q_0''}|^2 \approx u_0(|\rho_{q_0'}|^2 +|\rho_{q_0''}|^2)^2 + 2 u(\alpha) |\rho_{q_0'}|^2 |\rho_{q_0''}|^2 $. Hence, the energy goes up, and splitting is not favored. Indeed, the crystallization simulations starting from random initial conditions show the extinction behavior: large Bragg peak eats up its smaller neighbors, leaving in the end only a small number of spots that correspond to a (q)crystal.

Nontrivial angular dependence of $u$ can lead to nontrivial states. It can originate from electron-mediated interionic interaction, possibly in combination with the featureless local interaction that was just mentioned.

In 2D, highly symmetric states contain angles $\alpha = \pi n/N$, with $N = 1, 2, 3, 5$ corresponding to stripe, square, hexagonal, Penrose lattices.

Based on numerics in 2D for the electron-mediated $u(\alpha)$, even with additional local repulsion, QC states do not seem to be stable.

In 3D, situation is more complex:

  • cubuc lattice, $N = 3$. 2 neighbors with $\alpha = \pi/2$;
  • BCC lattice (FCC reciprocal), $N = 6$. 4 neighbors with $\alpha = \pi/3$, 1 with $\alpha = \pi/2$;
  • FCC lattice (BCC reciprocal), $N = 4$. 3 neighbors with $\alpha = \cos^{-1} (1/3)$
  • icosahedral-A (iQC) in momentum space, $N = 6$. 5 neighbors with $\alpha \approx 63.4^o$
  • icosahedral-B: momenta are the edges of icosahedron (favored by cubic interaction which we neglect). $N = 15$.
  • icosahedral-C: same as icosahedral, but with one vector missing (has $D_5$ symmetry); $N = 5$. 4 neighbors at icosahedral angles
  • dodecahedral in momentum space, $N = 10$, 3 neighbors with $\alpha_1\approx 41.8^o$, 6 neighbors with $\alpha_2\approx 70.5^o$.
  • HCP. Choice A: same as 2D hexagonal, $N = 3$; Choice B: $N = 6$, two $\alpha_1$, two $\alpha_2$, and one $\alpha_3$ neighbors.
  • smectic or stripe, $N = 1$. no neighbors
  • columnar, $N = 2$. 1 neighbor with arbitrary $\alpha$
  • rhombohedral, $N = 3$. 2 neighbors with same angles $\alpha$

Based on numerics in 3D for the electron-mediated $u(\alpha)$, including additional local repulsion, iQC IS stable. Other stables states are FCC, smectic, columnar, and rhombohedral. BCC is very close in energy to iQC, where that one is stable, but higher for the cases that we considered. The stability diagram depends on temperatures, just as $u$ does.