Symmetrization and reduction of effort

If the billiard has symmetry, then all its quantum eigenstates will have definite parities under the various symmetry operations [174]. If a basis set is chosen which does not share the problem symmetry, then all the eigenstates of the various classes will be found. However it is very beneficial to instead symmetrize the basis set, restricting the problem to that of finding only eigenstates in a single symmetry class. This can be repeated for all symmetry classes if desired.

Why is this beneficial? Consider a billiard with a single plane (or line) of reflection, $\mathcal{R}$. States fall into the odd or even classes. An unsymmetrized basis approach requires consideration of the full boundary area ${\mathsf{A}}$ and the full domain volume ${\mathsf{V}}$, leading to a basis set size $N$ proportional to ${\mathsf{A}}$. Consideration of only the odd class (by using an odd-symmetrized basis set) requires matching of the BCs on only half of the area (no matching is required on $\mathcal{R}$), and therefore $N$ is halved. If the effort per state found scales like $N^3$ ($N^2$) with the sweep (scaling) method, then symmetrization has increased efficiency by a factor 8 (4). Also the level density of states is halved, making the states easier to distinguish (see Section 5.5.3), raising the maximum $k$ reachable.

Generally the symmetrization of the basis functions means that they take $n$ times as long to evaluate (where $n$ is the symmetry factor). This has an effect on the speed of filling matrices, and of evaluating the final eigenfunctions (and gradients) in position-space.

For the stadium example common in this thesis, a fourfold symmetry is present ($C_{4v}$): reflection about both the x and y axes. Therefore all basis functions will undergo the following symmetrization:

$$\phi_i(x,y) \ \longleftarrow \ \frac{1}{4}\left[ \phi_i(x,... ..._i(-x,y) + p_y \phi_i(x,-y) + p_x p_y \phi_i(-x,-y) \right] ,$$ (J.8)

where the parities are $p_x, p_y = \pm 1$. The choice of the odd-odd symmetry class corresponds to $p_x = p_y = -1$. In the case of RPWs, such symmetrization results in only a single basis function per direction vector. For instance, the odd-odd class gives

$$\bar{\phi}_i(\bar{{\mathbf r}}) \ = \ \sin (n_{x,i} \bar{x}) \sin (n_{y,i} \bar{y}) ,$$ (J.9)

where the directions $\theta_i$ need now only cover the range $[0,\pi/2]$. For an example see Fig. J.1. So it turns out that for RPWs in $C_{4v}$ only twice as much time is spent per basis function evaluation than for the unsymmetrized case (J.3). For EPWs no such reduction in symmetrization time occurs.