Tension matrix in a scaling eigenfunction basis

It is important to define a spatial rescaling of a wavefunction, corresponding to dilation (or contraction) about the (arbitrary) origin. If the wavefunction $\psi{({\mathbf r})}$ is a Helmholtz solution at wavenumber $k_0$, then the rescaled version $\psi(k,{\mathbf r})$ of the same wavefunction obeys the Helmholtz equation at a new wavenumber $k$:

$$\psi(k,{\mathbf r}) \; \equiv \; \psi(\frac{k}{k_0} {\mathbf r}).$$ (6.1)

The rescaled function value at a given point is found by looking at the original function at a point a fractional distance $k/k_0 - 1$ further from the origin. The derivative with respect to $k$ of the wavefunction value at a point ${\mathbf r}$ is

$$\left. \frac{\partial }{\partial k} \psi(k,{\mathbf r}) \ri... ...\alpha=1} \; = \; \frac{1}{k_0} {\mathbf r}\cdot\nabla \psi .$$ (6.2)

This means that for $k=k_0+\delta$ with $\delta$ small ( $\ll 1/{\mathsf{L}}$), this can be expanded in powers of $\delta$ as

$$\psi(k_0 + \delta,{\mathbf r}) \; = \; \left[ 1 + \frac{\de... ...dot\nabla + O(\delta^2) \cdots \right] \, \psi{({\mathbf r})}.$$ (6.3)

We will not need any higher terms for the basic method (they are given in Appendix I).

Figure: Scaling eigenfunction $\psi_\mu(k,{\mathbf r})$ vanishes on a rescaled boundary $\Gamma_\mu$. It is clear that when $\Gamma_\mu$ is much closer than a wavelength from $\Gamma$, the wavefunction can be linearized, giving a value on $\Gamma$ proportional to the eigenfunction normal derivative. Also the boundary coordinate ${\mathbf s}$ and local surface normal ${\mathbf n}{({\mathbf s})}$ are shown.

We now choose a basis $\psi_\mu(k,{\mathbf r})$ of the exact billiard eigenfunctions scaled as above to have a single wavenumber $k$; these are called scaling eigenfunctions [195]. The unscaled eigenfunctions are taken to have unit normalization in the billiard. The index $\mu = 1\cdots\infty$ labels the eigenfunctions in ascending order of their unscaled wavenumber $k_\mu$. The scaling eigenfunctions no longer go to zero on the original billiard boundary $\Gamma$, however they do go to zero on rescaled (dilated or contracted) versions of the boundary $\Gamma_\mu$ (see Fig. 6.1). Imagine that a state $\mu$ has a wavenumber shift $\delta_\mu \equiv k-k_\mu$ which is very small ( $\ll 1/{\mathsf{L}}$, the inverse system size). Then $\Gamma$ will be very close to $\Gamma_\mu$, and the value of $\psi_\mu(k,{\mathbf r})$ for ${\mathbf r} = {\mathbf r}{({\mathbf s})}\in\Gamma$ will be close to zero. The boundary coordinate ${\mathbf s}$ measures location on $\Gamma$. Applying (6.3), the difference from zero is given by

$$\psi_\mu(k,{\mathbf s}) \; = \; \frac{\delta_\mu}{k_\mu}r_n \partial_n \psi_\mu + O(\delta_\mu^2) \cdots$$ (6.4)

where $\partial_n \psi_\mu$ is the (unscaled) eigenstate normal derivative at location ${\mathbf s}$. Notice that the zeroth-order term vanished because of the BCs. (This expansion is continued in Appendix I). A general sum of scaling eigenfunctions

$$\psi(k,{\mathbf r}) \ = \ \sum_\mu \tilde{x}_\mu \psi_\mu(k,{\mathbf r})$$ (6.5)

is also a scaling wavefunction.

The tension (as defined in the previous chapter for Dirichlet BCs) of a general scaling wavefunction is

$$f(k) \; = \; \oint_\Gamma \!\! d{\mathbf s} \,w{({\mathbf s})}\, \vert\psi(k,{\mathbf s})\vert^2 ,$$ (6.6)

where $w{({\mathbf s})}$ is a boundary weighting function. If $\psi$ is now written in the form (6.5) then the tension is $\tilde{{\mathbf x}}^{{\mbox{\tiny T}}} \tilde{F} \tilde{{\mathbf x}}$. The matrix $\tilde{F}$ is the scaling eigenfunction representation of the tension quadratic form, and we will see that it has wonderful properties. For $\delta_\mu, \delta_\nu \ll 1$ we can substitute (6.4) as follows,

$$\tilde{F}_{\mu \nu}(k) \; =\; \oint_\Gamma \!\! d{\mathbf s} \,w{({\mathbf s})}\, \psi_\mu(k,{\mathbf s}) \psi_\nu(k,{\mathbf s})$$

$$= \frac{\delta_\mu \delta_\nu}{k_\mu k_\nu} \oint_\Gamma \!\! d{\ma... ...f s})} r_n^2 (\partial_n \psi_\mu) \partial_n \psi_\nu \ + \ O(\delta^3) \cdots$$

$$= 2 \delta_\mu \delta_\nu M_{\mu \nu} \ + \ O(\delta^3) \cdots ,$$ (6.7)

where $r_n$ is an abbreviation for ${\mathbf r}\cdot{\mathbf n}$ at the location ${\mathbf s}$. The final form has been written in terms of the boundary inner-product matrix of the eigenstate normal derivatives,

$$M_{\mu \nu} \; = \; \frac{1}{2k^2} \oint_\Gamma \!\! d{\mat... ...{({\mathbf s})}\, (\partial_n \psi_\mu) \partial_n \psi_\nu ,$$ (6.8)

where $D{({\mathbf s})}= r_n^2 w{({\mathbf s})}$ is the weight defining the inner-product. The approximation $k \approx k_\mu \approx k_\nu$ has been used, because $M$ has only very slow dependence on $k$. Note that all the fast dependence of $F$ on $k$ appears in the wavenumber shifts $\delta _\mu$ and $\delta_\nu$.

Figure: Image of the matrix $M \propto \partial {\mathcal{H}} / \partial x$ shown as a density plot of $\vert\partial {\mathcal{H}} / \partial x_{\mu\nu}\vert^2$, for the case of dilation (a special deformation) of the 2D quarter stadium billiard. Darker pixels correspond to larger values. The wide central white region (of width $\sim 1/{\mathsf{L}}$) corresponds to quasi-orthogonality: the quantum band profile vanishes like $\kappa^4$ as $\kappa\rightarrow0$, where $\kappa \equiv k_\mu - k_\nu$. The diagonal is unity, and is not included in the band profile. The matrix involves all 451 eigenstates falling in the wavenumber range $398 < k < 402$.