Form of tension minima and resulting accuracy in $k$

As discussed in Section 5.3.1 (also see Appendix J), the values $\epsilon_\mu$ of these minima at each $k_\mu$ is determined by the quality of the Helmholtz basis set. Around each minimum the tension is quadratic,

$$f(k) \; = \; \epsilon_\mu + \mbox{\small$\frac{1}{2}$}c_\mu (k-k_\mu)^2 ,$$ (5.23)

where the curvatures $c_\mu$ are very similar for all states $\mu$ (see Fig. 5.2). This happens because the lowest tension state of constant norm (that is, the state ${\mathbf x}_1$ returned by (5.14)) is simply the scaling eigenfunction $\mu$ with lowest tension (see Section 6.1.1). The tension expansion of this function about $k=k_\mu$ is quadratic with $c_\mu = 4$ exactly, independent of $\mu$, for a modified weight function $w{({\mathbf s})}$ (see Appendix I). Therefore with our choice $w{({\mathbf s})}=1$, a curvature estimate is $c_\mu \approx 4 \langle r_n^2\rangle / \langle r_n \rangle$ where the averages are taken over ${\mathbf s} \in \Gamma$, and $r_n \equiv {\mathbf r}\cdot {\mathbf n}$. If either there is little scarring or the billiard is close to spherical, then we expect this estimate to be good.

Eq.(5.23) defines a natural `error scale' for the $k_\mu$ found: a change $k$ that is much less than $(\epsilon_\mu/{\mathsf{L}})^{1/2}$ from a tension minimum causes only a small fractional change in tension. Therefore I will call $\Delta k(\epsilon_\mu) = (\epsilon_\mu/{\mathsf{L}})^{1/2}$ the `tension rounding error' of the state $\mu$, because it defines an uncertainty in $k_\mu$ due to the basis-dependent finite tension minima. In practice, the errors in $k_\mu$ found appear to be an order of magnitude better than $\Delta k(\epsilon_\mu)$, as shwon by the lines in Fig. 6.6.