Tension matrix expansion

The tension matrix (using the `dilation' weighting function $w(s) = 1/r_n$) in the scaling eigenfunction basis has elements

$$\tilde{F}_{\mu\nu}(k) \ = \ \oint \! \frac{ds}{r_n} \, \psi_\mu(k_\mu + \delta_\mu) \, \psi_\nu(k_\nu + \delta_\nu),$$ (I.19)

where the wavenumber shifts for the two states involved are $\delta_\mu \equiv k-k_\mu$ and $\delta_\nu \equiv k - k_\nu$. As usual I define the wavenumber difference $\kappa \equiv k_\mu - k_\nu = \delta_\nu - \delta_\mu$. Substitution of (I.18) gives a power series in the $\delta$'s, which we will study and make estimates for unknown boundary integrals to grasp the general structure. We will assume for simplicity that the billiard size ${\mathsf{L}}\approx 1$, and remain in $d=2$. We will ignore differences in $k_\mu$ and $k_\nu$ from $k$ when appropriate, and look only for the dominant off-diagonal terms.

The $O(\delta^2)$ term in $\tilde{F}$ comes from the lowest order in (I.18),

$$\tilde{F}^{(2)}_{\mu\nu}(k) \ = \ \frac{\delta_\mu \delta_\nu}{k_\mu k_\nu} \oint \! ds \,r_n \, \psi_\mu^n \psi_\nu^n$$ (I.20)

which we can write as $2\delta_\mu \delta_\nu M_{\mu\nu}$ (see Section 6.1.1), where $M$ is quasi-diagonal. The diagonal is $M_{\mu\mu} = 1$, and we have from the considerations of Section 6.1.2 small off-diagonal elements of size $M_{\mu\nu} \sim k^{-1/2} \kappa^2 O(1)$, for $d=2$.

The $O(\delta^3)$ term in $\tilde{F}$ is, after some rearrangement,

$$\tilde{F}^{(3)}_{\mu\nu}(k) \ = \ -\frac{2 \delta_\mu^3}{k}... ..._t \, (\psi_\nu^n \psi_\mu^{nt} - \psi_\mu^n \psi_\nu^{nt}) ,$$ (I.21)

where an integral of $r_n r_t \partial_t(\psi_\mu^n \psi_\nu^n)$ was performed by parts. This required use of $\partial_t(r_n r_t) = r_n + \alpha(r_n^2 - r_t^2)$, a different expression than found by VS. It can be proved easily using $\partial_t r_n = \alpha r_t$ and $\partial_t r_t = 1 - \alpha r_n$, which follow from (I.8) and (I.9). Remarkably, the $\alpha$ dependence then cancels out, giving the same $\tilde{F}^{(3)}$ diagonal term as that of VS. We believe that the integral in (I.21) does not have any quasi-orthogonal property, so can be estimated using random waves. The estimate gives $k^{5/2} O(1)$ for this integral, and shows that this term dominates over any off-diagonal contribution from the first term (involving $M$). The factor of $\kappa$ in this term means that there is a weak form of quasi-diagonality at this order. Importantly, for $\kappa \ll 1$ the off-diagonal $O(\delta^3)$ error renders the $O(\delta^2)$ error insignificant. Hence we expect the quasi-diagonality property of $M$ to play no role in errors in the scaling method.

The $O(\delta^4)$ and higher terms in $F$ become very messy. I believe that the dominant $O(\delta^4)$ terms, both on and off the diagonal, are

$$\tilde{F}^{(4)}_{\mu\nu}(k) \ = \ -\frac{\delta_\mu\delta_\... ...} \oint \! ds \, r_n^2 r_t^2 \, \psi_\mu^{nt} \psi_\nu^{nt} .$$ (I.22)

This can be seen by comparing powers of $k$ and using random-wave estimates. A random-wave estimate of the integrals then gives $\tilde{F}^{(4)} \approx \delta_\mu^4 O(1)$ on the diagonal and $\delta_\mu^4 k^{-1/2} O(1)$ off-diagonal. Note that the off-diagonal has no quasi-diagonality property at this or higher orders. For higher orders $O(\delta^m)$ for $m$ even, we expect $\tilde{F}^{(m)} \approx \delta_\mu^m O(1)$ on the diagonal and $O(\delta^m) k^{-1/2} O(1)$ off-diagonal. For $m$ odd, the leading diagonal terms are down by a factor of $k$ which renders them insignificant.

To summarize, the diagonal of the tension matrix has the form given in Eq.(6.27), and for $\delta_\mu, \delta_\nu \ll 1$ the dominant off-diagonal terms are

$$\tilde{F}_{\mu\nu}(k) \ \approx \ \frac{\delta_\mu\delta_\... ... O(1) + \delta_\mu\delta_\nu O(1) \rule{0in}{0.2in} \right] .$$ (I.23)

Here contributions from orders 3 and 4 were included because it may be that the 4th order (the lowest order with no quasi-diagonality, i.e. no powers of $\kappa$) contributes most to errors in the scaling method. It is important to note that the 2nd order term (due to off-diagonal strength of $M$) is negligible. It is clear that more research is needed on the properties of the higher-order terms, especially if an explanation of the tension error $\delta^6$ power-law growth (Section 6.3.1) is sought.