factored out.

For simplicity we henceforth consider the case $1/\lambda=0$. We shall employ the notation $$Z'(k,{{\mathbf L}})=\frac{\partial Z}{\partial k}(k,{{\mathbf L}})=\sum_{j=1}^v
L_j\sec^2 kL_j.$$

The eigenfunction corresponding to the $n^{\rm th}$ eigenvalue is found to be $$\psi_i^{(n)}(x)=A^{(n)}\frac{\cos k_n(x-L_i)}{\cos k_nL_i}.$$ The constant $A^{(n)}$ is determined by the normalisation $$\sum_{j=1}^v \int_0^{L_j} |\psi_j^{(n)}(x)|^2 {{\mathrm d}}x=1,
\label{norm:const}$$ to be $$A^{(n)}=\left(\frac{2}{\sum_{j=1}^v L_j\sec^2 k_nL_j}\right)^{\frac{1}{2}}.$$ The value distribution of the eigenfunctions is determined by these normalisation constants. For definiteness, we shall focus here on the maximum amplitude squared of the eigenfunctions on a single bond, $$\begin{aligned}
A_i(n,{{\mathbf L}};v)&:=&\sup_{x\in[0,L_i]}\{|\psi_{i}^{(n)}(x)|^2\}\\
&=&(A^{(n)}\sec k_nL_i)^2\\
&=&\frac{2\sec^2 k_nL_i}{\sum_{j=1}^v L_j\sec^2 k_nL_j}.\end{aligned}$$

We now state our main results.

\[thm:1\] For any fixed ${\bar{L}}> 0$ $$\lim_{k\to\infty} \frac{1}{(\Delta L)^v}
{\mathop{\rm meas}}\left\{ {{\mathbf L}}\in[{\bar{L}},{\bar{L}}+\Delta L]^v:\frac{1}{v}Z(k,{{\mathbf L}})<y
\right\}=\frac{1}{\pi}
\int_{-\infty}^y \frac{1}{1+x^2} {{\mathrm d}}x,$$ provided that $k\Delta L\to\infty$ as $k\to\infty$.

We emphasise that in theorem \[thm:1\] we do not require that $\Delta L\to 0$. In some later results we shall make this stipulation.

\[thm:2\] Suppose that the components of ${{\mathbf L}}$ are fixed and linearly independent over ${\mathbb Q}$. Then $$\lim_{K\to\infty} \frac{1}{K}{\mathop{\rm meas}}\left\{k\in[0,K]:\frac{1}{v}Z(k,{{\mathbf L}})<y
\right\}=\frac{1}{\pi}\int_{-\infty}^y\frac{1}{1+x^2}{{\mathrm d}}x.$$

Theorems \[thm:1\] and \[thm:2\] demonstrate the equivalence of taking a $k$-average and a bond-length average at large $k$ for the distribution of values taken by the function 