equation thus reads $\dot{{
  \mathbf{p}
}} = -L{
  \mathbf{p}
}$ and it leads to the temporal evolution of ${
  \mathbf{p}
}$ in the form $$\begin{aligned}
  {
  \mathbf{p}
}(t) = \exp(-Lt) {
  \mathbf{p}
}_0,\end{aligned}$$ with ${
  \mathbf{p}
}_0$ being the initial condition. In our case, the initial distribution ${
  \mathbf{p}
}_0$ will always be completely localized, i.e. ${
  \mathbf{p}
}_0^2 = 1$.

If we average over all these localized initial conditions, then $Y_2 = {
  \mathbf{p}
}_0^T \exp(-L^T t) \exp(-L t) {
  \mathbf{p}
}_0$ becomes the trace $$\begin{aligned}
 \langle Y_2\rangle_{p_0} &= \frac{1}{N} ~{
  \mathrm{trace}\left\{ e^{-L^T t} e^{-L t} \right\}
}\label{equ:pTpSingularValue}\\
   &= \frac{1}{N} \sum_{j=1}^N \sigma_j(t),\nonumber    \end{aligned}$$ with $\sigma_j(t)$ being the $j$-th eigenvalue of the time-dependent operator $\mathcal{O}(t) = e^{-L^T t} e^{-L t}$. In what follows we will however use a more convenient representation and describe the process in the basis of the eigenfunctions of the Laplacian $L$ where the spectral decomposition is time-independent.

If $\mathcal{L}$ is the Laplacian for a simple random walk on a ring of size $N$ (i.e. a Toeplitz matrix with $1$ on the main diagonal, minus one half on the two adjacent diagonals and in the upper right and lower left corners, and all other elements equal to 