I've been working on a physics problem that has led me to make the following numerical observation. Let $L$ be an $N\times N$ real-symmetric tridiagonal matrix whose diagonal entries are zero. This matrix is fully specified by the entries of its first superdiagonal, which we denote as $c_{1},c_{2},\cdots c_{N}$.
Now consider the exponential of $L$ acting on a vector, $\vec{O}(a,b) = \exp\left[i(a+ib)L\right]\vec{e_{1}}$, where $\vec{e_{1}} =\left(1,0,0,\cdots, 0\right)$ and $a$ and $b$ are real numbers. The vector $\vec{O}(a,b)$ can be expressed as
$$\vec{O}(a,b) = \sum_{n=1}^{N} r_{n}e^{i\theta_{n}} \vec{e_{n}}$$
for real numbers $r_{n}, \theta_{n}$; these are functions of $a, b$, and the matrix elements of $L$, but I have suppressed this dependence for brevity. Now for my observation: for all choices of $L, a,$ and $b$ that I have considered, the phases $\theta_{n}$ vary linearly with $n$: $$\theta_{n} = \theta_{0} + n\alpha$$
for some real numbers $\theta_{0}$ and $\alpha$. This numerical observation is not perfect - there can be small deviations from it, particularly for small $n$, but it seems to be highly accurate and robust against variations in the parameters that define the problem. For example, if we take the $c_{i}$ to be random numbers drawn uniformly from the interval $[0,1]$, the phases vary linearly.
Since my numerics indicate that this property is generic, I would like to find a proof of it, perhaps by computing the asymptotic (large n) properties of $r_{n}$ and $\theta_{n}$ for an arbitrary choice of the $c_{i}$. I would also settle for any kind of intuitive argument for why this property is generic - I have no idea why this "should" be the case.