Let $L$ be the Laplacian matrix of a simple, connected graph, and $\mathcal{P}\_j$ the projector into the vertex $v\_j$, represented by the appropriate canonical basis vector $(0,...,1,...,0)^T$. Given the positive real parameters $t$ and $\lambda$, consider the functions $$P\_j(n,t,\lambda)=v\_j^T e^{-it(\lambda L-\mathcal{P}\_j)}s, $$ where $s=\frac{1}{\sqrt{n}}(1,...,1)^T$ is the normalized uniform vector.


My objective is to show that the condition $P\_i=P\_j$ implies the degrees of the two vertices $v\_i$ and $v\_j$ are the same. Does anyone have any tips on how to accomplish this?



Showing two vertices have same degree under a certain condition
I'll take $\lambda=1$ and use $E\_j$ for $\mathcal{P}\_j$. The $k$-th time derivative of $e^{-it(L-E\_j)}s$ at $t=0$ is
\[
(-i(L-E\_j))^k s.
\]
Now $(L-E\_j)s = -v\_j$ (because $Ls=0$) and, noting that $E\_j=v\_jv\_j^T$, we have
\[
(L-E\_j)^2s = -(L-E\_j)v\_j = -Lv\_j +v\_j.
\]
Therefore
\[
v\_j^T(L-E\_j)^2s = -v\_j^T Lv\_j + 1 = - L\_{j,j} + 1.
\]
Since $L\_{j,j}$ is the degree of vertex $j$, it follows that if $P\_i=P\_j$, then $i$ and $j$ have the same degree.


[I am curious as to where this problem is coming from?]


