So given $\Psi(x,t_0) = \Sigma_n a_n \psi_n(x)$ we multiply both sides by $\psi_m(x)$ and integrate and sum over all $m$ to get: $a_m = \int_{-\infty}^{\infty}\Psi(x,t_0)\psi_m(x)dx$

Sorry, it's explained wrong. We multiply both sides by $\psi_m(x)$ and integrate. The sum on the right will pick out exactly the case where $n=m$ and so there is no additional summing over $m$. The final answer is correct though. $a_m = \int_{-\infty}^{\infty}\Psi(x,t_0)\psi_m(x)dx$

@enumaris yes indeed one then exploits orthogonality to compute this. I spent some time last night attempting just this and going over this on the whiteboard. These are wave packets no?

will be preparing for an interview today. It is supposed to be a technical discussion by phone about php hehehe . I have the interview tomorrow at 2 pm.

@enumaris will be available sometime 7 pm Pacific time to discuss the physics of the spectrum?

probably not at that time

I'll be eating dinner and doing other stuff