Time evolution operator to find expectation value

Question

I have a state $\Psi (x,0) = \sum_{n=0}^{\infty} c_{n}u_n(x)$ and want to find the expectation value of any observable A at time t, $\langle \Psi(t)|\hat{A}|\Psi(t)\rangle$.

I know that I should apply the time evolution operator to find $\Psi(t)$ but am not sure how to find the expectation value when I'm not dealing with an eigenstate. I'm looking for the answer as a function of matrix elements $A_{mn} = \langle u_{m}|\hat{A}|u_{n}\rangle$.

Could anyone give me some guidance? I'm a QM newbie if you haven't noticed!

Answer 1

The key here is to notice that at any time $t$, one can expand the full time-dependent state $\Psi(x,t)$ at that time in terms of a basis $u_n$; $$ |\Psi(t)\rangle = \sum_n c_n(t)|u_n\rangle $$ The basic idea here is that the time-dependence of the state is being encoded in the expansion coefficients $c_n(t)$. Having done this, notice now that $$ \langle\Psi(t) |\hat A|\Psi(t)\rangle = \sum_{m,n}c_m^*(t)c_n(t)\langle u_m|\hat A|u_n\rangle = \sum_{m,n}c_m^*(t)A_{mn}c_n(t) $$
It is then possible to show (using the Schrodinger equation) that if the states $u_n$ are eigenstates of the Hamiltonian with corresponding eigenvalues $E_n$, then the coefficient $c_n(t)$ satisfy a simple differential equation in $t$ with a simple solution, and you should be able to put this all together to get what you're looking for.