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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!

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Hi KatieC25, and (slightly belated) welcome to Physics Stack Exchange! What have you tried to get the expectation value? Could you perhaps edit into the question an example of how you would start the process and show at which point you get stuck? –  David Z May 4 '13 at 21:25

1 Answer 1

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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.

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Thanks for helping me out (again!) –  KatieC25 May 5 '13 at 20:06
    
Sure thing @KatieC25 ! –  joshphysics May 5 '13 at 20:07

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