- #1
Logan Rudd
- 15
- 0
I know when the initial state (##\Psi(x,0)##) is given, ##\frac{d<x>}{dt} \not=<p>##. I thought you can only apply Ehrenfest's theorem when ##\Psi## is a function of x and t, however it seems like you can also apply it to the time-independent part (##\psi(x)##) by itself as well. Can someone explain to me why Ehrenfest's theorem is valid for ##\psi(x)##?
Thanks!
Thanks!
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