Thread: Ehrenfest's Theorem

I know when the initial state ($\Psi (x,0)$) is given, $\frac{d\langle x\rangle}{dt} \not= $ $\langle p\rangle$. 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 $\psi (x)$, and or why it is not valid for $\Psi (x,0)$. This is my first time posting, not sure how the latex coding will look.

Thanks!

The latex doesn't seem to work.

Latex here needs to be put between [tex] and [/tex] tags, rather than between $ signs.

So, ([tex]\Psi (x,0)[/tex]) becomes ()

and [tex]\frac{d\langle x\rangle}{dt} \not= [/tex] [tex]\langle p\rangle[/tex]

becomes

So, my attempt to fix your post results in...

Originally Posted by iNV9O7

I know when the initial state () is given, . I thought you can only apply Ehrenfest's theorem when is a function of and , however it seems like you can also apply it to the time-independent part () by itself as well. Can someone explain to me why Ehrenfest's theorem is valid for , and or why it is not valid for .

Hello, welcome to the forum, I would think it is because need not be a solution of the Shroedinger equation. I like to think of it an a boundry condition in time. So p(x,0) may have time dependence whereas p(x,t) won't for the solutions.