1. 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!

2. The latex doesn't seem to work.

3. Latex here needs to be put between $$and$$ tags, rather than between \$ signs.

So, ($$\Psi (x,0)$$) becomes ()

and $$\frac{d\langle x\rangle}{dt} \not=$$ $$\langle p\rangle$$

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 .

4. 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.

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