# Ehrenfest's Theorem

• 10-03-2014, 08:31 PM
iNV9O7
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!
• 10-03-2014, 08:49 PM
Jilan
The latex doesn't seem to work.
• 10-03-2014, 10:23 PM
SpeedFreek
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...

Quote:

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 .

• 10-04-2014, 07:09 AM
Jilan
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.