# Ehrenfest’s theorem

Starting from the definition $<p> =\int dx\Psi^*(x,t)\hat{p}\Psi(x,t)$ where $\hat{p}=-i\hbar(\partial /\partial x)$, and ignoring all terms that do not involve a factor $V(x)$, show that Ehrenfest second theorem is valid: $\frac{d\left \langle p \right \rangle}{d t}=\left \langle -\frac{\partial V}{\partial x} \right \rangle$ (Note the terms without the factor V all cancel but it’s not necessary for you to show this).

First you need to know the general time derivative of the average value of an operator A:

$\frac{d}{d t}\left \langle \Psi |A| \Psi \right \rangle=\left \langle \frac{d\Psi}{d t} |A| \Psi \right \rangle+ \langle\Psi \left |\frac{d A}{d t}\right |\Psi \rangle +\langle \Psi|A|\frac{d\Psi}{d t} \rangle$

Then use time dependent Schrodinger equation:

$\frac{d\Psi}{d t}=\frac{1}{i\hbar}\hat H\Psi$

$\frac{d}{d t}\left \langle \Psi |A| \Psi \right \rangle=\frac{1}{i\hbar}\left \langle H\Psi|A|\Psi \right \rangle+ \langle\Psi \left |\frac{d A}{d t}\right |\Psi \rangle -\frac{1}{i\hbar}\langle \Psi|A|\frac{d\Psi}{d t} \rangle$

$\frac{d}{d t}\left \langle \Psi |A| \Psi \right \rangle=\frac{1}{i\hbar}\left \langle \Psi|[H,A]|\Psi \right \rangle+ \langle\Psi \left |\frac{d A}{d t}\right |\Psi \rangle$

Now replace $\hat A =\hat p =-i\hbar*(d/d x)$

$\frac{d}{d t}\left \langle \hat p \right \rangle=\frac{d}{d t}\left \langle \Psi|p|\Psi \right \rangle=\frac{1}{i\hbar}\left \langle \Psi|[H,p]|\Psi \right \rangle +\left \langle \Psi|\frac{d \hat p}{d t}|\Psi \right \rangle=\frac{1}{i\hbar}\left \langle \Psi|[H,p]|\Psi \right \rangle$

Disregarding other terms we have

$[H,p] =[V(x),p]=\left[V(x),-i\hbar\frac{d}{d x} \right ]$

Now compute the remaining last term of the time derivative

$\frac{d}{d t}\left \langle p \right \rangle=-V\left \langle \Psi|\frac{d}{d x}|\Psi \right \rangle -\left \langle \Psi|\frac{d V}{d x}|Psi \right \rangle=C+\left \langle -\frac{d V}{d x} \right \rangle$

Just because

$\Psi =A*exp[i(k x-\omega*t)]$

and the first term above is a constant $C$

$V\left \langle \Psi|\frac{d V}{d x}|\Psi \right \rangle=-ikV\left \langle \Psi|\Psi \right \rangle= C$