# [Q]Time deviation of expectation value

1. Nov 14, 2008

### good_phy

Hi,

You know famous equation, $\frac{d<A>}{dt} = <\frac{i}{\hbar}[\hat{H},\hat{A}] + \frac{\partial\hat{H}}{\partial t} >$

But liboff said if $\frac{\partial \hat{A} }{\partial t} = 0$ then, $\frac{d<\hat{A}>}{dt} = 0$

this is the proof

$\frac{d<A>}{dt} = \frac{i}{\hbar}<\varphi_{n}|[\hat{H},\hat{A}]\varphi_{n}> = \frac{i}{\hbar}<\varphi_{n}|(\hat{H}\hat{A}-\hat{A}\hat{H})\varphi_{n}>$
$=\frac{i}{\hbar}(<\hat{H}\varphi_{n}|\hat{A}\varphi_{n}> - <\varphi|\hat{A}\hat{H}\varphi_{n}>)$
$\frac{i}{\hbar}E_{n}(<\varphi_{n}|\hat{A}\varphi_{n}> - <\varphi_{n}|\hat{A}\varphi_{n}>) = 0$

If it is right, we can conclude time deviation of expectation value of certain operator is zero if and only if corresponding operator is not depending on time, no matter what value of [H,A]
is!

is it right? i can't accept this theorm.

2. Nov 14, 2008

### weejee

This theorem is only valid for eigenstates of H. d<A>/dt is generally nonzero for non-stationary states unless A commutes with H.

3. Nov 14, 2008

### newbee

The state vector is a function of time in the Schodinger picture, as opposed to the Heisenberg picture, appears to be what you are using. In general the state vector will then be evolving as a superposition of eigenstates.