- #1
good_phy
- 45
- 0
Hi,
You know famous equation, [itex] \frac{d<A>}{dt} = <\frac{i}{\hbar}[\hat{H},\hat{A}] + \frac{\partial\hat{H}}{\partial t} >[/itex]
But liboff said if [itex] \frac{\partial \hat{A} }{\partial t} = 0 [/itex] then, [itex] \frac{d<\hat{A}>}{dt} = 0 [/itex]
this is the proof
[itex] \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}> [/itex]
[itex]=\frac{i}{\hbar}(<\hat{H}\varphi_{n}|\hat{A}\varphi_{n}> - <\varphi|\hat{A}\hat{H}\varphi_{n}>) [/itex]
[itex] \frac{i}{\hbar}E_{n}(<\varphi_{n}|\hat{A}\varphi_{n}> - <\varphi_{n}|\hat{A}\varphi_{n}>) = 0 [/itex]
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.
You know famous equation, [itex] \frac{d<A>}{dt} = <\frac{i}{\hbar}[\hat{H},\hat{A}] + \frac{\partial\hat{H}}{\partial t} >[/itex]
But liboff said if [itex] \frac{\partial \hat{A} }{\partial t} = 0 [/itex] then, [itex] \frac{d<\hat{A}>}{dt} = 0 [/itex]
this is the proof
[itex] \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}> [/itex]
[itex]=\frac{i}{\hbar}(<\hat{H}\varphi_{n}|\hat{A}\varphi_{n}> - <\varphi|\hat{A}\hat{H}\varphi_{n}>) [/itex]
[itex] \frac{i}{\hbar}E_{n}(<\varphi_{n}|\hat{A}\varphi_{n}> - <\varphi_{n}|\hat{A}\varphi_{n}>) = 0 [/itex]
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.