Are you asking about how quantum mechanical expectation values evolve with time? If so, then it evolves according to the differential equation
\frac{d}{dt}<{\psi}|O|{\psi}> = \frac{i}{\hbar}<{\psi}|[H,O]|{\psi}> + <{\psi}|\frac{{\partial}O}{{\partial}t}|{\psi}>
With O being a Hermitian operator.
i don't know what happened to my original post, but I am having an issue with the following problem:
Show that:
\frac{d}{dt}<x^2> =\frac{1}{m}(< xp_x> +<p_xx>)
for a three dimensional wave packet.
relevant equations:
Ehrenfest Theorem:(1)
i\hbar\frac{d}{dt}<O>=<[O,H]>+i\hbar<\frac{\partial }{\partial t}O>
where O is an operator
(2)
\frac{d}{dt}\int_{V}d^3r\psi ^*O\psi
I tried using both ways illustrated above and I arrived at the same answer:
\frac{-i\hbar}{2m}\int_{V}d^3r\psi ^*[\bigtriangledown ^2(x^2\psi)-x^2\bigtriangledown ^2\psi]
=\frac{-i\hbar}{2m}\int_{V}d^3r\psi ^*[\frac{\partial }{\partial x}
(x^2\psi'+2x\psi)-x^2\frac{\partial^2 }{\partial x^2}\psi]
=\frac{-i\hbar}{2m}\int_{V}d^3r\psi ^*[
2\psi+2x\frac{\partial }{\partial x}\psi+2x\frac{\partial }{\partial x}\psi+(x^2-x^2)\frac{\partial^2 }{\partial x^2}\psi]
=\frac{1}{m}\int_{V}d^3r[\psi ^*(-i\hbar)\psi+\psi^*x(-i\hbar\frac{\partial }{\partial x})\psi+\psi^*(-i\hbar\frac{\partial }{\partial x})x\psi]
=\frac{1}{m}(<xp_x>+<p_xx>)-\frac{i\hbar}{m}
Am I doing anything wrong? Where does the extra term come from, and does it mean anything?