Time Derivative of Expectation Values *

We wish to compute the time derivative of the expectation value of an operator \bgroup\color{black}$A$\egroup in the state \bgroup\color{black}$\psi$\egroup. Thinking about the integral, this has three terms.

{d\over dt}\left\langle\psi\left\vert A\right\vert\psi\right\r...
...\vert{\partial A\over\partial t}\right\vert\psi\right\rangle \\

This is an important general result for the time derivative of expectation values.

\bgroup\color{black}$\displaystyle {d\over dt}\left\langle\psi\left\vert A\right...
...le\psi\left\vert{\partial A\over\partial t}\right\vert\psi\right\rangle $\egroup
which becomes simple if the operator itself does not explicitly depend on time.

\begin{displaymath}\bgroup\color{black}{d\over dt}\left\langle\psi\left\vert A\r...
...t\langle\psi\left\vert[H,A]\right\vert\psi\right\rangle \egroup\end{displaymath}

Expectation values of operators that commute with the Hamiltonian are constants of the motion.

We can apply this to verify that the expectation value of \bgroup\color{black}$x$\egroup behaves as we would expect for a classical particle.

\bgroup\color{black}$\displaystyle {d\left\langle x\right\rangle\over dt}={i\ove...
...p^2\over 2m},x\right]\right\rangle
=\left\langle{p\over m}\right\rangle $\egroup
This is a good result. This is called the Ehrenfest Theorem.

For momentum,

\bgroup\color{black}$\displaystyle {d\langle p\rangle\over dt}={i\over\hbar}\lef...
...er dx}\right]\right\rangle
=-\left\langle{dV(x)\over dx} \right\rangle $\egroup
which Mr. Newton told us a long time ago.

Jim Branson 2013-04-22