Ehrenfest’s theorem
Generalized Ehrenfest’s theorem:
\[\boxed{ \frac{d}{dt}\expct{\hat{Q}} = \frac{i}{\hbar}\expct{\com{ \hat{\mathcal{H}} }{ \hat{Q} }} + \Expct{\frac{\partial \hat{Q}}{\partial t}} }\]
The equation is the same as the Heisenberg image, but let’s derive it from the Schrödinger image.
\[ \begin{align*} \frac{d}{dt}\expct{\hat{Q}} &= \frac{d}{dt}\brktop{\psi}{\hat{Q}}{\psi} \nl &= \Brktop{\frac{\partial\psi}{\partial t}}{\hat{Q}}{\psi} + \Brktop{\psi}{\frac{\partial \hat{Q}}{\partial t}}{\psi} + \Brktop{\psi}{\hat{Q}}{\frac{\partial\psi}{\partial t}} \end{align*} \]
Using the Schrödinger equation $i\hbar\dfrac{\partial}{\partial t}\ket{\psi}=\hat{\mathcal{H}}\ket{\psi}$,
\[ \frac{d}{dt}\expct{\hat{Q}} = -\frac{1}{i\hbar}\brkt{ \hat{\mathcal{H}}\psi }{ \hat{Q}\psi } + \Expct{\frac{\partial \hat{Q}}{\partial t}} + \frac{1}{i\hbar}\brkt{ \psi }{ \hat{Q}\hat{\mathcal{H}}\psi } \]
\[ \therefore \frac{d}{dt}\expct{\hat{Q}} = \frac{i}{\hbar}\expct{\com{ \hat{\mathcal{H}} }{ \hat{Q} }} + \Expct{\frac{\partial \hat{Q}}{\partial t}} \]
Examples
$ m\dfrac{d\expct{ \b{r} }}{dt} = \expct{\b{p}} $
$ \dfrac{d\expct{ \b{p} }}{dt} = -\Expct{ \nabla V(\b{r}) } $
Time-energy uncertainty relation
If $\hat{Q}$ is not explicit for $t$, by the Ehrenfest’s theorem;
- $ \sigma_\mathcal{H}\sigma_Q \ge \dfrac{1}{2}\abs{\expct{ \com{\hat{\mathcal{H}}}{\hat{Q}} }} = \dfrac{\hbar}{2}\abs{ \dfrac{d\expct{\hat{Q}}}{dt} } $
Here we define $\Delta E \coloneqq \sigma_\mathcal{H}$ and $\Delta t \coloneqq \dfrac{\sigma_Q}{\abs{ d\expct{\hat{Q}} / dt }} $.
Then we get, time time-energy uncertainty relation:
- $ \Delta E\Delta t \ge \dfrac{\hbar}{2} $
The ‘uncertainty’ in time is expressed as the average time taken, starting in state $\ket{\psi}$, for the expectation of some arbitrary observable $\hat{Q}$ to change by its standard deviation. This is reasonable as a definition for time uncertainty because it gives the shortest time scale on which we will be able to notice changes by using $\hat{Q}$ in state $\ket{\psi}$.