Ehrenfest's Theorem
- Quantum Mechanics 1.9
· 3 min 
from 「Modern Quantum Mechanics」: Sakurai, J. J.
About the Ehrenfest’s Theorem.
Ehrenfest’s theorem
Generalized Ehrenfest’s theorem:
\[ \odv{ }{t}\expct{\hat{Q}} = \frac{i}{\hbar}\expct{\comm{ \hat{H} }{ \hat{Q} }} + \Expct{\pdv{\hat{Q}}{t}} \]
The equation is the same as the Heisenberg image, but let’s derive it from the Schrödinger image.
\[ \begin{align*} \odv{ }{t}\expct{\hat{Q}} &= \odv{ }{t}\brktop{\psi}{\hat{Q}}{\psi} \nl &= \Brktop{\pdv{\psi}{t}}{\hat{Q}}{\psi} + \Brktop{\psi}{\pdv{\hat{Q}}{t}}{\psi} + \Brktop{\psi}{\hat{Q}}{\pdv{\psi}{t}} \end{align*} \]
Using the Schrödinger equation $\dps i\hbar\pdv{ }{t}\ket{\psi}=\hat{H}\ket{\psi}$,
\[ \odv{ }{t}\expct{\hat{Q}} = -\frac{1}{i\hbar}\brkt{ \hat{H}\psi }{ \hat{Q}\psi } + \Expct{\pdv{\hat{Q}}{t}} + \frac{1}{i\hbar}\brkt{ \psi }{ \hat{Q}\hat{H}\psi } \]
\[ \therefore \odv{ }{t}\expct{\hat{Q}} = \frac{i}{\hbar}\expct{\comm{ \hat{H} }{ \hat{Q} }} + \Expct{\pdv{\hat{Q}}{t}} \]
Examples
$\dps m\odv{\expct{\b{r}}}{t} = \expct{\b{p}} $
$\dps \odv{\expct{\b{p}}}{t} = -\Expct{ \nabla V(\b{r}) } $
Time-energy uncertainty relation
If $\hat{Q}$ is not explicit for $t$, by the Ehrenfest’s theorem;
- $\dps \sigma_H\sigma_Q \ge \frac{1}{2}\abs{\expct{ \comm{\hat{H}}{\hat{Q}} }} = \frac{\hbar}{2}\abs{ \odv{\expct{\hat{Q}}}{t} } $
Here we define $\Delta E \coloneqq \sigma_H$ and $\Delta t \coloneqq \dfrac{\sigma_Q}{\abs{ \odvi{\expct{\hat{Q}}}{t} }} $.
Then we get, time time-energy uncertainty relation:
- $\dps \Delta E\Delta t \ge \frac{\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 $\ket{\psi}$.
