the question I am trying to answer is:

Show that for any not explicitly time-dependent operator $\hat A$,

$$\frac{d}{dt} \langle \hat A \rangle = \frac{i}{\hbar} \langle \psi | [\hat H, \hat A]| \psi \rangle .$$ Use this relation to show that $\langle \hat A \rangle$ does not change whenever the system is in a stationary state.

I know how to show the relation, using the product rule on $\frac{d}{dt} \langle \psi | \hat A | \psi \rangle$, and using the Schrodinger equation to evaluate each term.

$$\frac{d}{dt} \langle \psi | \hat A | \psi \rangle = (\frac{d}{dt} \langle \psi |) \hat A | \psi \rangle + \langle \psi | \frac{d}{dt}\hat A | \psi \rangle + \langle \psi | \hat A \frac{d}{dt}| \psi \rangle$$

$$\frac{d}{dt}| \psi \rangle = - \frac{i}{\hbar} \hat H | \psi \rangle$$

$$ \frac{d}{dt} \langle \psi | = \frac{i}{\hbar} \langle \psi | \hat H$$

$$\frac{d}{dt} \hat A = 0$$

$$\frac{d}{dt} \langle \psi | \hat A | \psi \rangle = \frac{i}{\hbar} (\langle \psi | \hat H \hat A | \psi \rangle - \langle \psi | \hat A \hat H | \psi \rangle) = \frac{i}{\hbar} \langle \psi | [\hat H, \hat A] | \psi \rangle$$

I am unsure of how to show that $\frac{d}{dt} \langle \hat A \rangle = 0$ for the second part of the question.

Clearly, if $\hat A$ commutes with the hamiltonian, then this will be true since $\langle \psi | 0 | \psi \rangle = 0$. But I don't think I can assume this to be the case.

Thanks, I hope that makes sense.


  • $\begingroup$ I have tried to improve my question. Please let me know if you think it needs further work. $\endgroup$ – jm22b May 4 '17 at 12:57

For a stationary state, $\vert\psi\rangle=e^{-iEt/\hbar}\vert E\rangle$ with $\vert E\rangle$ an eigenstate of $\hat H$ with eigenvalue $E$. Then $$ \langle \psi \vert \hat H\hat A\vert \psi\rangle= e^{iEt/\hbar}E \langle E \vert \hat A\vert E \rangle e^{-iEt/\hbar} =E \langle E \vert \hat A\vert E \rangle $$ which is time-independent. A similar result follows for $\langle \psi \vert \hat A \hat H\vert \psi\rangle$. The difference between the two is $0$.


The key information here is that the system is in a stationary state. This implies that the system is in an eigenstate of the Hamiltonian. $$ \hat{H} |\psi\rangle = E |\psi\rangle$$ Plugging this into your expression pretty much gives you the answer.

  • $\begingroup$ I understand that $\hat A \hat H | \psi \rangle = \hat A E | \psi \rangle$, but I don't know about $\hat H \hat A | \psi \rangle$ $\endgroup$ – jm22b May 4 '17 at 13:07
  • 1
    $\begingroup$ $\hat{H}$ is hermitian. Apply it to the left! $\endgroup$ – noah May 4 '17 at 13:09

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