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I'm a bit embarrassed because there seems to be an obvious thing that I'm missing, but I can't see what it is.

Consider $[\mathcal{H},A]=B$, where $A$ and $B$ are some operators and $\mathcal{H}$ is the Hamiltonian. Let's look at the expectation value of $\mathcal{H}A$, and let's use energy eigenstates.

$$\langle \psi|\mathcal{H}A|\psi\rangle=\langle \psi|B+A\mathcal{H}|\psi\rangle=\langle \psi|B|\psi\rangle+\langle \psi|A\mathcal{H}|\psi\rangle=\langle B\rangle+E\ \langle A\rangle.$$

But, can't I also write

$$\langle \psi|\mathcal{H}A|\psi\rangle=\left(\langle \psi|\mathcal{H}\right) \left(A|\psi\rangle\right)=E\ \langle \psi|A|\psi\rangle=E\ \langle A\rangle,$$

where I've made use of $\mathcal{H}| \psi \rangle=E\ |\psi\rangle \Rightarrow \langle \psi|\mathcal{H}=E\ \langle \psi|?$

I don't see what's wrong with my argument, but I feel like something is wrong because I get different answers.

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  • $\begingroup$ This is all meaningless. There is no way to expect that the domains of A and B contain the eigenvectors of H. $\endgroup$
    – DanielC
    Commented Sep 14, 2019 at 20:18
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    $\begingroup$ @DanielC That is a bit harsh, given that we could perfectly well postulate that we're in a finite-dimensional space of states where there are no domain issues. $\endgroup$
    – ACuriousMind ♦
    Commented Sep 14, 2019 at 20:54

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Let's assume both $A$ and $B$ are Hermitian operators, since otherwise trying to take their expectation value doesn't make sense at all.

  1. If $H$ and $A$ do not commute, then their product $HA$ is not an observable, since $(HA)^\dagger = A^\dagger H^\dagger = AH \neq HA$, so $HA$ is not Hermitian. So it is questionable what you're trying to compute here in the first place from a physical viewpoint.

  2. Your argument is not wrong, it simply shows that $\langle B\rangle = 0$ for all eigenstates of $H$. In light of Ehrenfest's theorem ($\frac{\mathrm{d}}{\mathrm{d}t}\langle A\rangle \propto \langle [H,A]\rangle$ for not explicitly time-dependent $A$), this is not surprising: The eigenstates of the Hamiltonian are stationary states, so their expectation value for the commutator of an observable with the Hamiltonian needs to be zero, otherwise the expectation value (and thus the state) would not be stationary.

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  • $\begingroup$ Ok, so if my argument isn't wrong, then let's consider the following: At $t=0$, $\langle A \rangle=a_0$ is known. What is the time dependence of $\langle A \rangle$? On the one hand, $\langle \psi(0)|U(t)^{\dagger}AU(t)|\psi(0)\rangle = a_0$, since we can apply my argument above to each term if we expand the temporal unitary operator. On the other hand, $d/dt \langle A \rangle \propto \langle [H,A] \rangle$. So, if my argument above is correct, then the expectation value doesn't change and we should have $[H,A]=0$, but the problem says $A$ and $H$ don't commute. How do we resolve this? $\endgroup$
    – Ptheguy
    Commented Sep 14, 2019 at 20:36
  • $\begingroup$ @Ptheguy I do not understand your question. $\langle [H,A]\rangle = 0 $ for all eigenstates of $H$ does not imply $[H,A] = 0$. $\endgroup$
    – ACuriousMind ♦
    Commented Sep 14, 2019 at 20:41
  • $\begingroup$ I see, thank you for that clarification. That was one point I was missing. Still, if my argument in the original post is correct, then I get that the expectation value of an observable $A$ is time independent. That is, I arrive at $\langle \psi(t)|A|\psi(t)\rangle = \langle \psi(0)|A|\psi(0)\rangle$. Is this correct? Wouldn't the expectation value change over time since the states are evolving? $\endgroup$
    – Ptheguy
    Commented Sep 14, 2019 at 20:48
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    $\begingroup$ @Ptheguy Yes - which is entirely correct for an eigenstate of the Hamiltonian, since these states are eigenstates of the time evolution and do not change (except for a phase)! You seem to be confusing yourself by thinking that this applies to all states - it does not, you explicitly restricted yourself to eigenstates of the Hamiltonian in your question. $\endgroup$
    – ACuriousMind ♦
    Commented Sep 14, 2019 at 20:49
  • $\begingroup$ YES! Thank you. This now makes sense again. $\endgroup$
    – Ptheguy
    Commented Sep 14, 2019 at 20:50
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$$\left\langle B\right\rangle =\left\langle \left[H,A\right]\right\rangle=E\left\langle A\right\rangle-\left\langle A\right\rangle E=0$$

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  • $\begingroup$ I mean, this does answer the question but really would be a lot better with a sentence or two explaining why it does. $\endgroup$
    – jacob1729
    Commented Sep 14, 2019 at 22:36

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