Expectation value of operators with non-zero Hamiltonian commutators 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.
 A: Let's assume both $A$ and $B$ are Hermitian operators, since otherwise trying to take their expectation value doesn't make sense at all.


*

*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.

*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.
A: $$\left\langle B\right\rangle =\left\langle \left[H,A\right]\right\rangle=E\left\langle A\right\rangle-\left\langle A\right\rangle E=0$$
