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Suppose we have

\begin{equation} \langle[H,N]\rangle=0 \tag{1} \end{equation} where both $H$ and $N$ are hermitian. Under which assumption I can claim that then $$[H,N]=0~?\tag{2}$$

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  • $\begingroup$ If you are assuming that $N$ itself is independent of time, then, Ehrenfest's theorem / Heisenberg's equation of motion shows that the expectation value of of $N$ will be time independent if $<[H,N]> = 0$. $\endgroup$ – Dr. Ikjyot Singh Kohli Oct 27 '17 at 17:49
  • $\begingroup$ Can you make use of the same argument for the commutation? $\endgroup$ – Gbp Oct 27 '17 at 18:09
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    $\begingroup$ Does this vanishing expectation value hold for all states or just for a specific one? $\endgroup$ – ACuriousMind Oct 27 '17 at 18:38
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    $\begingroup$ If the statement is supposed to hold for all vectors (or a dense subset), then this question is a duplicate of physics.stackexchange.com/q/355883/50583 $\endgroup$ – ACuriousMind Oct 28 '17 at 10:49
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Let us see what this means: assuming that the commutator is properly defined, $\langle \psi,[H,N]\psi\rangle =0$ simply states that the image (range/codomain) of the commutator is orthogonal to its domain. If the domain of the commutator is dense everywhere in a complex separable Hilbert space, then the only vector orthogonal to every vector in the domain is the 0 vector. The only operator which sends an arbitrary vector into the 0 vector is the 0 operator. Thus a necessary condition to have [H,N] =0 is that the domain of the commutator be dense everywhere. In particular, the two operators could be bounded.

EDIT:

Statement: "If the domain of the commutator is dense everywhere in a complex separable Hilbert space, then its orthogonal complement is the 0 vector".

PROOF: The statement reads. "If $\bar{D_A} = \mathcal{H}$, then $ D_A ^{\perp} = \{0\}$". This follows trivially form the weak continuity of the scalar product. We prove that $D_A^\perp \perp \bar{D_A}$. Let $(s_n)\in D_A$ a sequence of vectors in $D_A$ which converges to $s\in\bar{D_A}$. Then $\forall x\in D_A^\perp, \langle x,s\rangle = \langle x,\lim_n s_n\rangle = \lim_n \langle x,s_n\rangle =0$. Therefore we have shown that $D_A^\perp \perp \mathcal{H}\Rightarrow D_A^\perp =\{0\} $.

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    $\begingroup$ The result is correct but the proof is not complete: " then the only vector orthogonal to every vector in the domain is the 0 vector". Here you cannot directly apply this statement. You first have to take advantage of the polarisation identity... $\endgroup$ – Valter Moretti Oct 27 '17 at 19:09
  • $\begingroup$ In a complex Hilbert space $\langle x |A x\rangle =0$ for $x$ varying in a dense domain implies $A=0$. The result is generally false in a real Hilbert space... $\endgroup$ – Valter Moretti Oct 27 '17 at 19:11

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