# If the mean of the commutator is zero, then what we can say about the commutator itself?

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 $$[H,N]=0~?\tag{2}$$

• 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 $<[H,N]> = 0$. Oct 27, 2017 at 17:49
• Can you make use of the same argument for the commutation?
– Gbp
Oct 27, 2017 at 18:09
• Does this vanishing expectation value hold for all states or just for a specific one? Oct 27, 2017 at 18:38
• 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 Oct 28, 2017 at 10:49
• The answer to the title question is, basically not much. A sufficient condition would be to show that the operator $i [H,N]$ is non-negative definite.
– lcv
Feb 4, 2020 at 12:18

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 $D_A^\perp \perp \mathcal{H}\Rightarrow D_A^\perp =\{0\}$.

• 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... Oct 27, 2017 at 19:09
• 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... Oct 27, 2017 at 19:11
• This is an old post but this answer clearly does not answer the current question. The average of an (anti-hermitian) operator can be zero without the operator being zero.
– lcv
Feb 4, 2020 at 12:15
• The question as it stands right now is 100% addressed by the answer. Feb 4, 2020 at 18:20