Whether $[A,B]=0$ if $\langle[A,B]\rangle=0$ for all states in the Hilbert space?

Question

Let $\hat{A}$ and $\hat{B}$ are two self-adjoint operators in quantum mechanics corresponding to two dynamical variables $A$ and $B$. If $$\langle[\hat{A},\hat{B}]\rangle\equiv \langle\psi|[\hat{A},\hat{B}]|\psi\rangle=0, \quad \forall\psi\in \mathcal{H}$$ where $\mathcal{H}$ is the Hilbert space of the problem, can we say that $[\hat{A},\hat{B}]=0$?

Answer 1

I'm just fleshing out the fine answer in the comment. Assuming what you call "dynamical variables" $\hat A, ~~\hat B$, are hermitian operators, their commutator is antihermitean, and so $$ i[\hat A, \hat B] \equiv \hat M $$ is hermitean. Whence unitarily diagonalizable, $\hat M= U^\dagger \hat D U$, to some diagonal operator $\hat D$.

If its expectation value in any orthonormal basis $|\psi _n\rangle$ vanishes, it must vanish itself, diagonal element by element, and hence $\hat M$ vanishes in the orthonormal basis $U^\dagger |\psi _n\rangle$; hence your commutator vanishes.

Answer 2

Let me just mention that, in quantum mechanics, observables are oftentimes unbounded self-adjoint (or Hermitian, in the physics terminology) operators on a Hilbert space $\mathscr{H}$. Unbounded operators can't be defined in the whole Hilbert space, so these are actually densely defined. In this case, one has to be a bit more careful when defining the commutator $[A,B] = AB - BA$ since their common domain need not to be dense and could even be empty!

That being said, suppose for simplicity we have $A$ and $B$ bounded and let $T = AB-BA$. Then, $$\|T\| = \sup\{\langle \psi, T\psi\rangle: \|\psi\| = 1\}$$ implies that $\|T\| = 0$, so that $T = 0$.

If $A$ and $B$ are unbounded but have a common dense domain, that is, if $AB - BA$ is densely defined, the same result follows but we can't use the same proof as before. Instead, suppose $T = AB-BA$ is densely defined, with domain $D(T) \subset \mathscr{H}$. Suppose $\langle \psi, T\psi\rangle = 0$ holds for every $\psi \in D(T)$. Suppose further that there exist some $\varphi \in D(T)$ such that $T\varphi \neq 0$. Because $D(T)$ is dense in $\mathscr{H}$, there exists a sequence $\{\psi_{n}\}_{n\in \mathbb{N}}$ of elements of $D(T)$ such that $\psi_{n} \to T\varphi$. But then: $$\|T\varphi\|^{2} = \langle T\varphi,T\varphi\rangle = \lim_{n\to \infty}\langle \psi_{n},T\varphi\rangle = 0$$ from where it follows that $T\varphi = 0$, a contradiction!

Answer 3

In general, if $T$ is an operator in $\mathcal H$ with dense domain $D(T)$ and $\langle \xi,T\xi\rangle=0$ for all $\xi\in D(T)$, then $T=0$ on $D(T)$. This follows simply by polarization. Indeed, $$ \langle \xi,T\eta\rangle=\frac 1 4 \sum_{k=0}^3 (-i)^k\langle \xi+i^k \eta,T(\xi+i^k \eta)\rangle=0 $$ for $\xi,\eta\in D(T)$. Since $D(T)$ is dense, $T\eta=0$ for all $\eta\in D(T)$.