I have a quantum operator $A.$ It's expectation value is constant respect to time. I mean
$$\langle \psi(t)|A|\psi(t)\rangle$$
is a constant values. If I know $|\psi(t)\rangle$ is not an eigenstate of Hamiltonian, can we say $A$ commutes with Hamiltonian $[A,H]=0\; ?$
Best Answer
No, the only thing you can conclude is that $\langle\psi|[H,A]|\psi \rangle =0$. Example, for some real constants $a,b$ and for a particle described in $L^2(\mathbb R^3)$ $A= aL_x$, $H=bL_z$, $$|\psi\rangle = |\phi(r)\rangle \otimes|l=0,m_z=0\rangle\:.$$ In this case $[H,A] \neq 0$ but $\langle \psi(t)|A|\psi(t) \rangle =0$ for every $t \in \mathbb R$ since $e^{-itbL_z}|0,0_z\rangle = |0,0_z\rangle$ and $|l=0,m_z=0\rangle = |l=0, m_x=0\rangle$.