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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 $[A,H]=0\; ?$

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  • $\begingroup$ Welcome to Physics Stack Exchange. This is a site for physics questions and answers, but we try to stay away from questions which can be solved with a little direct effort. What have you tried so far? What conceptual issue is making you stuck? $\endgroup$
    – DanielSank
    Nov 27 '15 at 6:50
  • $\begingroup$ Thanks so much for your guidance. In quantum litterateurs has been mentioned each operator commuting with Hamiltonian is Constant of motion. What about if we have the expectation values be constant!!!! it will be hard to say!!! Please assume an operator does not commute with Hamiltonian ([H,A]!=0), but the expectation value of the commutation relation (<\psi|[H,A]|\psi>=0) will be zero if |\psi> be an eigenvector of Hamiltonian. this case confuses me $\endgroup$ Nov 27 '15 at 7:15
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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 $|l=0,m_z=0\rangle = |l=0, m_x=0\rangle$.

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  • $\begingroup$ If we assume |\psi(t)>=e^{itL_{y}}|\psi> (|\psi(t)>=e^{itH}|\psi>) I am not sure about <\psi(t)|L_{x}|\psi(t)> be equal to zero!!! $\endgroup$ Nov 27 '15 at 8:27
  • $\begingroup$ You can replace $z$ for $x$, since $|0,0\rangle$ is isotropic, i.e., $L_i |0,0\rangle =0$ for $i=x,y,x$. $\endgroup$ Nov 27 '15 at 8:34

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