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I have been reading Modern Quantum Mechanics by J.J.Sakurai. Under the chapter Quantum Dynamics, the author says if an observable $A$ initially commutes with the Hamiltonian operator $H$, then it remain so for all later times. Also, he mentions that this is how an observable commuting with $H$ remains a constant of motion.

Question: What's special about $H$? Are the above statements valid if we used some other operator than $H$? For example, can we say that if the observables $A$ and $B$ are compatible, then they remain so for all other time?

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What's special about $H$ is that it generates time translations. This means that operators evolve, by postulate, through Heisenberg Equations of motion $$ i\frac{\mathrm d}{\mathrm dt}\mathcal O(t)=[H,\mathcal O] $$ modulo an explicit time dependence. Therefore, if an operator commutes with the Hamiltonian, $[H,\mathcal O]=0$ we automatically conclude that it is time-independent.

Can we say that if the observables $A$ and $B$ are compatible, then they remain so for all other time?

Let $[A(0),B(0)]=0$; then $$ [A(t),B(t)]=[U^\dagger(t)A(0)U(t),U^\dagger(t)B(0)U(t)]=U^\dagger(t)[A(0),B(0)]U(t)=0 $$ where $U(t)=\exp[-iHt]$ is the unitary time evolution operator. This proves that the answer is yes: if two operators commute at $t=0$, they commute for any $t$.

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  • $\begingroup$ Is this valid if H explicitly depends on time? $\endgroup$ – UKH May 14 '16 at 16:19
  • $\begingroup$ @Unnikrishnan in general, no; see this post for time evolution for time-dependent Hamiltonian (and, given the expression for $U$, try to think if an operator $\mathcal O(t)$ would in general commute with it) $\endgroup$ – AccidentalFourierTransform May 14 '16 at 16:31

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