If Aop commutes with H, A is a constant of motion - prove

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hi i was reading a text book and this statement puzzled me. it stated that [tex]\frac{d}{dt}<A>=\frac{1}{ih}[A_{op},H] if \frac{\partial A_{op}}{ \partial t}=0 [/tex].

i was wanting to prove this and hence show that if Aop commutes A is a constant of motion and can be a good quantum number.

I get that: H is the hamiltonian expressed as follows: [tex](H\phi)=i\hbar\frac{\partial \phi}{\partial t}[/tex] , <A> is the expectation value: [tex]<A>=\int{\phi^*A_{op}\phi d\tau} [/tex] and [A_{op},H] is the commutator [tex] (A_{op}H-HA_{op}) [/tex] and this equals zero when it commutes. However i cant put it together to get the above equation. can someone show me how to do it please?
 
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  • #2
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Would it help if I observed that it's implicit that it really means the expectation value of the commutator ?
 
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Fredrik
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<A> is the expectation value: [tex]<A>=\int{\phi^*A_{op}\phi d\tau} [/tex]
[itex]d\tau[/itex]? It should be [itex]d^3x[/itex], and things get easier if you write down how these things depend on time.

[tex]\langle A\rangle_{\phi(t)}=\langle\phi(t),A\phi(t)\rangle=\langle e^{-iHt}\phi,Ae^{-iHt}\phi\rangle=\langle\phi,e^{iHt}Ae^{-iHt}\phi\rangle[/tex]

I'm using units such that [itex]\hbar=1[/itex].
 

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