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

[lavster]

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

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: $$(H\phi)=i\hbar\frac{\partial \phi}{\partial t}$$ , <A> is the expectation value: $$<A>=\int{\phi^*A_{op}\phi d\tau}$$ and [A_{op},H] is the commutator $$(A_{op}H-HA_{op})$$ and this equals zero when it commutes. However i can't put it together to get the above equation. can someone show me how to do it please?

 

[muppet]

Would it help if I observed that it's implicit that it really means the expectation value of the commutator ?

 

[Fredrik]

<A> is the expectation value: $$<A>=\int{\phi^*A_{op}\phi d\tau}$$

$d\tau$? It should be $d^3x$, and things get easier if you write down how these things depend on time.

$$\langle A\rangle_{\phi(t)}=\langle\phi(t),A\phi(t)\rangle=\langle e^{-iHt}\phi,Ae^{-iHt}\phi\rangle=\langle\phi,e^{iHt}Ae^{-iHt}\phi\rangle$$

I'm using units such that $\hbar=1$.