- #1

maximus123

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**Hi everyone, my problem is this**

Using Dirac notation show that

[itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|[itex]\hat{A}[/itex]|[itex]\varphi[/itex]> = [itex]\frac{i}{\hbar}[/itex]<[itex]\varphi[/itex]|[[itex]\hat{H}[/itex],[itex]\hat{A}[/itex]]|[itex]\varphi[/itex]>

where A does not explicitly depend on t

Using Dirac notation show that

[itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|[itex]\hat{A}[/itex]|[itex]\varphi[/itex]> = [itex]\frac{i}{\hbar}[/itex]<[itex]\varphi[/itex]|[[itex]\hat{H}[/itex],[itex]\hat{A}[/itex]]|[itex]\varphi[/itex]>

where A does not explicitly depend on t

**I am given as a hint that the hamiltonian operator in Dirac notation is**

and that the conjugate expression is

*i*[itex]\hbar[/itex][itex]\frac{d}{dt}[/itex]|[itex]\varphi[/itex]>=[itex]\hat{H}[/itex]|[itex]\varphi[/itex]>and that the conjugate expression is

*-i*[itex]\hbar[/itex][itex]\frac{d}{dt}[/itex]<[itex]\varphi[/itex]|=<[itex]\varphi[/itex]|[itex]\hat{H}[/itex]**I'm a bit stumped, I know that the left hand side of the equation is the time derivative of the expectation value of operator A but as for the right hand side. I tried to expand the commutator as normal as if it were acting on the function (ket) |phi> but this came out to be zero which if true would surely make the whole RHS zero, which makes me think I don't understand what the rhs means. Any help would be appreciated.**