- #1
Niles
- 1,866
- 0
Hi all.
I have a question which arose from the answer of a homework problem. A particle is in the state given by
[tex]
\left| \psi \right\rangle = \frac{1}{{\sqrt 3 }}\left[ {\left| \psi \right\rangle _1 + \left| \psi \right\rangle _2 + \left| \psi \right\rangle _3 } \right],
[/tex]
where [itex]{\left| \psi \right\rangle }_i[/itex] is a stationary state of the Hamiltonian. Finding the expectation value of an operator Q in this state, it turns out that <Q> is time-dependent (in fact it oscillates). I was wondering how it is possible to have a time-dependent expectation value? If there was a time-dependent perturbation, then I would be convinced, since the state [itex]{\left| \psi \right\rangle }[/itex] would change over time, but in this case [itex]{\left| \psi \right\rangle }[/itex] doesn't change.
I have a question which arose from the answer of a homework problem. A particle is in the state given by
[tex]
\left| \psi \right\rangle = \frac{1}{{\sqrt 3 }}\left[ {\left| \psi \right\rangle _1 + \left| \psi \right\rangle _2 + \left| \psi \right\rangle _3 } \right],
[/tex]
where [itex]{\left| \psi \right\rangle }_i[/itex] is a stationary state of the Hamiltonian. Finding the expectation value of an operator Q in this state, it turns out that <Q> is time-dependent (in fact it oscillates). I was wondering how it is possible to have a time-dependent expectation value? If there was a time-dependent perturbation, then I would be convinced, since the state [itex]{\left| \psi \right\rangle }[/itex] would change over time, but in this case [itex]{\left| \psi \right\rangle }[/itex] doesn't change.