1
$\begingroup$

In Weinberg books The quantum field Theory he says that if a state is an energy eigenstate, then it cannot be localized in time. I am not understanding this statement. For example if $\Psi$ is energy eigenstate of free particle, can't we always localize it in time?

$\endgroup$
4
$\begingroup$

In the Schrödinger picture if $\left|\Psi\right>$ is an energy eigenstate, then $e^{-iHt}\left|\Psi\right> = e^{-iEt}\left|\Psi\right>$. That is, time-evolving the state gives back the same state (sinces states are only defined up to phases). So the state can't change with time and in particular could not be localized in time.

I'm not sure why you would think a free-particle state could be localized in time. Maybe you are understanding "localized in time" in a different sense than it is being used here. Localized in time should mean something like the state being zero outside a finite interval of time.

That is, localized in time means $\left|\Psi(t)\right> \neq 0$ if $t_1 < t < t_2$ and $\left|\Psi(t)\right> = 0$ if $t < t_1$ or $t > t_2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ what do you men by localized in time should mean something like the state being zero outside a finite interval of time $\endgroup$ – amilton moreira Nov 21 '19 at 19:16
  • $\begingroup$ edited to address the comment $\endgroup$ – d_b Nov 21 '19 at 19:23
  • $\begingroup$ Can you give me you a reference about Localization in time $\endgroup$ – amilton moreira Nov 21 '19 at 19:26
0
$\begingroup$

Apply the operator of an observable to a wave function that is a pure state of the Hamiltonian and then multiply by the complex conjugate of the wave function, you'll eliminate the time dependence of the probability of getting a given value of the observable.

Further, by The Ehrenfest Theorem, if an observable and its square both commute with the Hamiltonian, then the uncertainty in the observable in the observable is zero for all time suggesting it stays constant.

With no ability to change, no duration of time is meaningful.

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

Maybe have a look at the uncertainty principle, if a state's energy is known ('localized'), then you cannot simultaneously be localized in time.

Another nice thing to look into are stationary states, which are energy eigenstates: https://en.wikipedia.org/wiki/Stationary_state

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is not an informed answer. As there is no time operator, the usual derivation for the uncertainty principle doesn't work, nor does the interpretation (what is a standard deviation in the "time distribution"?). There are uncertainty principle-like relations which hold between some kind of time and some kind of energy, but these are not just a case of the usual uncertainty principle. $\endgroup$ – doublefelix Nov 22 '19 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.