1
$\begingroup$

I have an observable $O$ with operator $\hat{O}$. $\Psi_1$ is a wave function in an energy eigenstate, and $\psi_1$ is the corresponding spatial wave function. $E$ is the corresponding energy.

It is proved that the expectation value of $O$ is invariant over time with the following expressions:$$\langle O\rangle=\int\Psi_1^*\hat{O}\Psi_1~dx=\int\psi_1^*e^{iEt/\hbar}\hat{O}(\psi_1e^{-iEt/\hbar})~dx\rightarrow\int\psi_1^*\hat{O}\psi_1~dx$$

The last expression there is independent of time. But I used a right arrow because I don't see why it is equal to the other terms. I guess that the two exponential terms are supposed to cancel out by multiplying to 1. But why are you allowed to pull the exponential out of the operator's effect without any alteration? Why do we think that the operator must act upon the spatial wave function only? A counterexample of an operator for which this shouldn't be allowed is an energy operator $$\hat{E}=i\hbar\frac{\partial}{\partial t}.$$

Is there something about `normal' operators not doing such operations?


For instance, I think could define a linear Hermitian (but decidedly stupid and useless) operator $$\hat{t}\Psi=t\Psi$$ which would violate that proof for the in-variance of the expectation value over time (it just multiplies the wave function by time, the way a positon operator multiplies by position). What resolves this issue?

$\endgroup$
3
$\begingroup$
  • $\hat O$ is a linear operator. That means that $\hat O(\alpha \psi) = \alpha \hat O(\psi)$ for all scalar $\alpha$, and the factor $e^{-iEt / \hbar}$ is scalar.

  • "$\mathrm i\hbar \partial_t$" is not an operator on the Hilbert space, because time is a parameter in quantum mechanics, see e.g. here.

  • If $\hat O = \hat O(t)$ has an explicit time-dependence, like in your example at the end, then $\langle \hat O(t) \rangle$ is not time-independent of course. However, when talking about observables, we usually assume that they are not explicitly time-dependent.
    ("Explicitly time-dependent" here means that the operator $\hat O$ itself changes as a function of time.)

$\endgroup$

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.