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I don't understand at all what the time operators are in quantum mechanics. I thought that given a wave function, because it's a function of time, we could simple put in any time in the future to find it's time evolution. What is this time operator, why is it unitary, and how does it work?

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marked as duplicate by John Rennie, Abhimanyu Pallavi Sudhir, Qmechanic Nov 7 '13 at 11:31

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There is no time operator in quantum mechanics. At least, there's no nontrivial time operator. You could have an operator whose action is just to multiply a function by $t$, but time is a parameter in QM, so the operator will never do anything more complicated than that. Its eigenfunctions wouldn't be terribly useful either because they would just be delta functions in time; they don't obey the Schroedinger equation.

There is, however, a time evolution operator, $U(t_f,t_i)$ (so it's really an operator-valued function of two variables). Given a quantum state $\lvert\psi\rangle$, then $U(t_f,t_i)\lvert\psi\rangle$ is the state you would get at time $t_f$ from solving the Schroedinger equation with $\lvert\psi\rangle$ as your initial condition at time $t_i$. In other words, if $\lvert\Psi(t)\rangle$ is a quantum state-valued function of time, then if you take

$$i\hbar\frac{\partial}{\partial t}\lvert\Psi(t)\rangle = H\lvert\Psi(t)\rangle$$

as a given, you have

$$U(t_f,t_i)\lvert\Psi(t_i)\rangle = \lvert\Psi(t_f)\rangle$$

You can show from this that

$$U(t_f,t_i) = e^{iH(t_f - t_i)/\hbar}$$

and given that $H$ is hermitian, $U$ will be unitary.

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  • $\begingroup$ It seems like this operator is the normal quantity we add onto the schrodinger equation when we make it time dependent. Is it just more convenient to write it as an operator? $\endgroup$ – user24082 Nov 7 '13 at 15:34