# Expectation Value of Unitary Time Evolution Operator in Quantum Mechanics

Does the expression $\langle \Psi_i|U(t)|\Psi_i\rangle$ have a specific meaning, where $U(T)$ is the unitary time evolution operator of $\Psi$, and $\Psi_i$ is the initial state of $\Psi$?

If so, could you please explain that meaning and also please provide any clear references to this that I could read?

## 1 Answer

To talk about $\langle \psi_i \vert U(t) \vert \psi_i \rangle$ as the "expectation value of the time evolution operator" is probably the least insightful way to talk about this quantity. Since $U(t) = \exp(-\mathrm{i}Ht)$ for time-independent Hamiltonians, if you want to look at expectation values, you could as well look at that of $H$ directly. Note that, in particular, the time evolution operator is not self-adjoint, since $U(t)^\dagger = U(-t)\neq U(t)$, so it is not an observable, and speaking of its "expectation value" is physically meaningless.

The physical meaning is clearer when you realize that $\lvert \langle \psi_i \vert U(t)\vert \psi_i\rangle\rvert^2$ is the probability (assuming the states are normalized) to find a system that was at time $t_0 = 0$ in the state $\lvert \psi_i \angle$ again in the same state after the time $t$ has passed, hence $\langle \psi_i \vert U(t) \vert \psi_i \rangle$ is the "transition amplitude for a state into itself". This is probably most interesting when the $\lvert \psi_i\rangle$ are eigenstates of an operator that is being measured.