quantum mechanics - Why is time-evolution operator unitary?

When we shift the system's time from $t=0$ to $t = t$, we can define the following operator $\hat{U}$.

$$\hat{U} = e^{- i \hat{H} t / \hbar} \, .\tag{1}$$

So many (as far as I read, almost all of) documents assume $\hat{H}$ is Hamiltonian and $\hat{H} = \hat{H}^\dagger$ to prove that $\hat{U}$ is unitary.

I don't understand the reason why we can say $\hat{H}$ in (eq.1) is Hamiltonian. I believe $\hat{H}$ in $(1)$ is just an operator at this time and there is no reasonable context to conclude $\hat{H}$ here is nothing else but Hamiltonian we know.

Could anyone please tell me the reason?

Answer

1st point of view:

If you accept the Schrödinger equation $$ \mathrm i\hbar\, \partial_t \psi = \hat H \psi $$ with self-adjoint $\hat H$, then your equation 1 follows directly and $\hat U$ is unitary.

2nd point of view:

Time evolution must have the following properties:

• $\hat U$ must be norm-preserving so that probability is conserved.


• $\hat U$ should be invertible so that information is conserved.

Those two properties together imply that $\hat U$ is unitary. If you add the fact that $\hat U(t)$ should be a group, your equation 1 follows and it implies Schrödinger's equation with self-adjoint $\hat H$.