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?
 A: 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 $\hat H$.
A: The assumption is that the wave function is a probability amplitude. In particular, it's a vector that is normalized. In Dirac's notation, this is the statement:
$$\langle \psi |\psi\rangle = 1.$$
This can be made more concrete with:
$$\begin{align}
\mathrm{ordinary\ vectors\ } &\sum_{i} \psi^\star_i \psi_i = 1, \\
\mathrm{wave\ functions\ } &\int \psi^\star(x) \psi(x) \operatorname{d}x = 1,\ \mathrm{or} \\
\mathrm{even\ wave\ functionals\ } & \int \left[\mathcal{D}\phi(x)\right] \Psi^\star[\phi(x)] \Psi[\phi(x)] = 1.
\end{align}$$
Dont' worry if that last one is cryptic - it's for when you're dealing with quantum field theory.
The important point is that the wave function is confined to exist in only a part of the vector space; like how unit vectors are confined to lie on the surface of a sphere. Transformations that respect this constraint are called unitary. Thus that constraint means that every allowed transformation of $|\psi\rangle$ is unitary. Rotations, spatial translations, reflections, etc, all must respect the requirement that the wave function remains normalized.
The rest follows from the requirement that the time translation operation is a continuous change in $|\psi\rangle$ and that quantum mechanics maps on to classical mechanics on average (see: the correspondence principle). That means that $\hat{H}$, the generator of time translations in quantum mechanics, has to correspond with the generator of time translations in classical mechanics, the Hamiltonian.
There is one exception I know of to the unitarity requirement. That is time reflections. Time refletion is anti-unitary. For details, see the Wikipedia article on $T$-symmetry.
A: An intuitive approach would be to notice that the adjoint $U^{\dagger}(t)$ is the same as $U(-t)$. 
Thus, if $U(t)|\psi(0) \rangle  = |\psi(t)\rangle$
And $U^{\dagger}(t)|\psi(0) \rangle = |\psi(-t) \rangle$
Then $U^{\dagger}(t)U(t) |\psi(0)\rangle = |\psi(0)\rangle$
Meaning $U^{\dagger}U = I$, the requirement for a unitary operator.
