How do we know that the operator $U(t) = e^{-itH/\hbar}$ corresponds to time?

Question

By Stone's theorem on one-parameter unitary groups we know that there is a one-to-one correspondence between strongly continuous one-parameter unitary groups and self-adjoint operators. Hence, if $H$ (here the Hamiltonian) is a self-adjoint operator, there is a corresponding one-parameter unitary group of the form $U(t) = e^{-itH/\hbar}$. The mathematics is clear. What I am confused about is the physics. How do we know that the parameter $t \in \mathbb{R}$ corresponds to time and not some other parameter? Is this simply taken as a postulate?

Note I have already seen the answers on this post Why does time evolution operator have the form $U(t) = e^{-itH}$? but the questions seem to address more of the mathematics and not the physics.

Answer 1

Ths Schrodinger equation in general form reads $$i\hbar \dfrac{d}{dt}|\psi\rangle=H|\psi\rangle\tag{1}$$

where $H$ is the Hamiltonian of the system. This equation can be solved in terms of one time evolution operator which is exactly the one you ask about. The reason the parameter in it is time is because the operator in the exponent, $H$, is the Hamiltonian operator so that $|\psi(t)\rangle=U(t)|\psi_0\rangle$ solves (1).

So, asking why that parameter is time is tantamount to asking why the Schrodinger equation should govern the time evolution of a system. But then, a point to appreciate is that the Schrodinger equation is one of the postulates of Quantum Mechanics. As such it is one of the axioms defining the theory and not something we can prove from something else. It is what we postulate because we have derived predictions from it which were observed in experiment.

Nevertheless, one may still motivate the Schrodinger equation. In particular, one way is by taking inspiration from Classical Mechanics. The Schrodinger equation is exactly the statement that the Hamiltonian is the generator of time translations. But the same is true in Classical Mechanics. So one may postulate the quantum version by demading that, as in the classical theory, the Hamiltonian still generates time translation.

Answer 2

Because Hamiltonian is (by definition)

$H = ih\dot{S}S^+$

where operator S(t) determines evolution of the system over time:

$\psi(x,t) = S(t) \psi(x,0)$

$S(0) =1$