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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.

Qmechanic
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CBBAM
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  • Related : On the expression of evolution operator for a time-independent Hamiltonian in quantum mechanics. – Frobenius Dec 18 '22 at 10:27
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    Well, in classical mechanics the Hamiltonian is the generator of time translations; so you can take this as an "analogy" or motivation (same holds e.g. for the momentum operator: it is the generator of spatial translations). – Tobias Fünke Dec 18 '22 at 10:38
  • Note that the question title does not really fit the question itself. – Tobias Fünke Dec 18 '22 at 10:47
  • In the above it is defined that the wavefunction is a function of time. If this formula doesn't describe correctly what happens over time, then it contravenes the Schrodinger equation, and the solution formula is therefore wrong. Not all wavefunctions in general can be summed up neatly with a nice time dependent expression for its time evolution. – James Dec 18 '22 at 10:57
  • If you plug the solution $\psi(x,t) = e^{-i t H/\hbar}\psi_0(x)$ into the Schrödinger equation, you will see that it solves the equation. So the $t$ in both equations match up. If you agree that the Schrödinger Equation provides the time evolution for the state, then you know that this $t$ is really a time parameter. – doublefelix Dec 18 '22 at 12:21
  • @doublefelix So in short $t$ only corresponds to time in $U$ because $U$ is a solution of the Schordinger equation? – CBBAM Dec 18 '22 at 18:13

2 Answers2

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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.

Gold
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Because Hamiltonian is (by definition)

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

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

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

$S(0) =1$

Murod Abdukhakimov
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  • Considering the Schrodinger equation has the Hamiltonian as part of its definition, this seems rather circular. – CBBAM Dec 18 '22 at 18:11
  • Where do you find the hamiltonian defined like this? I have never seen this definition. – d_b Dec 18 '22 at 18:47
  • @d_b His equation is $$i\hbar \dfrac{dS}{dt} =HS$$ which is the equation for the time evolution operator. So it is just the Schrodinger equation. So what he seems to be saying is that the Hamiltonian is by definition the generator of time translations, which is right. Inverting the equation to write $H = i\hbar \dot{S} S^\dagger$ is just one unconventional way of expressing that. – Gold Dec 18 '22 at 19:02