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Disclaimer: The question has been answered in related posts, example this one which addresses LSZ reduction formula without explaining the minus sign issue.

We consider a unitary operator $U(t+dt,t):\mathcal{H}\to\mathcal{H}$ which acts on states $|a;t\rangle$ to give $|a;t+dt\rangle$. From classical mechanics the Hamiltonian generates time-evolution, we propose that $U(t+dt,t)=1-\frac{i}{\hbar}Hdt$ which satisfies the desired properties 1). conservation of probability 2). transitivity of time evolution by composition.

Well but $1+\frac{i}{\hbar}Hdt$ does the job perfectly as well. Clearly as put in all textbooks $e^{\frac{i}{\hbar}Ht}|a;t=0\rangle$ will not evolve the state to $|a;t\rangle$ when $H$ is time-independent, it is $e^{-\frac{i}{\hbar}Ht}|a;0\rangle=|a;t\rangle$. But then how do we know which sign to pick in the first place?

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Rescy_
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    mb. It has to satisfy Schrodinger's equation ofc... – Rescy_ Dec 14 '22 at 17:39
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  • It's not trivial, but it is a duplicate. – hft Dec 14 '22 at 17:53
  • See also the Luboš Motl answer here: https://physics.stackexchange.com/q/9557/ – hft Dec 14 '22 at 17:53
  • I do not think it's a duplicate in the sense you are suggesting. The unitary time-evolution operator comes after the Schrodinger's equation, in quantum mechanics settings. And the reason for the minus to be here is not the same as in the Schrodinger's equation as well. – Rescy_ Dec 14 '22 at 17:56
  • LAST comment: to be more specific, once we decide the form of the Schrodinger's equation and claim that our evolution operator should be analogous to classical case: we would need $H$ to play the role, then the sign is fixed because the evolved state has to satisfy Schrodinger's equation. – Rescy_ Dec 14 '22 at 17:58

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