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In response to asmaier's question, qmechanic showed why the propagator must be $\exp(cS)$. That made perfect sense. But can it also be shown that $c$ is imaginary? I believe it follows from infinitesimal canonical transformation logic, since time is a canonical transformation, but I can't work it out.

Qmechanic
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JoeJeffrey
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    if c is not purely imaginary, the propagation will not be unitary, meaning that probabilities will not add up to one. – naturallyInconsistent Sep 13 '23 at 15:38
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    For largely the same reasons Fourier transforms have imaginary exponents. – J.G. Sep 13 '23 at 15:47
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    Do you know how to derive the path integral (Feynman’s argument)? The $e^{iS}$ in the path integral is coming from the $e^{-iH t}$ in the time-evolution operator. As a side note, the Euclidean time-evolution operator is $e^{-H t}$ which gives $e^{- S}$ in the path integral instead of $e^{iS}$ (if you don’t know anything about Euclidean field theory, ignore the last comment completely!) – Prahar Sep 13 '23 at 16:03

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TL;DR: That the coefficient $c$ is imaginary is tied to unitary time-evolution, cf. above comment by naturallyInconsistent.

  1. Here is one check: Show that the path integral $$\psi(x_f,t_f)~\sim~K(x_f,t_f;x_i,t_i)~=\int_{x(t_i)=x_i}^{x(t_f)=x_f}\!{\cal D}x~e^{cS[x]}$$ satisfies the TDSE, cf. e.g. this Phys.SE post. This calculation is sensitive to the correct normalization of the coefficient $c=\frac{i}{\hbar}$.

  2. Alternatively, go with the standard textbook derivation of the path integral $$K(x_f,t_f;x_i,t_i)=\langle x_f,t_f;x_i,t_i\rangle$$ from the operator formalism, as Prahar suggests in above comment. See e.g. eq. (7) in my Phys.SE answer here.

Qmechanic
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