Why must the propagator exponent be imaginary?

Question

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.

Answer 1

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.