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For harmonic oscillators the prefactor for the semiclassical propagator is $Fe^{iS}$ where $$F=\sqrt {m\omega/{2πi\hbar\sin(ωt)}}$$ and $$S={m\omega[(x_0^2+x_1^2)\cos(\omega t)-2x_0x_1]}/{2\sin(\omega t)}.$$

Suppose there is a wavefunction $\psi(0)$. I want to calculate $$\int dx_0<x_1|U(0,0)|x_0><x_0|\psi(0)> ,$$ which is exact and gives $\psi(x_1,0)$.

Now if $<x_1|U(0,0)|x_0>$ were to be substituted with $Fe^{iS/\hbar}$, at $t=0$, both the exponent and the prefactor $F$ blow up, how could it converge to $\psi(x_1,0)$?

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The propagator must behave like Dirac's delta at t=0, and Dirac's delta has to have explosive nature by definition!

You should not evaluate the delta and then integrate, instead integrate away the discontinuity in limits.

First note that as $t\rightarrow 0$ your propagator evaluates to zero for $x_0 \neq x_1$ because of the destructive phases coming from $iS$. Therefore, the only relevant part of $\psi(x)$ is $\psi(x_1)$, which, assuming the function is smooth at the vicinity, you can take it out of the integral and ignore.

Then, you can carry out the integration with respect to $x_1$, which I believe should bring down a factor of $sin(\omega t)$, which prevents the blowout.

Hope this helps.

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