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I found something that I'm confused with when calculating the propagator of harmonic oscillator. Using the energy representation, the propagator of a quantum harmonic oscillator can be expressed as : $$ K(x_f,t_f;x_i,t_i)=\sum_ne^{-i\omega T(n+\frac{1}{2})}\psi_n(x_f)\psi_n^*(x_i)\tag{1} $$ where the $\psi_n$s are the wave functions of energy eigenstates, and $T=t_f-t_i$. Let $T^\prime=T+\frac{2\pi}{\omega}$, then $K(T^\prime)=-K$.
After some calculation, the final result of this propagator yields: $$ \begin{aligned} &K\left(x_f, t_f ; x_i, t_i\right) \\ &=\left[\frac{m \omega}{2 \pi \hbar i \sin \omega T}\right]^{1 / 2} e^{\frac{i}{\hbar} \frac{m \omega}{2 \sin \omega T}\left[\left(x_f^2+x_i^2\right) \cos \omega T-2 x_f x_i\right]} \end{aligned}\tag{2} $$ while $K(T^\prime)=K$ this time, rather than $K(T^\prime)=-K$. How did that happen?

Qmechanic
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CrisPhy
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1 Answers1

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Good observation. OP's eq. (2) should be amended with a metaplectic correction/Maslov index: There is a caustic at every half-period, which leads to a phase factor $\exp\left(-\frac{i\pi}{2}\right)$ jump, cf. e.g. my Phys.SE answer here.

Qmechanic
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