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I have computed the path integral for the harmonic oscillator, in the time sliced formalism. I have been able to reproduce the Maslov correction that appears every half-period. Now I want to compute the path integral for a harmonic oscillator, but now with a frequency $ω^2$ that is purely imaginary. What I get in this case is that there is no Maslov correction, since an imaginary $ω^2$ does not change branch in any of the $N$ Fresnel integral. Is this right?

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Sounds right. The Maslov index changes when the classical path has a turning point. If the frequency is pure imaginary, $\omega=i\,w$, $w\in\mathbb{R}$, then the classical EOM is $\ddot{x}=w\,x$. For this EOM, the solution is $x(t)=A\exp(t\surd w)$, which has no turning point.

Edit: now that I think further, one $could$ cook up initial conditions such that you have at most one turning point (e.g. a solution $x(t)=A\exp(-t\surd w)+B\exp(t\surd w)$ where $A>B$). Another way to check your Maslov result, which I think is a lot more straightforward than looking at the branch cuts of the amplitude, is to compute the eigenvalues of the operator formed by the second variation of the Lagrangian.

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  • $\begingroup$ Thank you for your answer. Could you provide a reference where the Maslov index is obtained by the means you point out in your answer? $\endgroup$
    – Manu
    Aug 14, 2017 at 22:54
  • $\begingroup$ Oof. Good question. I learned it years ago. In searching for a reference, I did find a reference for my comment about turning points: see Kleinert's Path Integrals book, page 128. Ah! Found it! It's in Littlejohn's beautiful notes on path integrals from his grad QM class at Berkeley: bohr.physics.berkeley.edu/classes/221/1011/notes/pathint.pdf. See esp. p. 20, 21, and 23. $\endgroup$ Aug 15, 2017 at 1:39

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