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I'm redoing the calculations of "Point Canonical Transformations in the path integral", by Gervais and Jevicki; while doing so I stumbled in integrals like $$ \int \mathrm{d}t \, \Delta_F^3(t) = -\frac{1}{12} \frac{i}{\omega^4}, \\ \int \mathrm{d}t \, \dot{\Delta}_F^2(t) \Delta_F(t) = \frac{1}{12} \frac{i}{\omega^2} $$ where $$ \Delta_F(t) = \int \frac{\mathrm{d}\nu}{2\pi} \, e^{i\nu t} \frac{i}{\nu^2-\omega^2+i \varepsilon}. $$

I tried various methods, without success. For instance, I integrated explicitly $\Delta_F(t)$ and after some calculations I found $\Delta_F(t) = \frac{1}{2\omega} \cos(\omega t)$; but this expression, inserted in the previous ones, does not make the integrals converge. Maybe in these integrals it's important to integrate in $\mathrm{d}t$ before doing the $\mathrm{d}\nu$ integration; but when trying to do so what I obtain is terribly complicated.

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    $\begingroup$ Full reference is J.-L. Gervais and A. Jevicki, Nucl. Phys. B110 (1976) 93. The pdf file is here. $\endgroup$
    – Qmechanic
    Commented Dec 28, 2013 at 11:32
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    $\begingroup$ Yes, for the first calculus, integrate first on $t$, and you will have to calculate terms like $\int d \nu_1 d\nu_2 \frac{1}{\nu_1 \pm (\omega-i\epsilon)}\frac{1}{\nu_2 \pm (\omega-i\epsilon)}\frac{1}{\nu_1 + \nu_2 \pm (\omega-i\epsilon)}$. You may choose paths wich are closing in the upper half-space, for instance, and apply $2$ times the residue theorem. $\endgroup$
    – Trimok
    Commented Dec 28, 2013 at 11:48
  • $\begingroup$ @Trimok: One of my approaches was the one you're telling me to follow. The only thing different is that I did an error on calculating the denominator in the third fraction of $\int d \nu_1 d\nu_2 \frac{1}{\nu_1 \pm (\omega-i\epsilon)}\frac{1}{\nu_2 \pm (\omega-i\epsilon)}\frac{1}{\nu_1 + \nu_2 \pm (\omega-i\epsilon)}$ (wow, I did an error in integrating a dirac delta function! maybe I should leave physics). Thank you very much; if you post your comment as an answer I can mark this question as answered. :-) $\endgroup$
    – Alex A
    Commented Dec 28, 2013 at 14:06

1 Answer 1

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The Feynman propagator for the QM harmonic oscillator reads

$$\tag{A} \Delta_F(t)~=~ \int \frac{\mathrm{d}\nu}{2\pi}~ e^{i\nu t} \frac{i}{\nu^2-\omega^2}~=~ \frac{1}{2\omega} \sum_{\pm}\theta(\pm t)e^{\mp i\omega t}~=~\Delta_F(-t), $$

where $\theta$ denotes the Heaviside step function, and $\omega>0$ is the characteristic angular frequency of the harmonic oscillator. The time derivative is

$$\tag{B} \dot{\Delta}_F(t)~=~ \frac{1}{2i} \sum_{\pm}\pm\theta(\pm t)e^{\mp i\omega t}, $$

where the two contributions in eq. (B) proportional to the delta function $\delta(t)$ have cancelled out. Hence

$$\tag{C} \Delta^3_F(t)~=~ \frac{1}{8\omega^3} \sum_{\pm}\theta(\pm t)e^{\mp 3i\omega t}, $$

and

$$\tag{D} \dot{\Delta}^2_F(t)\Delta_F(t)~=~ \frac{-1}{8\omega} \sum_{\pm}\theta(\pm t)e^{\mp 3i\omega t}. $$

It should be stressed that the Feynman $i\epsilon$ prescription is implicit understood in the above eqs. (A-D). Here it means that we should substitute

$$\tag{E} \omega \to \omega-i\epsilon $$

everywhere in eq. (A-D). This implies that the $t=\pm \infty$ boundary terms in an $\int_{\mathbb{R}} \mathrm{d}t$ integration are exponentially suppressed. The piecewise exponentials in eqs. (C-D) are readily integrated wrt. $t$. The only non-zero contributions come from the kinks at $t=0$. We arrive at the formulas (2.17) of Ref. 1:

$$\tag{2.17a} \int_{\mathbb{R}} \mathrm{d}t~\Delta^3_F(t)~=~ \frac{1}{12i\omega^4}, $$

and

$$\tag{2.17b} \int_{\mathbb{R}} \mathrm{d}t~\dot{\Delta}^2_F(t)\Delta_F(t)~=~ \frac{i}{12\omega^2}. $$

References:

  1. J.-L. Gervais and A. Jevicki, Point Canonical Transformations in Path Integral, Nucl. Phys. B110 (1976) 93. The pdf file is here.
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