Free particle propagator - Shankar POQM Eqn. 5.1.10, p. 153

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The forum discussion centers on the evaluation of the free particle propagator as presented in Shankar's "Principles of Quantum Mechanics," specifically Equation 5.1.10 on page 153. The integral preceding this equation is identified as non-traditional due to the presence of a purely imaginary coefficient, leading to oscillatory behavior that deviates from the standard Gaussian integral form. The discussion highlights the necessity of using Fresnel integrals to resolve the integral of the oscillatory function, indicating that Shankar's approach may overlook critical mathematical nuances.

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Can someone help with what must be a simple math issue that I'm stuck on. Shankar ("Principles of Quantum Mechanics" p. 153) evaluates the propagator for a free particle in Equation 5.1.10. A scan of the chapter is available here:

http://isites.harvard.edu/fs/docs/icb.topic1294975.files/Shankar%20-%20path%20integrals.pdf

The integral which precedes Eqn. 5.1.10 is not a traditional Gaussian of the form exp(-a(x+b)^2)
with Re(a)>0. Instead the integrand (after completing the square) is of the above form, but with Re(a)=0, i.e., a is a purely imaginary number. Therefore, the familiar closed form expression (which Shankar references in Appendix A.2) does not apply. The integrand oscillates and is not absolutely integrable; it may be integrable and expressible in elementary terms as shown in Eqn. 5.1.10, but that does not seem to follow from the traditional Gaussian integral of exp{-a(x+b)^2} with Re(a)>0.

To be more precise, completing the square explains the factor exp{ im(x - x')^2 / 2ћt } in Eqn 5.5.10. What remains is simply to evaluate (after a few changes of variable) the integral of exp(-iq^2)dq, a purely oscillatory function.
 
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Aha, OK I got it. Just need to make some variable changes, *but* eventually use the Fresnel integrals ##\int_0^∞ \text{cos}(x^2)dx = \int_0^∞ \text{sin}(x^2)dx = \sqrt{\pi/8}##. I think Shankar took a big liberty here.
 

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