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

  1. Nov 28, 2015 #1
    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 - path integrals.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.
    Last edited: Nov 29, 2015
  2. jcsd
  3. Nov 29, 2015 #2
    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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