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Integrating the product of a real and a complex exponential

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\Psi(x,t) = \int^{\infty}_{-\infty} C(p)\Psi_{p}(x,t) dp[/itex]

    is a solution to the Schroedinger equation for a free particle, where

    [itex]\Psi_{p}(x,t) = Ae^{i(px-Ept)/\hbar}[/itex].

    For the case [itex]C(p) = e^{-(p-p_{0})^{2}/\sigma}[/itex]

    where [itex]\sigma[/itex] is a real constant, compute the wavefunction at time t=0.

    2. Relevant equations

    [itex]\int^{\infty}_{-\infty} e^{-αp^{2}+βp} = \sqrt{\frac{\pi}{α}}e^{\frac{β^{2}}{4\alpha}}[/itex]

    where α is a positive real constant and β may be complex.

    2. The attempt at a solution

    This is the first part of one questions on a set of QM problems I've been given. I've made no progress with this part, because I don't know how to integrate the product of an exponential to a real number and an exponential to an imaginary number.
     
  2. jcsd
  3. Mar 17, 2013 #2

    vela

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    What's keeping you from plugging everything in and using the relevant equation/hint?
     
  4. Mar 17, 2013 #3
    I can't isolate the [itex]p^{2}[/itex] term.

    When I take the treat the integrand as an exponential to a complex power, I wind up with

    [itex]e^{-\frac{(p^{2}-2pp_{0}+p_{0}^{2})}{\sigma}+\frac{ix}{\hbar}p}[/itex]
     
  5. Mar 17, 2013 #4

    vela

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    I guess I'm still kind of confused about where you're getting stuck. This is just basic algebra.
    $$-\frac{p^2 - 2pp_0 + p_0^2}{\sigma}+\frac{ix}{\hbar}p = -\frac{1}{\sigma}p^2 + \left(\frac{2p_0}{\sigma}+\frac{ix}{\hbar}\right) p - p_0^2.$$ You may, however, find it easier to change variables in the integral first using the substitution ##p' = p-p_0##, and then deal with the exponentials.
     
  6. Mar 17, 2013 #5
    Ahh sweet, that's cleared things up a little - many thanks!
     
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