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Probability wave function is still in ground state after imparting momentum

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data
    An interaction occurs so that an instantaneous force acts on a particle imparting a momentum ## p_{0} = \hbar k_{0}## to the ground state SHO wave function. Find the probability that the system is still in its ground state.

    2. Relevant equations
    ##\psi _{0} =\left( \frac{mw}{\hbar\pi} \right )^\frac{1}{4} e^{-mwx^{2}/2\hbar} ##





    3. The attempt at a solution

    ##\Psi(x)=\psi_{0}e^{ik_{0}x}##
    This wave function gives a <p> =##\hbar k_{0}##
    Im confused whether this is the correct fourier transform to do.
    ##c(k)=\frac{1}{2\pi}^{0.5}\int_{-\infty}^{\infty}e^{-ikx}\psi_{0}e^{ik_{0}x}dx##
    And then solving for when the wavenumber of the ground state using E0=0.5* hbar * w.
    Therefore
    ##c(k)^2 ## = Probability

    Is this correct?
     
  2. jcsd
  3. Nov 21, 2016 #2

    Simon Bridge

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    You want to know the probability of a particular result of a measurement of energy.
     
  4. Nov 21, 2016 #3
    Ah right. So would this be the probability of the system still being in the ground state?
    ##\left | \int_{-\infty}^{\infty} \psi_{0} \psi_{0} e^{ik_{o}x}dx\right |^2=P##
     
  5. Nov 21, 2016 #4

    Simon Bridge

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    Well done - left out a star, but ##\psi_0## is real so....
     
  6. Nov 21, 2016 #5
    Thank you. Do you by any chance know any way to solve that integral by hand? I solved it on mathematica and it gave me ##P=e^\frac{-{k_{0}}^{2}\hbar}{4mw}## but I have no idea how to solve it by hand. Thanks again.
     
  7. Nov 21, 2016 #6

    Simon Bridge

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    Hint: integrate by parts. (That's normal for quantum.)
    You may be able to shortcut using ##\int_\infty \psi_0^2\; dx = 1##
     
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