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Expectaion value momentum

  1. Nov 29, 2007 #1
    1. The problem statement, all variables and given/known data
    find expectation value of the momentum fro

    SI(x,t) = (1/aPI)^1/4 exp((-x^2)/2a)


    2. Relevant equations
    pSI(x,t) = -ih dsi/dx
    <SI*[p]SI> = integral SI*pSI dx

    3. The attempt at a solution

    just need help with the integral
     
  2. jcsd
  3. Nov 29, 2007 #2
    I'm having a bit of trouble following your notation, but I'm guessing you're trying to calculate:

    [tex]<P> = \int_{-\infty}^\infty \psi(p) p \psi^*(p) dp[/tex]

    where:

    [tex]\psi(p) = \int_{-\infty}^\infty \psi(x) e^{\frac{\imath p x}{\hbar}} dx [/tex]

    Which integral do you need help with?

    --------
    Assaf
    Physically Incorrect
     
  4. Nov 29, 2007 #3
    yes its the first integral
    Im following a soloution to a similar question in my notes and it says something about odd/even function? thanks
     
  5. Nov 29, 2007 #4
    Have you calculated [tex]\psi(p)[/tex] yet? If so, what was your result? If not, let's start with that.
     
  6. Nov 29, 2007 #5
    hope you get the notation lol
    hopefully this is right

    SI(p) = ih(1/aPI)^1/4 ((x/a)exp ((-x^2)/2a)))


    what do you use to get the notation on the forum?
     
  7. Nov 29, 2007 #6
    When you use the operator p on Psi in the integral, a factor of x will pop out when you take the derivative of Psi. Then you'll have something like x times e^(-x^2) and then you should use the properties of even and odd functions to simplify the integral greatly.

    Hope that helps
     
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