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Schrodinger equation, potential energy function

  1. Apr 8, 2013 #1
    A particle of mass m is in the state
    [tex] \Psi (x,t) = A e^{-a[(mx^2 / \hbar ) + it]}[/tex]


    Find A
    For what potential energy function V(x) does [tex]\Psi[/tex] satisfy the Schrodinger equation?


    Do I just re-arrange for A? (Sorry if I seem really dumb). I'm not really getting this.
     
  2. jcsd
  3. Apr 8, 2013 #2

    DrClaude

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    Staff: Mentor

    To find ##A## means to normalize the function. You want
    $$
    \int_{-\infty}^{\infty} \left| \psi(x,t) \right|^2 dx = 1
    $$

    For the second part, you need to plug the wave function in the Schrödinger equation to find ##V(x)##. My guess is that you will need to use both the time-dependent and the time-independent Schrödinger equations.
     
  4. Apr 8, 2013 #3
    Here's my attempt, which took over 2 hours >.<

    I hope the final answers are right, but am I going about it the right way?
    Sorry the photos are blurry.
     

    Attached Files:

  5. Apr 8, 2013 #4

    DrClaude

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    For the normalization, you have to integrate ##| \psi(x,t) |^2 = \psi(x,t)^* \psi(x,t)##. Time should disappear for the final result for ##A##.
     
  6. Apr 8, 2013 #5

    DrClaude

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    I've check the calculation of ##V(x)## and you got it right!
     
  7. Apr 8, 2013 #6
    Sorry the picture is really blurry. My final answer does have time in it:

    [tex]A = (\dfrac{2ame^{4ait}}{\pi \hbar})^{1/4}[/tex]
     
  8. Apr 8, 2013 #7

    DrClaude

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    And that is not correct. I think you forgot to take the complex conjugate.
     
  9. Apr 8, 2013 #8
    If I do that wouldn't everything cancel out before I even integrate it?

    [tex]\int_{- \infty}^{\infty} A A^* e^{-a[mx^2 / \hbar + it]} e^{a[mx^2 / \hbar + it]} dx = 1[/tex]

    And so that would just become:

    [tex]\int_{- \infty}^{\infty} A A^*dx = 1[/tex]

    ??
     
  10. Apr 8, 2013 #9

    DrClaude

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    No, because only the imaginary part changes sign:
    $$
    \left( A e^{-a [m x^2/\hbar + i t]} \right)^* = A e^{-a [m x^2/\hbar - i t]}
    $$
     
  11. Apr 8, 2013 #10
    Ah yes, that makes sense.

    After a bit of working out again, I still think I'm missing something...
    Even if I use $$
    \left( A e^{-a [m x^2/\hbar + i t]} \right)^* = A e^{-a [m x^2/\hbar - i t]}
    $$
    my integral becomes:

    [tex]\int_{-\infty}^{\infty} A e^{-a [m x^2/\hbar + i t]} A e^{-a [m x^2/\hbar - i t]} dx = 1[/tex]

    Which becomes:

    [tex]\int_{-\infty}^{\infty} A^2 e^{-a m x^2/\hbar} e^{ -a i t} e^{-a m x^2/\hbar} e^{a i t} dx = 1[/tex]

    If I evaluate the integral, I get the final result of [tex]A = \displaystyle \left(\frac{2am}{ \pi \hbar}\right)^{1/4}[/tex]

    And then the time cancels out. So how would I get time left in the final result?
     
    Last edited: Apr 8, 2013
  12. Apr 9, 2013 #11

    DrClaude

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    That is the correct answer. You might have misunderstood what I wrote earlier, but time should disappear from the final answer.
     
  13. Apr 10, 2013 #12
    Great, thank you!
    Is there a 'thanks' or 'like' button on these forums? I can't seem to find it.
     
  14. Apr 10, 2013 #13

    DrClaude

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    Staff: Mentor

    You're welcome!

    Not that I know of.
     
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