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Momentum conversion

  1. Aug 11, 2007 #1
    Why does
    [tex] 1/\sqrt{2\pi\hbar }\int p-hat/2m * \psi(x) exp(-ipx/\hbar)*dx = p^2/2m \phi(p) [/tex], where phi(p) is that wavefunction in momentum space
    ?

    I understand why [tex] 1/\sqrt{2\pi\hbar }\int \psi(x) exp(-ipx/\hbar)*dx = \phi(p) [/tex].

    by the way, how do you make hats in latex

    EDIT: It should be [tex] 1/\sqrt{2\pi\hbar }\int p-hat^2/2m * \psi(x) exp(-ipx/\hbar)*dx = p^2/2m \phi(p) [/tex]
     
    Last edited: Aug 11, 2007
  2. jcsd
  3. Aug 11, 2007 #2

    malawi_glenn

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    try

    [tex]\hat{p}[/tex]

    And have done the p operator on the wave function?
     
  4. Aug 11, 2007 #3
    So, I should apply the opetor then do IBT two times to get rid of the second derivative? Aren't you left with a psi' and a psi if you do that?
     
  5. Aug 11, 2007 #4
    [tex] -i\hbar \frac{d \psi}{dx} exp(-ipx/\hbar)/(2m) [/tex]
    and if you apply it twice you get

    [tex] \frac{\hbar^2 d\psi} {dx^2} exp(-ipx/\hbar)/(2m) [/tex]
     
    Last edited: Aug 11, 2007
  6. Aug 11, 2007 #5

    malawi_glenn

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  7. Aug 11, 2007 #6
  8. Aug 11, 2007 #7

    malawi_glenn

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    Ah, ok that explains why i got zero from my first calculation;)

    Now check at the tips given in that thread I gave you, read all posts, you shall find how to apply the p-operator when it is "squared", and when it is applied to a function that is a product of two functions. Then after that, perform the Fourier-transformation.
     
  9. Aug 11, 2007 #8

    malawi_glenn

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    I probably gave you the wrong thread for showing you how to apply the p-operator to an arbitrary function. The operator is always going as far to the right as possible.

    So:

    [tex] -i\hbar \frac{d }{dx} (\psi (x)exp(-ipx/\hbar))/(2m) [/tex]

    use product rule for differentiation.
     
  10. Aug 11, 2007 #9
    I am pretty sure it the operator only applies to psi here. Perhaps I made another mistake in the initial post but I am not sure. The original equation which I did not show was:

    [tex]\hat{p}/2m \psi(x) - g\delta(x) \psi(x) = -|E|\psi(x)[/tex]

    my book then multiplies [tex]1/\sqrt(2\pi\hbar)exp( - ipx/\hbar)[/tex] to both sides and then integrates.

    Since you usually multiply things on the left, I think that you would not apply the operator to the exponential. I rearranged it because I did not think it made a difference. I thought an operator took as its argument only the function immediately following it. Was that wrong?
     
  11. Aug 11, 2007 #10

    malawi_glenn

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    Iam just stating what is usual in Quantum mechanics. But try both and see. Then write some clearer and nicer forumulas so can we see if it is good and maybe some more guys can come and help.

    This operator is a differential operator, and therefore one must be careful, we do not multiply operators, we operate with them =P
     
  12. Aug 11, 2007 #11
    [tex] \frac{1}{\sqrt{ 2\pi\hbar}}\int \frac{\hat{p}}{2m} \psi(x) e^{-ipx/\hbar}dx = \frac{\hbar^2}{2m\sqrt{ 2\pi\hbar}} \int \frac {d^2}{dx^2}\ \psi(x) e^{-ipx/\hbar}dx [/tex]

    If I integrate this by parts twice I get an expression with a psi'(x) and a psi(x) which I do not know how to get rid of.
     
  13. Aug 11, 2007 #12

    olgranpappy

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    ummm... no, on both counts. But anyways, this whole thread is confused, so let me just sort it out for you.

    We can write psi (a function of x) as a Fourier transform like this:
    [tex]
    \psi(x)=\int \frac{dp}{2\pi}e^{ipx}\phi(p)
    [/tex]

    Then

    [tex]
    \frac{d}{dx}\psi(x)=\frac{d}{dx}\int \frac{dp}{2\pi}e^{ipx}\phi(p)
    =\int \frac{dp}{2\pi}\left(\frac{d}{dx}e^{ipx}\right)\phi(p)
    =\int \frac{dp}{2\pi}ipe^{ipx}\phi(p)
    [/tex]

    And

    [tex]
    \frac{d^2}{dx^2}\psi(x)=\int \frac{dp}{2\pi}(ip)^2e^{ipx}\phi(p)\;.
    [/tex]
     
  14. Aug 11, 2007 #13
    The only problem is that i^2.
    Should the exponential be exp( -ipx)? We want to show that

    [tex] \frac{1}{\sqrt{ 2\pi\hbar}}\int \frac{\hat{p}}{2m} \psi(x) e^{-ipx/\hbar}dx = p^2/2m [/tex]

    and there is no sign change there.
     
    Last edited: Aug 11, 2007
  15. Aug 11, 2007 #14

    olgranpappy

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    That is there because I acted with [tex]\frac{d}{dx}[/tex] and
    not [tex]-i\frac{d}{dx}[/tex] which is the momentum operator in position space.

    This depends on your Fourier transform conventions and does not change the physics. Typically the Fourier transform conventions are specified in the way I specified them--A function of x (In this case \psi(x)) is written with the plus sign in the exponential of the Fourier kernel and a function of p is written with the minus sign in the exponential.

    I doubt very much that anyone in their right mind has asked you to show that the above is true... the above makes no sense.

    I'm sorry, but it just doesn't make sense.

    I have a feeling that the problem comes from how you are thinking about Fourier transforms. But, keep at it and you will understand eventually. Good luck.
     
  16. Aug 11, 2007 #15
    I see. Thanks.
     
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