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A complex integral from the text

  1. Sep 28, 2005 #1

    quasar987

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    In my book QM book, Gasiorowicz says that

    [tex]\int_{-\infty}^{\infty} e^{i(p-p')x/\hbar}dx = 2\pi \hbar \delta(p-p')[/tex]

    Where does that come from? I mean, set i(p-p')/h = K. Then the solution is

    [tex]\frac{\hbar e^{i(p-p')x/\hbar}}{i(p-p')}[/tex]

    and evaluate at infinity, it doesn't exists as the limit of cos and sin at infinity do not exist.

    There must be something in the fact that the integral is complex but I haven't studies complex analysis yet so go easy on me plz.
     
  2. jcsd
  3. Sep 29, 2005 #2

    quasar987

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    What, no one?
     
  4. Oct 1, 2005 #3

    George Jones

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    This takes a bit of work. Are you familiar with Fourier transforms?

    Regards,
    George
     
  5. Oct 1, 2005 #4

    quasar987

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    Just the basics. Forget it then. How about this one?

    [tex]\int_{-\infty}^{\infty}e^{2ikx}dx = 0[/tex]
     
  6. Oct 1, 2005 #5

    George Jones

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    This isn't consistent with your first post in this thread. To see this, set [itex]p'=0[/itex], [itex]\hbar =1[/itex], and [itex]x = 2u[/itex] in the integral in your first post.

    Regards,
    George
     
  7. Oct 1, 2005 #6

    quasar987

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    Well they are respectively equation 3-40 and 3-54 in Gasiorowizc. :frown:
     
  8. Oct 1, 2005 #7

    George Jones

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    It's okay if [itex]k \neq 0[/itex].

    [tex]
    \int_{-\infty}^{\infty}e^{2ikx}dx = \frac{1}{2} \int_{-\infty}^{\infty}e^{iku}du = \pi \delta (k)
    [/tex]

    This equals zero if [itex]k \neq 0[/itex].

    Regards,
    George
     
  9. Oct 1, 2005 #8

    quasar987

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    K! k is proportional to energy so k = 0 is kind of a trivial case. All good.


    :frown: --> :smile:
     
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