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<q|p>=exp(ip.x) why?

  1. Dec 16, 2004 #1
    Can anyone explain why?


  2. jcsd
  3. Dec 16, 2004 #2
    Depends on the nature of your function to begin with.. more detail needed.. probably due to orthogonality condition
  4. Dec 16, 2004 #3


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    What is [itex] <q| [/itex],an eigenstate of the momentum operator,or of the position operator (actually of their adjoints,but,because they are selfadjoint,you can think that way as well)?
    If it's for the momentum operator,then [tex] |p> [/itex] is an eigenstate as well,and the normalization condition reads:
    [tex] <q|p> =2\pi\hbar \delta(q-p) [/tex]
    ,but if it's for the coordinate operator,then the scalar product is zero,since one of them is an eigenvector from a Hilbert space and the other is a linear functional over another Hilbert space.

    So,anyway,what u've written there is wrong.

  5. Dec 16, 2004 #4
    correction <x|p>=exp(ip.x)

    sorry I meant <x|p>=exp(ip.x)

    and I think I have it


    <x|P|p> =p<x|p>=-id/dx<x|p>

    It's a long time since I did QM, and I started reading Prof Zee's Quantum field theory in a nut shell. He quotes this very early on without proof and I couldn't see where it came from. Thanks for comments.
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