Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

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

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

    thanks
     
  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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper


    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.

    Daniel.
     
  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

    P|p>=p|p>

    <x|P|p> =p<x|p>=-id/dx<x|p>
    solution
    <x|p>=exp(ipx)

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: <q|p>=exp(ip.x) why?
  1. Eigenvalues and [p,q]=i (Replies: 10)

  2. Expected value of Q(x,p) (Replies: 11)

  3. Why x and p (Replies: 1)

Loading...