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

Quantization on momentum space

  1. Aug 23, 2006 #1
    I have this doubt..quantization in momentum space using G(p) as the Fourier transform of the wave function was not common (at least when i studied Q. Physics) my doubt is, if we have that:

    [tex] x |G(p)>=i \hbar \frac{ \partial G(p)}{\partial p} [/tex]

    But..what would happen if we apply:

    [tex] \dot x |G(p)> = ? [/tex] here the "dot" means time derivative...

    [tex] G(p)= \int_{-\infty}^{\infty}dx \psi (x) e^{i\omega t -ipx} [/tex]
     
  2. jcsd
  3. Aug 23, 2006 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The position operator X isn't time varying; it doesn't make sense to ask for its time derivative! Also, there's no reason here to restrict yourself to stationary states.


    Anyways, you know that:

    [tex]
    | p \rangle = \int_{-\infty}^{+\infty} e^{-ipx} |x \rangle \, dx
    [/tex]

    (but I might have a sign wrong, or a constant multiple missing)

    So, you can always transform your state:

    [tex]
    \int_{-\infty}^{+\infty} G(p) |p \rangle \, dp
    [/tex]

    into

    [tex]
    \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty} G(p) e^{-ipx} \, dp \right) | x \rangle \, dx
    [/tex]

    if you really need to. (e.g. if you only know how to apply an operator to position states)
     
    Last edited: Aug 23, 2006
  4. Aug 30, 2006 #3

    reilly

    User Avatar
    Science Advisor

    It's been well known, at least since the 1930s, that dx/dt is a perfectly respectable operator in QM -- provided one is working in the Heisenberg or interaction representations (or pictures as they are sometimes called). This is pretty standard stuff, and thus is explained in countless books and papers. Dirac, in his Quantum Mechanics, discusses the dx/dt operator in section 30. He also gives an extensive discussion of the momentum represention in section 23.(For a non-relativistic free particle, dx/dt = p/m in the Heisenberg rep.)

    Regards,
    Reilly Atkinson
     
  5. Aug 30, 2006 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    He's working in the Schrödinger picture, though -- G(p) is a time-varying state.
     
  6. Aug 31, 2006 #5

    reilly

    User Avatar
    Science Advisor

    Right you are, as is often the case.. Reilly
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Quantization on momentum space
  1. Is Space Quantized? (Replies: 4)

Loading...