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Momentum operator

  1. May 9, 2006 #1

    Lee

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    How would I go about finding eigenfunctions and eigenvalues for the momentum operatro p?

    Is it just a case of stating Ae^i(kx-wt) is a eigenfunction and kh/2Pi is a eigen value? would (2/a)¹/²sin(kx-wt) also be a eigenfunction?

    PS. This is not homework, but revision.
     
  2. jcsd
  3. May 9, 2006 #2

    nrqed

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    Teh eigenfunctions are the solutions of the differential equation [itex] -i \hbar {\partial \over \partial x} f(x,t) = k f(x,t) [/itex]. The solution is clearly any function of time multiplied by [itex] e^{i k x} [/itex].
    There is nothing else you can say about the function of time. And there is nothing you can say about the eigenvalue "k". It could be any real number.

    As for sin(kx-wt), apply the momentum operator on this. Do you get a constant times the function back? No! So it is NOT an eigenstate of the momentum operator.
     
    Last edited: May 9, 2006
  4. May 9, 2006 #3

    Lee

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    Does the constant k have to be real? I would presume not but a real eigenvalue is probably more usefull than a complex one.
     
  5. May 9, 2006 #4

    nrqed

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    It has to be (otherwise measuring the momentum of a particle could give a complex result!! Which is physically nonsensical). The mathematical reason is that the momentum operator is hermitian, as are all observables, so its eigenvalues are real (things are actually a bit more tricky than this but that is usually the way it is presented in introductory QM courses).
     
    Last edited: May 9, 2006
  6. May 9, 2006 #5

    Lee

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    Along the same lines, what is ment by the periodc boundry coundition?
     
  7. May 10, 2006 #6

    dextercioby

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    It means the geometry of the system is in such way as to insure that the wavefunction obeys certain symmetry property. That's what "periodic" refers to: the wavefunction is a periodic function.

    As with "bounday", it definitely refers to the fact that the physical system under discussion is confined to a finite spatial extension, therefore exitibing "boundary conditions" .

    Daniel.
     
  8. May 10, 2006 #7

    dextercioby

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    That's actually false. Under special conditions, the momentum operator is selfadjoint. Selfadjoint operators do indeed have a completely real spectrum.

    Daniel.
     
  9. May 10, 2006 #8

    nrqed

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    I know that. This the reason for the statement next after.
    I di dnot think it would help the OP to get into that. But if you think it is something that needs to be gotten into at this level of introductory QM then why don't you explain carefully to the OP the distinction between hermitian and self-adjoint operators and show why it matters?:devil:

    EDIT: And I suggest that you e-mail David Griffiths and tell him to rewrite his book if you are so insistent on making the distinction at the introductory QM level.
     
    Last edited: May 10, 2006
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