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[Q]Is physical quantity depending on coordinate on which we solve problem?

  1. Oct 16, 2008 #1
    Hi.

    I have a question whether eigenvalue(corresponding physical quantity) is up to coordinate

    we establish to solve certain problem.

    For example, I have state function combined by momentum eigen states such as [itex] \Psi = \sqrt{\frac{1}{2}}ae^{ikx} + \sqrt{\frac{1}{2}}ae^{-ikx} [/itex]

    a is normalization factor of each eigenstate and [itex] \sqrt{\frac{1}{2}} [/itex] means

    probability each eigenstate is chosen after momentum measurement is the same.

    I think expactation value of momentum of that state is 0 because momentum we can get

    from each eigenstate is the same with opposite sign and probability is the same, according

    to [tex]<a> = \sum p_{i}a_{i}[/tex], [itex] p_{i} [/itex] is probability [itex] a_{i} [/itex] can be obtained.

    I attempted to get expectation value with firsthand calculation with [itex] <\Psi|\stackrel{{^}}{p}\Psi >[/itex]

    I establish integral range from 0 to a, [0,a] and i get fianl integral form [tex] \sqrt{\frac{1}{2}}\hbar ka\int_0^{a}2isin(2kx)dx [/itex]

    but its calculation is not zero except for case in which integral range is [-a/2,a/2] or
    [itex][-\infty,\infty] [/itex]

    i can not accept physical quantity is up to on what coordinate we solve problem.

    and i have a very important question, orthogonality [tex] <\varphi_{k} | \varphi_{k^{'}} >=0[/tex] only if integral range is [itex][-\infty,\infty] [/itex] ?

    Please remove my confusing.

    Thank you for pay attention to my question
     
  2. jcsd
  3. Oct 18, 2008 #2
    You can calculate expectation value by [itex] <\Psi|\stackrel{{^}}{p}\Psi >[/itex] only
    if the wavefunction is normalized. In your case the function is not normalized no matter what normalization constant you choose, because [itex] <\Psi|\Psi >[/itex]=infinity.
    So you tried to calculate the momentum of an infinite number of particles. I think that the fact that you got a finite value (even if you would extend the lenght of the interval towards infinity) proves that the momentum per 1 particle is in fact 0.
     
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