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Coordinate representation of the momentum operator

  1. Jun 15, 2012 #1
    The position operator in coordinate representation is:
    Xab=aδ(a-b)
    this is diagonal as expected

    The momentum operator turns out to look like
    Pab=-ih∂aδ(a-b)

    Now, this is not supposed to be diagonal because it does not commute with X.
    However it looks pretty diagonal to me.

    What am I missing?
     
  2. jcsd
  3. Jun 15, 2012 #2
    The momentum operator is not what you wrote here. It is equal to:
    [tex]P_{ab} = e^{i(a-b)}[/tex]
    with some scaling factors.
     
  4. Jun 15, 2012 #3
    What exactly are you saying?

    Anyone familiar with quantum mechanics in dirac notation?
     
  5. Jun 15, 2012 #4
    I don't see how it looks diagonal. For something to be diagonal, it needs to be multiplied by a delta function. An example of a diagonal matrix is:

    Oab=δ(a-b) f(a,b)

    for arbitrary function f(a,b).

    The delta function is essentially the identity matrix, and f(a,b) are the diagonal components.
     
  6. Jun 15, 2012 #5
    From what I understood on distributions the support of the derivative of a distribution is contained within the support of the original distribution.

    Therefore ∂aδ(a-b) should be zero wherever δ(a-b) is.
     
  7. Jun 15, 2012 #6

    fzero

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    Distributions are supported on functions ##f(a)##, not on the domain of the functions themselves. Given a function ##f(a)##, we have ##\delta[f]=f(0)## and ## \delta'[f]=f'(0)##. These are generally not equal. The same argument can be made for delta functions which are shifted away from the origin.

    ##\delta'## is as independent of ##\delta## as an ordinary function is from its derivative. There's no unitary operator that can map one to another (for all test functions ##f##).
     
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