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Abs. value of momentum in QM

  1. Sep 26, 2012 #1
    Hi,

    I am looking for the operator representing the absolute value of a particle's momentum. In other words: the square root of the laplacian (preferably in 3 dimensions, but 2 would also be fine).

    I am aware, that integral formulas exist for this operator, but is it possible to express it in terms of first order differential operators?
     
  2. jcsd
  3. Sep 26, 2012 #2

    tom.stoer

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    How should the solution look like?

    Something like an a-representation

    [tex]\sqrt{\hat{p}^2_a} \, \psi(a) = c\partial_a\,\psi(a)[/tex]

    with

    [tex]\psi(a) = \langle a|\psi\rangle[/tex]

    (a is neither the x- nor the p-rep. for which we know the expressions; so you are looking for a new a-basis)
     
  4. Sep 27, 2012 #3
    Interesting.But this a-basic representation is seldom used.Isn't it.
     
  5. Sep 27, 2012 #4

    tom.stoer

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    One remark: |p| is a postive operator, whereas a differential operator isn't positive i.e. has negative eigenvalues, therefore the above mentioned idea will not work.
     
    Last edited: Sep 27, 2012
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