Abs. value of momentum in QM

  • Thread starter csopi
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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)
 
  • #3
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Interesting.But this a-basic representation is seldom used.Isn't it.
 
  • #4
tom.stoer
Science Advisor
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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.
 
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