What is the Operator for Absolute Value of Momentum in Quantum Mechanics?

[csopi]

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?

 

[tom.stoer]

How should the solution look like?

Something like an a-representation

$$\sqrt{\hat{p}^2_a} \, \psi(a) = c\partial_a\,\psi(a)$$

with

$$\psi(a) = \langle a|\psi\rangle$$

(a is neither the x- nor the p-rep. for which we know the expressions; so you are looking for a new a-basis)

 

[zje2009]

Interesting.But this a-basic representation is seldom used.Isn't it.

 

[tom.stoer]

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.