Abs. value of momentum in QM

1. Sep 26, 2012

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?

2. Sep 26, 2012

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)

3. Sep 27, 2012

zje2009

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

4. Sep 27, 2012

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.

Last edited: Sep 27, 2012