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

  • Context: Graduate 
  • Thread starter Thread starter csopi
  • Start date Start date
  • Tags Tags
    Momentum Qm Value
Click For Summary

Discussion Overview

The discussion centers on identifying the operator that represents the absolute value of a particle's momentum in quantum mechanics. Participants explore the possibility of expressing this operator in terms of first-order differential operators, particularly in the context of different representations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant seeks an operator for the absolute value of momentum, specifically the square root of the Laplacian in 2 or 3 dimensions, and questions the feasibility of expressing it using first-order differential operators.
  • Another participant suggests a form of the solution involving an a-representation, indicating a desire for a new basis that is neither the position nor momentum representation.
  • A third participant notes that the a-basis representation is rarely used, implying a potential limitation in its application.
  • A later reply points out that the absolute value of momentum is a positive operator, while a differential operator may not be positive due to having negative eigenvalues, suggesting a fundamental issue with the proposed approach.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of representing the absolute value of momentum with first-order differential operators, with some highlighting theoretical challenges and others proposing alternative representations. No consensus is reached.

Contextual Notes

Participants acknowledge limitations related to the positivity of operators and the rarity of certain representations, which may affect the discussion's conclusions.

csopi
Messages
81
Reaction score
2
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?
 
Physics news on Phys.org
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)
 
Interesting.But this a-basic representation is seldom used.Isn't it.
 
One remark: |p| is a positive operator, whereas a differential operator isn't positive i.e. has negative eigenvalues, therefore the above mentioned idea will not work.
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K