Role of Angular Momentum in Defining Vector Operator ##\mathbf{V}##

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Discussion Overview

The discussion revolves around the role of the angular momentum operator ##\mathbf{J}## in defining a vector operator ##\mathbf{V}##, particularly in the context of quantum mechanics. Participants explore the implications of using total angular momentum versus orbital angular momentum in the definition of vector operators, considering both spin and orbital contributions.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant defines a vector operator ##\mathbf{V}## in relation to the angular momentum operator ##\mathbf{J}## and questions whether ##\mathbf{J}## should represent the total angular momentum, including spin.
  • Another participant agrees with the initial definition, suggesting that the total angular momentum is indeed relevant.
  • A different viewpoint argues that using orbital angular momentum ##\mathbf{L}## instead of total angular momentum ##\mathbf{J}## does not invalidate the commutation relation involving position operators.
  • Another participant contends that if ##\mathbf{V}## is defined in terms of spin ##\mathbf{S}## or total angular momentum ##\mathbf{J}##, then it is necessary to include both orbital and spin contributions to accurately describe the system.

Areas of Agreement / Disagreement

Participants express differing views on whether the total angular momentum must be used in defining the vector operator ##\mathbf{V}##. There is no consensus on the necessity of including spin in the definition of ##\mathbf{J}##.

Contextual Notes

The discussion highlights the complexity of angular momentum in quantum mechanics and the potential implications of different definitions on the behavior of vector operators. Specific assumptions about the system and the definitions of angular momentum are not fully explored.

blue_leaf77
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A vector operator ##\mathbf{V}## is defined as one satisfying the following property:

## [V_i,J_j] = i\hbar \epsilon_{ijk}V_k##

where ##\mathbf{J}## is an angular momentum operator. My question is what is the role of ##\mathbf{J}##, does it have to be the total angular momentum from all angular momenta appearing in the system, i.e. if we don't consider spin then ##\mathbf{J}=\mathbf{L}##, if we consider it then ##\mathbf{J} = \mathbf{L}+\mathbf{S}##?
 
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Yes, exactly.
 
But using L instead of the total angular momentum J even in the presence of spin doesn't prevent the relation ##[x_i,L_j] = i\hbar \epsilon_{ijk}x_k## to prevail.
 
Of course, but if you take V=S or V=J, you will need J=L+S to rotate also the spin part.
 

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