Orbital Angular Momentum Origin

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Discussion Overview

The discussion centers on the origin of orbital angular momentum in quantum mechanics, particularly how it relates to classical mechanics and the implications of choosing different origins. Participants explore the calculation of angular momentum and torque in quantum systems, especially in the context of atomic structures.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions how the dependence of angular momentum on the choice of origin in classical mechanics translates to quantum mechanics, specifically regarding eigenvalues and calculations at different points.
  • Another participant suggests that angular momentum in quantum mechanics is typically discussed in the rest frame of the center of mass, making it independent of position.
  • A third participant notes that in the context of electron orbitals, the choice of an arbitrary axis does not significantly impact problem-solving, implying that the axis is understood within the context of atomic structure.
  • A later reply expresses a desire for clarification on these points, indicating uncertainty about the explicit statements regarding angular momentum origins.

Areas of Agreement / Disagreement

Participants express varying views on the relevance of the choice of origin for angular momentum in quantum mechanics. While some suggest that it becomes independent of position in certain frames, others question the necessity of specifying an axis, indicating that the discussion remains unresolved.

Contextual Notes

Limitations include the potential dependence on definitions of angular momentum and the context of measurement. The discussion does not resolve how to handle angular momentum calculations at arbitrary points or the implications for different systems.

widderjoos
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We know from classical mechanics that angular momentum L = r \times p depends on your choice of origin. My question is: How does this work quantum mechanically? We know we get certain eigenvalues, but does this apply only in a certain choice of origin? How do we calculate angular momentum at some other point? I had a similar problem concerning torque on a magnetic dipole, \tau = \mu \times B = r \times F. About what point do we measure the moment arm?
Do we just assume our origin is at the "center" of the orbit?

Thanks for the help.
 
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widderjoos said:
We know from classical mechanics that angular momentum L = r \times p depends on your choice of origin. My question is: How does this work quantum mechanically? We know we get certain eigenvalues, but does this apply only in a certain choice of origin? How do we calculate angular momentum at some other point? I had a similar problem concerning torque on a magnetic dipole, \tau = \mu \times B = r \times F. About what point do we measure the moment arm?
Do we just assume our origin is at the "center" of the orbit?

Thanks for the help.

Angular momentum depends on position like in classical mechanics. However, when speaking of L e.g. for an atom, one always refers to the angular momentum of the atom in the rest frame of its center of mass so that L becomes independent of position. In principle it would be more appropriate to talk of a contribution to the spin of the compound particle than of angular momentum.
 
In quantum mechanics, orbital angular momentum usually describes electron orbitals, where orbitals are located in an atom. When we regarding to an atom, I don't think choosing an arbitrary axis, say the tree trunk outside, would mean any thing to solving problems.

Therefore, it has no necessity to specify the axis, since we all know what we are referring to
 
Yeah, that's what I was thinking, but just wanted to make sure since I couldn't find it explicitly stated anywhere. Thanks!
 

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