How do I check if the canonical angular momentum is conserved?

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Discussion Overview

The discussion centers on the conservation of canonical angular momentum in the context of a purely magnetic Hamiltonian involving a vector potential. Participants explore the implications of the Hamiltonian formulation and the definitions of angular momentum in this framework.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a Hamiltonian of the form $$ H = \dfrac{1}{2m} (\vec{p} - q\vec{A}) $$ and questions the conservation of $$ \vec{L} = \vec{r} \times \vec{p} $$.
  • Another participant challenges the correctness of the Hamiltonian, noting that it should be a scalar and suggesting the computation of the Poisson bracket between $$ H $$ and $$ \vec{L} $$ as a method to investigate conservation.
  • There is a suggestion that the proper canonical angular momentum might be $$ \partial {\mathcal L}/\partial\dot\phi $$ instead of the ordinary orbital angular momentum.
  • Some participants agree that the term $$ (\vec{p} - q \vec{A}) $$ should be squared in the Hamiltonian formulation.
  • Clarifications are made regarding the distinction between canonical angular momentum and "naive" angular momentum.

Areas of Agreement / Disagreement

Participants express disagreement regarding the formulation of the Hamiltonian and the definitions of angular momentum. There is no consensus on the correct approach to determine conservation.

Contextual Notes

Participants note potential limitations in the Hamiltonian's formulation and the definitions used, which may affect the conclusions drawn about conservation.

phos19
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Specifically given a purely magnetic hamiltonian with some associated vector potential :
$$ H = \dfrac{1}{2m} (\vec{p} - q\vec{A}) $$

How can I deduce if $$ \vec{L} = \vec{r} \times \vec{p}$$ is conserved? ( $$\vec{p} = \dfrac{\partial L}{\partial x'}$$, i.e. the momentum is canonical)
 
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phos19 said:
$$ H = \dfrac{1}{2m} (\vec{p} - q\vec{A}) $$
Are you sure this is the correct Hamiltonian? Normally ##H## is a scalar here, but your RHS is a vector.
phos19 said:
How can I deduce if ## \vec{L} = \vec{r} \times \vec{p}\,## is conserved?
When you've figured out a correct Hamiltonian, you could try computing the Poisson bracket between ##H## and ##\vec{L}## ?

Btw, maybe you really want the proper canonical angular momentum ##\partial {\mathcal L}/\partial\dot\phi##, rather the ordinary orbital angular momentum?
 
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The ## (\vec p - q \vec A)## should be squared
 
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malawi_glenn said:
The ## (\vec p - q \vec A)## should be squared
yes, my mistake
 
strangerep said:
Are you sure this is the correct Hamiltonian? Normally ##H## is a scalar here, but your RHS is a vector.

When you've figured out a correct Hamiltonian, you could try computing the Poisson bracket between ##H## and ##\vec{L}## ?

Btw, maybe you really want the proper canonical angular momentum ##\partial {\mathcal L}/\partial\dot\phi##, rather the ordinary orbital angular momentum?

strangerep said:
Are you sure this is the correct Hamiltonian? Normally ##H## is a scalar here, but your RHS is a vector.

When you've figured out a correct Hamiltonian, you could try computing the Poisson bracket between ##H## and ##\vec{L}## ?

Btw, maybe you really want the proper canonical angular momentum ##\partial {\mathcal L}/\partial\dot\phi##, rather the ordinary orbital angular momentum?
yes ##H## is supposed to be squared. Here ##\vec{L}## is the canonical angular momentum, not the "naive" angular momentum.
 

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