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

AI Thread Summary
The discussion centers on the conservation of angular momentum in a system described by a magnetic Hamiltonian. Participants question the formulation of the Hamiltonian, noting that it should be a scalar rather than a vector. To determine if angular momentum is conserved, they suggest computing the Poisson bracket between the Hamiltonian and the angular momentum vector. There is also a clarification that the focus should be on the canonical angular momentum rather than the ordinary orbital angular momentum. The conversation emphasizes the importance of correctly defining the Hamiltonian to analyze the system's dynamics effectively.
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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