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

[phos19]

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)

 

[strangerep]

$$ H = \dfrac{1}{2m} (\vec{p} - q\vec{A}) $$

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

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?

 

[malawi_glenn]

The $(\vec p - q \vec A)$ should be squared

 

[phos19]

The $(\vec p - q \vec A)$ should be squared

yes, my mistake

 

[phos19]

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?

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.