Charged particle in a purely radial magnetic field, is the canonical angular momentum conserved?

Question

Let $ \vec{B} = k \dfrac{\vec{u_r}}{r^2}$ (assuming magnetic monopoles exist) and let $q$ be a charged particle. The associated hamiltonian is $H = \dfrac{(\vec{p} - q \vec{A})^2}{2m}$ and the canonical momentum is given by $ \vec{p} = m \vec{v} + q \vec{A}$ with an acceptable $ \vec{A} = \dfrac{b}{r} \tan(\theta/2) \vec{u_{\phi}}$.

How to know if $ \vec{L} = \vec{r} \times \vec{p}$ (the canonical angular momentum, not the standard angular momentum) is conserved? Is there a simple way to test this? (I've tried using Poisson brackets but it is a mess).

Answer 1

Since your problem is invariant by rotation, you should have conservation of angular momentum. Only usually, since $p=mv$ the two alternatives: $$ L_v=r\times mv\\ L_p=r\times p $$ are equivalent. Only in this case they are not and it turns out that $L_v$ is the correct one. This can be guessed from gauge invariance: $L_v$ is invariant ($v$ is) while $L_p$ is not.

More formally, if you have a valid trajectory $r,v$, rotation invariance says that it is the new trajectory $Rr,Rv$ which is again a valid trajectory for any rotation $R$. The rotation on phase space therefore needs to rotate $r,v$ by the same amount. This is generated by $L_v$ since from the formula: $$ v=p-qA $$ you can check that you still have the correct bracket: $$ \{r_i,v_j\}=\delta_{ij} $$ so the same reasoning for proving that $L_p$ generates the same rotation for $r,p$ applies.

Hope this helps