Quantum angular momentum of a particle in an homogeneous magnetic field

Question

In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ The last term is weird to my eyes!

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ How should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field (I'm using the symmetric gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$), I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian \begin{align} H &= \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} \\[2ex] &= \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{1}{2} \, m \omega^2 (x^2 + y^2) + i \hbar \omega \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), \tag{9} \end{align} I get this commutator: $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}) \equiv i \hbar \omega \, (\vec{\mathrm{r}} \cdot \vec{\mathrm{\pi}} + \vec{\mathrm{\pi}} \cdot \vec{\mathrm{r}}). $$ This thing is weird! I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?). How should I consider these commutators? I have the impression that they may be related to the arbitrary choice of axes origin, while the particle may evolve around any point in the $(x, y)$ plane.

Answer 1

Here I try to give some results that do not rely on the B field being uniform, or a particular choice of gauge. Starting from the definitions of the quantum operators $$ \vec{M} = \vec{r} \times \vec{\pi} \\ H = \frac{1}{2m} \vec{\pi}^2 $$

Work in the Heisenberg picture, the torque equation is $$ \frac{d}{dt}\vec{M} = \vec{r} \times \frac{d}{dt}\vec{\pi} $$

Using Heisenberg's equation of motion, $$ [\vec{M}, H] = \vec{r} \times [\vec{\pi}, H] $$

You were asking why the commutator $[\vec{M}, H]$ is non-zero. In general, it is not, as $[\vec{\pi}, H]$ is non-zero. Below are some equations which might be helpful for your analysis. $$ \frac{1}{i\hbar} [\vec{r}, H] = \frac{1}{m} \vec{\pi} \\ \frac{1}{i\hbar} [\vec{\pi}, H] = \frac{q}{2m} (\vec{\pi}\times\vec{B} - \vec{B}\times\vec{\pi}) $$

Answer 2

I think I got it! The commutator (10) is the quantum version of the classical time derivative of the angular momentum (classical expressions below): $$\tag{A} \vec{\mathrm{M}} = \vec{\mathrm{r}} \times m \vec{\mathrm{v}}. $$ So, using the magnetic force $\vec{\mathrm{F}} = q \, \vec{\mathrm{v}} \times \vec{\mathrm{B}}$: $$ \frac{d \vec{\mathrm{M}}}{d t} = \vec{\mathrm{r}} \times q (\vec{\mathrm{v}} \times \vec{\mathrm{B}}) = q (\vec{\mathrm{r}} \cdot \vec{\mathrm{B}}) \vec{\mathrm{v}} - q (\vec{\mathrm{r}} \cdot \vec{\mathrm{v}}) \vec{\mathrm{B}}. $$ With a magnetic field oriented along the $z$ direction and considering the classical circular motion in the $(x, y)$ plane around any point (not necessarily the origin), we get $\vec{\mathrm{r}} \cdot \vec{\mathrm{B}} = 0$ and $$\tag{B} \frac{d M_z}{d t} = - \frac{q B}{m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{v}}) = 2 \omega \, (\vec{\mathrm{r}} \cdot m \vec{\mathrm{v}}), $$ where $\omega = - q B / 2 m$. This is the classical version of (10), since $m \, \vec{\mathrm{v}} \equiv \vec{\pi}$ (which is the same as $\vec{\mathrm{p}} - q \vec{\mathrm{A}}$). If the axes origin is selected at the center of the circular motion, then $\vec{\mathrm{r}} \cdot m \vec{\mathrm{v}} = 0$ and the angular momentum $M_z$ is trivialy conserved. But in QM, we can't cancel (10) because the particle has a probability to be anywhere in the plane and it isn't just moving in a circle around the origin.

About (6), well, I'm not sure yet, but classically we could let $\vec{\mathrm{r}} \cdot \vec{\mathrm{B}} = 0$.