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:
$$\tag{6}
[ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \,  \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}).
$$
So I need to confirm this.  Is (6) correct?

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:
$$\tag{7}
[L_z, H] = 0, \qquad [M_z, H] \ne 0.
$$
Is this right and 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 get
$$\tag{8}
M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2).
$$
So, for the Hamiltonian (I'm using the gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$)
$$\tag{9}
H = \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} = \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{q^2 B^2}{8 m} (x^2 + y^2) + i \hbar \frac{q B}{2 m} \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr),
$$
I get
$$\tag{10}
[M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}).
$$
This thing is weird to me.  Is (10) correct too?  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?), if these relations are correct (I believe they are).