What's the coordinate of the Landau tube?

The Hamiltonian of a three-dimensional electron gas in a static magnetic field $vec{B}$ is :
$$hat{H}=frac{(hat{vec{p}}+ehat{vec{A}})^2}{2m}$$

If choosing $vec{B}=Bvec{e_z}$ and using Landau gauge : $vec{A}=(0,Bx,0)$, we can rewrite the Hamiltonian in a quasi harmonic oscillator form:

$$hat{H}=frac{hat{p}_x^2}{2m}+frac{1}{2}momega_c^2left(hat{x}+frac{hat{p}_y}{momega_c}right)^2+frac{hat{p}_z^2}{2m}$$
where $omega_c$ is the cyclotron frequency.

For the first two terms, they are just in harmonic oscillator form, so the eigen-energy corresponding to this part is $(n+frac{1}{2})hbaromega_c$. For the last term, since the magnetic field doesn’t break the translation symmetry along z-direction (i.e., $[hat{H},hat{p}_z]=0$), the eigen-energy corresponding to this part is $(hbar k_z)^2/2m$.

Hence we achieve the eigen-energy of the Hamiltonian with two quantum number:

$$E_{nk_z}=left(n+frac{1}{2}right)hbaromega_c+frac{(hbar k_z)^2}{2m}$$and use landau tube to describe the degeneracy for each eigen-energy:

We can find that for a fixed $n$ and $k_z$, the states of the same energy are just denoted by a circle with centre axis along $k_z$ axis and satisfying $frac{hbar^2 (k_x^2+ k_y^2)}{2m}=(n+frac{1}{2})hbaromega_c$.

Here is my question : Are the coordinates of Landau tube ($k_x$, $k_y$ and $k_z$) corresponding to three components of mechanical momentums ($hat{vec{pi}}=hat{vec{p}}+ehat{vec{A}}$) or canonical momentums($hat{vec{p}}$)?

If they are mechanical, why can we use both $k_x$ and $k_y$ to label a state when $[hat{vec{pi_x}},hat{vec{pi_y}}]neq0$?

If they are canonical, why can we easily find the states of same energy by satisfying $frac{hbar^2 (k_x^2+ k_y^2)}{2m}=left(n+frac{1}{2}right)hbaromega_c$ when the form of Hamiltonian is not simple $frac{hat{p}_x^2+ hat{p}_y^2+hat{p}_z^2}{2m}$?