quantum mechanics - For a constant magnetic field, is there a gauge with both canonical momenta conserved?

To describe a constant magnetic field $\mathbf B=(0,0,B)$ (ignoring the motion along the $z$ dimension) within hamiltonian (or quantum) mechanics, one needs to choose a gauge. One common gauge is the symmetric gauge, in which the vector potential is $\mathbf A=\frac12(-By,Bx,0)$ and the hamiltonian is $$ H=\frac{1}{2m}\left(p_x+\frac{eB}{2m}y\right)^2 + \frac{1}{2m}\left(p_y-\frac{eB}{2m}x\right)^2. $$ Similarly, one can also choose the Landau gauge, which breaks the symmetry to take a vector potential of the form $\mathbf A=(0,Bx,0)$, giving the hamiltonian as $$ H=\frac{1}{2m}p_x^2 + \frac{1}{2m}\left(p_y-\frac{eB}{m}x\right)^2. $$

It can be shown quite easily that the problem has two conserved quantities, $$ x_0= x+\frac{mc}{eB}v_y \quad\text{and}\quad y_0 =y-\frac{m c}{eB}v_x, $$ which give the center of the classical circular motion. As I showed in a previous answer, the Landau gauge manages to rotate its axes within phase space in a way that sets the canonical momentum $p_y$ axis along one of these two conserved quantities (specifically, $x_0$), so it is conserved. (This conservation can easily be seen from the fact that $H$ does not depend on $y$.) This is not particularly problematic, since the canonical momentum is now related to the (non-conserved) kinematic momentum via $mv_y=p_y+\frac{eB}{c}x$.

However, this is a bit ugly since you break the symmetry, and you only get one conserved component of the canonical momentum, in a problem that is manifestly rotation-symmetric.

My question, then, is whether there exists a gauge transformation that will make both canonical momenta into conserved quantities. If not, then, conversely, is it possible to show that this is impossible?

Answer

The condition that the $p_i$ be conserved is equivalent to $\{H,p_i\} = 0$ for the Hamiltonian $H = (p-A)^2$, where I've dropped all constants for convenience. A straightforward computation yields $$ \{H,p_i\} = -2\sum_j \frac{\partial A_k}{\partial x^i}(p_k - A_k)$$ and $p_k =A_k(x)$ is impossible since this is an off-shell equation where $p$ and $x$ are independent. Conservation of the canonical momentum $p_i$ is therefore equivalent to $$ \frac{\partial A}{\partial x^i} = 0,$$ or, in other words, $A(x,y,z) = A(z)$ if we want both $p_x$ and $p_y$ to be conserved. But the magnetic field of such a vector potential is $$ B = \nabla\times A = \begin{pmatrix} -\partial_z A_y \\ \partial_z A_x \\ 0\end{pmatrix},$$ which can obviously never be equal to $B = (0,0,B)$. Therefore, a gauge in which both canonical momenta are conserved is impossible.