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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.