Consider the Hamiltonian of a charged particle of charge $q$ in a uniform magnetic field $\textbf{B}=B\hat{\textbf{z}}$ is given by $$H=\frac{(\textbf{p}-q\textbf{A})^2}{2m}$$ where $\textbf{p}$ is the canonical momentum given by $\textbf{p}=m\textbf{v}+q\textbf{A}$.

What can we say about the classical symmetries of this Hamiltonian and the associated conservation laws?

In particular, since $\textbf{B}$ is constant, then $\textbf{A}$ is necessarily lineraly dependent on $x$ and/or $y$. For example, a possible choice of $\textbf{A}$ is given by $\textbf{A}=(-\frac{1}{2}By+\frac{1}{2}Bx,0)$. Therefore, the Hamiltonian is not invariant under coordinate shifts $x\to x+x_0$ and/or $y\to y+y_0$. Hence, the system is not translation invariant along $x$ and $y$ directions. And we do not expect the canonical momenta $p_x$ and $p_y$ to be constants of motion. Is it not in contradiction with the answer here?

  • $\begingroup$ Where do you see a contradiction to any of the answers in the other question? Please elaborate on what you mean. $\endgroup$ – ACuriousMind Apr 6 '17 at 14:08
  • $\begingroup$ @ACuriousMind The maximum voted answer says "On the classical level, the system is translation invariant in both x- and y-direction". It's not. Isn't it? $\endgroup$ – SRS Apr 6 '17 at 19:34

It depends on what you mean by "the system" being invariant here. If you mean the Hamiltonian with a specific choice of $A$, then that's not translation invariant. However, the equations of motion do not depend on the choice of $A$, and they are translation-invariant in both directions. You can argue that, classically, the equations of motion are what matters, but when you quantize you have no choice but to choose a Hamiltonian.

Symmetries of the equation of motion need not necessarily lift to the Hamiltonian, as we can plainly see here and as is discussed more generally here by Qmechanic. One can show that there is no way to make both canonical momenta conserved for any choice of $A$, see this answer of mine.

This particular situation is also discussed at length and excellently by Emilio Pisanty in this answer. Summarizing, this Hamiltonian system (regardless of the choice of $A$), has two conserved quantities, namely the $(x,y)$-coordinate of the center of the circle the moving particle describes. As usual in Hamiltonian mechanics, conserved quantitites generate symmetries. The choice of $A$ is unphysical and hence cannot change these quantities, but what the choice of $A$ does is "rotate" the canonical momenta $p_x,p_y$. So the Hamiltonian is translation invariant in one direction exactly when one of the canonical momenta aligns with one of the conserved quantities, but, in general, the transformation generated by them will not be translation.

This underscores that the canonical momentum of Hamiltonian mechanics is, in general, a purely fictitious quantity. There is no general physical interpretation for it and transformations that do not change the physics, such as the choice of $A$, may change it from a physically meaningful conserved quantity to...just something else that describes the a point in phase space.

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  • $\begingroup$ Since the Hamiltonian (as well as the Lagrangian) is not translation invaraint, the canonically conjugate momenta $p_x=mv_x+qA_x$ and $p_y=mv_y+qA_y$ are not conserved. Can we say anything about $\pi_x=mv_x$ and $\pi_y=mv_y$ (which are not canonically conjugate to $x$ and $y$)? @ACuriousMind $\endgroup$ – SRS Apr 7 '17 at 5:59
  • $\begingroup$ @SRS What do you want to say about them? The particle moves in a circle in the x-y plane, so they're obviously not conserved. $\endgroup$ – ACuriousMind Apr 7 '17 at 10:36

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