Physically intuitive explanation for Hamiltonian of charged particle in EM field I've recently been looking at the quantum mechanical description of a charged particle in an EM field and have come across this classical Hamiltonian: $$H = \cfrac{(\mathbf p - q\mathbf A)^2}{2m} + q\phi \, .$$
I can interpret the second term as the electric potential energy of the charged particle due to the $E$-field. Is there a similar, physically intuitive, way to understand why there is a correction of $-q\mathbf A$ to the kinematic momentum $\mathbf p$?
I have managed to obtain this Hamiltonian from the Lagrangian for the system, so it's just the intuition that I'm struggling with.
 A: Yes, $${\bf v}~=~\frac{{\bf p} - q{\bf A}}{m}$$ is the velocity of the point charge, so the Hamiltonian $$H~=~\frac{m}{2}{\bf v}^2+q\phi$$ is nothing but the mechanical energy. Also note that the canonical/conjugate momentum $${\bf p}~=~\frac{\partial L}{\partial {\bf v}}~=~m{\bf v}+q{\bf A}$$ is different from the kinetic momentum $m{\bf v}$!
A: You might also be wondering where the $q \mathbf{A}$ comes from in
$$ \mathbf{p} = m\mathbf{v} + q \mathbf{A}.$$
One way to think of this is Noether's theorem, which says that total momentum must be conserved if your system has a translational symmetry. If you have a constant magnetic field in the z direction,
$$ \mathbf{B} = (0, 0, B)$$
then the Lorentz force law says
$$
\frac{d}{dt} \mathbf{v} = \frac{q}{m}\mathbf{v} \times \mathbf{B}.
$$
If we confine the particle to move in the x-y plane, $\mathbf{v} = (v_x, v_y, 0)$, and
$$ (\dot v_x, \dot v_y, 0) =  \frac{q}{m} B(v_y, -v_x, 0).$$
This could be solved by, say,
$$
v_x = v \sin\left(\frac{q B}{m} t \right)
$$
$$
v_y = v \cos\left(\frac{q B}{m} t \right).
$$
But note that the mechanical momentum, $m \mathbf{v}$ is not conserved, even though our system has translational symmetry! What's going on?
The answer is that momentum (in some direction) is conserved, but we have to modify our definition of momentum. The position of the particle will be
$$
x = -\frac{mv}{q B} \cos \left(\frac{q B}{m} t\right)
$$
$$
y = \frac{mv}{q B} \sin \left(\frac{q B}{m} t\right).
$$
The vector potential $\mathbf{A}$ could be any field that satisfies
$$
\mathbf{B} = \nabla \times \mathbf{A}.
$$
One such example is
$$
\mathbf{A} = B (-y, 0, 0).
$$
Note that the vector potential is not uniquely defined because we could add the gradient of any function to it and it would still satisfy $\mathbf{B} = \nabla \times \mathbf{A}$. But with the $\mathbf{A}$ we have chosen here, it is symmetric under translations in the x direction but NOT in the y direction.
So now, the total momentum in the x direction is
$$ p_x = m v_x + q A_x = m v \sin\left(\frac{q B}{m} t \right) + q (- B ) \frac{mv}{q B} \sin \left(\frac{q B}{m} t\right) = 0$$
so it is conserved. However, the total momentum $p_y$ is NOT conserved, because $\mathbf{A}$ is not symmetric under $y$ translations.
For an other, yet perfectly acceptable choice of $\mathbf{A}$,
$$
\mathbf{A} = B(0, x, 0)
$$
then $p_y$ would be conserved but $p_x$ would not be.
So, at the end of the day, this modification to the momentum allows for it to be conserved when $\mathbf{A}$ has a symmetry, but there's still an ambiguity because it matters what you take $\mathbf{A}$ to be.
Edit: Noether's theorem is actually very simple in this case. It's actually just the Euler Lagrange equation. For a position $q_i$ (when $i = 1,2,3$ we have $q_i$ is $x,y,z$) the Euler Lagrange equation of motion is just
$$
\frac{d}{dt} \Big( \frac{\partial L}{\partial \dot q_i} \Big) - \frac{\partial L}{\partial q_i} = 0.
$$
Namely, when we have a translational symmetry in the $q_i$ direction
$$
 \frac{\partial L}{\partial q_i} = 0
$$
(when $L$ doesn't depend on $q_i$) the Euler Lagrange equation implies we have momentum conservation in the $i$th direction:
$$
\frac{d}{dt} \Big( \frac{\partial L}{\partial \dot q_i} \Big) = \dot p_i = 0.
$$
