# 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.

• It is not a good idea to call $\mathbf p$ kinematic momentum. Instead, it is usual to call it canonical momentum. The reason for this is in classical Hamiltonian mechanics, $\mathbf p$ is not equal to $m\mathbf v$, but it is equal to $m\mathbf v + q\mathbf A$. Commented Jul 19, 2022 at 2:17

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 $$m{\bf v}$$!

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: $$\frac{d}{dt} \Big( \frac{\partial L}{\partial \dot q_i} \Big) = \dot p_i = 0.$$

• In the given field and coordinate system, kinematic momentum components $mv_x,mv_y$ are not conserved, and neither are canonical momentum components $p_x,p_y$, in general. The canonical momenta $p_x,p_y$ are only conserved in special cases where the vector potential is chosen conveniently to produce such conservation. So its an artifact of gauge choice, physical situation allows both $p_x,p_y$ being conserved and not being conserved. Commented Jul 19, 2022 at 2:14
• > "$m\mathbf v$ is not conserved, even though our system has translational symmetry! What's going on?" -- Here you are, I think, confusing symmetry of the magnetic field $\mathbf B$ wrt translations in $x$ or $y$ direction, with symmetry of the Lagrangian with respect to translations in $x$ or $y$ direction. The former is present, but the latter is not present except in special cases you pointed out (vector potential being constant in direction $x$, or $y$). The Noether theorem is concerned with symmetries of action, or Lagrangian, thus here vector p. A. Not with symmetries of $\mathbf B$. Commented Jul 19, 2022 at 2:20
• $\mathbf{A}$ appears in the action, and if $\mathbf{A}$ has a translational symmetry then the corresponding momentum is conserved. I agree that the symmetries of $\mathbf{A}$ are gauge dependent. OP's question was of the form "what is intuition for this definition of momentum" and I decided to approach this question from the perspective of Noether's theorem. My answer explicitly only refers to symmetries of $\mathbf{A}$, not $\mathbf{B}$. Commented Jul 20, 2022 at 1:59
• My point is the answer is misleading, because the relation between canonical momentum and kinetic momentum is true irrespective of whether the vector potential is chosen by us with a particular symmetry implying conservation, or not. Commented Jul 20, 2022 at 3:21