I recently came across the fact that, for a charged particle in an electromagnetic field, the generalized momentum differs from the kinetic momentum. Specifically, the generalized momentum is given by $$p=mv+qA,$$ whereas the kinetic momentum is just $$p_{kin}=mv.$$ I understand the mathematical origin of this distinction from the Lagrangian and Hamiltonian formulations of mechanics. However, I'm wondering if there is a deeper physical interpretation of this difference-perhaps in terms of energy, work, or some other fundamental concept. Why does the electromagnetic potential $A$ appear in the momentum, and what does that mean physically?
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Here is a perhaps more fundamental way to look at momentum. In quantum mechanics $P^{\mu}=(P_x,P_y,P_z.E)$ are the generators of space-time translations by $X_{\mu}=(x,y,z,ct)$. $$ |Translated\ Object>=e^{iX_{\mu}P^{\mu}}|Object> $$ where Object is any chunk of stuff (eg: an electron) and the ket $|Object>$ is a vector in Hilbert space that stands for the Object. Now, you wrote: $$ P^{\mu}=p^{\mu}+A^{\mu}Q $$ where $p^{\mu}$ is the mechanical momentum, you called mv. These $p^{\mu}$ commute with each other, are generators in the Poincare group, and just multiply $|Object>$ by a complex number $e^{iX_{\mu}p^{\mu}}$ when a translation is done. But wait, experimentally more things can happen to an Object than this when you translate it! For example, if there are other charged particles around, their " vector potential fields $A^{\mu}$ " will cause the charge generator Q to be done to the Object as you translate resulting in a phase change if the Object is charged. Hence the second term $A^{\mu}Q$ is added to do this. When $A^{\mu}(x)$ is different in different locations, the transformations (rotations and boosts) caused by electromagnetic fields $\vec{E},\vec{B}$ (=derivatives of $A^{\mu}(x)$) are done to the Object as it is translated. Q is called a generator in the "Gauge group" (in this case the group U(1)), and $A^{\mu}$ is called the "Gauge field".
There are even more gauge transformations ( Phys Stack Answer) added to $P^{\mu}$ to patch it up so it does all the things we experimentally see translations do.
Gary Godfrey
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Just as the electric scalar potential can be thought of as potential energy per unit charge, one interpretation of the magnetic vector potential is potential momentum per unit charge. Then the canonical momentum of a point charge is a combination of its linear and "potential" momentum, $\mathbf{p}_\text{can}=\mathbf{p} + q\mathbf{A}$. The justification for this form is outlined below.
In classical mechanics, the force on a particle is its rate of change of momentum. If the force is conservative, $\boldsymbol{\nabla}\times\mathbf{F}=\mathbf{0}$, then we can represent it as the gradient of some potential energy,
\begin{equation} \mathbf{F} = \frac{d\mathbf{p}}{dt} = -\boldsymbol{\nabla}U\,. \end{equation}
In electromagnetism, the force on a charged particle is given by the Lorentz force law,
\begin{equation} \mathbf{F} = q\,(\mathbf{E} + \mathbf{v}\times\mathbf{B})\,. \end{equation}
Our goal is to express this force as the gradient of some potential as in classical mechanics. Rewriting the Lorentz force law in terms of scalar and vector potentials,
\begin{equation} \mathbf{F} = q\,\bigg(-\boldsymbol{\nabla}\phi - \frac{\partial\mathbf{A}}{\partial t} + \mathbf{v}\times(\boldsymbol{\nabla}\times\mathbf{A})\bigg)\,. \end{equation}
With the help of one product rule,
\begin{equation} \boldsymbol{\nabla}(\mathbf{v}\cdot\mathbf{A}) = \mathbf{v}\times(\boldsymbol{\nabla}\times\mathbf{A}) + (\mathbf{v}\cdot\boldsymbol{\nabla})\,\mathbf{A}\,, \end{equation}
and defining the convective (total) derivative of the vector potential,
\begin{equation} \frac{d\mathbf{A}}{dt} = \frac{\partial\mathbf{A}}{\partial t} + (\mathbf{v}\cdot\boldsymbol{\nabla})\,\mathbf{A} \end{equation}
the Lorentz force law can be rewritten as
\begin{equation} \frac{d}{dt}(\mathbf{p} + q\mathbf{A}) = -\boldsymbol{\nabla}[q\,(\phi - \mathbf{v}\cdot\mathbf{A})]\,. \end{equation}
The similarity of this expression and Newton's second law invites us to define the canonical momentum for a charged particle in an electromagnetic field as $\mathbf{p}_\text{can} = \mathbf{p} + q\mathbf{A}$. A similar calculation allows us to express the energy of the charged particle as
\begin{equation} \frac{d}{dt}\bigg(\frac{p\,^2}{2m} + q\phi\bigg) = \frac{\partial}{\partial t}[q\,(\phi - \mathbf{v}\cdot\mathbf{A})]\,. \end{equation}
Notice the potential on the right hand side is the same as that of the analogous momentum equation.
Aiden
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