# Gauge ivariance and canonical versus kinetic momenta for a charged particle in an EM field

I all, I am struggling to grasp the notion of gauge invariant when talking about an object like the canonical momenta $$\frac{\partial L}{\partial \dot{q}_i}$$ or kinetic momenta $$m\dot{q}_i$$.

I am very comfortable with gauge theory in the field theory context, starting with a Lagrangian, requiring its invariance under a local symmetry, partial $$\rightarrow$$ covariant derivatives and the corresponding transformation of the gauge field connections,etc. But showing an object like a momentum is gauge invariant is new for me.

I am looking into the difference between the canonical and kinetic momenta in the case of a charged particle in an EM field, described by the standard Lagrangian

$$$$L = \frac{1}{2} m\dot{r}^2 - q \phi + q \dot{r}\cdot A$$$$

The canonical momenta are $$\vec{p}_c=m\dot{\vec{r}} +q\vec{A}$$ and the kinetic are just $$\vec{p}_k=m\dot{\vec{r}}$$.

I am trying to figure out how to explicit show that $$\vec{p}_c$$ are not gauge-invariant (presumably under the U(1) symmetry of EM?) whereas $$\vec{p}_k$$ are. I know this to be the case by ACuriousMind's answer here, Emilio Pisanty's answer here, and the following section of Wikipedia's minimal coupling article.

Any tips are appreciated! :)

The canonical momentum is always (in position "$$q$$" basis) given by $$-i\hbar\partial_q$$ so the mapping of the commutator to the Poisson bracket $$[q,p]=i\hbar \leftrightarrow \{q,p\}=1$$ stays true. The nice covariant object however involves the covaraint derivative $$\nabla_q$$ as $$-i\hbar\nabla_q =-i\hbar\left(\partial_q-\frac{i}{\hbar}qA\right)= p-qA$$ which in your example represents the gauge invariant $$m\dot q$$. On its own $$\partial_q$$ does not map nicely under gauge transformations $$|x\rangle \to e^{i\Lambda(x)} |x\rangle$$, which translate to $$\psi(x)=\langle x|\psi\rangle \to e^{-i\Lambda(x)} \psi(x)$$.