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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

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

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! :)

Lopey Tall
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1 Answers1

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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 $\psi(x)=\langle x|\psi\rangle \to e^{-i\Lambda(x)} \psi(x)$.

mike stone
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