0
$\begingroup$

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! :)

$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thank you very much for the pointer to QM! This and the wiki article (en.wikipedia.org/wiki/…) were exactly what i needed. Cheers! $\endgroup$
    – Lopey Tall
    Jul 31, 2020 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.