In the presence of a magnetic field, the canonical momentum of a charge particle changes from $p_{i}\equiv mv^i$ to $\pi_{i}\equiv p_{i}+eA_{i}$, where $e$ is the charge of the particle.

  1. It is possible to define another kind of canonical momentum $\tilde{\pi}_{i}=p_{i}-eA_{i}$?

  2. How can you show that this new definition of momentum is not gauge-invariant?


1 Answer 1


Just as $qV$ is potential energy, $q \vec{A}$ could easily be considered to be 'potential momentum', while ordinary momentum could be considered 'kinetic momentum'.
$${\vec p}_{total}={\vec p}_{kinetic}+{\vec p}_{potential}$$

While this 'total momentum' is not gauge invariant, conservation of total momentum is gauge invariant, and that is the critical item.

  • $\begingroup$ Ah, I see! So, under a gauge transformation, $\tilde{\pi} \rightarrow \tilde{\pi}-e\nabla\chi$ for a gauge transformation $A\rightarrow A+\nabla\chi$, which is why $\tilde{\pi}$ is not gauge invariant? $\endgroup$ Nov 14, 2016 at 7:46
  • $\begingroup$ That's right. The gauge choice is time invariant and so the time derivative will cause the extra gauge term to drop out when you look at conservation of momentum. $\endgroup$
    – David Elm
    Dec 6, 2016 at 9:14

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