# Conjugate momentum is not gauge invariant

The conjugate momentum of a charged particle moving in a uniform magnetic field is given by

$$\vec p=m\vec v+q \vec A$$

This expression is not unique because $\vec A$ is not unique. $\vec A$ is not a measurable quantity. But $\vec p$ is a phase space variable and if it is not unique the prediction of future of the system is not unique either. Isn't this a problem?

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It is not a problem. In fact that trajectory in phase space is not uniquely determined by equation of motion and initial/boundary condition is the meaning of "gauge symmetry is in fact not a symmetry but redundancy in our description" –  Jia Yiyang May 3 '14 at 14:15
@Roopam, what you want is that your equations of motion are gauge invariant. –  Nick May 3 '14 at 17:55

As you say, $\mathbf p$ here is the canonical momentum (conjugate to $\mathbf x$, i.e. the generator of translations). This is different from the kinematical momentum $$\Pi=m \mathbf v=\mathbf p-q \mathbf A$$ which is the physical quantity involved in any observable physical prediction, and is gauge invariant.