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Consider the Dirac action $S=\int d^4x\bar{\psi}(x)(i\not\partial-m)\psi(x)$. Since there are no time derivative of $\bar{\psi}$, we get the constraint that its canonical momenta vanishes. This constraint is of course first class. Does this mean that the Dirac equation has a gauge symmetry? Why do we usually not care about it when doing the canonical quantization of the Dirac field? Thanks!

Qmechanic
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Ivan Burbano
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    If you think about the phase space as containing momenta associated to both $\psi$ and $\bar{\psi}$, then the two constraints mentioned below are $\bar{\pi} = 0 $ and $\pi - i\bar{\psi}\gamma^0= 0$. The Poisson bracket between the constraints is then proportional to the Poisson bracket between $\bar{\psi}$ and $\bar{\pi}$, and is therefore nonzero. – Bruno De Souza Leão Apr 27 '20 at 23:42
  • Got it! Muito obrigado! – Ivan Burbano Apr 28 '20 at 00:26

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The are 2 primary constraints rather than 1. (Recall that there is also 1 primary constraint for the complex conjugate field). The 2 primary constraints do not (super)Poisson-commute, so they are second-class rather than first-class constraints. And therefore no gauge symmetry. See e.g. my related Phys.SE answer here.

Qmechanic
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  • I don't see the second primary constraint since the action does have time derivatives with respect to $\psi$. Funnily enough, I think that my confusion has to do with the question you linked. In particular, using the more symmetric action $\frac{i}{2}\bar{\psi}\not\partial\psi+\frac{i}{2}\bar{\psi}\overleftarrow{\not\partial}\psi+m\bar{\psi}\psi$ it is clear that both momenta do not vanish and we have the two second class primary constraints. – Ivan Burbano Apr 27 '20 at 20:43