In the Dirac equation, the only thing about the gamma matrices that is "fixed" is the anticommutation rule:
[itex]\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}[/itex]
We can get an equivalent equation by taking a unitary matrix [itex]U[/itex] and defining new spinors and gamma-matrices via:
[itex]\gamma'^\mu = U \gamma^\mu U^{-1}[/itex]
[itex]\psi' = U \psi[/itex]
[itex]\bar{\psi'} = \bar{\psi} U^{-1}[/itex]
(Actually, it occurs to me now that [itex]U[/itex] doesn't need to be unitary. But if it's not unitary, we need to define [itex]\bar{\psi'} = \psi'^\dagger (U U^\dagger)^{-1} \gamma'^0[/itex], rather than [itex]\bar{\psi'} = \psi'^\dagger \gamma'^0[/itex])
My question is whether this freedom to choose a representation is a gauge symmetry. Is there a corresponding gauge field so that we are free to choose [itex]U(x^\mu)[/itex] differently at every point, if we make the corresponding change to the gauge field?
[itex]\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}[/itex]
We can get an equivalent equation by taking a unitary matrix [itex]U[/itex] and defining new spinors and gamma-matrices via:
[itex]\gamma'^\mu = U \gamma^\mu U^{-1}[/itex]
[itex]\psi' = U \psi[/itex]
[itex]\bar{\psi'} = \bar{\psi} U^{-1}[/itex]
(Actually, it occurs to me now that [itex]U[/itex] doesn't need to be unitary. But if it's not unitary, we need to define [itex]\bar{\psi'} = \psi'^\dagger (U U^\dagger)^{-1} \gamma'^0[/itex], rather than [itex]\bar{\psi'} = \psi'^\dagger \gamma'^0[/itex])
My question is whether this freedom to choose a representation is a gauge symmetry. Is there a corresponding gauge field so that we are free to choose [itex]U(x^\mu)[/itex] differently at every point, if we make the corresponding change to the gauge field?
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