# Is Choice of Spinor Representation a Gauge Symmetry?

Staff Emeritus

## Main Question or Discussion Point

In the Dirac equation, the only thing about the gamma matrices that is "fixed" is the anticommutation rule:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$

We can get an equivalent equation by taking a unitary matrix $U$ and defining new spinors and gamma-matrices via:

$\gamma'^\mu = U \gamma^\mu U^{-1}$
$\psi' = U \psi$
$\bar{\psi'} = \bar{\psi} U^{-1}$

(Actually, it occurs to me now that $U$ doesn't need to be unitary. But if it's not unitary, we need to define $\bar{\psi'} = \psi'^\dagger (U U^\dagger)^{-1} \gamma'^0$, rather than $\bar{\psi'} = \psi'^\dagger \gamma'^0$)

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 $U(x^\mu)$ differently at every point, if we make the corresponding change to the gauge field?

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WannabeNewton
No. It is no more a gauge symmetry than the ability to express the electric and magnetic fields in terms of cartesian basis vectors or spherical polar basis vectors.

Staff Emeritus
To me, the choice of the matrix $U$ at each point seems like a generalization of the choice of the phase $e^{i \phi}$ at each point. That's the special case where $U = e^{i \phi} I$. The choice of phase is the gauge symmetry associated with electromagnetic interactions. I was wondering if there was a more general gauge symmetry that involved more complicated choices of $U$.