# Is Choice of Spinor Representation a Gauge Symmetry?

1. Jul 13, 2014

### stevendaryl

Staff Emeritus
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?

2. Jul 13, 2014

### 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.

3. Jul 13, 2014

### stevendaryl

Staff Emeritus
Well, the choice of a different basis at each point in spacetime IS a gauge symmetry, isn't it? Can't GR be described in those terms?

4. Jul 13, 2014

### stevendaryl

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 $U$.