Under a discrete parity transformation, how does a non-abelian gauge field $A^a_{\mu}(x)$ transform? Is it possible to get mixing between the colors? How about the fermion $\psi_n(x)$ which is coupled to the gauge field? Let's say they transform under some representation of the gauged Lie group, with generators $(t_a)_{nm}$, does the fermion mix its $n$ index under a parity transformation?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The gauge field $A_\mu$ transforms as a covector (here $A_\mu = T^a A^a_\mu$ is the full connection matrix). This means that $A_\mu$ transforms in the same way that partial derivatives $\partial_\mu$ transform. This is most easily seen by looking at the covariant derivative $D_\mu = \partial_\mu + A_\mu$. The covariant derivative transforms under coordinate changes $x \rightarrow y$ as
$\frac{D}{dy^\mu} = \frac{dx^\nu}{dy^\mu} \frac{D}{dx^\nu}$
Or, written another way,
$D_\mu \rightarrow \frac{dx^\nu}{dy^\mu} D_\nu$
This implies that under a coordinate change, $A_\mu$ transforms the same way,
$A_\mu \rightarrow \frac{dx^\nu}{dy^\mu} A_\nu$
So under a reflection, there is one component $x^i$ that transforms to $x^i \rightarrow -x^i$ all others staying same. So this means that
$A_i \rightarrow -A_i$ (i.e $A^a_i \rightarrow -A^a_i$)
and all other components stay the same. Note that this a purely geometric statement having nothing to do with quantizing the theory, and comes from viewing $A_\mu$ as a connection on a vector bundle (for example, see this other StackExchange post).
Fermions transform as usual, $\psi \rightarrow \gamma^0 \psi$ under parity, which corresponds to switching the left and right components of the fermi field.
