Why is the Yang-Mills Comparator unitary?

Question

In chapter 15.2 of Peskin, the comparator is defined, as some object $U\left(y,\,x\right)$ which transforms as: $$ U\left(y,\,x\right) \mapsto V\left(y\right) U\left(y,\,x\right) \left[V\left(x\right)\dagger\right] $$

where $V\left(x\right)\in SU\left(2\right)^{\mathbb{R}^4}$.

This object is introduced mainly in order to be able to define the covariant derivative, but its main defining property is that the Fermion field transforms then as:

$$ U\left(y,\,x\right)\psi\left(x\right)\mapsto V\left(y\right)U\left(y,\,x\right)\psi\left(x\right)$$ so we basically have a way to make the Fermion field transform as if it were at $y$ instead of at $x$.

At some stage Peskin states it is reasonable to that this comparator be unitary. My question is why? What breaks down, and where, if we don't assume that?

This post imported from StackExchange Physics at 2014-05-04 11:19 (UCT), posted by SE-user Psycho_pr

Answer 1

If the comparator was not unitary, you could not expand it in terms of Hermitian generators of $SU(2)$, which is required in order to construct the non-Abelian covariant derivative.

This post imported from StackExchange Physics at 2014-05-04 11:19 (UCT), posted by SE-user Frederic Brünner