Commutator of a quark current

Question

In Quantum Chromodynamics, when we take the limit in which the u, d and s quarks have no mass, there exists a global symmetry $G \equiv SU(3)_L \otimes SU(3)_R$ in flavour space. The corresponding Noether currents to this symmetry are given by

$$ J_X^{a,\mu} = \bar{q}_X \gamma^\mu T^a q_X \quad \text{where} \quad X = L,R $$

In the current, $q_X$ is the left- or right-handed quark field, $\gamma^\mu$ are the Dirac matrices and $T^a$ are the generators of $SU(3) in the fundamental representation.

And y have to prove that the Noether charges $Q_X^a = \int d^3x J_X^{a,0}(x)$ satisfy the commutation relation

$$ \left[Q_X^a, Q_Y^b \right] = i \delta_{XY} f^{abc} Q_X^c $$

The problem is that in order to demonstrate the commutation relation I know that I have to start by using

$$ \left[q(x) \mathcal{O} q(x) . q^\dagger (y) \mathcal{O}' q(y) \right] = q^\dagger (x) \left[ \mathcal{O} , \mathcal{O}' \right] q(x) \delta^{(4)}(x-y) $$

and I have no clue of how to demonstrate that relation, except for that I need the anticommutation relations of the quark fields.

Thanks for your attention. I’m looking forward to your reply.

Answer 1

Just expand the commutator using the identity

\begin{equation} [AB,CD] = -AC\{D,B\}+A\{B,C\}D-C\{A,D\}B+\{C,A\}DB. \end{equation}

It should be easy with that. But take into account that you are only interested in the commutation relations of the fields, not with the ones of the operator $\mathcal{O}$ (it has nothing to do with color, flavor and so on), so write down the matrix indices and take the operator out of the commutator.