The term quadratic in $g_s$ vanished because we are considering transformations infinitesimal in $\alpha$ here, and only linear terms are considered.

To see that $T^a$ and $\gamma^\mu$ commute with each other we can write the kinetic terms expliciting the indices:
$$ \tag{1} \bar \psi (i \gamma^\mu \partial_\mu - m ) \psi
\equiv \bar \psi_{\alpha,i} (i \gamma^\mu_{\alpha\beta} \delta_{ij} \partial_\mu - m \delta_{\alpha \beta} \delta_{ij}) \psi_{\beta,j},$$
where the $i,j$ indices are *flavour indices*, and come from the fact the spinor fields $\psi$ transform under some representation (in this case the fundamental) of the gauge group (in this case $SU(3)$).
The $\delta_{ij}$ and $\delta_{\alpha \beta}$ terms are there as the components of the identity matrices acting respectively on the gauge group and on the spin degrees of freedom.
Another way to write **(1)**, summing over the index $i$ and using the $\delta_{ij}$ term, is:
$$ \tag{1'} \bar \psi_{\alpha, j} ( i \gamma^\mu_{\alpha \beta} \partial_\mu - m \delta_{\alpha \beta}) \psi_{\beta,j} = 0. $$
The physical reason for the kinetic operator $(i \! \not\! \partial-m) $ to be diagonal on the flavour indices is that during propagation the flavour does not change (e.g. a quark down does not decide by itself to become a bottom, if not interacting with something else).
Yet another way to write **(1)**, with which you may or may not be more familiar is
$$ \tag{1''} i \bar \psi_\alpha^j \gamma^\mu_{\alpha\beta} \partial_\mu \psi_\beta^j
- m \bar \psi_{\alpha}^j \psi_{\alpha}^j
\equiv i \bar \psi \gamma^\mu \partial_\mu \psi - m \bar \psi \psi = 0. $$

Expliciting the indices the gauge transformation has the form:
$$ \tag{2} \psi_{\alpha,i}(x)
\rightarrow (e^{ i g_s \alpha^a(x) T^a} )_{ij} \psi_{\alpha,j}(x),$$
which keeping only terms linear in $\alpha$ becomes:
$$ \tag{3} \psi_{\alpha,i}(x) \rightarrow ( \delta_{ij} + ig_s \alpha^a(x) T^a_{ij}) \psi_{\alpha,j}(x)
\equiv \psi_{\alpha,i}(x) + ig_s \alpha^a(x) T^a_{ij} \psi_{\alpha,j}(x), $$
where it is important to notice how the generators $T^a$ act on the flavour indices $i,j$, *not* on the spinor indices $\alpha,\beta$.
Once you have made explicit all the sum involved in the indices, all the object you are left with are (eventually complex) numbers, hence *they all commute* (with the exception of the spinor fields of course, which are Grassman numbers.

How does this apply to your calculation?

- Your fourth term is neglected because of order $\mathcal{O}(\alpha^2)$.
- In your second term the derivative acts
*both* on $\alpha$ and on $\psi$, giving the term (I'm no longer expliciting indices here for brevity):
$$ \tag{4} \bar \psi ( i \gamma^\mu \partial_\mu - m) ig_s \alpha^a T^a \psi
= -g_s (\partial_\mu \alpha^a) \bar \psi \gamma^\mu T^a \psi
+ i g_s \alpha^a \bar \psi ( i \gamma^\mu \partial_\mu - m ) T^a \psi $$
- The term involving both the $\gamma$ and the generators $T^a$ (and the derivative $\partial_\mu$ acting on $\psi$) has the form:
$$ \tag{5} (i)^2g \alpha^a \bar \psi_{\alpha,i} (\gamma^\mu_{\alpha\beta} T^a_{ij} - T^a_{ij} \gamma^\mu_{\alpha\beta}) \partial_\mu \psi_{\beta,j} = 0, $$
where as I said above $\gamma^\mu_{\alpha\beta}, T^a_{ij} \in \mathbb{C}$ hence they commute.

See also the wikipedia article on the gauge covariant derivative for similar calculations.

This post imported from StackExchange Physics at 2015-01-18 13:49 (UTC), posted by SE-user glance