$SU(3)$ gauge invariance in QCD

Question

In QCD, the gauge-invariant Lagrangian under the transformation

$$ \psi \to \psi' = e^{ig T^a \theta^a(x)} \psi$$

is written as:

$$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psi - \frac{1}{4}G^a_{\mu\nu}G_a^{\mu\nu}$$

where the covariant derivative is:

$$D_\mu = \partial_\mu - ig T^a G^a_\mu$$

and the field strength tensor is defined as:

$$ G^a_{\mu\nu} = \partial_\mu G^a_\nu - \partial_\nu G^a_\mu + g f_{abc} G^b_\mu G^c_\nu $$

If I impose the gauge-invariance, I find that the gauge field transforms as:

$$ G^a_\mu \to G'^a_\mu = G^a_\mu + \partial_\mu \theta^a $$

Am I correct? I think I am, but if I look at how the field strength transforms, I expect it to remain invariant, but instead I find an extra term:

$$ G^a_{\mu\nu} \to G^a_{\mu\nu} + g_s f_{abc}(\partial_\mu \theta^b \partial_\nu \theta^c + \partial_\mu \theta^b G^c_\nu + \partial_\nu \theta^c G^b_\mu) $$

Does this term vanish? If does then Why? Or am I totally wrong on the transformation of the gauge field?

Answer 1

The Gauge field transforms as $$ G_{\mu}^{i}\frac{\lambda^{i}}{2} \rightarrow G_{\mu}' = uG_{\mu}u^{-1}+\frac{i}{g}u\partial_{\mu}u^{-1} $$ with $u \in SU(3)$ such that $$ u^{-1}(x)=\exp (-i\alpha^i \lambda^i /2) $$ this can be expanded in a series for infinitesimal transformations.

EDIT:

expanding $u^{-1}$ $$ u^{-1} \approx 1 -i\alpha^i \lambda^i/2 + \mathcal{O}(\alpha^2) $$ you can preform this expansion for both $u$ and it's inverse to first order and make sure to keep everything to first order in $\alpha$, it will include a commutator.