Gauge invariance of Yang-Mills lagrangian

Question

I am trying to show gauge invarince of the Yang-Mills lagrangian

$$\mathcal{L}= -\frac{1}{4}F_{\mu \nu }^{a}F^{\mu \nu ,a}+\sum_{i,j}^{N}\overline{\psi}_{i} (\delta _{ij}i\partial_{\alpha}\gamma^{\alpha } -\delta _{ij}m+gA_{\alpha }^{a}\gamma^{ \alpha } T^{a}_{ij})\psi_{j},$$ by rewriting it in terms of the covariant derivative $D_{\mu}=\partial_{\mu}-igA^{a}_{\mu}T^{a},$ for which I know that $F_{\mu \nu }=\frac{i}{g}[D_{\mu},D_{\nu}],$ (where $F_{\mu \nu }=F_{\mu \nu }^{a}T^{a}$) and that it transforms as $D_{\mu} \rightarrow U(x)D_{\mu}U^{-1}(x)$ under the gauge transformation. I am stuck with the following two questions:

• When evaluating the transformation of the first term, I have seen the idendity $$-\frac{1}{4}F_{\mu \nu }^{a}F^{\mu \nu ,a}=-\frac{1}{2}F_{\mu \nu }^{a}F^{\mu \nu ,b}\text{tr}[T^{a}T^{b}]=-\frac{1}{2} \text{tr} [F_{\mu \nu }F^{\mu \nu}]$$ been used, but I dont understand the second equality. The components of the Yang-Mills field tensor are matrices, so how does one justify including them in the trace? (It is understood that that $T^{a}$ matrices has been normalized so that $\text{tr}[T^{a}T^{b}]=\frac{1}{2}\delta^{ab}$ by the way.)

• For the second term of the lagrangian I have seen the equality $$\sum_{i,j}^{N}\overline{\psi}_{i} (\delta _{ij}i \partial_{\alpha}\gamma^{\alpha }-\delta _{ij}m+gA_{\alpha }^{a}\gamma^{ \alpha } T^{a}_{ij})\psi_{j} =\sum_{i,j}^{N}\overline{\psi}_{i} ( i D_{ij, \alpha}\gamma^{\alpha }-\delta _{ij}m)\psi_{j},$$ been used, but I don't understand how this is true unless $gA_{\alpha }^{a}\gamma^{ \alpha } T^{a}_{ij}=0$ for $i\neq j$. I am very eager to know why this equality holds?

Answer 1

In the first point you're wrong in the normalisation condition which is $$\text{Tr}(T^aT^b) = \frac{1}{2}\delta^{ab}$$ and it couldn't be $\delta_{ij}$ since you're tracing over the indices $$\text{Tr}(T^aT^b) =T^a_{ij}T^b_{ji}$$ With this the first result is trivial.

The second point just comes from the definition of the covariant derivative $D_\mu = \partial_\mu-igA_\mu^aT^a$ in which the internal indices are understood. If you write them down you would get $$(D_{\mu})_{ij} =\partial_\mu\delta_{ij}-igA_\mu^aT^a_{ij} $$ in fact $$i\bar\psi_i (D_\mu)_{ij}\gamma^\mu\psi_j = i\bar\psi_i\left(\partial_\mu\gamma^\mu\delta_{ij}-igA_\mu^a\gamma^\mu T^a_{ij}\right)\psi_j = \bar\psi_i\delta_{ij}i\gamma^\mu\partial_\mu\psi_j+g\bar\psi_i\gamma^\mu A_\mu^aT^a_{ij}\psi_j$$ which is exactly what you have in the second equality