Normalisation of QCD Lagrangian

Question

In QCD, and more generally in representations of $\mathfrak{su}(N)$, there is a freedom to choose the normalisation of the generators, $$ \mathrm{Tr} \, \left[R(T^a) R(T^b)\right] = T_R \delta^{ab}.\tag{1} $$

I am trying to work out the implications of this for the kinetic term for the gluon field in the QCD Lagrangian density. I've looked in various textbooks and notes, and every source I can find either glosses over this entirely, or gets the details trivially wrong (e.g. eqn. 33 in these or eqn. 4.66 in these).

Conventionally physicists choose $ T_F = \frac{1}{2}$ and write $$ \mathcal{L}_\mathrm{kin} = - \frac{1}{4} F_{\mu\nu}^a F^{\mu\nu}{}^a,\tag{2} $$ where the field strength tensor has been expanded into components $$ \mathbf{F}_{\mu\nu} = \sum_a F_{\mu\nu}^a T^a.\tag{3} $$

It seems clear to me that $$ \mathrm{Tr} \, \left[\mathbf{F}_{\mu\nu} \mathbf{F}^{\mu\nu} \right] = T_F \; F_{\mu\nu}^a F^{\mu\nu}{}^a,\tag{4} $$ and so the correct, Lie-algebra-normalisation-convention-independent expression for the Lagrangian density should be $$ \mathcal{L}_\mathrm{kin} = - \frac{1}{2} \mathrm{Tr} \, \left[\mathbf{F}_{\mu\nu} \mathbf{F}^{\mu\nu} \right] = - \frac{1}{2} T_F \; F_{\mu\nu}^a F^{\mu\nu}{}^a,\tag{5} $$ which restores the usual result in the conventional case.

Is this correct, and is there a good source for it? (ie one that could be cited in an academic work).

Why is the convention-specific component expansion used preferentially in the literature?

Answer 1

I'm not sure there is anything to be said about your "is this correct" question. Of course it is correct. You may expand F in any representation you wish, beyond the fundamental, but why should you? Presumably your text (try Schwartz, if P&S irks you) explains to you that, at the end of the day, when the trace is taken, $$ {1\over T_R}\mathrm{Tr} \, \left[\mathbf{F}_{\mu\nu} \mathbf{F}^{\mu\nu} \right] = F_{\mu\nu}^a F^{\mu\nu}{}^a, $$ in any representation, which is immaterial, since this quantity is actually in the adjoint. Again, the free indices a are adjoint indices, 8 in QCD, regardless of the dandified superfluous representations you wished to express the left-hand-side in. So your left hand side provides equal traces for 137000 different representations, provided you specify them clearly. I need not demonstrate to you that the r.h.s is, really, the l.h.s in the adjoint if you perversely chose to formulate it that way!

Again, adjoint action in any representation group elements U, yields the very same answer, in that representation, $$\mathbf{F}_{\mu\nu} \to U\mathbf{F}_{\mu\nu}U^{-1}\approx F_{\mu\nu}^b T^b+ i\theta^aF_{\mu\nu}^b[T^a,T^b]+ O(\theta^2)\\ = F_{\mu\nu}^b T^b-f^{cab}\theta^aF_{\mu\nu}^bT^c + O(\theta^2)= T^c (F_{\mu\nu}^c -f^{cab}\theta^aF_{\mu\nu}^b) + O(\theta^2). $$ The commutator cares not about the representation, by definition.

Why is the convention-specific component expansion used preferentially in the literature?

One normally couples the gluons to quarks, which are in the Fundamental of color, so your U group elements are in the fundamental, easy to combine gluon and quark entities through. However, if you further needed to couple novel fermions in the sextet color representation, as, of course, several model-builders do, then you'd have to use Us in the 6, etc, and, if you asked for trouble, you'd use the dandified matrix field strength, suitably normalized as specified, to chase down your local details.

Answer 2

The standard normalization of the kinetic term in components is OP's eq. (2)

$$ -\frac{1}{4}F_{\mu\nu}^{ a}F^{\mu\nu a}~=~\underbrace{\frac{1}{2} \sum_{i=1}^3\dot{A}^a_i\dot{A}^a_i}_{\text{kinetic term}}+\ldots, $$

cf. e.g. this Phys.SE post. How this is written with a trace depends on the author's normalization of the trace.