# Normalisation of QCD Lagrangian

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?

• You are positive Peskin & Schroeder, or Mathew Schwarz, etc..., the mainstream texts, don't cove this? I mean, this is an optional arbitrary normalization of the coupling, or the action. Have you calculated beta functions or anomalies? Feb 21, 2021 at 1:04
• Yes, I am sure. The books I have looked in (those plus others, including standard QCD textbooks) generally make reference to the trace notation only to prove gauge invariance of the component form, if at all, for which getting the normalisation right is unimportant. As for your question - yes, I have calculated in QCD, but using the usual conventions of $T^a = \frac{1}{2} \lambda^a$. I'm interested in what happens if we adopt different conventions, which we are of course free to do, and the difference cannot be physical; and it seems likely that there is a convention-independent formulation.
– JCW
Feb 21, 2021 at 11:05
• This definitely doesn't correspond to our freedom to scale the action without affecting the equations of motion, since other terms of the QCD Lagrangian don't contain $T_F$ (and in the SM Lagrangian other terms have no QCD content at all). It can probably be consistently compensated for by a redefinition of $g$, but it's unclear why that would make it a bad question or trivial - then we're just compensating for the freedom to fix one convention differently by fixing another differently.
– JCW
Feb 21, 2021 at 11:34
• I really sympathise with this question. Physics should be about concepts, but since nobody states their conventions clearly, I often find myself spending more time chasing down signs and factors of two. When you see something that looks wrong, you are never sure if the authors made a mistake or if you are just misunderstanding their conventions... Feb 21, 2021 at 15:33
• I also sympathize. I very often find physics notation and texts are written to make it easy to do computations, and don't discuss subtleties like this because, from a computational point of view, why would you make any choice of representation but the easiest to work with? But especially when you are learning, you often want to understand why an apparent ambiguity in the formalism doesn't affect any physical results, and it's not always obvious or discussed explicitly. Feb 21, 2021 at 15:56

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.

• To be clear, I am not suggesting using any representation other than the fundamental. The freedom I am talking about isn't to choose a representation, but to choose its normalisation. Working entirely within the fundamental representation, $T_F = \frac{1}{2}$ is a choice. Both P&S and Schwartz, and other textbooks I have consulted, acknowledge that it is a choice, but not what the expression in question should be if a different choice is made. Your identity agrees with mine; I agree that it is correct. That is not my question. I will endeavour to clarify my question.
– JCW
Feb 21, 2021 at 16:03
• So we all agree, the choice exists, and any well meaning reader of the literature should not think about it twice. It is elementary group theory, and such texts (Georgi?) cover it. It's hard to see if you are shaking your fist at the literature, or you are asking for references in the "I dare you" mode.... Feb 21, 2021 at 16:19
• To illustrate with an example, the book by Elvang and Huang (arxiv.org/abs/1308.1697) chooses $T_F = 1$, so the generators are trace-orthonormal. Eq 2.63 states the term of the Lagrangian as they claim it should be, in a convention-independent way, $-\frac{1}{4} \mathrm{Tr} \, \mathbf{F}_{\mu\nu} \mathbf{F}_{\mu\nu}$. Yet it's clear that this only gives the familiar $-\frac{1}{4}$ factor for their choice of $T_F$, and if you used another you'd get something else; yet elsewhere other authors claim that $-\frac{1}{4}$ arises if you use their conventions instead. They can't all be right!
– JCW
Feb 21, 2021 at 16:21
• Ugh... but they further redefine their structure constants, and their couplings! to match. If you believed they are inconsistent, which I doubt, you might write to them. This is standard procedure in the field, to spend/waste some due diligence on squaring conventions, if inclined... Feb 21, 2021 at 16:27
• I did notice their definition of $f^{abc}$, but I can't see where that comes into it - shouldn't it only change the factor accompanying $A_\mu^bA_\nu^c$ in $F^a_{\mu\nu}$, not the overall factor? Some of my queries are resolved by a set of notes I've just found, edu.itp.phys.ethz.ch/fs13/qft2/QFT2FS13Notes.pdf, which keep full generality throughout (eg eq 4.18 for $\mathcal{L}$, culminating in eqs 4.77-4.81). If you know of a book/paper with a similar level of generality, I would appreciate a reference. At a meta-level, I'm curious to know why this more general formulation is so rare.
– JCW
Feb 21, 2021 at 18:55

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.

• Ok, but isn't the fundamental object here the Lie-algebra-valued field strength tensor $\mathbf{F}_{\mu\nu}$, rather than its components in a specific basis? Why doesn't the factor accompanying this term in components change according to the basis we expand in? To be concrete, if I scale the generators of the fundamental rep in two different ways, to give $\{T^a\}$ with $T_F = K$ and $\{t^a\}$ with $T_F = k$, ie $$T^a = t^a \sqrt{\frac{K}{k}},$$ is it not true that $$F^a_{\mu\nu} \equiv {f}^a_{\mu\nu} \sqrt{\frac{k}{K}}?$$ Then they have to come with different factors to agree.
– JCW
Feb 21, 2021 at 17:29
• The obvious analogy is with two orthogonal bases of a vector space, $\{\mathbf{e}_i\}$ and $\{\mathbf{e}'_i\}$, normalised $$\mathbf{e}_i\cdot\mathbf{e}_j=k\,\delta_{ij}$$ and $$\mathbf{e}'_i\cdot\mathbf{e}'_j=k' \,\delta_{ij}.$$ Any expression involving an inner product has an induced dependence on this normalisation when expressed in component notation, $$\left\lVert\mathbf{a}\right\rVert^2=\sum_{i,j}a_ia_j\left(\mathbf{e}_i\cdot\mathbf{e}_j \right)=k\,a_ia_i=k'\,a'_ia'_i,$$ and as a result you couldn't say, eg, $$-\frac{1}{4} a_i a_i \equiv -\frac{1}{4} a'_ia'_i,$$ for any $k,k'$.
– JCW
Feb 21, 2021 at 18:16