Connection between QCD Lagrangian and the strong running coupling constant

I know the QCD Lagrangian as well as the running coupling constant for the strong force.

But how are they connected? The Lagrangian should contain the coupling constant, shouldn't it?

• $\alpha_S=g^2/4\pi$ – user154997 Sep 7 '17 at 11:20
• Thank you! But where do I see g² in return in the Lagrangian? edit: The gluon fields fields strength tensor do that, right? – Ben Sep 7 '17 at 12:28
• Look at the definition of $G_{\mu\nu}^a$ on the Wikipedia page you quoted: second equation below this point – user154997 Sep 7 '17 at 13:05
• And then, of course, $D_\mu=\partial_\mu + igA_\mu$ which Wikipedia did not bother to repeat on that page as far as I can tell. – user154997 Sep 7 '17 at 13:07
• @LucJ.Bourhis You should make that comment an answer. – ACuriousMind Sep 10 '17 at 4:32

The Wikipedia page you quoted gives the QCD Lagrangian in its most compact form, using the convention that there is an implicit sum over any repeated indices,

$$\mathcal{L}_\text{QCD}=\bar{\psi}_i\big(\text i(\gamma^\mu D_\mu)_{ij}-m\delta_{ij}\big)\psi_j - \frac{1}{4}G^a_{\mu\nu}G^{\mu\nu,a},$$

where the gluonic tensor $G^a_{\mu\nu}$ reads

$$G^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+gf^{abc}A^b_\mu A^c_\nu \tag{1},$$

where $A^a_\mu$ is the gluon field. In those formula $\mu$ and $\nu$ are the spacetime indices whereas $a$, $b$, $c$ on one hand and $i$, $j$ on the other hand are colour indices, representing two different representations of $SU(3)$. By that I mean that $\psi_i$ represents a quark of colour $i$ whereas $A^a_\mu$ represents a gluon of colour $a$. That wikipedia page did not bother to remind readers of the definition of the covariant derivative $D$, which, I am sure, can be found on their page about Yang-Mills gauge theories. Here it is:

$$D^\mu_{ij}=\partial^\mu\delta_{ij}+\text ig(T_a)_{ij}A^\mu_a \tag{2}$$

where the $T_a$'s are generators of that representation of $SU(3)$ (the i in front of $g$ is the imaginary number, not the index $i$, just to be clear!). I am probably too terse I know but the alternative would be to write many pages, and better people than me have already done that: I suggest you check at least Wikipedia or much better, a good book on quantum field theory, especially about that question of representations.

This is the same $g$ appearing in equations (1) and (2) and it embodies the strength of QCD coupling. Then $\alpha_S$ is defined as

$$\alpha_S=\frac{g^2}{4\pi}.$$

This answers your question. But since I went that far, I'll address the questions which naturally come next: why using $g^2$ in that definition? And why $4\pi$?

Let's look at an example: the process $q\bar{q}\to ggg$ (the $g$'s stands for gluons here, not for the coupling constant $g$ in eqn. (1) and (2), just to make sure!). Here are two Feynman diagrams to consider at the lowest order in $g$:

The $qqg$ vertices come from the term featuring $\text ig\bar{\psi}_i \psi_j A_a^\mu$ in the lagrangian: note the factor $g$. The $ggg$ vertices come from the terms featuring three fields $A_a^\mu$: it too has a factor $g$. This is true for the other type of vertices not appearing on those diagrams: each vertex brings a factor $g$. As a result, the transition amplitude represented by those diagrams is proportional to $g^3$. But then we measure cross-sections, which are proportional to the modulus square of the amplitude. So the cross-section for that process, taking into account only those lowest order diagrams, will be proportional to $g^6$, i.e. $\alpha_S^3$.

What happens if we add more vertices, or equivalently go to a higher order in $g$? One will necessarily add 2 vertices at least, e.g.,

Thus we get an order $g^5$. So the amplitude will be the sum of contributions of order $g^3$ and order $g^5$ and when we square we will get contributions of order $g^6$ and $g^8$, keeping only the next leading order, i.e. of order $\alpha_S^3$ and $\alpha_S^4$. So that answer the question about using $g^2$.

As for the $4\pi$, it is a historical fluke as far as I know. In QED, you would have the electron charge $e$ instead of $g$ and it was customary since the early days of Quantum Mechanics to use $\alpha=e^2/4\pi$. So as not to introduce a spurious factor in the comparison of $\alpha$ and $\alpha_S$, a $4\pi$ was used for the latter too.