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?

  • 1
    $\begingroup$ $\alpha_S=g^2/4\pi$ $\endgroup$ – user154997 Sep 7 '17 at 11:20
  • $\begingroup$ Thank you! But where do I see g² in return in the Lagrangian? edit: The gluon fields fields strength tensor do that, right? $\endgroup$ – Ben Sep 7 '17 at 12:28
  • $\begingroup$ Look at the definition of $G_{\mu\nu}^a$ on the Wikipedia page you quoted: second equation below this point $\endgroup$ – user154997 Sep 7 '17 at 13:05
  • $\begingroup$ 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. $\endgroup$ – user154997 Sep 7 '17 at 13:07
  • $\begingroup$ @LucJ.Bourhis You should make that comment an answer. $\endgroup$ – 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


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 $g$:

enter image description here

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.,

enter image description here

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.