# How does the underlying symmetry of QCD imply the allowance of a 4-gluon vertex?

Quantum chromodynamics allows for a four-gluon vertex such as this, in a diagram

Such a vertex would never be allowed in quantum electrodynamics, which has an underlying U(1) gauge symmetry.

I know that quantum chromodynamics has an underlying SU(3) gauge symmetry, but how does this imply that a four-gluon vertex like what is shown above would be allowed?

• QCD also has a three-gluon vertex. Vertices with three and four gauge bosons arise whenever the gauge group is non-abelian (i.e., group transformations are non-commutative). The N in SU(N) doesn’t determine how many gluons can come together at a vertex. $SU(5)$ would only have 3- and 4-gluon vertices. – G. Smith Aug 20 '19 at 16:35
• Ulsss, what is your level of understanding? Are you (somewhat) familiar with Yang- Mills theories? – Alfred Centauri Aug 21 '19 at 0:00
• @AlfredCentauri The book I'm following has covered the construction of a Yang-Mills theory but not really much beyond that. I just found them interesting so I wanted to do some more, but there's only one chapter on them in the book. – Featherball Aug 21 '19 at 12:21

Let's compare the following train of thought for $$U(1)$$ vs $$SU(3)$$:

### U(1)

1. $$U(1)$$ is an Abelian group.
2. This means, the commutator of two group elements is zero.
3. The field strength tensor that appears in the Lagrangian, $$F^{\mu\nu}$$, contains such a commutator: $$F_{\mu\nu}^{QED} = \partial_\mu A_\nu - \partial_\nu A_\mu + \text{i} g [A_\mu, A_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu$$
4. In the Lagrangian, we have a term like $$F_{\mu\nu} F^{\mu\nu}$$.
5. This means that at most, we can have terms quadratic in $$A_\mu$$, which leads to the photon propagator and the fermion-photon interaction.

### SU(3)

1. $$SU(3)$$ is a non-Abelian group.
2. This means, the commutator of two group elements is not zero.
3. The field strength tensor that appears in the Lagrangian, $$F^{\mu\nu}$$, contains such a commutator: $$F_{\mu\nu}^{QCD} = \partial_\mu A_\nu - \partial_\nu A_\mu + \text{i} g [A_\mu, A_\nu]$$
4. In the Lagrangian, we have a term like $$F_{\mu\nu} F^{\mu\nu}$$.
5. This means that at most, we can have terms quartic in $$A_\mu$$. This means, $$A^4$$, $$A^3$$, $$A^2$$, $$A$$. Terms linear in $$A$$ lead to quark-gluon interaction, terms quadratic in $$A$$ lead to gluon propagators and the terms with $$A^3$$ and $$A^4$$ lead to the three-gluon and four-gluon interaction terms.

And this is how the underlying symmetry of QCD implies the allowance of a 4-gluon vertex!

The QCD Lagrangian is,

$$\mathcal{L}=\sum_{\alpha} \overline{\psi}_{q, a}\left(i \gamma^{\mu} \partial_{\mu} \delta_{a b}-g_{s} \gamma^{\mu} t_{a b}^{C} \mathcal{A}_{\mu}^{C}-m_{q} \delta_{a b}\right) \psi_{q, b}-\frac{1}{4} F_{\mu \nu}^{A} F^{A \mu \nu}$$.

The field strength tensor is given by, $$F_{\mu \nu}^{A}=\partial_{\mu} \mathcal{A}_{\nu}^{A}-\partial_{\nu} \mathcal{A}_{\mu}^{A}-g_{s} f_{A B C} \mathcal{A}_{\mu}^{B} \mathcal{A}_{\nu}^{C}$$,

where the $$A_\mu^A$$ terms represent the gluon field. It is then clear that $$\frac{1}{4} F_{\mu \nu}^{A} F^{A \mu \nu}$$ involves a product of four gluon field operators. In the $$U(1)$$ case, the field strength tensor has no $$-g_{s} f_{A B C} \mathcal{A}_{\mu}^{B} \mathcal{A}_{\nu}^{C}$$ term, and so we do not get photon-photon interactions.

For more on this, see the PDG's review of QCD.