# How does one derive the QCD Lagrangian?

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How  does one read the Quantum Chromodynamics Lagrangian (which is similiar to the QED Lagrangian)? What does the rank 3 tensor, $H^{\mu\nu\rho}$ represent? What is the fermion term and which is the gluon term? I suspect that the latter term $\bar\psi(i\gamma^\mu\nabla_\mu -m)\psi$ is the fermion term, as it seems to be the Lagrangian of a Dirac field.

Where can I see a derivation of this Lagrangian?

This post imported from StackExchange Physics at 2014-03-06 11:10 (UCT), posted by SE-user user34039

edited Apr 19, 2015
QFT textbook will explain the meaning of the symbols and tensors. It is not really a derivation to find these Lagragnians. It is postulated by educated guess and experimental evidence (e.g. parity non-conservation)

This post imported from StackExchange Physics at 2014-03-06 11:10 (UCT), posted by SE-user user26143

I don't think this is gradute level. Even if granting it is, this question does not seem to contain much thought. So -1.

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Quantum Chromodynamics is a Quantum Field Theory that describes Quarks, Gluons, and their interactions through the Strong Force. It is a strongly-coupled theory, which means that there is the need of Renormalisation (unless you believe in Vladimir Kalitvanski).

## Free Quarks

Free Quarks clearly should obey the Free (i.e. Potential-less) Dirac Equation: $\left( i\hbar {{\gamma }^{\mu }}{{\partial }_{\mu }}-m{{c}_{0}} \right)\psi =0$

Applying the Euler-Lagrange Equations, we see that the Lagrangian Density would then be: $\mathsf{\mathcal{L}}={{c}_{0}}\bar{\psi }\left( i\hbar {{\gamma }^{\mu }}{{\partial }_{\mu }}-m{{c}_{0}} \right)\psi$

Sidenote: This is the Lagrangian Density for a Free Quark. For the entire Quantum Chromodynamics Lagrangian Density, one also needs to find the Lagrangian Density \ for only Gluons, and the Lagrangian Density for Quark-Gluon interaction.

This Lagrangian Density ensures invariance under $SU(3)$ transformations. I.e. transforming the quark field $\psi$ as $\psi \mapsto Q\psi$ where $Q\in SU\left( 3 \right)$does not change the Lagrangian Density.

## Quark Confinement

In experiment, Quarks have never been observed as free, they are always interacting through the Strong Force, with Gluons and other Quarks. This phenomenon is known as Quark Confinement.

## Gluons and the strong force

Let us now introduce 8 Gluon Potentials $A_{a}^{\mu }$ where $a$ goes from 1 to 8. The Quarks will no longer be free. The interaction between the Quarks and theGluons is known as the Strong Force. The Lagrangian Density due to the Strong Force is then given by the following expression: $\mathcal L=-\hbar {{g}_{QCD}}{{\gamma }^{\mu }}A_{a}^{\mu }\frac{{{\lambda }^{a}}}{2}$

The Gluons themselves are Yang-Mills Fields, and have the Yang-Mills Lagrangian Density. This is as follows: $\mathcal L=-\frac{1}{4}{{H}^{\mu \nu \rho }}{{H}_{\mu \nu \rho }}$

With these Gluon Fields, the covariant derivative can be defined as follows: $\mathcal L={{\nabla }_{\mu }}_{\operatorname{QCD}}={{\partial }_{\mu }}+i{{g}_{QCD}}A_{a}^{\mu }\frac{{{\lambda }^{a}}}{2}$

Which invites correct comparisons with General Relativity, including bundle curvatures, etc.

## Total Lagrangian Density

Adding up the three individual Lagrangian Densityies previously discussed, one obtains the total Lagrangian Density of Quantum Chrodynamics: Note that we do not immediately observe the interaction term, but this is merely because we have quietly replaced $\partial$ with $\nabla$.

## Gauge Invariance

Quantum Chromodynamics is then invariant under $SU(3)$ transformations if the Gluon Potentials simultaneously transform as: $A_{a}^{\mu }\mapsto {{U}^{\dagger }}\left( A_{a}^{\mu }+\frac{i}{{{g}_{QCD}}}{{\partial }_{\mu }} \right)U$

This is a Gauge Transformation, and therefore, Quantum Chromodynamics is a Gauge Theory of with a Gauge Group of $SU(3)$.

answered Apr 5, 2014 by (1,975 points)
edited Apr 19, 2015
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Most of the standard model action is determined by the requirements of renormalizability, Poincare symmetry, and the assumption of appropriate (partially broken) internal symmetry groups.

Renormalizability implies that the total degree of the Lagrangian is at most four, with Bosonic and Fermionic fields being counted as of degree 1 and 1.5, respectively.

Poincare symmetry enforces that the fields come in groups that can be arranged in terms of scalars, spinors, or vectors (higher order tensors are not renormalizable, and vectors only if they are generators of a gauge group), and restricts the possible combinations of the fields, leading to the combinations also found in the standard model.

The internal symmetry group assumed groups the fields into bigger entities, composed of generators of representations of the group.

Apart from that, the number and type of fields is almost unconstrained (with a restriction due to anomaly cancellation). The standard model chooses these in the simplest way compatible with experiment.

Gravitation (which is a symmetric tensor field, hence has spin 2) is not included, as it violates the renormalizability assumption.

answered Apr 12, 2014 by (14,437 points)

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