# Why is $SU(3)$ chosen as the gauge group in QCD?

Why is $SU(3)$ chosen as the gauge group. Why not $U(3)$? Why does it even have to be unitary?

• If you like this question you may also enjoy reading this post. Apr 16, 2014 at 12:27

Let me first make a general remark about internal symmetry groups, unrelated to our problem of the correct symmetry group for QCD.

The symmetry must act on Hilbert space as a unitary operator for the conservation of probability.

Now let us turn to the strong interaction. The most important experimental facts were that

1. Observed hadron spectrum was understood as that of bound states of quarks.
2. The SLAC experiment found that in high energy, deep inelastic scattering, the bound quarks behaved as if they were weakly coupled.
3. The measured decay rate of $\pi\to\gamma\gamma$ was nine times greater than expected.

In theoretical language, we thus require the properties of confinement (hadrons) and asymptotic freedom (coupling gets weaker at high energies).

A theoretical result is that only Yang-Mills $SU(N)$ gauge theories exhibit asymptotic freedom.

The question is now, what is the value of $N$? Well, experimental fact number three helps us. If the quarks transform in the fundamental triplet of $SU(3)$, the decay rate is enhanced by $3^2$.

• Ok, but why do we want to conserve probability. i thought that only has to hold in a non-relativistic theory Apr 16, 2014 at 13:14
• No, for a good symmetry, the probability of seeing something should be invariant under the action of tht symmetry. If the symmetry were not unitary, probabilities could be made arbitrarily big or small by successive symmetry transformations. unbounded probabilities greater than 1 make no sense. Apr 16, 2014 at 13:19
• Suggestion to the answer (v2): Mention $SU(N_1)\times\ldots \times SU(N_r)$. Apr 16, 2014 at 13:31
• @innisfree Can you give a reference explaining why only $SU(N)$ exhibit asymptotic freedom? Apr 16, 2014 at 17:35
• Cheng and Li, gauge theory of elementary particles @JSchwinger Apr 16, 2014 at 18:19

Proving, that $$N_c=3$$ is not enough to verify, that the transformations are governed by $$SU(3)$$. Imagine a SU(2) color triplet $$\phi \to \phi^\prime=W\phi$$. A meson would be color neutral, because $$\phi^\dagger \phi \to \phi^{\dagger\prime} \phi^\prime=\phi^{\dagger} W^\dagger W \phi$$ and $$W$$ is unitary (by definition). The problem with that transformation is, that we need to choose a representation of $$SU(2)$$ acting in the 3D color-space. Here one can choose real $$W$$s, which leads to a problem. $$\phi^\ast$$ and $$\phi$$ transform identically, allowing for color neutral $$qq$$ or $$\bar{q}qq$$ particles. Checking, that this is not true for $$SU(3)$$, is an exercise in patience and using explicit $$\lambda$$-matrices. This only leaves the question of Groups $$G(N>3)$$. As long as it has a subgroup, meeting our criteria, it can be used. But then we just introduce a Group of operations, that we do not need to explain the states we observe.

• I'm not sure this answers the (nearly six-year-old) question.
– d_b
Mar 22, 2020 at 17:13