# Why do quarks tend to bond in groups of three?

Why do quarks tend to bond in groups of three?

I understand why they might in bond in groups of three to create a net charge of zero (i.e. neutrons), but I don't understand why they form protons and not, for example, commonly form in groups of four. I have done background research on this question, and did not find any information. The wikipedia article on baryons (https://en.wikipedia.org/wiki/Baryon) was not helpful either.

To summarize, why are mesons and baryons more common than other hadrons?

• A state of four quarks is called a tetraquark. Also, it's not really true that they "tend to bond in groups of three", there's plenty of mesons, they just are all unstable. – ACuriousMind Nov 20 '15 at 22:56
• Fair enough, but why do baryons such as the proton essentially not decay? Is this not an indicator of a strong bond? – Mason Nov 20 '15 at 23:00
• Not being able to decay is not really an indication of a "strong bond" but rather that there is no decay process which could happen within the confines of the known conservation laws. – ACuriousMind Nov 20 '15 at 23:02
• Alright, but why is most common matter made of mesons and baryons? – Mason Nov 20 '15 at 23:05
• Mason, common matter must be made of something. Are you asking why stable particles are stable? – Alfred Centauri Nov 20 '15 at 23:24

Here is one way to think about it...

Anything made of quarks must have net color neutrality.

Quarks come in the following 'colors', RED, BLUE, and GREEN and all three combine to be color neutral

Antiquarks come in the following 'colors', ANTIRED, ANTIBLUE, and ANTIGREEN and all three combine to be color neutral

Colors and their anticolors combine to be color neutral (i.e. RED and ANTIRED combine to be color neutral)

Hence...
One quark cannot be color neutral
Two quarks cannot be color neutral
One quark and one antiquark can be color neutral [MESON]
3 quarks can be color neutral [BARYON]
Two quarks and two antiquarks can be color neutral [TETRAQUARK]
4 quarks and one antiquark can be color neutral (three colors + color/anticolor) [PENTAQUARK]

You can see that singular quarks and diquarks are not possible and that baryons and mesons are simpler in structure, hence, more common.

Briefly, a hadron has to be color singlet ${\bf 1}$ under the $SU(3)_C$ color gauge group, due to color confinement. Examples:

• A single quark $q$ transforms in the fundamental representation ${\bf 3}$ of $SU(3)_C$, and is hence not allowed. See also related Phys.SE post here.

• A diquark $qq$ belongs to the tensor representation ${\bf 3}^{\otimes 2}:={\bf 3}\otimes{\bf 3}\cong\bar{\bf 3}\oplus{\bf 6}_S$, which we have decomposed in irreps. This contains no singlet ${\bf 1}$, and is hence not allowed. See also related Phys.SE posts here and here about $SU(3)$ tensor representations.

• In a meson, the quark-antiquark-pair $q\bar{q}$ belongs to ${\bf 3}\otimes\bar{\bf 3}\cong{\bf 1}\oplus{\bf 8}_M$, which contains a singlet ${\bf 1}$, and is hence allowed.

• The third tensor product is ${\bf 3}^{\otimes 3}={\bf 3}\otimes{\bf 3}\otimes{\bf 3}\cong {\bf 1}\oplus 2\cdot{\bf 8}_M\oplus{\bf 10}_S$. In a baryon, the three quarks $qqq$ form a totally antisymmetric representation $\wedge^3 {\bf 3}\cong {\bf 1}$ of $SU(3)_C$, which is isomorphic to a singlet ${\bf 1}$, and is hence allowed. See also related Phys.SE post here. The lightest baryon, the proton, is stable in the standard model due to baryon/quark number conservation. (However, see the hypothetical proton decay.)

• The tensor product ${\bf 3}\otimes{\bf 3}\otimes\bar{\bf 3}\cong 2\cdot{\bf 3}\oplus {\bf 6}_M\oplus {\bf 15}_M$ contains no singlet ${\bf 1}$, so the combination $qq\bar{q}$ is not allowed.

• The fourth tensor product is ${\bf 3}^{\otimes 4}\cong 3\cdot{\bf 3}\oplus 2\cdot{\bf 6}_M\oplus 3\cdot{\bf 15}_M\oplus{\bf 15}_S$. This contains no singlet ${\bf 1}$, so four quarks $qqqq$ are not allowed.

• A "molecule" of mesons and baryons, such e.g. a tetraquark $q\bar{q}q\bar{q}$ or a pentaquark $qqqq\bar{q}$, is also allowed, but is obviously heavier. See also related Phys.SE posts here, here, and here.

As one can see the number of quarks minus the number of anti-quarks should be divisible by 3.

References:

1. G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 10. The pdf file is available here.
• What is meant by the subscripts $M$ and $S$? – Robin Ekman Nov 23 '15 at 1:29
• $S$=totally symmetric (Young tableaux with only one row); $M$=mixed symmetry (Young tableaux with more than one row and more than one column); For $SU(3)$ there is effectively only one totally antisymmetric Young tableaux (two boxes in a column): $\wedge^2 {\bf 3}\cong\bar{\bf 3}$. – Qmechanic Nov 23 '15 at 1:34