Chapter 12 Baryons and mesons

12.1 Introduction

The goal of this lecture is to understand the allowed baryons and mesons observed in nature. We will focus on the lightest $u,d,s$ quark states, since these are most prevalent.

Quarks have spin 1/2, so mesons, consisting of 2 quarks, can have a total spin of 0 or 1. Baryons have three quarks, so adding a third quark to a quark pair means the total spin can either be 1/2 or 3/2. Protons (uud) and neutrons (udd) are examples of baryons with spin 1/2. However, not all baryons with spin 1/2 are seen in nature – in particular, the $ddd$, $uuu$ and $sss$ baryons are not observed.

12.2 Pauli exclusion principle

The reason for this is the Pauli Exclusion Principle (PEP), which forbids more than 1 fermion from occupying the same quantum state. An implication of the PEP is that for any two fermions the overall wavefunction picks up a minus sign (it is anti-symmetric) under particle exchange,

$$\psi ({𝐫}_1,{𝐫}_2)=-\psi ({𝐫}_2,{𝐫}_1).$$ (12.1)

It is this requirement that the overall wavefunction is anti-symmetric for fermions which restricts the allowed baryon states.

12.3 Wavefunction symmetry

The most stable systems of quarks arise when they have no net angular momentum. This preferred configuration is called an ‘s-state’, with $\mathrm{\ell }=0$. In our discussion, we will assume the baryons and mesons are in this state.

The overall wavefunction of the baryons made up of three quarks (fermions) is

$$𝚿={\mathrm{\Psi }}_{\mathrm{spin}}{\mathrm{\Psi }}_{\mathrm{colour}}{\mathrm{\Psi }}_{\mathrm{flavour}}{\mathrm{\Psi }}_{\mathrm{space}}(𝐫),$$ (12.2)

that is the product of four terms. Since baryons have half-integer spin they are fermions and so the PEP requires the overall wavefunction to be anti-symmetric. One way of getting an anti-symmetric total wavefunction is multiplying 3 symmetric wavefunctions by an anti-symmetric wavefunction. This is indeed how nature works:

1. The spin wavefunctions are symmetric on exchange of spins.

2. The space wavefunction for the lowest energy state is also symmetric on interchanging particles (overall s-state).

3. The flavour wavefunctions are also symmetric on interchange of particles. For example, the neutron, whose flavour is written crudely as $udd$, has the symmetric flavour wavefunction

   $${\mathrm{\Psi }}_{\mathrm{flavour}}=|u,d,d⟩+|d,u,d⟩+|d,d,u⟩,$$ (12.3)

   where $|u,d,d⟩$ is shorthand for the first quark being in the up state, the second quark being in the down state, and the third quark being in the down state.

4. The colour wavefunction is completely antisymmetric on interchange. For baryons it is of the form

   ${\mathrm{\Psi }}_{\mathrm{colour}}=|r,b,g⟩-|r,g,b⟩+|b,g,r⟩$ (12.4)

   $-|b,r,g⟩+|g,r,b⟩-|g,b,r⟩,$

   where $|r,b,g⟩$ is shorthand for the first quark being in the r state, the second quark being in the b state, and the third quark being in the g state. Colour is the only fully antisymmetric wavefunction for the three colours. Since there is only one such state it is called the colour singlet. The colour singlet state has the lowest energy. All known baryons are colour singlets of this form. Other colour states correspond to a repulsive interaction between quarks and so do not lead to bound states.

12.4 Baryons of total spin $3/2$, for $\mathrm{\ell }=0$

Before discussing baryons let us recall the allowed spins for a quark pair. These can have either spin 0 or 1. The $s=0$ state only has one z-component of spin so is a singlet state. It has a normalized wavefunction

$${\mathrm{\Psi }}_{\mathrm{spin}}=\frac{1}{\sqrt{2}}\left(|\uparrow ,\downarrow ⟩-|\downarrow ,\uparrow ⟩\right).$$ (12.5)

The $s=1$ state has three z-components of spin so is a triplet state. It has normalized wavefunctions

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\uparrow ⟩,$ (12.6)

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ ${\displaystyle \frac{1}{\sqrt{2}}}\left(|\uparrow ,\downarrow ⟩+|\downarrow ,\uparrow ⟩\right),$

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\downarrow ,\downarrow ⟩.$

In order to get a spin 3/2 baryon we add a quark to the 2 quark $s=1$ state. We won’t be concerned with the normalization factors in the spin wavefunctions so will neglect them – we are only interested in the symmetry. These states are

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\uparrow ,\uparrow ⟩,$ (12.7)

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\uparrow ,\downarrow ⟩+|\uparrow ,\downarrow ,\uparrow ⟩+|\downarrow ,\uparrow ,\uparrow ⟩,$

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\downarrow ,\downarrow ⟩+|\downarrow ,\downarrow ,\uparrow ⟩+|\downarrow ,\uparrow ,\downarrow ⟩,$

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\downarrow ,\downarrow ,\downarrow ⟩.$

Notice all the spin wavefunctions are fully symmetric on exchange of spins, which is what was required by the PEP – it told us the spin wavefunction needed to be symmetric. The allowed particles can then be arranged as in Fig. 12.1.

$S=0$	${\mathrm{\Delta }}^{-}$		${\mathrm{\Delta }}^0$		${\mathrm{\Delta }}^{+}$		${\mathrm{\Delta }}^{++}$	1232MeV
	(ddd)		(udd)		(uud)		(uuu)
$S=-1$		${\mathrm{\Sigma }}^{\ast -}$		${\mathrm{\Sigma }}^{\ast 0}$		${\mathrm{\Sigma }}^{\ast +}$		1385MeV
		(dds)		(dus)		(uus)
$S=-2$			${\mathrm{\Xi }}^{\ast -}$		${\mathrm{\Xi }}^{\ast 0}$			1533MeV
			(dss)		(uss)
$S=-3$				${\mathrm{\Omega }}^{-}$				1672.5MeV
				(sss)

Table 12.1: Baryons of total spin $3/2$, for $\mathrm{\ell }=0$.

• Notice that the mass difference between states of different $S$ is about 150 MeV/c².

• The particle ${\mathrm{\Delta }}^{+}$ has the same quark structure ($uud$) as the proton (the baryon of spin 1/2) with a spin quantum number of 3/2. It is an excited state. Similarly ${\mathrm{\Xi }}^{\ast -}$ is an excited state of ${\mathrm{\Xi }}^{-}$. It is composed of the same combination of quarks.

• Note that the ${\mathrm{\Omega }}^{-}$ particle has three strange quarks of spin 1/2 so it seems that we have three identical particles of the same flavour, spin and orbital state. However, the three quarks have different colours so they are not identical. Pauli’s exclusion principle is not violated.

• Originally the table was drawn up by Gell-Mann before the ${\mathrm{\Omega }}^{-}$ was discovered. Its mass was guessed from the masses of the other particles by assuming that the mass would be about 150 MeV/c² more than the ${\mathrm{\Xi }}^{\ast -}$ particles. The confirmation of the existence of the ${\mathrm{\Omega }}^{-}$ was an early triumph for the quark model of Gell-Mann.

• The same sort of table can be drawn up for any three quarks, for example the $u$, $d$ and $c$ quarks.

• The presence of the ${\mathrm{\Delta }}^{++}$ particle implies that there are three $u$ quarks which must be in a symmetric state, ${\mathrm{\Psi }}_{\mathrm{flavour}}=|u,u,u⟩$. It then follows that the colour wavefunction must be antisymmetric. That was why colour was first introduced.

12.5 Baryons of total spin 1/2, for $\mathrm{\ell }=0$

In order to get a spin 1/2 baryon we can either add a quark to the 2 quark $s=1$ state or the 2 quark $s=0$ state. Adding the third quark to the pair with $s=0$ gives the spin wave functions (again ignoring normalization factors)

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\downarrow ,\uparrow ⟩-|\downarrow ,\uparrow ,\uparrow ⟩,$ (12.8)

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\downarrow ,\downarrow ⟩-|\downarrow ,\uparrow ,\downarrow ⟩.$

Note that the wavefunctions are not symmetric.

Adding the third quark to the pair with $s=1$ gives the spin wavefunctions

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\uparrow ,\downarrow ⟩-|\uparrow ,\downarrow ,\uparrow ⟩-|\downarrow ,\uparrow ,\uparrow ⟩,$ (12.9)

${\mathrm{\Psi }}_{\mathrm{spin}}$ $=$ $|\uparrow ,\downarrow ,\downarrow ⟩+|\downarrow ,\uparrow ,\downarrow ⟩-|\downarrow ,\downarrow ,\uparrow ⟩.$

Note that the wavefunctions are not fully symmetric, but they do have partial symmetry on 2 of the quarks. Therefore for any quark composition of the form $aab$, where $a\ne b$, the $aa$ pair must be must be in the $s=1$ state. Hence for the six combinations

$$uud,uus,ddu,dds,ssu,ssd,$$ (12.10)

the $aa$ pair is in the $s=1$ state.

When we have three identical quarks (e.g. $uuu$) there is no fully symmetric state under all 3 particle exchanges, and therefore violates the PEP. This is why we do not see the combinations $uuu$, $ddd$ and $sss$ in Fig. 12.3.

$S=0$		$n$		$p$		939MeV
		(ddu)		(uud)
$S=-1$	${\mathrm{\Sigma }}^{-}$		${\mathrm{\Sigma }}^0$		${\mathrm{\Sigma }}^{+}$	1190MeV
	(dds)		(dus)		(uus)
$S=-2$		${\mathrm{\Xi }}^{-}$		${\mathrm{\Xi }}^0$		1320MeV
		(dss)		(uss)

Table 12.2: Baryons of total spin 1/2, for $\mathrm{\ell }=0$.

$S=1$		$K^0$		$K^{+}$		495MeV
		(d$\overline{\mathrm{s}}$)		(u$\overline{\mathrm{s}}$)
$S=0$	${\pi }^{-}$		$\eta ,{\eta }^{\prime },{\pi }^0$		${\pi }^{+}$
	(d$\overline{\mathrm{u}}$)				(u$\overline{\mathrm{d}}$)	136MeV
$S=-1$		$K^{-}$		$K^0$
		(s$\overline{\mathrm{u}}$)		(s$\overline{\mathrm{d}}$)		495MeV

Table 12.3: Mesons of spin 0 and $\mathrm{\ell }=0$.

The masses in MeV/c² that are quoted are approximate, for there is a small splitting between the energies in any row. In part this is due to the mass difference between the $u$ and $d$ quark masses, in part it is due to a spin-spin interaction. For example Griffiths gives the masses to be:

• $p=939.280\mathrm{MeV}/{\mathrm{c}}^2$

• $n=939.573\mathrm{MeV}/{\mathrm{c}}^2$

• ${\mathrm{\Sigma }}^{+}=1189.36\mathrm{MeV}/{\mathrm{c}}^2$

• ${\mathrm{\Sigma }}^0=1192.5\mathrm{MeV}/{\mathrm{c}}^2$

• ${\mathrm{\Sigma }}^{-}=1197.34\mathrm{MeV}/{\mathrm{c}}^2$

• ${\mathrm{\Xi }}^0=1314.9\mathrm{MeV}/{\mathrm{c}}^2$

• ${\mathrm{\Xi }}^{-}=1321.3\mathrm{MeV}/{\mathrm{c}}^2$

12.6 Mesons of spin 0 and $\mathrm{\ell }=0$

For simplicity, let’s again consider mesons made of $u$, $d$ and $s$ quarks only, together with their antiparticles. These consist of one quark and one antiquark, where the antiquark is denoted with a bar over the top. The quark anti-quark pair are bound together by the strong force with exchange of gluons. They are similar to the bound state of an electron and an anti-electron, making positronium. But the quarks cannot be pulled apart. We can have $s=0$ (called pseudo-scalar) and $s=1$ (called vector) mesons. The latter is an excited state of the former.

In Fig. 12.3 we give the particles with $s=0$, which are arranged in a hexagon.

• Note that the $\eta$, the ${\eta }^{\prime }$ and the ${\pi }^0$ are different linear combinations of the states $u\overline{u}$, $d\overline{d}$ and $s\overline{s}$, so we cannot identify them as simply as the rest. These three particles have different masses: the ${\pi }^0(=u\overline{u}-d\overline{d})$ has a mass of 134.96 MeV/c², the $\eta (=u\overline{u}+d\overline{d}-2s\overline{s})$ a mass of 548 MeV/c² and the ${\eta }^{\prime }(=u\overline{u}+d\overline{d}+s\overline{s})$ a mass of 958 MeV/c².

• There are excited states of mesons with $\mathrm{\ell }=1,2,3$ and so on with a set of spectroscopic notation that would gladden any spectroscopist’s eye. In all over 200 excited states of baryons and mesons have been seen.

There are so many hadrons known today that no-one can remember all their names, let alone their properties. We don’t expect students to remember all the atomic energy levels, let alone give them names. Similarly it is sufficient to be able to work out the spin and other quantum numbers by coupling the quarks together. This turns out to describe the pattern of hadrons precisely and provides compelling evidence for the quark model of hadrons.