---
title: Baryon Multiplets, Spin, and the Color Puzzle
module: The Quark Model
moduleNumber: 4
lessonNumber: 3
order: 403
summary: >
  Baryons as three-quark states, with a wavefunction factored into space, spin,
  flavor, and color. The spin-3/2 Δ⁺⁺ = uuu forces a totally symmetric state that
  the Pauli principle forbids, and the resolution is an antisymmetric color factor —
  the first evidence for color. The octet and decuplet spin content, and baryon
  magnetic moments as a quantitative test of the model.
topics: [The Quark Model]
draft: false
sources:
  - book: Griffiths
    ref: "Ch. 5 — Bound States, §5.7–5.8 Baryons and the color argument"
  - book: Thomson
    ref: "Ch. 9 — The Quark Model, §9.5"
  - book: Tong
    ref: "The Standard Model (Cambridge Part III), §3.3.3 Baryons"
---

A baryon is a bound state of three quarks. The flavor $SU(3)$ decomposition
$\mathbf 3 \otimes \mathbf 3 \otimes \mathbf 3 = \mathbf 1 \oplus \mathbf 8 \oplus
\mathbf 8 \oplus \mathbf{10}$ from the [Eightfold-Way lesson](/particle-physics/quark-model/eightfold-way-su3)
fixes the flavor multiplets, but it leaves open which spin accompanies each. Fixing
that requires the full three-quark wavefunction and the Pauli principle, and the
answer forced a new quantum number into the theory: color.[^grif-baryons]

## The four-part wavefunction

Quarks are fermions, so the total wavefunction of a baryon must be antisymmetric
under exchange of any two of them. The wavefunction factors into four independent
pieces,

$$
\Psi = \psi_{\text{space}} \times \chi_{\text{spin}} \times \phi_{\text{flavor}}
       \times \xi_{\text{color}},
$$

and the exchange symmetry of the whole is the product of the symmetries of the
parts. For the ground-state baryons the spatial wavefunction has no orbital angular
momentum ($L = 0$) and is therefore symmetric. The requirement is then a statement
about the remaining three factors.

$$
% caption: The baryon wavefunction as a product of four exchange factors. For the
% L = 0 ground states the spatial part is symmetric, so the product of spin, flavor,
% and color must be antisymmetric to satisfy Fermi statistics.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \x/\lab/\sym in {0/space/symmetric, 2.5/spin/{}, 5/f\/lavor/{}}{
    \draw[thick] (\x-0.85,-0.5) rectangle (\x+0.85,0.6);
    \node at (\x,0.35) {\lab};
    \node[font=\scriptsize] at (\x,-0.2) {\sym};
  }
  \draw[acc, thick, fill=acc!10] (6.65,-0.5) rectangle (8.35,0.6);
  \node[acc] at (7.5,0.35) {color};
  \node[black, font=\scriptsize] at (7.5,-0.2) {antisym};
  \foreach \x in {1.65, 4.15, 6.65}
    \node[black] at (\x,0.05) {$\times$};
  \node[black, font=\scriptsize, align=center] at (3.75,-1.2)
    {total wavefunction must be antisymmetric};
\end{tikzpicture}
$$

## The Δ⁺⁺ puzzle

The sharpest case is the $\Delta^{++}$, a spin-$\tfrac32$ baryon of charge $+2$. Its
charge and spin fix it uniquely: three up quarks with all spins aligned. Examine the
three factors.

- **Flavor.** All three quarks are $u$, so $\phi_{\text{flavor}} = uuu$ is symmetric
  under any exchange.
- **Spin.** The spin-$\tfrac32$ state with $J_z = +\tfrac32$ has all three spins up,
  $\chi = {\uparrow\uparrow\uparrow}$, symmetric.
- **Space.** The ground state has $L = 0$, symmetric.

The product of these three is totally symmetric. For three identical spin-$\tfrac12$
fermions this is forbidden: the wavefunction must be antisymmetric, yet every
available factor is symmetric. Either the quark model is wrong, or a fourth,
antisymmetric factor exists.

$$
% caption: The statistics problem. The Delta-plus-plus is three up quarks with all
% spins aligned in the L = 0 ground state: symmetric in flavor, spin, and space. For
% identical fermions this totally symmetric state violates the Pauli principle,
% unless a further antisymmetric degree of freedom is present.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \x in {-1.6,0,1.6}{
    \fill (\x,0) circle (3pt);
    \draw[very thick, ->] (\x,-0.5) -- (\x,0.6);
    \node[below, font=\scriptsize] at (\x,-0.6) {u};
  }
  \node[black, font=\scriptsize] at (0,1.2) {three up quarks, spins aligned};
  \node[black, font=\scriptsize] at (0,-1.4) {symmetric in space, spin, f\/lavor};
  \node[acc, font=\scriptsize] at (0,-1.9) {Pauli principle violated};
\end{tikzpicture}
$$

## Color to the rescue

The resolution assigns each quark a further three-valued charge, **color**, taking
values labelled red, green, blue. A baryon is built as the totally antisymmetric
combination in the three color labels,

$$
\xi_{\text{color}} = \frac{1}{\sqrt 6}\,\epsilon_{abc}\,
   \left|a\right\rangle_1 \left|b\right\rangle_2 \left|c\right\rangle_3,
$$

with $\epsilon_{abc}$ the totally antisymmetric symbol on the three colors. This is
antisymmetric by construction, and it is a **color singlet** — invariant under the
color $SU(3)$. With $\xi_{\text{color}}$ antisymmetric and the space–spin–flavor
product symmetric, the total wavefunction is antisymmetric, and Fermi statistics is
satisfied. The $\Delta^{++}$ exists precisely because color exists.

$$
% caption: The antisymmetric color factor. The three quarks carry distinct color
% charges (red, green, blue) combined through the epsilon symbol into a color
% singlet. This factor is antisymmetric under exchange of any two quarks, so the
% otherwise-symmetric Delta-plus-plus wavefunction becomes antisymmetric overall.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \x/\c in {-1.6/R, 0/G, 1.6/B}{
    \draw[thick] (\x,0) circle (0.42);
    \node at (\x,0) {\c};
  }
  \node[black, font=\scriptsize] at (0,1.0) {distinct colors, antisymmetrized};
  \node[acc, font=\scriptsize] at (0,-1.1) {color singlet: total wavefunction antisymmetric};
\end{tikzpicture}
$$

The same color factor removes the fractional-charge embarrassment of the light
baryons and, applied to the meson case, explains why $q\bar q$ and $qqq$ are the
only simple combinations seen: they are the smallest color singlets. The
independent, quantitative confirmation of exactly three colors comes from the
$R$-ratio and the $\pi^0 \to \gamma\gamma$ rate, developed in the
[final lesson](/particle-physics/quark-model/color-confinement-exotics).

## Octet and decuplet spin content

With color fixed as antisymmetric, the space–spin–flavor product is symmetric, and
the allowed spins follow from how flavor and spin combine.

> **Decuplet, spin $\tfrac32$.** The decuplet flavor wavefunction is totally
> symmetric. To make space–spin–flavor symmetric with $L = 0$, the spin must also be
> totally symmetric, which for three spin-$\tfrac12$ quarks means $S = \tfrac32$. The
> ten decuplet baryons — the four $\Delta$, three $\Sigma^\ast$, two $\Xi^\ast$, and
> the $\Omega^-$ — all carry spin $\tfrac32$.

> **Octet, spin $\tfrac12$.** The octet flavor wavefunction has mixed symmetry. A
> mixed-symmetry flavor state paired with a mixed-symmetry spin state ($S = \tfrac12$)
> can be combined into a totally symmetric product. The eight octet baryons — the
> nucleons, three $\Sigma$, the $\Lambda$, and two $\Xi$ — all carry spin $\tfrac12$.

$$
% caption: Spin content of the two ground-state baryon multiplets. The flavor-
% symmetric decuplet (ten states) must carry symmetric spin, giving spin three-halves;
% the mixed-symmetry octet (eight states) carries spin one-half. The apex of the
% decuplet is the triple-strange Omega-minus.
\begin{tikzpicture}[font=\footnotesize, scale=0.95]
  \definecolor{acc}{HTML}{4A6FA5}
  % octet hexagon (left)
  \begin{scope}[shift={(-3.4,0)}]
    \draw (-1,0.9) -- (1,0.9) -- (2,0) -- (1,-0.9) -- (-1,-0.9) -- (-2,0) -- cycle;
    \foreach \x/\y in {-1/0.9, 1/0.9, -2/0, 2/0, -1/-0.9, 1/-0.9}
      \fill (\x,\y) circle (2pt);
    \draw[fill=white] (0,0) circle (2pt);
    \node[black, font=\scriptsize] at (0,-1.5) {octet: 8 states};
    \node[black, font=\scriptsize] at (0,-1.95) {spin one-half};
  \end{scope}
  % decuplet triangle (right)
  \begin{scope}[shift={(3.4,0.5)}]
    \foreach \x in {-1.5,-0.5,0.5,1.5} \fill (\x,0.6) circle (2pt);
    \foreach \x in {-1,0,1} \fill (\x,-0.2) circle (2pt);
    \foreach \x in {-0.5,0.5} \fill (\x,-1.0) circle (2pt);
    \draw[acc, very thick, fill=acc!12] (0,-1.8) circle (2.5pt);
    \node[acc, right, font=\scriptsize] at (0.2,-1.8) {sss};
    \node[black, font=\scriptsize] at (0,-2.5) {decuplet: 10 states};
    \node[black, font=\scriptsize] at (0,-2.95) {spin three-halves};
  \end{scope}
\end{tikzpicture}
$$

## Magnetic moments

The quark model makes a quantitative, parameter-light prediction for baryon
magnetic moments. Each quark is a point Dirac particle with moment
$\mu_q = q_q\hbar/(2 m_q)$ (in units where $g = 2$), pointing along its spin. A
baryon's moment is the expectation value of the sum $\sum_i \mu_i \sigma_{iz}$ in
its spin-flavor wavefunction.

The proton's spin-up state is the totally symmetric combination of two up quarks
and one down. Coupling the two like quarks to spin $1$ (they sit in the symmetric
pair) and adding the down quark to reach total $J_z = +\tfrac12$ gives, after
normalization,

$$
|p\!\uparrow\rangle = \tfrac{1}{\sqrt{18}}\Big(
   2\,u\!\uparrow u\!\uparrow d\!\downarrow
   - u\!\uparrow u\!\downarrow d\!\uparrow
   - u\!\downarrow u\!\uparrow d\!\uparrow
   + \text{permutations}\Big).
$$

Taking $\langle \sum_i \mu_i \sigma_{iz}\rangle$ in this state, the up quarks
contribute with net weight $+\tfrac43$ and the down with $-\tfrac13$, so the proton
(uud) moment is

$$
\mu_p = \frac{1}{3}\left(4\mu_u - \mu_d\right),
\qquad
\mu_n = \frac{1}{3}\left(4\mu_d - \mu_u\right),
$$

and, since $\mu_u/\mu_d = q_u/q_d = -2$ when $m_u \approx m_d$, the ratio is
predicted with no free parameters:

$$
\frac{\mu_n}{\mu_p} = \frac{4\mu_d - \mu_u}{4\mu_u - \mu_d}
   = \frac{4\mu_d + 2\mu_d}{-8\mu_d - \mu_d} = \frac{6\mu_d}{-9\mu_d} = -\frac{2}{3}.
$$

The measured ratio is $-1.913/2.793 = -0.685$, within a few percent of
$-\tfrac23$. Fixing the two light-quark moments from $\mu_p$ and the strange moment
from $\mu_\Lambda = \mu_s$, the model predicts the entire octet.[^grif-moments]

$$
% caption: Predicted versus measured magnetic moments for the spin-one-half baryon
% octet, in nuclear magnetons. Bars above the axis are positive moments, below are
% negative (the sign is shown by direction, magnitudes on the vertical scale). The
% proton and lambda are used to fix the quark moments; the rest are predictions,
% agreeing with data to about ten percent.
\begin{tikzpicture}[font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-0.4,0) -- (9.4,0);
  \draw[->, black] (0,-2.4) -- (0,3.4) node[above, font=\scriptsize] {moment magnitude};
  \foreach \y in {1,2} {
    \draw[black] (-0.1,\y) -- (0.1,\y) node[left=2pt, black, font=\scriptsize] {\y};
    \draw[black] (-0.1,-\y) -- (0.1,-\y) node[left=2pt, black, font=\scriptsize] {\y};
  }
  % pairs: x-center, predicted, measured (signed height), label-y
  % p: +2.79 / +2.79 ; n: -1.86 / -1.91 ; Lambda: -0.58/-0.61 ; Sigma+: +2.68/+2.42 ; Xi-: -0.47/-0.65
  \foreach \x/\pp/\mm/\ly/\lab in {1/2.79/2.79/-0.3/p, 3/-1.86/-1.91/0.3/n, 5/-0.58/-0.61/0.3/{Lambda}, 7/2.68/2.42/-0.3/{Sigma}, 9/-0.47/-0.65/0.3/{Xi}}{
    \draw[acc, fill=acc!18] (\x-0.42,0) rectangle (\x-0.04,\pp);
    \draw[black, fill=black!22] (\x+0.04,0) rectangle (\x+0.42,\mm);
    \node[black, font=\scriptsize] at (\x,\ly) {\lab};
  }
  \draw[acc, fill=acc!18] (6.6,-2.9) rectangle (6.9,-2.6);
  \node[black, right, font=\scriptsize] at (7.0,-2.75) {predicted};
  \draw[black, fill=black!22] (6.6,-3.3) rectangle (6.9,-3.0);
  \node[black, right, font=\scriptsize] at (7.0,-3.15) {measured};
\end{tikzpicture}
$$

The agreement is the model's most direct quantitative success at the level of
individual particles. It also carries a physical message: the moments come out
right only if the quarks behave as nearly free spin-$\tfrac12$ Dirac particles
inside the baryon, with effective (constituent) masses of a few hundred MeV. The
constituent mass is larger than the current-quark mass that enters the QCD
Lagrangian because it includes the energy of the surrounding gluon and
quark–antiquark field, a distinction developed in the QCD module. Where the model
misses — the $\Sigma^+$ prediction is high by about ten percent — the discrepancy
marks the limits of treating the baryon as three static constituent quarks.[^tong-baryonmass]

[^grif-baryons]: Griffiths, _Introduction to Elementary Particles_, §5.7–5.8, gives the three-quark wavefunction, the $\Delta^{++}$ statistics argument, and the color resolution. The spin assignments of the octet and decuplet follow the exchange-symmetry analysis there.

[^grif-moments]: Griffiths, §5.8, derives $\mu_p = (4\mu_u - \mu_d)/3$ and the octet magnetic moments; predicted and measured values are tabulated there, using constituent masses fit to $\mu_p$. Measured moments from the Particle Data Group, [pdg.lbl.gov](https://pdg.lbl.gov).

[^tong-baryonmass]: Tong, _The Standard Model_ (Cambridge Part III), §3.3.3, stresses that the baryon mass is set by the QCD scale $\Lambda_{\text{QCD}}$ rather than the bare quark masses, and that the constituent quark carries the energy of the surrounding strongly interacting fields.