Combining quarks into baryons

The combinatorics in this article are the daily language of hadron spectroscopy. Every entry in the Particle Data Group's baryon tables is labelled by an isospin multiplet, and the near-degeneracy of its members remains one of the cleanest tests that $m_u\approx m_d$. Lattice QCD collaborations now compute the tiny isospin-breaking splittings — the $1.29\ \mathrm{MeV}$ neutron–proton mass difference, resolved into quark-mass and electromagnetic parts — from first principles, turning what Heisenberg treated as an exact symmetry into a precision measurement of $m_d-m_u$. The very same $2\otimes2\otimes2=4\oplus2\oplus2$ counting, promoted to $SU(3)$ and then spin-flavour $SU(6)$, underlies quark-model predictions of baryon magnetic moments and the search for exotic pentaquark and dibaryon states. And the $\Delta\to N\pi$ ratios derived here still help disentangle overlapping resonances in pion-photoproduction experiments at JLab and MAMI, where isospin selection rules separate signal from background.

Picture the two light quarks as the two orientations of a single abstract arrow. The strong interaction does not care which way the arrow points — rotate $u$ into $d$ and every strong observable is unchanged — so $u$ and $d$ sit at the top and bottom of one isospin doublet, at $I_3=+\tfrac12$ and $I_3=-\tfrac12$. Everything that follows is the bookkeeping of stacking these arrows two and three at a time. When you combine two, the extreme states $uu$ and $dd$ have $I_3=\pm1$ and can only belong to a symmetric isospin-1 triplet; the ladder operator fills in the symmetric $I_3=0$ member, and the one leftover orthogonal combination is the antisymmetric isospin-0 singlet. That is the whole content of $2\otimes2=3\oplus1$, and the first figure shows it as a weight diagram.

Add a third quark and the eight product states $uuu,uud,udu,udd,duu,dud,ddu,ddd$ rearrange into $4\oplus2\oplus2$. The extreme states $uuu$ ($I_3=+\tfrac32$) and $ddd$ ($I_3=-\tfrac32$) can only live in a fully symmetric isospin-$\tfrac32$ quadruplet — and that quadruplet is not an abstraction: it is precisely the four members of the $\Delta(1232)$ resonance, $\Delta^{++}(uuu)$, $\Delta^{+}(uud)$, $\Delta^{0}(udd)$, $\Delta^{-}(ddd)$, degenerate at $1232\ \mathrm{MeV}$ up to tiny isospin-breaking splittings. The remaining four states pair into two mixed-symmetry isospin-$\tfrac12$ doublets. The second figure is the payoff: the quadruplet as a weight ladder, carrying the real mass and the isospin prediction that the $\Delta^{+}$ decays to $p\pi^{0}$ twice as often as to $n\pi^{+}$.

The starting point is the assignment that makes isospin work: the two light quarks are the two states of a single $SU(2)$ doublet, and the strong Hamiltonian commutes with the isospin operators, so $I$ and $I_3$ are good quantum numbers.

where: $I$ — total isospin, the length of the isospin vector; $I_3$ — its third component, tied to electric charge and quark content; $\hat I^{2}$ — the Casimir operator with eigenvalue $I(I+1)$; the $u,d$ doublet has $I=\tfrac12$. Reading: the algebra is a carbon copy of ordinary angular momentum, with $u\leftrightarrow{\uparrow}$ and $d\leftrightarrow{\downarrow}$, so every result you know for spin transfers verbatim. Combining isospins uses the ladder operators. $\hat I_+$ turns a $d$ into a $u$ (raising $I_3$ by one), $\hat I_-$ does the reverse, and on a many-quark state they act as a sum over the constituents, which is what makes "adding isospin" mechanical.

where: $\hat I_-^{(i)}$ — the lowering operator acting on the $i$-th quark only; the square-root factor is the standard angular-momentum normalization. Reading: to build a multiplet you place the top state by inspection, then apply $\hat I_-$ repeatedly; the coefficients come out automatically, and whatever states are left over are fixed by orthogonality. For two quarks the four product states split into a symmetric triplet and an antisymmetric singlet — the isospin content of the light diquark and, with $u,d$ replaced by $\uparrow,\downarrow$, of any two spin-$\tfrac12$ particles.

where: the triplet carries $I_3=+1,0,-1$ and is even under $1\leftrightarrow2$ exchange; the singlet carries $I_3=0$ and is odd. Reading: the two $I_3=0$ states are not $ud$ and $du$ separately — only their symmetric and antisymmetric combinations have definite isospin. Adding the third quark builds the eight-state decomposition that this whole article is about: one fully symmetric quadruplet and two mixed-symmetry doublets.

where: the subscripts $S,A$ label the behaviour under interchange of quarks 1 and 2 (symmetric, antisymmetric); neither doublet has any definite symmetry under $1\leftrightarrow3$ or $2\leftrightarrow3$. Reading: $8=4+2+2$, and only the quadruplet is totally symmetric — a fact that will decide which spin states pair with which flavour states when real baryons are built.

1 · The Δ⁺ branching ratio (fully worked)

Using isospin alone, predict the ratio of decay rates $\Gamma(\Delta^{+}\to p\pi^{0})$ to $\Gamma(\Delta^{+}\to n\pi^{+})$, and compare with the other members of the $\Delta$ multiplet.

Step 1. Fix the isospin labels. The decaying $\Delta^{+}$ is the $|I,I_3\rangle=|\tfrac32,+\tfrac12\rangle$ member. In the final state the pion carries $I=1$ and the nucleon $I=\tfrac12$, so the allowed total isospin is $\tfrac12$ or $\tfrac32$; conservation of isospin in the strong decay selects the $|\tfrac32,+\tfrac12\rangle$ piece.

Step 2. Read the two relevant Clebsch–Gordan coefficients from the $1\otimes\tfrac12$ table: $\langle1,0;\tfrac12,+\tfrac12|\tfrac32,+\tfrac12\rangle=\sqrt{\tfrac23}$ couples to $\pi^0 p$, and $\langle1,+1;\tfrac12,-\tfrac12|\tfrac32,+\tfrac12\rangle=\sqrt{\tfrac13}$ couples to $\pi^+ n$.

Step 3. Square and take the ratio, treating the phase space as common to both channels: $\dfrac{\Gamma(\Delta^{+}\to p\pi^{0})}{\Gamma(\Delta^{+}\to n\pi^{+})}=\dfrac{2/3}{1/3}=\mathbf{2}$. So the $\Delta^{+}$ decays to $p\pi^{0}$ twice as often as to $n\pi^{+}$.

Step 4. Cross-check against the whole multiplet. The same coefficients give the full pattern $\Gamma(\Delta^{++}\to p\pi^{+}):\Gamma(\Delta^{+}\to p\pi^{0}):\Gamma(\Delta^{+}\to n\pi^{+})=3:2:1$, so the fully charged $\Delta^{++}\to p\pi^{+}$ (a single channel, coefficient 1) is the strongest. Result: $\Gamma(\Delta^{+}\to p\pi^{0})=2\,\Gamma(\Delta^{+}\to n\pi^{+})$, with the $\Delta^{++}$ decaying at the top relative rate of 3 — all from $SU(2)$, no dynamics.

2 · Orthonormality and symmetry of the three-quark states

Take the three $I_3=+\tfrac12$ states in the basis ordered $(uud,udu,duu)$ and verify they form an orthonormal set with the advertised exchange symmetries.

Step 1. Write the coefficient vectors: quadruplet $\phi_{\rm q}=\tfrac1{\sqrt3}(1,1,1)$, mixed-symmetric $\phi_S=\tfrac1{\sqrt6}(2,-1,-1)$, mixed-antisymmetric $\phi_A=\tfrac1{\sqrt2}(0,1,-1)$.

Step 2. Check norms. $|\phi_{\rm q}|^2=\tfrac13(1+1+1)=1$; $|\phi_S|^2=\tfrac16(4+1+1)=1$; $|\phi_A|^2=\tfrac12(0+1+1)=1$. All normalized.

Step 3. Check the overlaps. $\phi_{\rm q}\cdot\phi_S\propto(2-1-1)=0$; $\phi_{\rm q}\cdot\phi_A\propto(0+1-1)=0$; $\phi_S\cdot\phi_A\propto(0-1+1)=0$. All three are mutually orthogonal, confirming an orthonormal triad.

Step 4. Check symmetry. Under $1\leftrightarrow2$ the basis maps $uud\to uud$, $udu\leftrightarrow duu$: $\phi_S\to\tfrac1{\sqrt6}(2,-1,-1)=+\phi_S$ (symmetric), $\phi_A\to\tfrac1{\sqrt2}(0,-1,1)=-\phi_A$ (antisymmetric). Under $1\leftrightarrow3$ ($uud\leftrightarrow duu$, $udu\to udu$), $\phi_A\to\tfrac1{\sqrt2}(-1,1,0)$, which is neither $\pm\phi_A$ — it is a mixture of $\phi_S$ and $\phi_A$. Result: the three states are orthonormal; $\phi_S,\phi_A$ have definite symmetry only under $1\leftrightarrow2$, the hallmark of mixed symmetry.

Isospin is one of those ideas that arrived a full generation before the objects it describes were understood. Its author reached for it to explain the nucleus, long before anyone had heard of a quark.

Texts. Thomson, Modern Particle Physics, §9.3 (combining quarks into baryons) and §9.3.1 (three-quark spin states), with Problem 9.3 on the $\Delta$ decay ratios · Griffiths, Introduction to Elementary Particles, 2nd ed., §5.8–5.9 (light quark multiplets and the quark model) · Halzen & Martin, Quarks and Leptons, Ch. 2 (symmetries and quarks) · Perkins, Introduction to High Energy Physics, Ch. 4 (isospin and $SU(2)$). For the group-theory backbone, Georgi, Lie Algebras in Particle Physics, Ch. 1–3. The isospin $SU(2)$ algebra and ladder-operator machinery are prerequisites assumed here. The step that assembles the mixed-symmetry flavour doublets $\phi_S,\phi_A$ with the spin doublets $\chi_S,\chi_A$ into a totally symmetric baryon wavefunction — folding in the antisymmetric colour singlet — is treated in the following article on baryon wavefunctions. The generalization $SU(2)\to SU(3)$ that turns the doublet into the $(u,d,s)$ triplet and the multiplets into the octet and decuplet is developed in the articles on $SU(3)$ flavour and meson multiplets. Three key results underpin this article: (1) $\hat I_-(uu)=ud+du$ with normalization $\sqrt2$; (2) the orthonormality of $\{\phi_{\rm q},\phi_S,\phi_A\}$ at $I_3=+\tfrac12$; (3) the $1\otimes\tfrac12$ Clebsch–Gordan coefficients $\sqrt{2/3},\sqrt{1/3}$ giving the $2:1$ decay ratio.

Prerequisites. Isospin · Symmetries in Quantum Mechanics · Non-Relativistic QM Review. Leads to. Baryon Wavefunctions · SU(3) Flavour · Meson Multiplets.

Ask the tutor: Ask me to walk through the ladder-operator construction of any multiplet, work out an isospin decay ratio, or show how $\phi_S$ and $\phi_A$ combine with the spin states into a symmetric proton.