# Why there are no $uuu$ and $ddd$ baryons with spin 1/2?

What is preventing $Δ^{++}$ and $Δ^-$ spin 3/2 baryons from going to a lower-energy state with spin 1/2 similar to that of protons and neutrons? I don't think the Pauli exclusion principle can prevent it because the quarks have different colors. The whole purpose of the quark color is to allow more than one quark to be in the same state. What's so special about protons and neutrons? What allows them to have lower energy compared to $Δ^+$ and $Δ^0$?

• Seems there are $\Delta$ resonances (of quark content "uuu" as well we "ddd") which have spin $J = 1/2$; including $\Delta(1620)1/2^-$, $\Delta(1750)1/2^+$, $\Delta(1900)1/2^-$ ... They just happen to be more massive than $\Delta(1232)3/2^+$. – user12262 Apr 25 '14 at 5:25
• @user12262: Better to use the PDG summary tables than PDGLive for this purpose; two of your particles haven't been seen for a while. – rob Apr 25 '14 at 6:14
• It looks like there is some reason why baryons and light nuclei with lower isospin are more stable. The Pauli exclusion principle alone cannot explain it, at least for baryons. Maybe two quarks can have the same color temporarily during gluon exchange, and then they would have to be in different energetic states if they have the same spin and isospin. That configuration would require more energy. – proski Apr 25 '14 at 14:37
• @proski, I'd put it the other way. Nature's preference for low isospin is the reason that we have two stable nucleons, and stable isotopes with mostly $N\approx Z$. – rob Apr 25 '14 at 15:21
• @rob: "two of your particles haven't been seen for a while." -- Touché. (Meanwhile I'm pondering whether to try and ask some more incisive question about "isospin"; or whether I should read more articles by Christian Wiesendanger.) "Better to use the PDG summary tables than PDGLive for this purpose" -- We may come to appreciate PDGLive for this purpose if and when if lives more up to its name and provides minute-by-minute live updated charts of "status" of particles; and not only those lame static stars ... – user12262 Apr 26 '14 at 6:43

You are correct to point out that there's no symmetry that forbids a state with isospin 3/2 and spin 1/2; in the nomenclature, this is also called a $\Delta$ resonance. The Particle Data Group lists two such particles, with mass 1620 MeV and 1910 MeV. They exist, but they are heavier than the spin-3/2 $\Delta$ at 1232 MeV.

The reason why is isospin, although the exclusion principle is involved.

From the standpoint of the strong nuclear interaction, you can sometimes treat the proton and the neutron as two states of the same particle, the "nucleon." In quantum mechanics, a system with two internally available states usually tends to follow the same mathematical rules as a spinor with angular momentum ℏ/2; this is the case for the nucleon. So the strong interaction operator that distinguishes between is a "rotation" in "isotope space," or isospin.

Isospin is a good quantum number for the ground states and excited states of many light nuclei. In heavy nuclei, where the energy due to electrostatic repulsion starts to compete with the nuclear binding energy, the symmetry between proton and neutron is broken and you can't assign a definite isospin to a particular state.

In isotope space the pion is a three-state triplet, obeying the same algebra as a spin-one system in angular momentum space. You can think of the $\pi^+$ and $\pi^-$ as the isotopic raising and lowering operators on the proton and the neutron.

Similarly, a $\Delta$ is a strongly-interacting particle with total isospin 3/2. The $\Delta$ has four projections onto the charge axis, corresponding to the four charge states: $\Delta^{++}, \Delta^+, \Delta^0, \Delta^-$. Historically I believe the existence of the $\Delta^{++}$ with spin 3/2 was a lynchpin in the argument for the existence of quark color. The $\Delta^{++}(1232)$ has spin 3/2, so its spin wavefunction is symmetric under exchange; its isospin wavefunction, for the same reason, is symmetric under exchange; therefore there must be another degree of freedom with three states so that the quark wavefunction can be antisymmetric.

So why is a spin-1/2 $\Delta$ heavier than the lightest spin-3/2 $\Delta$? You can compare with the case of the deuteron. Nucleons don't have the color degree of freedom, so exchange symmetry — the exclusion principle — requires that a two-nucleon system with spin 0 must have isospin 1, and vice-versa. Isospin symmetry tells us that a proton-neutron pair with spin 0 should have roughly the same energy as a diproton or a dineutron. Since neither of those systems is bound, we expect to find the deuteron with isospin 0 and spin 1. Which it has. Apparently, in baryons and light nuclei, total isospin contributes more to the total energy of a system than does total angular momentum.

## Correction

$\Delta(1620) 1/2^-$ is actually pretty well settled. (Thanks to rob.)

Actually, the Pauli exclusion principle can explain why there are no (uuu,ddd,sss) spin-1/2 ground states.

In baryons, quarks have four degree of freedom: orbital, spin, flavor, color. As you already know, the quarks' total wave functions should be anti-symmetric.

1. If we have uuu(or ddd, or sss) then the flavor part is symmetric;
2. Since we assume they are ground states,the orbital part is also symmetric;
3. For baryons(three-quark bound state), the color part is always anti-symmetric;

So we conclude that the spin part must be symmetric. It is the case for spin-3/2 states (baryon decuplet) not for spin-1/2 baryons (they have mixed symmetry). That's why uuu spin-1/2 ground state doesn't exist.

However, if the spin-1/2 uuu state is excited, then the orbital part of the wave function may also have mixed-symmetry. The orbital-spin can combine together (by group theory calculations) to be symmetric, the Pauli exclusion principle can't forbid this.

The only thing left is to find these state on experiments. This is not easy. As already pointed by user12262, there are only signs of the existence of $\Delta$ spin-1/2 states at present.

• Actually the $\Delta(1620) S_{31}$, with $I(J^P)=\frac32(\frac12^-)$, is pretty well-established. I'm surprised by your statement that the color part of the wavefunction must always be antisymmetric; can you elaborate? – rob Sep 28 '14 at 14:06
• @rob I only search on the PDG website; it says $\Delta(1620)$ is a four-star particle. I mean it is not absolutely settled. As for your question, if we assume that the baryons are color singlet (color neutral). The quarks color part lie in SU(3) fundamental representation, using Littlewood–Richardson rule, $3\otimes 3 \otimes 3=10\oplus 8\oplus \bar{8}\oplus 1$, we can find that only the last one have dimension 1(color neutral) and it is antisymmetric. In other words, for baryons, color neutral= color siglet = antisymmetric. – luyuwuli Oct 10 '14 at 3:31
• Hmmm, the proton and neutron are also four-star particles. Your comment about SU(3) representation is helpful, thanks. – rob Oct 10 '14 at 4:29
• @rob Oops, I was so wrong about four-star particle. Thanks a lot. However, I'm also wondering why PDGLive doesn't mark stars for most of the confirmed mesons. – luyuwuli Oct 10 '14 at 7:06
• @Paul You can refer to the baryon summary table at PDGLive, although I don't know why mesons don't have this star notation. – luyuwuli May 31 '17 at 8:40