Why does the $\Lambda^0$ exist?

Question

In looking at the particle charts I see 3 Sigma baryons with very similar masses and other similar properties, which makes sense since they all have one strange quark plus two others from the (up,down) group. As I understand things, up and down have similar small masses but differ in charge, so if you take a strange quark and two of the (up,down) group you can get 3 different particles similar in properties except for the charges from up/down - uus, uds, dds. So far so good.

However, there is also a Lambda-0, which is also uds, like the sigma-0. As far as I can tell, it is exactly the same as the sigma-0 (spin etc.) except it is less massive. (in fact sigma-0 decays to it). So what is going on here? What are the quarks doing differently, and what prevents them from doing the same to form a lambda-plus? (I assume the Fermi stats prevent two up quarks in such a thing, but if so, why can the sigma-plus exist?) What changes when the sigma-0 decays to lambda-0? Trying to find the answer results in adding and subtracting the quarks and multiplying by the square root of two - huh?

Possibly related question in reverse - Take a quark and an antiquark from the (up,down) group and you get three pions, plus, minus and 0. Plus and minus are understandable, u-dbar and d-ubar. But why not two pi-0 mesons? U-ubar and d-dbar? Again they are adding quarks and multiplying by the square root of 2 and voila, a pi-0! Huh again?

Answer 1

Indeed, as indicated, @Qmechanic answered your questions pre-emptively, but you might be unfamiliar with the language and fail to understand the powerful logic of the respective quark wave functions of the Λ and the Σ⁰ upon inspection. The SU(6) wave functions, already implicit to Zweig and Gell-Mann when they put the quark model together, should be in your text, or else Fayyazuddin & Riazuddin, etc.

Inspect them: $$ |\Lambda \uparrow\rangle = \frac{1}{\sqrt{12}} \Bigl ( |( u\downarrow d\uparrow -d\downarrow u\uparrow +d\uparrow u\downarrow -u\uparrow d\downarrow ) ~ s\uparrow\rangle \\ + | u\downarrow s\uparrow d\uparrow -d\downarrow s\uparrow u\uparrow +d\uparrow s\uparrow u\downarrow -u\uparrow s\uparrow d\downarrow \rangle \\ + |s\uparrow ( u\downarrow d\uparrow -d\downarrow u\uparrow +d\uparrow u\downarrow -u\uparrow d\downarrow)\rangle \Bigr ),\\ |\Sigma^0 \uparrow\rangle = \frac{1}{6} \Bigl ( 2|( u\uparrow d\uparrow +d\uparrow u\uparrow)~ s\downarrow\rangle +2|s\downarrow ( u\uparrow d\uparrow +d\uparrow u\uparrow) \rangle \\ -|( u\downarrow d\uparrow +d\downarrow u\uparrow +d\uparrow u\downarrow +u\uparrow d\downarrow ) ~ s\uparrow\rangle -|s\uparrow ( u\downarrow d\uparrow +d\downarrow u\uparrow +d\uparrow u\downarrow +u\uparrow d\downarrow ) \rangle \\ +2 | u\uparrow s\downarrow d\uparrow +d\uparrow s\downarrow u\uparrow \rangle - | u\downarrow s\uparrow d\uparrow +d\downarrow s\uparrow u\uparrow +d\uparrow s\uparrow u\downarrow +u\uparrow s\uparrow d\downarrow \rangle \Bigr ). $$

I have grouped the pieces of the wave functions together so the spin-isospin symmetry of the u,d diquark is more apparent. Recall, these are anticommuting fermions, but since the implicit color labels fully antisymmetrize them, the space-flavor-spin part you see is symmetric.

The order of the sequencing denotes the space-wavefunction--confirm its complete symmetry; and so your question focuses on the isospin-spin pieces.

Note that, in the isoscalar Λ, the net spin (up) is carried by the s, while the light diquark is aways in an isosinglet-spin-singlet state, $(ud-du)(\uparrow \downarrow-\downarrow \uparrow)$, so jointly symmetric, despite the antisymmetry of each factor.

By contrast, in the isovector Σ⁰, the isotriplet-spin-triplet combinations $(ud+du)\uparrow\uparrow$, and $(ud+du)(\uparrow \downarrow+\downarrow \uparrow)$ are both individually and hence jointly symmetric. You may recall this in reassuring yourself of the mutual orthogonality of the Λ and the Σ⁰, which it is highly advisable to check.