# Why does the $\Lambda^0$ exist?

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?

• Possible duplicate: physics.stackexchange.com/q/248661/2451 – Qmechanic Oct 3 '17 at 22:08
• Isospin seems to be a fancy way of stating the number of different states produced by swapping up and down quarks for each other, so lambda with isospin 0 (only one combo) and sigma w isospin 1 (3 combos) really tells me nothing about what's going on. – Madman Oct 3 '17 at 22:58

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.

• I don't understand all the math and notations in Cosmas' response, but he seems to hit on something I thought of last night. The strange quark is rather different from up and down and this breaks symmetry that up and down seem to have with each other. I was thinking, what if in the lambda, the up and down have the same spin with the strange spin opposite. Because of Fermi rules, the only way that is possible is if the two non-strange quarks are not the same - an up and a down. In the Sigmas, up and down have antiparallel spin, so they both can be the same, allowing 3 combinations. Am I close? – Madman Oct 5 '17 at 19:36
• These are product wavefunctions of the hadrons with all the allowable symmetries. Well, you observe that 4 of the 18 terms in the Σ do have the spin of the s opposite to that of the Σ. Once you appreciate the logic of the symmetries, the answer is pegged, like sudoku. If the s were degenerate with d and the u much heavier, you'd have completely different states, a U-spin triplet and a U-spin singlet, with very different, but analogous, wave functions. – Cosmas Zachos Oct 5 '17 at 22:56
• "If the s were degenerate with d and the u much heavier, you'd have completely different states, a U-spin triplet and a U-spin singlet, with very different, but analogous, wave functions" – Madman Oct 9 '17 at 14:49
• "If the s were degenerate with d and the u much heavier, you'd have completely different states, a U-spin triplet and a U-spin singlet, with very different, but analogous, wave functions" If that were the case, we'd have the neutron, the sigma-0 and the xi-0 all with similar masses to each other and probably considered a family, correct? Also, I don't think you intended this, but if d and s were very similar, all particles in this family would have the same charge,would the family pretty much degenerate into a single particle? I don't quite get the concept of particle states combining into 1. – Madman Oct 9 '17 at 14:56
• No, as you see from the entire wave functions, above, not the mere daft quark content, you'd have 4 neutral states: the n, Ξ 0, and two orthogonal uds, states, but not the Σ and Λ, since now you'd have to rotate those so as to consist of a U-triplet central state and a U-singlet. See WP. I think you could master quark wave functions from a good book. – Cosmas Zachos Oct 9 '17 at 16:11