# $\Lambda^o \Sigma^o$ mass difference

1. Aug 27, 2011

### fermi

I have a couple of questions about $\Lambda^o \Sigma^o$ mass difference:

(1) Does anyone know a reference to a publication (excluding publications with lattice calculations) where they calculate this difference? (Please don't get me wrong, lattice calculations are perfectly good way of computing, but that is not what I look for at the moment. I want something that will not only get the right numerical value, but also give me some intuitive idea why the numerical value is what it is.)

(2) Is there an intuitive explanation for the fact that there is any difference at all? $\Lambda^o$ and $\Sigma^o$ have the same quark content: uds. The observation that $\Sigma^o$ is part of an isospin triplett while $\Lambda^o$ is a singlet does not explain it. Why should the quantum state (ud+du)s be more massive than (ud-du)s? Since we are firmly convinced that the quarks are held together by gluons, and the gluons are completely flavor blind, coming up with an intuitive explanation for different uds masses must be difficult. (Also keep in mind that both Hyperons have the same ${J}^{PC}$ quantum numbers.) Even so, there must be one. Can anyone tell me what it is?

2. Aug 27, 2011

### clem

The two nucleon quarks in the Sigma are in the spin 1 state and in the spin 0 state for the Lambda. It is this spin difference that makes them different.
The spin.spin force between two quarks is attractive in the spin 0 state and about three times larger and repulsive in the spin one state. It just comes from adding two spin 1/2.
This spin difference is what makes the Delta heavier than the nucleon.
There are many quark potential models that correlate these forces and get the Sigma-Lambda difference, as well as other hadron mass differences.

3. Aug 27, 2011

### clem

Incidentally, Fermi did not know about quarks.

4. Aug 27, 2011

Staff Emeritus
Why not? It explains why the deuteron is lighter than it's isotriplet partner (which isn't even bound).

5. Aug 27, 2011

### fermi

The force that binds nucleons is different. Even though it too is based on gluons (so we assume) it is more akin to the Van der Waals forces in atoms and molecules: short ranged, and much weaker than the corresponding first order force. The Deutoron binding energy is only a couple of MeVs, compared to the nucleon binding energy of 300Mev per quark roughly. One is a first order effect, the other is a third order effect. If the mass splitting of $\Sigma - \Lambda$ was in the order of a few Mev, this would be a different issue. But 75 Mev difference is a first order effect.

But that's not all. The presence of the strange quark re-symmetrizes the problem. In Deutoron, the two nucleons are in I=0, L=0, and s=1 state to make the Fermion wave function antisymmetric. In $\Lambda$ too, the u and d quarks are in I = 0 and s=1 state, and the s quark spin is opposite (so that ${J}^{P}$ comes out at $(1/2)^{+}$. ) (The analogy actually breaks down a bit here: the two nucleons in Deutoron are very non-relativistic, and they are well described by the Schrodinger wave function and the quantum numbers I, L, and S. On the other hand, the u and d quarks inside a baryon are relativistic and L, and S are not good Quantum numbers, only the total Angular Momentum J is.) Anyway, in $\Sigma$ the ud quarks are in I=1 and s=0 state (the same as neutron). But here comes the s quark. It needs to be in s=1 state either with u or d quark (but not both.) So, whatever the u and d quarks are doing in $\Lambda$, the u and s (or d and s) quarks are doing the same thing in $\Sigma$. The first order gluon effects should not distinguish between d and s.

6. Aug 27, 2011

### fermi

Ahh well, we all have to learn new tricks, don't we? Getting old is no excuse for not doing so.

7. Aug 28, 2011

### Hans de Vries

David Griffits' "Introduction to Elementary Particles" has a section on the
Baryon masses using elementary spin-spin coupling. ( Chapter 5.10 )

Regards, Hans

8. Aug 28, 2011

### fermi

Thank you for the reference, I will check it out.