Baryon singlet representation for SU(3) flavour symmetry

[Pietjuh]

Hi there!

As most people already might know, we can decompose the 27 dimensional representation for the baryons under SU(3) flavour symmetry as 27 = 10 + 8 + 8 + 1. I can find a lot of information about the particles that lie in the decuplet and in the octet, but nothing about which particle is associated to the singlet representation. Can anyone give me some information about this? :)

Thanks in advance!

 

[hamster143]

the singlet particle is uds (lambda) and it does not exist in the ground state (angular momentum zero) due to Fermi statistics.

 

[Vanadium 50]

Hamster is right that the SU(3) baryon flavor singlet is an isospin 0 uds combination. He's also right that it doesn't exist: however, it still possesses (or would possess) angular momentum of 1/2, because the three quarks each have spin 1/2. I think he means there is no orbital angular momentum.

There is an excited SU(3) singlet. It must have T=0, J=3/2, and a quick look in the PDG suggests to me that the best candidate is the Lambda(1890).

 

[Meir Achuz]

There is a uds spin 1/2 state that is the Lambda particle of the baryon octet.
It does exist in the octet ground state.

 

[petergreat]

Hamster is right that the SU(3) baryon flavor singlet is an isospin 0 uds combination. He's also right that it doesn't exist

Could you explain to me why it doesn't exist? I've also been wondering about this question.

 

[Vanadium 50]

It's a fermion so it's total wave function has to be antisymmetric. If I make it symmetric in color, under SU(3) and spatially, it's total wave function isn't antisymmetric.

 

[Meir Achuz]

Could you explain to me why it doesn't exist? I've also been wondering about this question.

The spin addition 1/2+1/2+1/2=1/2 leads to a spin state of mixed symmetry (not completely symmetric or antisymmetric). This means the flavor state must also be mixed, which requires a flavor octet, and not a singlet.