# Why do we have only 4 SU(2) total spin and isospin 3/2 baryons?

In the context of SU(2), we find that for a quark triplet $qqq$ there are 4 wavefunctions of total Isospin $3/2$, ranging for $I_3$ going from $-3/2$ to $3/2$. Furthermore, these wavefunctions are completely symmetric with respect to the exchange of any two quarks.

Since spins also obey the same laws, we also have 4 totally symmetric wavefunctions for the addition of three $1/2$ spins.

Now, knowing the total wavefunction of a bound $qqq$ state must be totally antisymmetric, we find that for $L=0$ ground state baryons, $\phi_{flavour}\chi_{spin}$ must be totally symmetric.

Knowing that, it seems to me we would be able to find 16 different combinations that are totally symmetric. My book (Modern Particle Physics, Thomson) however seems to tell me otherwise, implying there are only 4 such baryons ($\Delta^-,\Delta^0,\Delta^+,\Delta^{++}$). Is this there a reason or there indeed exists 12 more baryons of total isospin and spin 3/2 ?

Also, is there a systematic way of combining mixed symmetry wavefunctions to yield total symmetric ones ? Again, in the Thomson, he just happens to see that some combination of mixed symmetry wavefunctions yields a totally symmetric one.

• ? Each Delta has 4 spin states: 4x4=16.... – Cosmas Zachos Dec 22 '16 at 22:52
• Oh right, it was only that... – Frotaur Dec 23 '16 at 2:48