We considered earlier how 3 quarks, with spin (s) ½ and isospin (I) ½, can combine to make a total wavefunction (space, spin, isospin and colour parts) which is antisymmetric under the interchange of two particles, as required for fermions. For the lowest lying (l = 0) baryons, we saw that the product of the spin and isospin wavefunctions must then be symmetric, and the two solutions are the Δ, with I =  and s = , and the nucleons, with I = ½ and s = ½. There were a few simplifications in this account, so here is a more rigorous explanation. (This more detailed approach is not required for the examinations, but understanding it will help your appreciation of the role of symmetry in quark systems.)

First we will look as the combination of two objects with spin ½ and isospin ½. Since two quarks do not form a bound state, it is helpful to consider combinations of two nucleons as these have similar quantum numbers. In this case, since there is no colour part to the total wavefunction, the product of spin and isospin states must be antisymmetric. We will denote s_z = ½ by ↑ and s_z = −½ by ↓, and I₃ = ½ by p and I₃ = −½ by n.

First consider just the isospin. There are 4 combinations of two nucleons: pp, pn, np and nn. The first and last are obviously symmetric under interchange; the other 2 do not have a defined symmetry. We can produce the following combinations:
pp, ¹/_√2 (pn + np), nn, ¹/_√2 (pn − np). The first 3 are symmetric, corresponding respectively to (I, I₃) = (1, 1), (1, 0), (1, −1). The last is antisymmetric, (I, I₃) = (0, 0). Similarly the spin states are ↑↑, ¹/_√2 (↑↓ + ↓↑), ↓↓, ¹/_√2 (↑↓ − ↓↑), corresponding to (s, s_z) = (1, 1), (1, 0), (1, −1), (0, 0) respectively. (The ¹/_√2 factors are simply the normalisations.)

The allowed states are those where spin and isospin have opposite symmetry; the antisymmetric isospin state and symmetric spin states (s = 1) give the bound deuteron, while the complementary states are only allowed for free pairs of protons and neutrons. (This is discussed further in the Nuclear Physics course next semester.)

We now return to the quarks, denoting I₃ = ½ by u and I₃ = −½ by d. Since the colour part of the wavefunction is always antisymmetric, the combined isospin×spin must be symmetric. First we will consider symmetric isospin combinations of the three quarks.
uuu is (I, I₃) = (, ) so must be the Δ⁺⁺.
uud is I₃ = ½; forming combinations which are symmetric under 2-quark permutations we get ¹/_√3 (uud + udu + duu). This must be (I, I₃) = (, ½) so the Δ⁺.
Similarly ¹/_√3 (udd + dud + ddu) is (I, I₃) = (, −½) so the Δ⁰.
ddd is (I, I₃) = (, −) so the Δ⁻.

Note: the first and last state must be I = . There must therefore be symmetric states with I =  and intermediate I₃ values, and these are the only possibilities!

The spin states for the Δ must also be symmetric, and we can write these down in an analogous way:
(s, s_z) = (, ) is ↑↑↑.
(s, s_z) = (, ½) is ¹/_√3 (↑↑↓ + ↑↓↑ + ↓↑↑).
(s, s_z) = (, −½) is ¹/_√3 (↑↓↓ + ↓↑↓ + ↓↓↑).
(s, s_z) = (, −) is ↓↓↓.

It was implied earlier that the nucleon states were antisymmetric in isospin and antisymmetric in spin. In fact, it is impossible to write down 3-quark states that are antisymmetric under the interchange of all pairs! (Try it!) However, it is possible to use "partially antisymmetric" isospin and spin states as a basis to find the required states which are symmetric overall.

First consider one pair of quarks. Ignoring the normalisation factors, (u↑d↓ − d↑u↓) is antisymmetric under exchange of isospin labels; (u↑d↓ − u↓d↑) is antisymmetric under exchange of spin labels; so (u↑d↓ − u↓d↑ − d↑u↓ + d↓u↑) is antisymmetric under exchange of isospin or spin labels but symmetric overall (i.e. when both the spin and isospin properties of a pair of particles are swapped). This combination has I = 0 and s = 0.

We can now combine this state with a third quark u↑ and symmetrise to find the combined state representing a proton (I₃ = ½) with spin s_z = ½. Simply adding a u↑ on the left gives:
u↑u↑d↓ − u↑u↓d↑ − u↑d↑u↓ + u↑d↓u↑. This must be made symmetric under interchange of pairs of quarks. Permuting first and second adds on + u↑u↑d↓ − u↓u↑d↑ − d↑u↑u↓ + d↓u↑u↑, while permuting first and third adds on + d↓u↑u↑ − d↑u↓u↑ − u↓d↑u↑ + u↑d↓u↑. (Note that interchange of 2^nd and 3^rd quarks was already considered above.)

Gathering terms together, and including the normalisation factor (equal to the reciprocal of the sum of the squares of the coefficients) gives

p↑ =	1/√18 (2u↑u↑d↓ − u↑u↓d↑ − u↓u↑d↑       + 2u↑d↓u↑ − u↑d↑u↓ − u↓d↑u↑       + 2d↓u↑u↑ − d↑u↓u↑ − d↑u↑u↓).

The spin-down state of the proton and the spin states of the neutron can be found in the same way.