PHY304 | Particle Physics | Dr C N Booth |

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

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*, *I*_{3}) = (1, 1), (1, 0), (1, −1).*I*, *I*_{3}) = (0, 0).^{1}/_{√2} (↑↓ + ↓↑),^{1}/_{√2} (↑↓ − ↓↑),*s*, *s _{z}*) = (1, 1), (1, 0), (1, −1), (0, 0)

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 *I*_{3} = ½ by u and
*I*_{3} = −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*, *I*_{3}) =
(, )^{++}.

uud is *I*_{3} = ½; forming combinations which are symmetric under 2-quark
permutations we get
^{1}/_{√3} (uud + udu + duu).*I*, *I*_{3}) =
(, ½)^{+}.

Similarly ^{1}/_{√3} (udd + dud + ddu)*I*, *I*_{3}) =
(, −½)^{0}.

ddd is *I*, *I*_{3}) =
(, −)^{−}.

Note: the first and last state *must* be
*I* = .
There must therefore be symmetric states with
*I* =
and intermediate *I*_{3} 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}*) =
(, )

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,
*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 (*I*_{3} = 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

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.