# What is Diquark?

1. Aug 22, 2006

### BuckeyePhysicist

Is a diquark $$[q_1q_2]$$ a totally antisymmetric state in color, flavor and spin space,

i.e. a color antitriplet $$\overline{3}_c$$, flavor antitriplet $$\overline{3}_f$$

and spin singlet?

So if it has u, d flavors, how to write it explicitely?

$$|ud -du \rangle \otimes \frac{1}{\sqrt{N_c}}|cc> \otimes \frac{1}{\sqrt{2}}(|\uparrow\downarrow \rangle - |\downarrow\uparrow \rangle)$$ ???

Last edited: Aug 22, 2006
2. Aug 23, 2006

### Meir Achuz

1. A diquark need not be symmetrized or antisymmetrized if the two quarks are different flavors.
2. If the two flavors are the same, the product of space, spin, and color must be antisymmetric.
3. For a diquark L=0 is usually assumed, so the product spinXcolor must be antisymmetric.
4. This allows for a spin one or a spin zero diquark.
5. Many people also symmetrize or antisymmetrize with respect to flavor for different quarks in the diquark. This does not produce different states, but permits the distinction into flavor multiplets.
6. The diquark you give is one example. You could also have spin one. Any combination with total antisymmetry would work. This includes color or flavor sextets, depending on the spin.
7. "Totally antisymmetric" means the product, not antisymmetry in each space.

3. Aug 23, 2006

### BuckeyePhysicist

Thanks for the helpful answer. Could you at least give me one concrete example? If I want a totally antisymmetric diquark which has flavor content u and d, how it should look like?

4. Aug 24, 2006

### Meir Achuz

There are many possible states. If color is symmetric, like RR (or 5 other symmetric combinations), then either:
1. I=1, S=0, or 2. I=0, S=1, (24 states of u and d).

If color is one of the 3 antisymmetric combnations, then:
3. I=1, S=1, or 4. I=0, S=0, (12 states of u and d).

5. Aug 29, 2006

### arivero

A interesting play I did once if that if one considers that top quark is not able to bind, then the extant five quarks can be combined in a way to be partly supersymmetric to the existing six quarks, except that there are three exotic +4/3 diquarks. If we could get rid of them, the degrees of freedom between bosons (diquarks) and fermions (quarks) would to a nice supersymetric match.

6. Aug 29, 2006