# What if colour didn't exist?

1. Jan 21, 2005

### babtridge

Hi there,
Could anybody please explain how electron-positron annihilation supports the conjecture that 3 colour states exist for quarks?
Plus in the Baryon 1/2 octet and 3/2 decuplet, what would be the corresponding set of states if colour did not exist, such that quarks of the same flavour were identical?

Thanks a lot in advance guys and gals.....

2. Jan 21, 2005

### misogynisticfeminist

uhhhh i don't know how leptons can prove that 3 colour states exists.

3. Jan 21, 2005

### marlon

This has nothing to do with the existence of the colour-quantum-number. This number was introduced because certain wavefunctions (like that of the $$\Delta^{++}$$)-particle) could exhibit "all equal" quantumnumbers. In the case of the above mentioned particle the wavefunction consists out of three up quarks + three spin up-numbers + three 1s-energy levels. This kind of wave function does not obey the Pauli-exclusion rule and because of this, the wavefunction must be antisymmetric. In order to obey this anti-symmetry, colours were introduced.

regards
marlon

Last edited: Jan 21, 2005
4. Jan 21, 2005

### anti_crank

Do a Feynman diagram as follows. Going in you have an electron and a positron, that annihilate into a virtual photon; the photon then turns into a quark-antiquark pair that go out.

The idea is that for high enough energies, this diagram contributes to the total electron-positron cross section. Let's assume only the light quarks contribute, then you can sum up the amplitudes for annihilation into u and b quark/antiquark pairs and calculate the fraction of the time quarks are produced as opposed to a photon pair or a muon-antimuon. If color exists, then there are 3 diagrams as far as the EM interaction is concerned for each quark flavor; one for each color. This increases the fraction of producing baryons or mesons compared to the no-color case.
I do not have the time to work this out now; the idea is that you are now looking for quark wavefunctions that are antisymmetric wrt flavor and spin instead of symmetric. The spatial wavefunction is always symmetric in the ground state, and the color singlet wavefunction is always antisymmetric if it is included.

Edit: if you have Griffiths' book on elementary particles, he touches on this somewhere in the chapter on building baryon wavefunctions.

5. Jan 24, 2005

### babtridge

Thanks for taking the time to help me out on this one...you've all made stuff a lot clearer.
Tekiteasy peeps!