# Spin S of 2 quarks

1. Feb 23, 2013

### Gregg

If two quarks are in a S=1 state, are they in either

$|\uparrow \uparrow \rangle$ or $|\downarrow \downarrow \rangle$ ?

2. Feb 23, 2013

### fzero

3. Feb 23, 2013

### Gregg

yes I forgot and the other, symm. $\frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \right)$ ...

I don't really know how to ask this question very well, but, it is how spin and color wave-functions are related to whether or not it is symmetric or not and where isospin comes into this.

If we have $\Psi_{\text{spin}} \Psi_{\text{flavor}}$ then it seems to me that to get a symmetric wave-function here we must have either both symmetric or both anti symmetric.

I have read something that suggest that the latter is not true and that in this case we actually have a mixture of symmetric and antisymmetric states.

Given two identical quarks $| q q u \rangle$, do you have as the symmetric color-spin wavefunction

$\frac{1}{\sqrt{3}} \left( |q^\uparrow q^\uparrow u^\uparrow \rangle+|q^\uparrow u^\uparrow q^\uparrow \rangle+|u^\uparrow q^\uparrow q^\uparrow \rangle \right)$ ?

Does the qq pair have to always have to be in the symmetric state ( s=1), why?

The above is just $\frac{1}{\sqrt{3}} \left( |q q u \rangle+ |q u q \rangle+ |u q q \rangle\right) \otimes | \uparrow \uparrow \uparrow \rangle$ isn't it?

and this refers to the S=3/2 state since all the spins are parallel, is there an analogous one for the all spin down?

So it seems that all there is left to do is the case of spin 1/2 where the up quark is opposing the pair, can we have

$| q^\uparrow q^\downarrow u^\downarrow \rangle$ states? would they appear due to the symmetric $\frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \right)$

Last edited: Feb 23, 2013
4. Feb 23, 2013

### Bill_K

Don't forget there is also ψcolor, and since the total color must be zero, ψcolor is always totally antisymmetric.

Since quarks are fermions, the remaining factors ψspinψspaceψflavor must together be symmetric.

5. Feb 23, 2013

### Gregg

Yes that was what I was attempting to get at, the color wavefunction is antisymmetric, assuming we are in l=0 we have the space wavefunction symmetric and now I am looking at the product of the spin and flavor which I require to be together symmetric and to see how many particles there are, but I think I have confused myself as to how to do this.

If it was obvious to me why the quark pair qq needed to be in the symmetric state I could model the states on the proton also,

$\chi_\text{p}(S=\frac{1}{2}, S_z=\frac{1}{2}) = \sqrt{\frac{2}{3}} \chi_{\text{uu}}(1,1) \chi_{\text{d}}( \frac{1}{2}, -\frac{1}{2} ) - \sqrt{\frac{1}{3}} \chi_{\text{uu}}(1,0) \chi_{\text{d}}( \frac{1}{2}, \frac{1}{2} )$

Where

$\chi_{\text{uu}}(1,0)$ is the $\frac{1}{\sqrt{2}} \left( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \right)$ state, where I have got the above $\chi_{\text{p}}$ from using ladder operator on the $\chi (S=3/2, S_z=3/2)$ (and Clebsch Gordon) ?

I think that if I could write out all of these wavefunctions in an exhaustive way then I would understand why this is all important a lot better!

Last edited: Feb 23, 2013
6. Feb 23, 2013

### Gregg

I think my questions should be:

How does isospin, I, affect the symmetry of the $\Psi_{\text{flavor}}$ wavefunction?

Spins of like quarks have to be in the S=1 state by symmetry, and for J=3/2 the spins have to be parallel. Doesn't a mix of anti-symmetric states in the spin and flavor states allow us to not use the symmetric S=1 state for the pair, qq?

How are different particles distinguished? The qq S=1 states are a triplet so for S=3/2, say, with an up quark's spin +1/2, there would be 3 different wave functions for qqu, when do different wavefunctions refer to different particles?

Last edited: Feb 23, 2013