# Psuedoscalar Mesons - why is there an eta and an eta prime?

1. Apr 10, 2013

### bayners123

The pseudoscalar mesons have $J^P = 0^-$

They form a nonet: for S = ±1, I (isospin) = 1/2 and so there are two particles for each value of strangeness. This account for 4 particles: the ground-state Kaons.
For S=0, I can be 0 or 1. I=1 gives a triplet: $\pi^\pm \mbox{ and } \pi^0$.

For S=0 and I = 0 however there are two particles: the $\eta \mbox{ and the } \eta^\prime$

As far as I can see the eta and the eta prime have exactly the same characteristics. My question is: why is there an eta prime? All the other mesons seem justified by the quark model. Why are there two $I^CJ^P = 0^+0^-$ particles and what distinguishes them?

2. Apr 10, 2013

### Bill_K

η is an isospin singlet, but part of the SU(3) octet, while η' is an SU(3) singlet. The three states with Y = 0, I3 = 0 are:

π0 = (uu - dd)/√2
η0 = (uu + dd - 2ss)/√6
η'0 = (uu + dd + ss)/√3

3. Apr 10, 2013

### bayners123

Hmm ok thanks. I think I need to understand group theory better.

4. Apr 10, 2013

### kloptok

The quark states are in the representation space of the $3$ representation of SU(3), and antiquarks in the $\bar{3}$. The mesons are states obtained by combining quarks and antiquarks, i.e. $|q\rangle \otimes|\bar{q}\rangle$ states. These states lie in the direct product of the $3$ and the $\bar{3}$ representation spaces. But this space separates into an octet and a singlet, $3\otimes\bar{3}=8\oplus 1$, so the mesons consist of the octet (the pions, the kaons and the $\eta$) and the singlet (the $\eta'$).

It is like in quantum mechanics when you combine two particles with spin 1/2 - the Hilbert space for the "total spin" states $|j,m\rangle$ (where here $j=0,1$ and $m=0$ for $j=0$ and $m=-1,0,1$ for $j=1$) in terms of group representations is the direct product space $2\otimes 2=3\oplus 1$, it separates into a singlet and a triplet (denoting representations by their dimensionalities). Note that spin 1/2 corresponds to the $2$ representation of the group SU(2) which is the proper group here (and thus not SU(3) like for the mesons).

If I remember correctly it is actually even a bit more complicated since the isospin singlet state in the SU(3) octet and the SU(3) singlet state mix with each other.

5. Apr 11, 2013

### bayners123

Brilliant, thanks. I think I understand that better now. The one that still confuses me is how the hadrons break down as
$$\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}=\mathbf{10}_S\oplus\mathbf{8}_M\oplus\mathbf{8}_M\oplus\mathbf{1}_A$$

Where does this breakdown come from? That's rhetorical by the way; I'll go read a group theory book :)

6. Apr 11, 2013