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

[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?

 

[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

 

[bayners123]

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

 

[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.

 

[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 :)

 

[Vanadium 50]

You need 27, and you have a purely symmetric state (singlet), a purely antisymmetric state (decuplet) and 16 states of mixed symmetry. You can only do that with two octets.

It's perhaps worth mentioning that while group theory is fine for counting states, the actual particles are more complicated: neither the eta nor eta-prime is a pure SU(3) flavor singlet. Additionally, the mass of the eta-prime is much larger than its quark content would suggest.