# How many mesons and baryons are there?

1. Oct 3, 2008

### CarlB

That is, how many different states are there assuming you distinguish two particles as different if they have different quantum numbers or different masses.

For example, the $$\Delta^-(1232),\Delta^0(1232),\Delta^+(1232),\Delta^{++}(1232)$$ are four different states. And these are all different from the four different charge states of the Delta(1600) states and the four Delta(1920) states. The other Delta resonances are not $$P_{33}$$, but just these give 12 states already.

There are a total of 22 mass multiplets called "Delta" which gives a total of 22x4 =88 states. But these are all particles. Double them for antiparticles and we're up to 176.

The nucleons (i.e. N and P) also have 22 mass multiplets but there are only 2 states in each (with charge 0 or +1, like the neutron and proton) so, counting anti-particles this gives 88 states.

So I'm up to 176+88 = 264 and I've only covered two letters, $$\Delta, N$$. Does anybody know how many there are in total?

2. Oct 4, 2008

Staff Emeritus
An infinite number. (Just like there are an infinite number of states of the hydrogen atom)

3. Oct 4, 2008

### hamster143

I'm not so sure about that. We haven't found any new delta resonances above 2950 MeV since 1978. Maybe there aren't any.

4. Oct 4, 2008

### humanino

I agree with Vanadium, in the sense that we have not observed an infinite number of excited states of hydrogen either. You run into difficulties such as "what is a state ?" when they are unstable, or too broad anyway.

5. Oct 4, 2008

### clem

If you call resonances with strong decays particles, then the number is ambiguous.
The number would be infinite if you consider all resonances as particles even if their width is so broad as to be effectively unobservable. That is why "We haven't found any new delta resonances above 2950 MeV".
I prefer to call only strongly stable quark bound states as particles.
Then you get a reasonable, but still large, number.

6. Oct 4, 2008

Staff Emeritus
That's not a bad choice, but it excludes the Delta, which was the example CarlB originally chose. So I don't think he's going along with that particular convention.

7. Oct 4, 2008

### arivero

If you ask instead "How many regge trajectories", then the number is finite: one for each quark content, and degenerate parity if you wish.

Furthermore, I am strongly in the side of not including the top quark between the possible quark contents.

8. Feb 25, 2010

### nelufar

Well, as I see from the earlier posts, people have been talking only about the baryons. What about how many mesons are there? If just for the sake of telling a number, what it should be? 100, 200, or 400? :uhh:

9. Feb 25, 2010

### Staff: Mentor

10. Feb 25, 2010

### arivero

Do you call an excited hydrogen atom a different particle than an hydrogen atom in the fundamental state?????

My only trouble could be parity. A 0+ and a 0- state with the same quark composition, is the same particle? Can we get from one to the other via some transition?

Last edited: Feb 25, 2010
11. Feb 25, 2010

### arivero

Ah, that is easy. The top quark does not hadronize, so there are only five pieces. A bit of group theory leaves out a neutral U(1) of U(5)-flavour, and you are left with SU(5) flavour.

12. Feb 25, 2010

### nelufar

Can you please elaborate on this. I am not so familiar with group theory. Thanks!

13. Feb 25, 2010

### arivero

Ok, you only have five quarks now: u d s c b. In principle, you have 25 combinations of quark/antiquark, simply build an square. But due to some construct in quantum mechanics, a combination of the 5 mesons in the diagonal of the square has different properties, it is the "singlet". The other 4 combinations, jointly with the other 20 pieces of this square, make a single object of group theory, and each of them is a meson.

I am not very sure if the extant singlet is another 25th meson, but I believe it is not. For the same token, we speak of the SU(3) octet, instead a nonet.

Still there is the question of the parity of combinations, I am not sure if you can consider the scalar and pseudoscalar versions of a combination as being states of the same particle: they have even the same spin, but different parity.