How many hadrons are there?

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  • #1
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Main Question or Discussion Point

I keep on hearing that there are hundreds of hadrons, or even more ambiguous a zoo of hadrons, but for some reason I've never seen an exact answer.

Given that there are 6 quarks, it seems you can form 6*6*6=216 baryons. Including anti-baryons, that would be 2*216=432 baryons. Also there are two sets, spin 1/2 and spin 3/2, so there are actually 864 baryons?

For mesons it seems you can form 6*6=36, including anti-mesons.

Now I know mesons come in spin 0 and spin 1, so that doubles it to 2*36=72

So are there 864+72=936 hadrons and anti-hadrons, which would imply just 936/2=468 hadrons?

Also why are spin 0 mesons different from spin 1 mesons, and spin 1/2 baryons different from spin 3/2 baryons? Aren't they the same particles, just with different total spin?
 

Answers and Replies

  • #2
blechman
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There are only so many quark combinations; even less than you say since the top quark never forms hadrons (it decays too quickly). But there are nearly limitless hadrons.

Each hadron has, in addition to quark content, various spins as you say (and not just 0,1, but also 2, 3, ...). The higher spin hadrons are heavier (as a rule) so I'm not sure what you mean by saying they're the same - they have different mass (compare the pion to the rho, for instance; or the proton to the Delta+)! Also, there is orbital angular momentum of the constituents, which can be anything from 0 to infinity, which is why the hadron spin can be arbitrarily high in principle.

If you want an "exact" number, go to the Particle Data Group and count all the particles. Be prepared to be there all day!

Hope that helps.
 
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  • #3
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That helped a lot. It bothered me because everyone says its hundreds or it's a zoo, and I just couldn't understand why people couldn't just say how many there are, since naively 6*6*6=... But I forgot about orbital angular momentum, so I guess you can go beyond 3/2 for baryons and 1 for mesons, so I guess theoretically it's limitless.

So I guess if you have really high energy, you can produce hadrons whose quarks have huge orbital angular momentum (and hence the hadrons would be very massive because of all the energy from the angular momentum), and hence a high overall spin for the hadron. But say you are searching for a really heavy fundamental particle. Could it be possible that one of the hadrons with high spin almost has the same mass as the particle you are looking for, so they would be impossible to distinguish?
 
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  • #4
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First, let's consider what goes into hadron states.
Quark flavors
Quark total spin
Quark orbital states
(leaving aside gluons and "glueball" states)

The OP discusses only flavors and spin states, and not orbital ones, and I shall do likewise.

In my discussions, I will assume n flavors for full generality.

Mesons

These are composed of a ordinary quark and an antiquark, so there are n2 possible flavor states and total spins 0 and 1.

That means n2 spin-0 mesons and the same number of spin-1 ones, for a total of 2n2.

Baryons

These are composed of three ordinary quarks; antibaryons parallel them with antiquarks.

Quarks obey Fermi-Dirac statistics, meaning that their total wavefunction will be antisymmetric. However, they are in a color-singlet state, making them antisymmetric in color. This means that they must be symmetric in flavor and spin combined.

Their possible spin states are 3/2 (completely symmetric) or 1/2 (partially symmetric). So their corresponding flavor states must be likewise. Working it out, there are n(n+1)(n+2)/6 completely symmetric flavor states and n(n2-1)/3 mixed-symmetry ones.

So there are n(n2-1)/3 spin-1/2 baryons and n(n+1)(n+2)/6 spin-3/2 ones.

There can be more than one mixed-symmetry flavor state; consider the lambda and the sigma-0 baryons. In general:
111 -- 0 states -- all the same flavor
112 -- 1 state -- two the same flavor, one different
123 -- 2 states -- all different flavors

This yields n2(n+1)/2 total baryons, the same number of antibaryons, and the total of these n2(n+1).

Total Hadrons

So the total number of meson and baryon states is n2(n+3)

Number of flavors, each spin of meson, spin-1/2 baryons, spin-3/2 baryons, total
1: 1 0 1 4
2: 4 2 4 20
3: 9 8 10 54
4: 16 20 20 112
5: 25 40 35 200
6: 36 70 56 324

There's a problem with 6: top quarks decay too fast to hadronize -- they decay directly into an up/strange/bottom and a W with a mean life of around 5*10-25 s, equivalent to about 10-16 m or 0.1 fermis.
 
  • #5
blechman
Science Advisor
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So I guess if you have really high energy, you can produce hadrons whose quarks have huge orbital angular momentum (and hence the hadrons would be very massive because of all the energy from the angular momentum), and hence a high overall spin for the hadron. But say you are searching for a really heavy fundamental particle. Could it be possible that one of the hadrons with high spin almost has the same mass as the particle you are looking for, so they would be impossible to distinguish?
Well, the problem is that after you reach around 10 GeV or so, you will stop forming "hadrons" -- if you produce quarks with that much energy, they will form jets instead! So while IN THE QUARK MODEL it looks like there are an infinite number of hadrons, in reality it only goes so far before you enter a new physical regime.

So if you're looking for a "fundamental" particle at 100 GeV, for example, it won't be a hadron!

That being said: some heavy (few GeV) hadron states (such as the dreaded X(3872)) are quite a confusion for the field - nobody really knows what they are...
 
  • #7
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The "resonances" there are orbital excited states, and there are a lot of such states now known. But from the ranges of masses of various sets of resonances, the maximum excitation energy a hadron can have is about 1.5 to 2 GeV.

Let's approach the maximum-mass question from a different direction. What is the maximum mass a hadron can have before its quarks become too short-lived to form hadrons?

The top quark is clearly too massive:
http://www-d0.fnal.gov/Run2Physics/WWW/results/prelim/TOP/T87/T87.pdf
Determination of the width of the top quark
The DØ Collaboration
URL http://www-d0.fnal.gov
(Dated: March 11, 2010)
Decay width: 2.1 +- 0.6 GeV
Mean life: 3 +- 1 * 10-25 s

That's because it decays into a physical (non-virtual) W.
Decay width:
Real W: m3/v2
Virtual W: m5/v4
v = Higgs-field vacuum value

Hadronization time -- hard to find any good numbers on that, but I'll assume that it's equivalent to a decay width of about the QCD energy scale: 200 MeV, corresponding to a mean life of 3.3*10-24 s.

A top quark extrapolated down to close to the W mass gives a decay width of about 300 MeV. Phase-space factors would reduce it even further, since all but the W will have relatively little energy. It clearly won't have any time to hadronize.


To extrapolate upwards, let's use the B+- meson, with mass 5.28 GeV and mean life 1.6*10-12 s, equivalent to a decay width of 4*10-10 MeV.

Dividing out quark-mixing effects gives a decay width of 2.4*10-7 MeV.

Extrapolating upward from that worst case gives a hadronization upper limit of 320 GeV, which is above the W mass.


Thus, heavy quarks will form heavy/light hadrons like the B meson all the way up to the W mass.


Let's now consider quarkonia. We can extrapolate upward from the upsilon(1S) b-b* meson.
Mass: 9.46 GeV
Width: 54 keV
It can decay like a QCD version of orthopositronium, the quark-antiquark pair annihilating into two gluons. Or it can decay into a virtual gluon, which makes a quark-antiquark pair.

The decay width ~ (mass)*(alphas)5
where alphas ~ 1/log(mass/QCD_scale)
Should be alphas(m)2 * alphas(m*alphas)3

The quark mass needs to reach a few TeV before this decay rate approaches the hadronization limit.

So let's consider weak decay vs. annihilation. Extrapolating upward for the worst case gives a quark mass of about 50 GeV, for the actual case of B mesons, 200 GeV.

So quarks will form quarkonia up to close to the W mass.


In summary, the heaviest possible hadron will have a mass a little less than the W mass. W's being virtual will slow down the quarks' decay enough to allow hadrons and quarkonia to form.
 
  • #8
arivero
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For mesons it seems you can form 6*6=36, including anti-mesons.
You do it in SU(6), so 35 instead of 36

Now I know mesons come in spin 0 and spin 1, so that doubles it to 2*36=72
Mesons come in any excitation state, so it is even worse than doubling. And you also have parity. Fortunately, the spin can be ordered using regge trajectories.

Also why are spin 0 mesons different from spin 1 mesons, and spin 1/2 baryons different from spin 3/2 baryons? Aren't they the same particles, just with different total spin?
It is a bit more complicated, for instance spin 1 is symmetric in the exchange of particle spins, and then you must look the right combination of symmetry and antisymmetry in the other "labels" of the particle, say color, flavour, etc. Because of it, people use group theory to put some order.
 
  • #9
arivero
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Let's approach the maximum-mass question from a different direction. What is the maximum mass a hadron can have before its quarks become too short-lived to form hadrons?

The top quark is clearly too massive:
http://www-d0.fnal.gov/Run2Physics/WWW/results/prelim/TOP/T87/T87.pdf
This is very important to me, because it means that you should classify mesons by doing the pairings inside SU(5) flavour. And amusingly, the 24 of SU(5) has 12 neutrals, 6 of charge +1 and 6 of charge -1. So Nature repeats herself: the MSSM also contains 12 scalars of charge 0, 6 +1 and 6 -1.
 
  • #10
tom.stoer
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You do it in SU(6), so 35 instead of 36
These are the pseudo-scalar mesons J=0 (like pions) only, but of course there are J=1 states (like rho) as well
 
  • #11
arivero
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These are the pseudo-scalar mesons J=0 (like pions) only, but of course there are J=1 states (like rho) as well
Yeah, in any case the moral is: use group theory.
 
  • #12
Vanadium 50
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This is all well and good, but it will give you the wrong answer. Toss the heavy quarks out, and just look at the light quarks, and you will discover there are too many mesons. There is no natural place for the a0(980) and f0(980) in qqbar picture, and there is at least one too many 1+ states.

I don't know what the right answer is (I doubt anyone does), but we know for certain that this approach will undercount.
 
  • #13
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I'll count meson states -- it's n2, not n2 - 1 mesons with each spin. I'll ignore different-flavor mesons, and concentrate on same-flavor ones, since that's what produces the difference. Ground states:

Spin 0:
pi0: ((uu*)-(dd*))/sqrt(2)
eta, eta': ((uu*)+(dd*))/sqrt(2), (ss*) (different mixtures)
etc-c: cc*
eta-b: bb*
Spin 1:
rho0: ((uu*)-(dd*))/sqrt(2)
omega, phi: ((uu*)+(dd*))/sqrt(2), (ss*) (different mixtures)
J/psi: cc*
upsilon: bb*

For each spin, 5 same-flavor mesons corresponding to the 5 hadronizable quark flavors.

Table of uu*,dd*,ss* mesons by value of I(G),J(PC)
S = 0, L even
0(+),0(-+) -- eta, eta'
0(+),2(-+) -- eta2
1(-),0(-+) -- pi
1(-),1(-+) -- pi1
1(-),2(-+) -- pi2
S = 1, L even
0(-),1(--) -- omega, phi
0(-),3(--) -- omega3, phi3
1(+),1(--) -- rho
1(+),3(--) -- rho3
S = 0, L odd
0(+),0(++) -- f0
0(+),1(++) -- f1
0(+),2(++) -- f2, f2'
0(+),4(++) -- f4
1(-),0(++) -- a0
1(-),1(++) -- a1
1(-),2(++) -- a2
1(-),4(++) -- a4
S = 1, L odd
0(-),1(+-) -- h1
1(+),1(+-) -- b1
Source: meson table in Particle Data Group - 2010 Summary Tables
 
  • #14
arivero
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I'll count meson states -- it's n2, not n2 - 1 mesons with each spin. I'll ignore different-flavor mesons, and concentrate on same-flavor ones, since that's what produces the difference. Ground states:

Spin 0:
pi0: ((uu*)-(dd*))/sqrt(2)
eta, eta': ((uu*)+(dd*))/sqrt(2), (ss*) (different mixtures)
etc-c: cc*
eta-b: bb*
Agreed, but I assume you are looking with detail http://pdg.lbl.gov/2010/reviews/rpp2010-rev-quark-model.pdf in order to keep in mind how it works. For u,d,s, your 3 x 3 decomposes in 8 x 1 and then you have uu*+dd*+ss* in the 1 (the "singlet") while uu*+dd*-2ss* is the eigenstate in the 8. Then these states mix to produce eta and eta'. Equations 14.4 to 14.16 in the review. It is important to keep them in mind because mass relationships are derived from these mixings. i

Ideally, with no mixing, all the particles in the octet should have one value of the mass, and the particle in the singled should have another different mass. One way to consistently avoid the mixing is to move the singlet out to infinite mass, in this sense one can think of n^2 -1 instead of n^2. But you are right, really the singlet is there.

What is important is that, it is always a singlet, the diagonal sum aa*+bb*+cc*+dd*+ee*+.... So if all the quarks were having the same mass, you should be almost in the same position, still having a n^2-1 plet for a mass eigenvalue, and an "eta" singlet with a different mass.

To abstract: it is important, when doing the classification, to consider the limit of equal masses for all the involved quarks, which is the restoration of global flavour symmetry.
 
  • #15
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Their possible spin states are 3/2 (completely symmetric) or 1/2 (partially symmetric). So their corresponding flavor states must be likewise. Working it out, there are n(n+1)(n+2)/6 completely symmetric flavor states and n(n2-1)/3 mixed-symmetry ones.

So there are n(n2-1)/3 spin-1/2 baryons and n(n+1)(n+2)/6 spin-3/2 ones.
Why is the product of a mixed symmetry state (spin) with another mixed symmetry state (flavor) altogether symmetric?

Say you symmetrize i and j, and then antisymmetrize i and k:

[tex]\psi=\psi_{ijk}+\psi_{jik}-\psi_{kji} -\psi_{jki} [/tex]

So let ijk=uud, then:

[tex]\psi=2uud-duu-udu [/tex]

Now swap the middle quark with the last quark to get:

[tex]\psi'=2udu-duu-uud [/tex]

Even if you multiply [tex]\psi'[/tex] by the correspondingly swapped spin part of the wavefunction and the color part of the wavefunction, you still won't get the original wavefunction times minus one, since 2udu-duu-uud is fundamentally different from 2uud-duu-udu.

Maybe I'm not understanding this correctly.
 
  • #16
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173
The trick here is to construct a mixed state that need not be decomposable into (flavors) * (spins). For an upward-spinning proton:

u+u+d-: 2
u+u-d+: -1
u-u+d+:-1
u+d+u-: -1
u+d-u+: 2
u-d+u+:-1
d+u+u-: -1
d+u-u+: -1
d-u+u+: 2

Downward spin and neutrons (udd) work similarly.

For 3 different flavors, there are two possible states. I'll do the lambda and sigma0 baryons, which are both uds. I'll ignore flavor permutations; the spins follow the flavors, like in my earlier example.

lambda
uds
++-: 0
+-+: 1
-++: -1

sigma0
uds
++-: 2
+-+: -1
-++: -1

The sigma+ (uus) and sigma- (dds) baryons work much like the proton.
 
  • #17
970
3
The trick here is to construct a mixed state that need not be decomposable into (flavors) * (spins). For an upward-spinning proton:

u+u+d-: 2
u+u-d+: -1
u-u+d+:-1
u+d+u-: -1
u+d-u+: 2
u-d+u+:-1
d+u+u-: -1
d+u-u+: -1
d-u+u+: 2
How did you derive this? The state is certainly symmetric, but how does one know that this doesn't come from the direct product of symmetric flavor states and symmetric spin states?

For a mixed state I only know this formula:

[tex] \psi=\psi_{ijk}+\psi_{jik}-\psi_{kji} -\psi_{jki} [/tex]

If I can't use a direct product, then I have to let each index i,j,k vary from 1 to 4, with 1 standing for u+, 2 for u-, 3 for d+, and 4 for d-.

If I choose 114, then I get:

[tex] \psi_{[114]}=\psi_{114}+\psi_{114}-\psi_{411} -\psi_{141}
=2(u+u+d-) + (d-u+u+)-(u+d-u+) [/tex]

If I choose 123, then I get:

[tex] \psi_{[123]}=\psi_{123}+\psi_{213}-\psi_{321} -\psi_{231}
=(u+u-d+) + (u-u+d+)-(d+u-u+)-(u-d+u+) [/tex]

But this seems to be all the combinations you can form with 2 u's and 1 d, and 2 +'s and 1 -, and if I add [tex]\psi_{[114]}+\psi_{[123]} [/tex], then the result is not symmetric.

So I have no clue how you got this state.
 
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  • #18
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There are many hadrons, due to dyslexic geeks watching Leia in Return of the Jedi.

Originally they were going to call leptons "softons", but Feynman wouldn't stop giggling, so Gell-Mann got angry and ruined that joke forever.
 

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