1. Lecture 8: Quarks I Meson & Baryon Multiplets
3-Quark Model & The Meson Nonets
Quarks and the Baryon Multiplets
Useful Sections in Martin & Shaw:
Chap 3, Section 6.2

2. 2
sheet 3
The figure below shows the cross section for the production
of pion pairs as a function of CM energy in e+e- annihilation.
Relate the FWHM of the resonance to the lifetime of the  .
Breit-Wigner:
 1
(E ER)2 + 2/4
max at E=ER
( 4/)
FWHM ~ 100 MeV
1/2 max when |E-ER| = /2
or FWHM/2 = /2
so, indeed,  = FWHM
Et ~ ħ
 = ħ   = 6.58x10-22 MeV s  100 MeV =6x10-24 s 2

3.   ee
Consider the following decay modes of the 
  ee
Explain which of these decay modes is forbidden and the
relative dominance of the other modes.
J P C




e
e


However, identical bosons must be
produced in indistinguishable states,
so wavefunction must be even
in terms of angular momentum.
Cannot get any of the quantum
numbers, so this mode is forbidden
Of the remaining modes, is a strong interaction coupling,
so this will dominate compared with EM coupling for ee&  3

4. Lecture 8: Quarks I Meson & Baryon Multiplets
3-Quark Model & The Meson Nonets
Quarks and the Baryon Multiplets
Useful Sections in Martin & Shaw:
Section 2.2, Section 6.2

5. Gell-Mann - Nishijima Formula thus, define ''Hypercharge" as Y  B + S
Meson Nonets
For ''pre-1974" hadrons, the following relationships were also observed
Q = I3 + (B+S)/2
Gell-Mann - Nishijima Formula
thus, define ''Hypercharge" as
Y  B + S
Mesons -1 1 Y -1 1 I3 Y

(892)

0
(896)
 
(769)
(769) 0
(782) 
(1019) 
 nonet 
(494)  0
(498) 
(140)
(135) 0
(547) 
(958) 
0 nonet
( SpinParity ) I3
Note the presence
of both particles
and antiparticles

6. Baryon Octet & Decuplet
Baryons -1 1 I3 Y p
(938) 
(1321) 0
(1315) n
(940) 
(1197)
0 (1193)
(1116) 
(1189)
1/2 octet
( SpinParity ) -1 1 I3 Y 
(1232)

(1535) 
*
(1387) 
(1672)
3/2+ decuplet 

*
(1383)
*
(1384)

(1532)
Note antiparticles
are not present

7. Inelastic Scattering: Evidence for Compositeness

8. 3-Quark Model
Consider a 3-component ''parton" model where the
constituents have the following quantum numbers: -1 1 I3 Y u d s -1 1 I3 Y s d u
''anti-quarks"
''quarks"

9. We can add quarks and anti-quarks quantum numbers together
Quarks and Mesons
Mesons are generally lighter than baryons, suggesting they contain fewer quarks
Also, the presence of anti-particles in the meson nonets suggests they might be composed of equal numbers of quarks and anti-quarks (so all possible combinations would yield both particles and anti-particles)
Further, if we assume quarks are fermions, the integer spins of mesons suggest quark-antiquark pairs
We can add quarks and anti-quarks quantum numbers together
graphically by appropriately shifting the coordinates of one ''triangle"
with respect to the other: Y -1 1 I3 s d u s u d u s d

10. ( ) ( ) Nice! But we still have some work to do...
Quarks and Isospin
Nice! But we still have some work to do...
While the central states certainly involve uu, dd and ss,
they can, in fact, be any set of orthogonal, linear combinations
Start with the pions: originally related by rotations in isospin space...
 now clear this refers to symmetry between u and d quarks
Isospin doublet u d
( )
+1/2
-1/2
So parameterize the
isospin rotation by:
(I3= 1/2) u  ucos  dsin
(I3= 1/2) d  usin  + dcos  2
Apply charge
conjugation:
(I3= 1/2) u ucos  dsin
(I3= 1/2) d usin + dcos  2
Note: top/bottom isospin members transform differently in each case  messy!
(I3= 1/2) (u (u) cos  + d sin
(I3= 1/2) d  d cos  usin   2
We can ''fix" this by
rewriting the latter as:
So the isospin pairs and transform the same way u d
( ) u

11.  dduu 
Pion Wave Functions
Thus, we rotate u  d and d  u
So, in terms of the wave functions, we will actually define
duandud
The 0 is a neutral ''half-way" state in the rotation.
We can get to a neutral state by rotating to either
dd or uufrom either the  or +, respectively.
So take the superposition:
spins
anti-parallel
 dduu 
A similar argument follows for the ’s of the 1- nonet,
but the quark spins must be parallel in that case.
Note from the nonets that spin interaction must play
a big role in determining masses!

12. dd uu
1 Middle Bits
Now look at the 1 nonet...
The mass of the  is very nearly the same as for the ’s,
suggesting it might be composed of similar quarks
Since
dd uu
spins parallel
We seek another orthogonal such combination, so
dduu
spins parallel
Which leaves
 ss
spins parallel

13. dduuss
Etas
Now look back at the 0 nonet...
The masses of the  and differ by ~400 MeV, suggesting
a different, heavier quark pair is involved. And we know from
the  that the s is heavy compared with either u or d quarks
The  differs by another ~400 MeV, suggesting that another
such pair is involved.
Indeed, if we try:
dduuss
Orthogonality
then requires:
dduuss
Warning: most texts talk about 1 and 8 , which are the SU(3)
states of group theory if the symmetry were perfect... it isn’t, so
these are not actually the physical states! The physical states are
usually explained by ''mixing" between these.

14. Baryons:  So try building 3-quark states Start with 2:
Building Baryons
Baryons:
Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions
Absence of anti-particles suggests there is not substantial anti-quark content
(note that m() ≠ m(+) so they are not anti-particles, and similarly for the * group)
 So try building
3-quark states
Start with 2:

15. Baryons: The baryon decuplet !!  So try building ddd ddu duu uuu
The Decuplet
Baryons:
Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions
Absence of anti-particles suggests there is not substantial anti-quark content
(note that m() ≠ m(+) so they are not anti-particles, and similarly for the * group)
 So try building
3-quark states
ddd ddu duu uuu
dds uus
dss uss
sss
uds
Now add a 3rd:
The baryon decuplet !!
and the  sealed the Nobel prize 

16. 0  But what about the octet?
Coping with the Octet -1 1 I3 Y p
(938) 
(1321) 0
(1315) n
(940) 
(1197)
0 (1193)
(1116) 
(1189)
But what about the octet?
It must have something to do with spin...
(in the decuplet they’re all parallel,
here one quark points the other way)
We can ''chop off the corners" by
artificially demanding that 3 identical
quarks must point in the same direction
J=1/2
But why 2 states in the middle?
ddd ddu duu uuu
dds uus
dss uss
sss
uds
ways of getting spin 1/2:
  
u d s
 
u d s
 
u d s
these ''look" pretty much
the same as far as the strong
force is concerned (Isospin) 0 
J=3/2

17. Charge: d + d + u = 0 u = -2d d + u + u = +1 d + 2(-2d) = +1 -3d = +1
Coping with the Octet -1 1 I3 Y p
(938) 
(1321) 0
(1315) n
(940) 
(1197)
0 (1193)
(1116) 
(1189)
Charge:
d + d + u = 0
u = -2d
d + u + u = +1
J=1/2
d + 2(-2d) = +1
ddd ddu duu uuu
dds uus
dss uss
sss
uds
-3d = +1
d = -1/3 & u= +2/3
u + d + s = 0
J=3/2
s = -1/3

18. ( ) What happened to the Pauli Exclusion Principle ???
Quark Questions -1 1 I3 Y p
(938) 
(1321) 0
(1315) n
(940) 
(1197)
0 (1193)
(1116) 
(1189)
So having 2 states in the centre isn’t strange...
but why there aren’t more states elsewhere ?!
  
u u s
 
i.e. why not
and
???
We can patch this up again by altering
the previous artificial criterion to:
J=1/2
The lowest energy state is
''Any pair of similar quarks must
be in identical spin states"
ddd ddu duu uuu
dds uus
dss uss
sss
uds
Not so crazy  lowest energy states of
simple, 2-particle systems tend to be
''s-wave" (symmetric under exchange)
( )
What happened to the
Pauli Exclusion Principle ???
Why are there no groupings suggesting
qq, qqq, qqqq, etc. ??
J=3/2
What holds these things together anyway ??