Transcript Lecture 8: Quarks I - University of Oxford

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
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
FWHM ~ 100 MeV
max at E=ER
( 4/)
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
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
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

e
e


Of the remaining modes, is a strong interaction coupling,
so this will dominate compared with EM coupling for ee& 
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
For ''pre-1974" hadrons, the following relationships were also observed
Q = I3 + (B+S)/2
Gell-Mann - Nishijima Formula
Mesons
Y
0
(498)

(140)
1
-1
( SpinParity ) 0 nonet
Y

(494)
(135) 0 (547)

(958)

(494)
thus, define ''Hypercharge" as
YB+S
0
(498)

(896)

(140)
I3

(769)
Note the presence
of both particles
and antiparticles
1

(892)
(769) 0 (782)

(1019)

(892)
-1
0
(896)
 nonet

(769)
I3
Baryons
Y
n
(940)
1
0 (1193)
(1116)

(1197)
-1

(1321)

(1232)

(1232)
p
(938)

(1189)
I3
Y
*
(1387)
0
(1315)
1

(1232)
*
I3
(1383)
*
(1384)
-1

(1535)

(1532)

(1672)
( SpinParity )
1/2
octet
Note antiparticles
are not present

(1232)
3/2+ decuplet
Inelastic Scattering: Evidence for Compositeness
Consider a 3-component ''parton" model where the
constituents have the following quantum numbers:
Y
Y
1
1
s
d
u
-1
1
I3
-1
1
u
d
s
-1
''quarks"
-1
''anti-quarks"
I3
•
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"
Y
with respect to the other:
ds
1
us
dd uu
ss
du
-1
su
-1
ud
1
sd
I3
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
(I3= 1/2)
u  ucos 2  dsin 2
d  usin 2 + dcos 2
(I3= 1/2)
u ucos   dsin 
(I3= 1/2)
So parameterize the
isospin rotation by:
Apply charge
conjugation:
(I3= 1/2)
Isospin doublet
2
2
d usin  + dcos 
2
2
()
()
u
d
+1/2
d
u
+1/2
Note: top/bottom isospin members transform differently in each case  messy!
(I3= 1/2)
We can ''fix" this by
rewriting the latter as:
So the isospin pairs
(I3= 1/2)
() ()
u
d
d  d cos 2 usin 
2
(u (u) cos 2 + d sin 2
d
and u transform the same way
-1/2
-1/2
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:
 dduu 
spins
anti-parallel
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!
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
Which leaves
 dduu
spins parallel
  ss
spins parallel
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.
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:
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
Now add a 3rd:
ddu
dds
duu
uus
uds
dss
uss
The baryon decuplet !!
and the  sealed the Nobel prize 
sss
uuu
Y
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)
n 1
(940)
But why 2 states in the middle?
0 (1193) 
(1116) (1189)

(1197)
We can ''chop off the corners" by
artificially demanding that 3 identical
quarks must point in the same direction

-1
I3
J=1/2
0
(1315)
(1321)
ddd
p
(938)
ddu
duu
uuu
ways of getting spin 1/2:

uds

uds

uds
dds
0
these ''look" pretty much
the same as far as the strong
force is concerned (Isospin)

uus
uds
dss
uss
J=3/2
sss
Y
Charge:
n 1
(940)
p
(938)
d+d+u= 0
u = -2d
0 (1193) 
(1116) (1189)

(1197)
d + u + u = +1

-1
ddd
J=1/2
0
(1315)
(1321)
d + 2(-2d) = +1
I3
ddu
duu
uuu
-3d = +1
d = -1/3 & u= +2/3
dds
dss
uss
u+d+s= 0
s = -1/3
uus
uds
J=3/2
sss
Y
So having 2 states in the centre isn’t strange...
but why there aren’t more states elsewhere ?!
i.e. why not
   and   ???
uus
uus
We can patch this up again by altering
the previous artificial criterion to:
The lowest energy state
''Any pair of similar quarks must
is be in identical spin states"
(
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. ??
What holds these things together anyway ??
n 1
(940)
0 (1193) 
(1116) (1189)

(1197)

-1
I3
J=1/2
0
(1315)
(1321)
ddd
p
(938)
ddu
dds
duu
uuu
uus
uds
dss
uss
J=3/2
sss