Lecture 8: Quarks I

• Meson & Baryon Multiplets

• 3-Quark Model & The Meson Nonets

• Quarks and the Baryon Multiplets

Chap 3, Section 6.2

Useful Sections in Martin & Shaw:

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 .2

sheet 3

Breit-Wigner:

1

(E ER)2 + 2/4

FWHM ~ 100 MeV

1/2 max when |E-ER| = /2

or FWHM/2 = /2

so, indeed, = FWHM

max at E=ER

( 4/)

= ħ = 6.58x10-22 MeV s 100 MeV =6x10-24 s

Et ~ ħ

Consider the following decay modes of the eeExplain which of these decay modes is forbidden and therelative dominance of the other modes.

Of the remaining modes, is a strong interaction coupling,

so this will dominate compared with EM coupling for ee&

J P C

ee

However, identical bosons must be produced in indistinguishable states,

so wavefunction must be evenin terms of angular momentum.

Cannot get any of the quantum numbers, so this mode is forbidden

Lecture 8: Quarks I

• Meson & Baryon Multiplets

• 3-Quark Model & The Meson Nonets

• Quarks and the Baryon Multiplets

Section 2.2, Section 6.2

Useful Sections in Martin & Shaw:

For ''pre-1974" hadrons, the following relationships were also observed

Q = I3 + (B+S)/2

Gell-Mann - Nishijima Formulathus, define ''Hypercharge" as Y B + S

Mesons

I3

-1

1

Y

(494)

(494)

0

(498)

0

(498)

(140)

(135) 0 (547) (958)

(140)

0 nonet( SpinParity )

-1

1

I3

Y

(892)

(892)

0

(896)

(896)

(769)

(769) 0 (782) (1019)

(769)

nonet

Note the presenceof both particlesand antiparticles

Baryons

-1

1

I3

Y

p

(938)

(1321)

0

(1315)

n

(940)

(1197)

0 (1193)

(1116) (1189)

1/2 octet( SpinParity )Note antiparticlesare not present

-1

1

I3

Y

(1232)

(1535)

(1232)

*

(1387)

(1672)

3/2+ decuplet

(1232)

(1232)

*

(1383) *

(1384)

(1532)

Inelastic Scattering: Evidence for Compositeness

Consider a 3-component ''parton" model where theconstituents have the following quantum numbers:

-1

1

I3

Y

1-1

ud

s

''quarks"

-1

1

I3

Y

1-1

s

du

''anti-quarks"

Y

-1

1

I3 1-1

We can add quarks and anti-quarks quantum numbers together graphically by appropriately shifting the coordinates of one ''triangle" with respect to the other:

sd

ud dd

su

duuu

us ds

ss

• 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

Nice! But we still have some work to do...

Start with the pions: originally related by rotations in isospin space... now clear this refers to symmetry between u and d quarks

So parameterize the isospin rotation by:

(I3= 1/2) u ucos dsin

(I3= 1/2) d usin + dcos

22

2

2

Apply charge conjugation:

(I3= 1/2) u ucos dsin

(I3= 1/2) d usin + dcos

2

2

2

2

Note: top/bottom isospin members transform differently in each case messy!

Isospin doublet

u d d u

( )( )

+1/2

-1/2

+1/2

-1/2

We can ''fix" this by rewriting the latter as:

(I3= 1/2) (u (u) cos + d sin

(I3= 1/2) d d cos usin 2

2

2

2

So the isospin pairs and transform the same way u d( )

d

u( )

While the central states certainly involve uu, dd and ss,they can, in fact, be any set of orthogonal, linear combinations

Thus, we rotate u d and d u

So, in terms of the wave functions, we will actually define

duandud

The 0 is a neutral ''half-way" state in the rotation.

dduu spinsanti-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 playa big role in determining masses!

We can get to a neutral state by rotating to eitherddoruufrom either the or +, respectively.

So take the superposition:

Now look at the 1 nonet...

Since dduu spins parallel

We seek another orthogonal such combination, so

dduu spins parallel

Which leavesss spins parallel

The mass of the is very nearly the same as for the ’s,suggesting it might be composed of similar quarks

Now look back at the 0 nonet...

The differs by another ~400 MeV, suggesting that anothersuch pair is involved.

Indeed, if we try:dduussOrthogonality then requires: dduuss

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, sothese are not actually the physical states! The physical states are usually explained by ''mixing" between these.

The masses of the and differ by ~400 MeV, suggesting a different, heavier quark pair is involved. And we know fromthe that the s is heavy compared with either u or d quarks

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

So try building 3-quark states

Start with 2:

(note that m() ≠ m(+) so they are not anti-particles, and similarly for the * group)

So try building 3-quark states

Now add a 3rd:

The baryon decuplet !!

and the sealed the Nobel prize

ddd ddu duu uuu

dds uus

dss uss

sss

uds

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)

-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" byartificially demanding that 3 identicalquarks must point in the same direction

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=1/2

J=3/2

-1

1

I3

Y

p

(938)

(1321)

0

(1315)

n

(940)

(1197)

0 (1193)(1116)

(1189)

ddd ddu duu uuu

dds uus

dss uss

sss

uds

J=1/2

J=3/2

Charge:d + d + u = 0

d + u + u = +1

u = -2d

d + 2(-2d) = +1

-3d = +1

d = -1/3 & u= +2/3

u + d + s = 0

s = -1/3

ddd ddu duu uuu

dds uus

dss uss

sss

uds

We can patch this up again by altering the previous artificial criterion to:

''Any pair of similar quarks must be in identical spin states"

What happened to the Pauli Exclusion Principle ???

Why are there no groupings suggestingqq, qqq, qqqq, etc. ??

What holds these things together anyway ??

Not so crazy lowest energy states of simple, 2-particle systems tend to be ''s-wave" (symmetric under exchange)( )

So having 2 states in the centre isn’t strange...but why there aren’t more states elsewhere ?!

J=1/2

J=3/2

u u s

u u s

i.e. why not and ???

-1

1

I3

Y

p

(938)

(1321)

0

(1315)

n

(940)

(1197)

0 (1193)(1116)

(1189)

The lowest energy state

is