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Over the years inquiring minds have asked: “Can we describe the known physics with just a few building blocks ?” Þ Historically the answer has been yes. Elements of Mendeleev’s Periodic Table (chemistry) nucleus of atom made of protons, neutrons

Quarks

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Richard Kass

Over the years inquiring minds have asked:

“Can we describe the known physics with just a few building blocks ?”

Þ Historically the answer has been yes.

Elements of Mendeleev’s Periodic Table (chemistry)

nucleus of atom made of protons, neutrons

proton and neutron really same “particle” (different isotopic spin)

By 1950’s there was evidence for many new particles beyond g, e, p, n

It was realized that even these new particles fit certain patterns:

pions: p+(140 MeV)p-(140 MeV)po(135 MeV)

kaons:k+(496 MeV)k-(496 MeV)ko(498 MeV)

Some sort of pattern was emerging, but ........... lots of questions

Þ If mass difference between proton neutrons, pions, and kaons is due

to electromagnetism then how come:

Mn > Mp and Mko > Mk+but Mp+ > Mpo

Lots of models concocted to try to explain why these particles exist:

Þ Model of Fermi and Yang (late 1940’s-early 50’s):

pion is composed of nucleons and anti-nucleons (used SU(2) symmetry)

note this model was proposed

before discovery of anti-proton !

With the discovery of new unstable particles (L, k) a new quantum number was invented:

Þ strangeness

Richard Kass

Gell-Mann, Nakano, Nishijima realized that electric charge (Q) of all particles could be related to isospin (3rd component), Baryon number (B) and Strangeness (S):

Q = I3 +(S + B)/2= I3 +Y/2

Coin the name hypercharge (Y) for (S+B)

Interesting patterns started to emerge when I3 was plotted vs. Y:

Particle Model of Sakata (mid 50’s):

used Q = I3 +(S + B)/2

assumed that all particles could be made from a combination of p,n, L

tried to use SU(3) symmetry

In this model:

This model obeys Fermi statistics and explains why:

Mn > Mp and Mko > Mk+ and Mp+ > Mpo

Unfortunately, the model had major problems….

Richard Kass

Problems with Sakata’s Model:

Why should the p, n, and L be the fundamental objects ?

why not pions and/or kaons

This model did not have the proper group structure for SU(3)

What do we mean by “group structure” ?

SU(n)= (nxn) Unitary matrices (MT*M=1) with determinant = 1 (=Special) and

n=simplest non-trivial matrix representation

Example: With 2 fundamental objects obeying SU(2) (e.g. n and p)

We can combine these objects using 1 quantum number (e.g. isospin)

Get 3 Isospin 1 states that are symmetric under interchange of n and p:

|11> =|1/2 1/2> |1/2 1/2>

|1-1> =|1/2 -1/2> |1/2 -1/2>

|10> = [1/Ö2](|1/2 1/2> |1/2 -1/2> + |1/2 -1/2> |1/2 1/2>)

Get 1 Isospin state that is anti-symmetric under interchange of n and p

|00> = [1/Ö2](|1/2 1/2> |1/2 -1/2> - |1/2 -1/2> |1/2 1/2>)

In group theory we have 2 multiplets, a 3 and a 1:

2 Ä 2 = 3 Å1

Back to Sakata's model:

For SU(3) there are 2 quantum numbers and the group structure is more complicated:

3 Ä 3 Ä 3 = 1 Å 8 Å 8 Å 10

Expect 4 multiplets (groups of similar particles) with either 1, 8, or 10 members.

Sakata’s model said that the p, n, and L were a multiplet which does not fit into the above scheme of known particles! (e.g. could not account for So, S+)

Richard Kass

“Three Quarks for Muster Mark”, J. Joyce, Finnegan’s Wake

Model was developed by: Gell-Mann, Zweig, Okubo, and Ne’eman (Salam)

Three fundamental building blocks 1960’s (p,n,l) Þ 1970’s (u,d,s)

mesons are bound states of a of quark and anti-quark:

Can make up "wavefunctions" by combing quarks:

baryons are bound state of 3 quarks:

proton = (uud), neutron = (udd), L= (uds)

anti-baryons are bound states of 3 anti-quarks:

These quark objects are:

point like

spin 1/2 fermions

parity = +1 (-1 for anti-quarks)

two quarks are in isospin doublet (u and d), s is an iso-singlet (=0)

Obey Q = I3 +1/2(S+B) = I3 +Y/2

Group Structure is SU(3)

For every quark there is an anti-quark

quarks feel all interactions (have mass, electric charge, etc)

Richard Kass

The additive quark quantum numbers are given below:

Quantum #udscbt

electric charge2/3-1/3-1/32/3-1/32/3

I31/2-1/20000

Strangeness00-1000

Charm000100

bottom0000-10

top000001

Baryon number1/31/3 1/31/3 1/31/3

Lepton number000000

Successes of 1960’s Quark Model:

Classify all known (in the early 1960’s) particles in terms of 3 building blocks

predict new particles (e.g. W-)

explain why certain particles don’t exist (e.g. baryons with S = +1)

explain mass splitting between meson and baryons

explain/predict magnetic moments of mesons and baryons

explain/predict scattering cross sections (e.g. spp/spp = 2/3)

Failures of the 1960's model:

No evidence for free quarks (fixed up by QCD)

Pauli principle violated (D++= uuu wavefunction is totally symmetric) (fixed up by color)

What holds quarks together in a proton ? (gluons! )

How many different types of quarks exist ? (6?)

Richard Kass

Dynamic Quark Model (mid 70’s to now!)

Theory of quark-quark interaction Þ QCD

includes gluons

Successes of QCD:

“Real” Field Theory i.e.

Gluons instead of photons

Color instead of electric charge

explains why no free quarks Þ confinement of quarks

calculate lifetimes of baryons, mesons

Failures/problems of the model:

Hard to do calculations in QCD (non-perturbative)

Polarization of hadrons (e.g. L’s) in high energy collisions

How many quarks are there ?

Historical note:

Original quark model assumed approximate SU(3) for the quarks.

Once charm quark was discovered SU(4) was considered.

But SU(4) is a badly “broken” symmetry.

Standard Model puts quarks in SU(2) doublet,

COLOR exact SU(3) symmetry.

Richard Kass

How do we "construct" baryons and mesons from quarks ?

Use SU(3) as the group (1960’s model)

This group has 8 generators (n2-1, n=3)

Each generator is a 3x3 linearly independent traceless hermitian matrix

Only 2 of the generators are diagonal Þ 2 quantum numbers

Hypercharge = Strangeness + Baryon number = Y

Isospin (I3)

In this model (1960’s) there are 3 quarks, which are the eigenvectors (3 row column vector)

of the two diagonal generators (Y and I3)

Baryons are made up of a bound state of 3 quarks

Mesons are a quark-antiquark bound state

The quarks are added together to form mesons and baryons using the rules of SU(3).

M&S P133-140

It is interesting to plot Y vs. I3 for quarks and anti-quarks:

Richard Kass

Making mesons with (orbital angular momentum L=0)

The properties of SU(3) tell us how many mesons to expect:

old Vs new

p«u

n«d

l«s

Thus we expect an octet with 8 particles and a singlet with 1 particle.

If SU(3) were a perfect

symmetry then all particles

in a multiplet would have

the same mass.

Richard Kass

Making Baryons (orbital angular momentum L=0).

Now must combine 3 quarks together:

Expect a singlet, 2 octets, and a decuplet (10 particles) Þ 27 objects total.

Octet with J=1/2:

Richard Kass

Baryon Decuplet (J=3/2)

Expect 10 states.

Prediction of the W- (mass =1672 MeV/c2, S=-3)

Use bubble chamber to find the event.

1969 Nobel Prize to Gell-Mann!

“Observation of a hyperon with strangeness minus 3”

PRL V12, 1964.

Richard Kass

Leptonic Decays of Vector Mesons

What is the experimental evidence that quarks have non-integer charge ?

Þ Both the mass splitting of baryons and mesons and baryon magnetic moments depend on (e/m) not e.

Some quark models with integer charge quarks (e.g. Han-Nambu) were also successful in explaining mass patterns of mesons and baryons.

Need a quantity that can be measured that depends only on electric charge !

Consider the vector mesons (V=r, w, f, y, U): quark-antiquark bound states with:

mass ¹ 0

electric charge = 0

orbital angular momentum (L) =0

spin = 1

charge parity (C) = -1

parity = -1

strangeness = charm = bottom=top = 0

These particles have the same quantum numbers as the photon.

The vector mesons can be produced by its coupling to a photon:

e+e-® g ® V e.g. : e+e-® g ® Y(1S) or y

The vector mesons can decay by its coupling to a photon:

V® g ® e+e- e.g. : r® g ® e+e- (BR=6x10-5) or y®g ® e+e- (BR=6.3x10-2)

Richard Kass

The decay rate (or partial width) for a vector meson to decay to leptons is:

The Van Royen-

Weisskopf Formula

In the above MV is the mass of the vector meson, the sum is over the amplitudes that make up the meson, Q is the charge of the quarks and y(0) is the wavefunction for the two quarks to overlap each other.

|SaiQi|2

GL(exp) |SaiQi|-2

quarks

meson

GL(exp)

If we assume that |y(o)|2/M2 is the same forr, w, f, (good assumption since masses are 770 MeV, 780 MeV, and 1020 MeV respectively) then:

expect:GL(r) : GL(w) : GL(f) = 9 : 1 : 2

measure: (8.8 ± 2.6) : 1 : (1.7 ± 0.4)

Good agreement!

Richard Kass

Once the charm quark was discovered SU(3) was extended to SU(4) !

Richard Kass

PDG listing of the

known mesons.

With the exception of

the hb, all ground state

mesons (L=0) have been

observed and are in good

agreement with the

quark model.

A search for the hb is

presently underway!

Richard Kass

Magnetic Moments of Baryons

The magnetic moment of a spin 1/2 point like object in Dirac Theory is:

m = (eh/2pmc)s = (eh/2pmc) s/2, (s = Pauli matrix)

The magnetic moment depends on the mass (m), spin (s), and electric charge (e) of a point like object.

From QED we know the magnetic moment of the leptons is responsible for the energy difference between the 13S1 and 11So states of positronium (e-e+):

13S1® 11So Energy splitting calculated = 203400±10 Mhz

measured = 203387±2 Mhz

If baryons (s =1/2, 3/2...) are made up of point like spin = 1/2 fermions (i.e. quarks!) then we should be able to go from quark magnetic moments to baryon magnetic moments.

Note: Long standing physics puzzle was the ratio of neutron and proton moments:

Experimentally: mp/mn»-3/2

In order to calculate m we need to know the wavefunction of the particle.

In the quark model the space, spin, and flavor (isotopic spin) part of the wavefunction is symmetric under the exchange of two quarks.

The color part of the wavefunction must be anti-symmetric to satisfy the Pauli Principle (remember the D++). Thus we have:

y = R(x,y,z) (Isotopic) (Spin) (Color)

Þ Since we are dealing with ground states (L=0), R(x,y,z) will be symmetric.

always anti-symmetric because

hadrons are colorless

Richard Kass

Þ Consider the spin of the proton. We must make a spin 1/2 object out of

3 spin 1/2 objects (proton = uud)

From table of Clebsch-Gordon coefficients we find:

Also we have: |1 1> = |1/2 1/2> |1/2 1/2>

For convenience, switch notation to “spin up” and “spin down”:

|1/2 1/2> = and|1/2-1/2> = ¯

Thus the spin part of the wavefunction can be written as:

Note: the above is symmetric under the interchange of the first two spins.

Consider the Isospin (flavor) part of the proton wavefunction.

Since Isospin must have the two u quarks in a symmetric (I=1) state this means that spin must also have the u quarks in a symmetric state.

This implies that in the 2¯ term in the spin function the two are the u quarks.

But in the other terms the u’s have opposite sz’s.

We need to make a symmetric spin and flavor (Isospin) proton wavefunction.

Richard Kass

We can write the symmetric spin and flavor (Isospin) proton wavefunction as:

The above wavefunction is symmetric under the interchange of any two quarks.

To calculate the magnetic moment of the proton we note that if m is the magnetic moment operator:

m=m1+m2+m3 Composite magnetic moment = sum of moments.

<u| m|u> = mu = magnet moment of u quark

<d| m|d> = md = magnet moment of d quark

<usz| m|usz> = mu|sz = (2e/3)(1/mu)(sz)(h/2pc), with sz = ±1/2

<dsz| m|dsz> = md|sz = (-e/3)(1/md)(sz)(h/2pc), with sz = ±1/2

<usz=1/2| m|usz=-1/2> =0, etc..

For the proton (uud) we have:

<yp|m|yp> = (1/18) [24mu,1/2 + 12 m d,-1/2 + 3 md,1/2 + 3 md,1/2]

<yp|m|yp> = (24/18)mu,1/2 - (6/18)md,1/2 using md,1/2 = - md,-1/2

<yp|m|yp> = (4/3)mu,1/2 - (1/3)md,1/2

For the neutron (udd) we find:

<yn|m|yn> = (4/3)md,1/2 - (1/3)mu,1/2

Richard Kass

Let’s assume that mu = md = m, then we find:

<yp| m|yp>=(4/3)(h/2pc) (1/2)(2e/3)(1/m)-(1/3)(h/2pc)(1/2)(-e/3)(1/m)

<yp| m|yp>=(he/4pmc) [1]

<yn| m|yn>=( 4/3)(h/2pc) (1/2)(-e/3)(1/m)-(1/3)(h/2pc)(1/2)(2e/3)(1/m)

<yn| m|yn>=( he/4pmc) [-2/3]

Thus we find:

In general, the magnetic moments calculated from the quark model are in

good agreement with the experimental data!

-1.46

Richard Kass

Try to look inside a proton (or neutron) by shooting high energy electrons and muons at it and see how they scatter.

Review of scatterings and differential cross section.

The cross section (s) gives the probability for a scattering to occur.

unit of cross section is area (barn=10-24 cm2)

differential cross section= ds/dW

number of scatters into a given amount of solid angle: dW=dfdcosq

Total amount of solid angle (W):

Cross section (s) and Impact parameter (b)

and relationship between ds and db:

ds =|bdbdf|

Solid angle: dW =|sinqdqdf|

Richard Kass

Hard Sphere scattering:

Two marbles of radius r and R with R>>r.

b=Rsin(a)=Rcos(q/2)

db= -1/2Rsin(q/2)dq

ds=|bdbdf|= [Rcos(q/2)][1/2Rsin(q/2)dq]df

ds=|bdbdf|= [R2][1/4][sin(q)dq] df

The differential cross section is:

The total cross section is:

This result should not be too surprising since any “small” (r) marble within this area

will scatter and any marble at larger radius will not.

Richard Kass

Rutherford Scattering:

A spin-less, point particle with initial kinetic energy E and electric charge e scatters off a stationary point-like target with electric charge also=e:

note: s = ¥which is not too surprising since the coloumb force is long range.

This formula can be derived using either classical mechanics or non-relativistic QM.

The quantum mechanics treatment usually uses the Born Approximation:

with f(q2) given by the Fourier transform of the scattering potential V:

Mott Scattering: A relativistic spin 1/2 point particle with mass m, initial momentum p and electric charge e scatters off a stationary point-like target with electric charge e:

stationary target has

M>>m

In the low energy limit, p<< mc2, this reduces to the Rutherford cross section.

Kinetic Energy = E=p2/2m

Richard Kass

Mott Scattering: A relativistic spin 1/2 point particle with mass m, initial momentum p

and electric charge e scatters off a stationary point-like target with electric charge e:

In the high energy limit p>>mc2 and E»p we have:

“Dirac” proton: The scattering of a relativistic electron with initial energy E and final energy E' by a heavy point-like spin 1/2 particle with finite mass M and electric charge e is:

scattering with recoil,

neglect mass of electron,

E >>me.

q2 is the electron four momentum transfer: (p¢-p)2 = -4EE'sin2(q/2)

The final electron energy E' depends on the scattering angle q:

Richard Kass

What happens if we don’t have a point-like target, i.e. there is some structure inside

the target?

The most common example is when the electric charge is spread out over space and

is not just a “point” charge.

Example: Scattering off of a charge distribution. The Rutherford cross section is modified to be:

with: E=E¢ and

The new term |F(q2)| is often called the form factor.

The form factor is related to Fourier transform of the charge distribution r(r) by:

usually

In this simple model we could learn about an unknown charge distribution (structure) by measuring how many scatters occur in an angular region and comparing this measurement with what is expected for a "point charge" (|F(q2)|2=1 (what's the charge distribution here?) and our favorite theoretical mode of the charge distribution.

Richard Kass

Electron-proton scattering: We assume that the electron is a point particle. The "target" is a proton which is assumed to have some "size" (structure). Consider the case where the scattering does not break the proton apart (elastic scattering). Here everything is "known" about the electron and photon part of the scattering process since we are using QED.

As shown in Griffiths (8.3) and many other textbooks we can describe the proton in terms of two (theoretically) unknown (but measurable) functions or "form factors", K1, K2:

This is known as the Rosenbluth formula (1950). This formula assumes that scattering takes place due to interactions that involve both the electric charge and the magnetic moment of the proton.

Thus by shooting electrons at protons at various energies and counting the number of electrons scattered into a given solid angle (dW =|sinqdqdf|) one can measure K1 and K2.

note:q2 is the electron four momentum transfer: (p¢-p)2 = -4EE'sin2(q/2), and:

Richard Kass

An extensive experimental program of electron nucleon (e.g. proton, neutron) scattering was carried out by Hofstadter (Nobel Prize 1961) and collaborators at Stanford. Here they measured the "size" of the proton by measuring the form factors. We can get information concerning the "size" of the charge distribution by noting that:

For a spherically symmetric charge distribution we have:

Hofstadter et al. measured the root mean square radii

of the proton charge to be:

McAllister and Hofstadter,

PR, V102, May 1, 1956.

Scattering of 188 MeV electrons

from protons and helium.