Transcript Lecture 8, Quarks

P780.02 Spring 2003 L8
Quarks
Richard Kass
Over the years inquiring minds have asked:
“Can we describe the known physics with just a few building blocks ?”
 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
 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:
 Model of Fermi and Yang (late 1940’s-early 50’s):
pion is composed of nucleons and anti-nucleons (used SU(2) symmetry)
p+ = pn, p- = np, po = pp - n n
With the discovery of new unstable particles (L, k) a new quantum
strangeness
note this model was proposed
before discovery of anti-proton
!
number
was invented:
P780.02 Spring 2003 L8
Quarks
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:
ko Y
1
p-
k+
p+
I3
1
po
-1
k-
-1
k
o
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
+
o
In this model: p+ = pn, p- = np, po = pp - n n, k = p L, k = n L
+
Baryons are made up of 3 at a time:  = Lpn
This model obeys Fermi statistics and explains why:
Mn > Mp and Mko > Mk+ and Mp+ > Mpo
Unfortunately, the model had major problems….
P780.02 Spring 2003 L8
Quarks
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/2](|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/2](|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  2 = 3 1
Back to Sakata's model:
For SU(3) there are 2 quantum numbers and the group structure is more complicated:
3  3  3 = 1  8  8  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 o, +)
P780.02 Spring 2003 L8
Early 1960’s Quarks
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)  1970’s (u,d,s)
mesons are bound states of a of quark and anti-quark:
Can make up "wavefunctions" by combing quarks:
+
o
p+ = ud, p- = du, po = 1 (uu - d d), k = ds, k = ds
2
baryons are bound state of 3 quarks:
proton = (uud), neutron = (udd), L= (uds)
anti-baryons are bound states of 3 anti-quarks:
p  uud n  udd L uds
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)
P780.02 Spring 2003 L8
Early 1960’s Quarks
Richard Kass
The additive quark quantum numbers are given below:
Quantum #
u
d
s
c b
t
electric charge 2/3
-1/3 -1/3 2/3 -1/3 2/3
I3
1/2
-1/2 0
0 0
0
Strangeness
0
0
-1
0 0
0
Charm
0
0
0
1 0
0
bottom
0
0
0
0 -1 0
top
0
0
0
0 0
1
Baryon number 1/3
1/3 1/3 1/3 1/3 1/3
Lepton number 0
0
0
0 0
0
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?)
P780.02 Spring 2003 L8
Dynamic Quarks
Richard Kass
Dynamic Quark Model (mid 70’s to now!)
Theory of quark-quark interaction  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  confinement of quarks
calculate cross sections, e.g.
e+e-  qq
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.
P780.02 Spring 2003 L8
Richard Kass
From Quarks to Particles
How do we "construct" baryons and mesons from quarks ?
M&S P133-140
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  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).
It is interesting to plot Y vs. I3 for quarks and anti-quarks:
Y 1
Y 1
2/3
d
1/3
u
I3
1
-1
s
s
-1
u
d
2/3
-1
-1
I3
1
P780.02 Spring 2003 L8
Making Mesons with Quarks
Making mesons with (orbital angular momentum L=0)
The properties of SU(3) tell us how many mesons to expect:
3 3  1 8
Thus we expect an octet with 8 particles and a singlet with 1 particle.
Richard Kass
old Vs new
pu
nd
ls
If SU(3) were a perfect
symmetry then all particles
in a multiplet would have
the same mass.
P780.02 Spring 2003 L8
Baryon Octet
Richard Kass
Making Baryons (orbital angular momentum L=0).
Now must combine 3 quarks together:
3  3  3  1  8  8  10
Expect a singlet, 2 octets, and a decuplet (10 particles) 27 objects total.
Octet with J=1/2:
P780.02 Spring 2003 L8
Baryon Decuplet
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
P780.02 Spring 2003 L8
Quarks and Vector Mesons
Richard Kass
Leptonic Decays of Vector Mesons
What is the experimental evidence that quarks have non-integer charge ?
 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  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)
P780.02 Spring 2003 L8
Richard Kass
Quarks and Vector Mesons
The decay rate (or partial width) for a vector meson to decay to leptons is:
g 
16p2em
M
2
V
|  ai Qi | y(0)
2
2
The Van RoyenWeisskopf Formula
i
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.
meson
r
1 ( uu - d d)
2
1 ( 2 -(2 3
w
1 ( uu + dd)
2
1 ( 2 +(- 1 ) )
3
2 3
f
ss
y
cc
(- 1 )
1)
3
2
)
2
2
=
1
2
7 Ke V
13 .2
=
1
18
0.8 Ke V
12 .8
=
1
9
1.3 Ke V
11 .8
=
4
9
4.7 Ke V
10 .5
=
1
9
1.2 Ke V
10 .6
3
(2 )
2
3
Y
L(exp) L(exp) |aiQi|-2
|aiQi|2
quarks
bb
(- 1 )
3
2
If we assume that |y(o)|2/M2 is the same for r,w,f, (good assumption since masses are 770
MeV, 780 MeV, and 1020 MeV respectively) then:
expect:
measure:
L(r) : L(w) : L(f) = 9 : 1 : 2
(8.8 ± 2.6) : 1 : (1.7 ± 0.4)
Good agreement!
P780.02 Spring 2003 L8
Four Quarks
Richard Kass
Once the charm quark was discovered SU(3) was extended to SU(4) !
P780.02 Spring 2003 L8
More Quarks
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!
P780.02 Spring 2003 L8
Magnet Moments of Baryons
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 =
measured =
203400±10 Mhz
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:
always anti-symmetric because
y = R(x,y,z) (Isotopic) (Spin) (Color) hadrons are colorless
 Since we are dealing with ground states (L=0), R(x,y,z) will be symmetric.
P780.02 Spring 2003 L8
Magnet Moments of Baryons
Richard Kass
 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:
2 |1 1> |1/2 -1/2> 1 |1 0> |1/2 1/2>
|1/2 1/2> =
3
3
Also we have: |1 1> = |1/2 1/2> |1/2 1/2>
|10> =
1
2
|1/2 1/2> |1/2 -1/2> +
1
2
|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:




1 / 2 1 / 2  2 / 3    - 1 / 3 1 / 2         1/ 6 2    -    -   
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.
P780.02 Spring 2003 L8
Magnet Moments of Baryons
Richard Kass
We can write the symmetric spin and flavor (Isospin) proton wavefunction as:
2u  u  d  2d  u  u  2u  d  u 

 p  1 / 18 

-u  d  u  -u  u  d  -u  u  d  - d  u  u  -u  d  u  -d  u  u 
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
<usz=1/2| m |usz=-1/2> =0, etc..
<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
For the proton (uud) we have:
<yp|m|yp> = (1/18) [24mu,1/2 + 12 md,-1/2 + 3 md,1/2 + 3 m d,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
P780.02 Spring 2003 L8
Magnet Moments of Baryons
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:
 yp |m | yp 
 yn | m | yn 

mp
mn
-
3
2
In general, the magnetic moments calculated from the quark model are in
good agreement with the experimental data!
-1.46
P780.02 Spring 2003 L8
Richard Kass
Are Quarks really inside the proton?
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):
1 2p
1
2p
-1 0
-1
0

 dW   d cos q  df  4 p
Cross section (s) and Impact parameter (b)
and relationship between ds and db:
ds =|bdbdf|
Solid angle: dW =|sinqdqdf|
P780.02 Spring 2003 L8
Richard Kass
Examples of scattering cross sections
Hard Sphere scattering:
Two marbles of radius r and R with R>>r.
b=Rsin()=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:
ds
bdbdf
( R 2 / 4) sin qdqdf R 2



dW sin qdqdf
sin qdqdf
4
The total cross section is:
ds
R2
2pR 2 1
2
s  
dW   sin qdqdf 
sin
q
d
q

p
R

dW
4
4 -1
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.
P780.02 Spring 2003 L8
Richard Kass
Examples of scattering cross sections
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:
2

ds 
e2



2
dW 4E sin ( q / 2) 
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:
ds
2 2
 f (q )
dW
2
with f(q ) given by the Fourier transform of the scattering potential V:
f (q 2 )   eiq r V(r )dr
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:
2
ds 


 (mc)2  p 2 cos2
  2 2
dW 2p sin (q / 2)  
q 
2 
stationary target has
M>>m
In the low energy limit, p<< mc2, this reduces to the Rutherford cross section.
Kinetic Energy = E=p2/2m
P780.02 Spring 2003 L8
Richard Kass
Examples of scattering cross sections
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:
2
ds 


 (mc)2  p 2 cos2 q 
  2 2
dW 2p sin (q / 2)  
2 
In the high energy limit p>>mc2 and Ep we have:
2
ds 

cos(
q
/
2)


 
2
dW 2Esin (q / 2) 
“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:
2
scattering with recoil,
2



ds   cos(q / 2)  E 
q


1
tan 2 (q / 2)
2
2

dW 2Esin (q / 2)  E 
 2M

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:
E
E 
2E 2
1
sin ( q / 2)
M
P780.02 Spring 2003 L8
Richard Kass
Examples of scattering cross sections
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:
2
2

ds 
e2
2

 F(q )

with: E=E and q 2  -4Esin 2 (q / 2)
2
dW 4E sin ( q / 2) 
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:
F(q2 )  r(r )eiqr d 3r
3
usually  r(r)d r  1
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.
P780.02 Spring 2003 L8
Richard Kass
Elastic electron proton scattering (1950’s)
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,
2
K2:


ds

E 

2
2
2
2

 

2K1(q )sin (q / 2)  K2 (q )cos (q / 2)
dW 4M p Esin 2 (q / 2)  E
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:
E 
E
1
2E 2
sin ( q / 2)
M
P780.02 Spring 2003 L8
Richard Kass
Elastic electron proton scattering (1950’s)
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:


(q  r )2
d r   r(r)1  iq  r   d 3r
2!


iqr 3
2
F(q )  r(r )e
For a spherically symmetric charge distribution we have:
q2 2
q2 2
F(q ) 1 r
 r r(r)dr  1 6
6
2
Hofstadter et al. measured the root mean square radii
of the proton charge to be:
r2
1/2
charge
 (0.74  0.24)  10 -13 cm
McAllister and Hofstadter,
PR, V102, May 1, 1956.
Scattering of 188 MeV electrons
from protons and helium.