1. P780.02 Spring 2002 L9Richard Kass Four Quarks Once the charm quark was discovered SU(3) was extended to SU(4) !

2. P780.02 Spring 2002 L9Richard Kass More Quarks PDG listing of the known mesons. With the exception of the  b, all ground state mesons (L=0) have been observed and are in good agreement with the quark model. A search for the  b is presently underway!

3. P780.02 Spring 2002 L9Richard Kass Magnet Moments of Baryons Magnetic Moments of Baryons The magnetic moment of a spin 1/2 point like object in Dirac Theory is:  = (eh/2  mc)s = (eh/2  mc)  /2, (  = 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 1 3 S 1 and 1 1 S o states of positronium (e - e + ): 1 3 S 1  1 1 S o 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: m p /m n  -3/2 In order to calculate  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  ++ ). Thus we have:  = R(x,y,z) (Isotopic) (Spin) (Color)  Since we are dealing with ground states (L=0), R(x,y,z) will be symmetric.

4. P780.02 Spring 2002 L9Richard Kass Magnet Moments of Baryons  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 s z ’s. We need to make a symmetric spin and flavor (Isospin) proton wavefunction.

5. P780.02 Spring 2002 L9Richard Kass Magnet Moments of Baryons 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  is the magnetic moment operator:  =  1 +  2 +  3 Composite magnetic moment = sum of moments. =  u = magnet moment of u quark =  d = magnet moment of d quark =  u |s z = (2e/3)(1/m u )(s z )(h/2  c), with s z = ±1/2 =  d |s z = (-e/3)(1/m d )(s z )(h/2  c), with s z = ±1/2 For the proton we have: = (1/18) [24  u,1/2 + 12  d,-1/2 + 3  d,1/2 + 3  d,1/2 ] = (24/18)  u,1/2 - (6/18)  d,1/2 using  d,1/2 = -  d,-1/2 = (4/3)  u,1/2 - (1/3)  d,1/2 For the neutron we find: = (4/3)  d,1/2 - (1/3)  u,1/2

6. P780.02 Spring 2002 L9Richard Kass Magnet Moments of Baryons Let’s assume that m u = m d = m, then we find: =(4/3)(h/2  c) (1/2)(2e/3)(1/m)-(1/3)(h/2  c)(1/2)(-e/3)(1/m) =(he/4  mc) [1] =( 4/3)(h/2  c) (1/2)(-e/3)(1/m)-(1/3)(h/2  c)(1/2)(2e/3)(1/m) =( he/4  mc) [-2/3] Thus we find: In general, the magnetic moments calculated from the quark model are in good agreement with the experimental data!

7. P780.02 Spring 2002 L9Richard 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 (  ) gives the probability for a scattering to occur. unit of cross section is area (barn=10 -24 cm 2 ) differential cross section= d  /d  number of scatters into a given amount of solid angle: d  =d  dcos  Total amount of solid angle (  ): Cross section (  ) and Impact parameter (b) and relationship between d  and db: d  =|bdbd  | Solid angle: d  =|sin  d  d  |

8. P780.02 Spring 2002 L9Richard Kass Examples of scattering cross sections Hard Sphere scattering: Two marbles of radius r and R with R>>r. b=Rsin(  )=Rcos(  /2) db= -1/2Rsin(  /2)d  d  =|bdbd  = [Rcos(  /2)][1/2Rsin(  /2)d  ]d  d  =|bdbd  = [R 2 ][1/4][sin(  )d  ] d  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.

9. P780.02 Spring 2002 L9Richard 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: note:  =  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(q 2 ) 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 : In the low energy limit, p<< mc 2, this reduces to the Rutherford cross section. Kinetic Energy = E=p 2 /2m stationary target has M>>m

10. P780.02 Spring 2002 L9Richard 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 : In the high energy limit p>>mc 2 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 >>m e. q 2 is the electron four momentum transfer: (p-p) 2 = -4EE'sin 2 (  /2) The final electron energy E' depends on the scattering angle  :

11. P780.02 Spring 2002 L9Richard 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: with: E=E and The new term |F(q 2 )| is often called the form factor. The form factor is related to Fourier transform of the charge distribution  (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(q 2 )| 2 =1 (what's the charge distribution here?) and our favorite theoretical mode of the charge distribution.

12. P780.02 Spring 2002 L9Richard 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", K 1, K 2 : 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 ( d  =|sin  d  d  | ) one can measure K 1 and K 2. note:q 2 is the electron four momentum transfer: (p-p) 2 = -4EE'sin 2 (  /2), and:

13. P780.02 Spring 2002 L9Richard 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: 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.