Particle Physics -I

1. Particle Physics -I • Nucleon-nucleon interaction; meson exchange; Exchange giving rise to the four forces of nature; properties of pion • Pion-nucleon interaction; Collision kinematics and conservation laws;Phase shift Analysis; More conservation laws; N*s and s; Photon-nucleon interaction

2. Book: K R Krane, Introductory Nuclear Physics, Wiley • Nucleon-Nucleon force -Chapter 4 • Intro/ the deuteron pp 80-81 • Spin and Parity pp 83-84 • Magnetic Dipole moment pp 84-85 • Electric Quadrupole moment p 85 • Properties of force pp 100-102 • Exchange force pp 108-112 • Isospin pp 388-389 • Properties of pions pp 656-665 • Pion-proton interaction pp 671-675 • Scattering theory - partial wave analysis pp 408-411 • Conservation of Baryon number p715

3. NUCLEONS Nucleon is the generic name given to the proton(p+) and the neutron(n0) - particles with essentially the same mass but different charges. We view the proton and the neutron as different CHARGE STATES of the nucleon.

4. Properties of Nucleons Property Unit Proton Neutron Charge|e|+|e| 0 Mass MeV/c2 938.3 939.6 Spin h/2 ½½  N+2.793 -1.913 Radius fm(10-15m) 1 1 Stability Stable >1032 y  ~ 11.1 min Decay Mode ----- n p + e- +e

5. Di- Nucleons Only ONE stable - the DEUTERON (D) Charge+|e| Mass(MeV/c2) 1875.7 = 938.3 + 939.6 - 2.2 Binding Energy 2.2 MeV Spin (h/2)1 = ½ + ½ ( N) 0.857 ~ +2.793 -1.913 Radius fm(10-15m) 2 ~ 1 +1

6. Deuteron Cross- section  ~  rD2 ~ 12.5 fm2 = 125 mb Deformation from circle ~ 3mb P N

7. What is the deuteron state? • From ~ p + n we infer that the proton and neutron are in an orbital angular momentum L=0 state most of the time • This would imply a spherical nucleus. From the fact there is a deformation this must mean for a small fraction of the time the proton and neutron are in a different orbital angular momentum state

8. Unstable Di - nucleons D* - Unbound excited state of Deuteron. Eex ~ + 70 KeV. Spin=0.  ~ 10-22 sec Di-proton - Unbound state of two protons. Spin=0.  ~ 10-22 sec Di-neutron - Unbound state of two neutrons. Spin=0.  ~ 10-22 sec

9. Evidence for existence of di-proton (2-p) Nn En(MeV) p + D  p + n + p p + D  n + (2-p)

10. Possible combinations of two S=½ Nucleons in L=0 state

11. States of two nucleons with L=0 Only the deuteron - 1 CHARGE STATE -exists with S=1, Sz = +1,0, -1 i.e. 1 CHARGE STATE, 3 possible spin orientations D*, 2-p and 2-n - 3 CHARGE STATES -exist with S=0, Sz =0 i.e. 3 CHARGE STATES, 1 spin orientation There is a certain symmetry here!

12. 2-p D* 2-n S=0 Sz = 0 D Sz = +1,0,-1 S=1

13. Introducing ISOSPIN Formally, Isospin is a vector in a ‘Charge Space’ orthogonal to normal space, whose orientation in ‘charge space’ defines the charge on the particle. As a vector it obeys the same rules as the Spin vector - hence its name.Other than that it is totally unrelated to spin. For the nucleon, the Isospin vector t = ½, tz = ½. The relation between charge and tz is: Q/e = tz + ½ For two nucleons Isospin T = t1 + t2, Tz = t1z + t2z. There are two possibilities: T=1, Tz = +1, 0, -1 T=0, Tz = 0. and Q/e = Tz + 1

14. 2-p D* 2-n S=0 T=1 Sz = 0 Tz = +1 Tz = 0 Tz = -1 D Sz = +1,0,-1 S=1 T=0, Tz = 0

15. COMPONENTS OF NUCLEON-NUCLEON FORCE • Spin dependent - D bound, D* unbound , only difference orientation of spins • Charge or Isospin Independent - 2-p, D*, 2-n have same mass • Range of force 1- 2 fm • Evidence: RD = 2fm • RA = 1.2A1/3 V= 4/3RA3 ~ A  V(Nucleon) • Attractive - nuclei exist • Repulsive core • Evidence: RD= 2fm R(3He) = 2fm R(4He) = 1.7 fm

16. D = 2  ( 2 Nucleons) (3He) = 1.33  (3 Nucleons) (4He) = 0.7  (4 nucleons) D 3He 4He Binding Energy per nucleon pair B/½A(A-1) 2.23 2.53 4.67

17. The -particle (4He nucleus) is the smallest and most strongly bound nucleus in the periodic table. The nucleons within it clearly overlap. Why doesn’t it become even smaller? The N-N interaction appears to have a Repulsive Core of range ~ 0.7 fm. The magnitude of Binding Energies (MeV) means that the force must be a STRONG interaction (~ 100 times stronger than the EM interaction). Tensor Component To explain the non-spherical deuteron the N-N force must have a component which is non-central ie depends on the orientation of spins relative to spatial position.

18. Deuteron r r

19. Force between two dipole magnets N S N S d1 d2 r N S V(r ) = -1/r3{3(d1.r)(d2.r) - d1.d2} where r is the unit vector d1 r N S d2

20. Deuteron r r The EM potential V(r ) = -1/r3{3(p.r)(n.r) - p. n} leads to a deformation which is of order b. A strong interaction potential is needed: 2spsnVT(r) V(r ) = VT(r) {3(sp.r)(sn.r) - sp.sn} to give a deformation of order mb. -spsnVT(r)

21. Meson Exchange What can give rise to the tensor potential part of the N-N interaction? The answer is thought to be the virtual exchange of mesons between the nucleons. What do we mean by this?

22. The 4 exchange forces Force Range Exchange particle Strength Gravitational  Graviton,g, 0 10-39 Electromagnetic  Photon, , 0 10-2 Strong < 2fm Pion, , 140 MeV/c2 1 Weak ~ 0 W-boson,W,81 GeV/c2 10-5

23. The Electromagnetic Interaction In the interaction between two charged particles (eg an electron and a proton) photons are exchanged between the two particles. However we do not observe any photons being exchanged. As long as the exchange is described by the Uncertainty Principle it can still occur even if it is unobservable. Et ~  R~ c t =  c/E If E = 0 (photon) the range R = . Also as E2 =p2c2 + m02 c4 and m0 = 0 then the change in momentum of the emitting particle will be p= E/c and F = p/ t =  c/R2

24. The relativistic relation between total energy, momentum and mass helps us to visualise what may be happening in this virtual ( ie real but non - observable) exchange of photons. We see that: E =  (p2c2 + m02c4) The negative solution implies that we can view the vacuum as a sea of negative energies: p e-   E=0 The photon interacts with the vacuum locally. Energy is then transmitted out in all directions. When another charged particle is encountered this is then transferred to the particle by a photon.

25. The Strong Interaction Meson Exchange Et ~  R~ c t =  c/E For the strong interaction the long range part of the interaction is produced by the exchange of the lightest meson the pion, . M ~ 140 MeV/c2 ie E  140 MeV R = 6.6 10-22 MeV sec  3 1023 fm sec-1/ 140 MeV ie R = 1.4 fm   E=0

26. Properties of Pions Pions exist in three charge states + , - and 0 Therefore the isospin of the PionT = 1 with orientations in charge space Tz= +1, -1 and 0 respectively. Q/e = Tz M+ = M- = 139.57 MeV/c2 M0 = 134.97 MeV/c2. The spin of each pion is S = 0. Another important property of the Pion is its PARITY

27. PARITY This is defined for wave-functions in a symmetric potential. E.g. The N-N potential is symmetric about the mid-point of the two nucleons: V(r) = V(-r). Then for the probability density of the two nucleons:(r)2 = (-r)2 and: (r) = P(-r) where P = 1 A wave-function has parity P = +1 if it is even under reflection ( i.e. a swap) of both the particles co-ordinates and parity P = -1 if it is odd under reflection. In a symmetric potential wave-functions with Angular Momentum L=0,2,4,… are Even Parity, those with L=1,3,5,… are Odd Parity. All Strong and EM potentials are symmetric and each energy state has a definite parity - either Even or Odd. Particles of finite size with internal structure held together by a symmetric interaction to bediscussed later - e.g. Nucleons, Pions - have INTRINSIC PARITIES. Parity is conserved in Strong or EM interactions between particles.

28. Intrinsic Parity of the - Consider the reaction AT REST between a - and a deuteron: - + D  n + n Spin (S) 01½½ Angular Momentumtum (L) LD=0Lnn? Now Total angular Momentum J is conserved in all interactions. Therefore for Total Angular Momentum(J) 1 = 1 What is the value of Lnn? As the neutrons are like particles with S= ½ they are subject to Pauli Principle. i.e. the overall wave-function must be anti-symmetric

29. The overall wave-function of two neutrons consists of two parts: nn = (L) (S) • a space wave-function produced by solution of Schrodinger’s equation in a potential Vnn (r) i.e. containing a Legendre polynomial dependent on angular momentum state L. Wavefunctions with L= 0,2, .. Are Symmetric - Even Parity. Those with L=1,3,.. Are Anti-symmetric - Odd Parity. • a spin wave-function dependent on the spin of the two neutrons, S. This too can be symmetric or anti-symmetric.

30. The Spin Wave-function, (S) Two neutrons can exist with their spins parallel: This is a Symmetric state S = 1, Sz= +1,0 0r -1 Or anti-parallel: This is an anti-symmetric state S=0, Sz =0

31. For two neutrons J= 1cannot occur if either: • L=0 S=1 or L=2 S=1- Symmetric Space Wavefunction, Symmetric spin Wavefunction ie Overall Symmetry • L=1, S=0 - Anti-Symmetric Space Wavefunction, Anti-Symmetric spin Wavefunction. ie Overall Symmetry The ONLY possibility is L=1 S=1 - Anti-Symmetric Space Wavefunction, Symmetric spin Wavefunction. ie Overall Anti-Symmetry.J can then have values J=2 • orimportantly J=1 J=0 L=1 S=1 L=1 S=1 S=1 L=1 J=1

32. So now, in the reaction AT REST between a - and a deuteron: - + D  n + n Spin (S) 01½½ Angular Momentum (L) LD = 0Lnn? Total Angular Momentum(J) 1 = 1 We see that the value of Lnn has to be Lnn =1.Thus in the reaction, an initial even parity WF (LD = 0 ) goes to an odd parity Lnn=1 WF. But parity is conserved in the Strong Interaction.We must thus consider Intrinsic Parities. Then: Pinitial = PPnPpP(LD = 0 ) = Pfinal= PnPn P (Lnn =1) The intrinsic parities of the nucleons cancel each other. So: PP(LD = 0 ) = P (Lnn =1) i.e. -1  +1 = -1 The INTRINSIC PARITY of the PION is NEGATIVE.Incidentally the INTRINSIC PARITY of the NUCLEON is POSITIVE - by convention.

33. Pion Exchange between Nucleons The negative intrinsic parity of the pion has important consequences when it is exchanged between nucleons. It gives rise to the TENSOR force. When the pion is emitted by a nucleon, as the nucleon continues to have positive parity, the pion must be emitted with L =1 - to cancel out its intrinsic negative parity. In order for this to happen, to conserve total angular momentum either: (a) the emitting and receiving nucleons must reverse their initial spin directions (b) the emitting and receiving nucleons must recoil with L=1. Depending on their initial spatial orientation (a) or (b) happens.

34. Lnp=0 Lnp=0 Pion transfer occurs by nucleons changing spin direction 0 L=1 Lnp=2 Lnp=0 Pion transfer by angular momentum change L=1 L=1 0 L=1

35. So in the loosely bound deuteron ONE PION exchange is the dominant process leading to the tensor force. In 3He, with nucleons closer together, TWO PION exchange is more likely. Here no nett angular momentum need be transferred between two nucleons - i.e. a central force 0 0 This has a range ~ 0.7 fm. In 4He, with the nucleons overlapping, even heavier mesons can be exchanged. The , and  mesons have masses between 600 and 900 MeV/c2, negative parity and S=1. When these are transferred (with L=1 to conserve parity) nucleons are forced to L=2 state -i.e . Must move apart, are effectively repelled.

36. Pion-Proton interaction We can create beams of pions by bombarding a heavy nucleus with a high energy proton beam from an accelerator. + 0 Beams of charged pions are formed using electric and magnetic fields. Proton beam T= 3000 MeV n Tungsten (W) target

37. Liquid hydrogen target Area AH, length tH,density H + beam Pion detectors • Number of scattered pions = N(IN) –N(OUT) = • N(IN)* (NAV*H*tH *AH )*(T /AH) • By analogy with throwing point-like darts at a board T would just be the cross-sectional area of a proton, but this analogy needs to be extended as • there is an interaction between the pion and the nucleon, predominantly the STRONG force with a certain strength and range, but there is also a small contribution from the Coulomb force. • So we extend the definition of T to: • THE TOTAL CROSS-SECTION FOR THE STRONG INTERACTION OF A PION WITH A PROTON • T is measured in barns (10-24 cm2) and, when divided by AH, is the probability for a pion to interact with a proton. It is energy dependent.

38. Pion-proton total cross-sections 195 600 900 1380 T  (MeV)

39. Peaks in the p cross-section are interpreted as resonances in the p system:  +p  ( p)*   +p • The lowest resonant state occurs in both the + p and the -p cross-sections at a pion incident laboratory energy of T = 195 MeV. • The width of this resonance is ~ 100 MeV. Using the Uncertainty Principle  E t ~ we find that the lifetime of the resonance is ~ 10-23 sec. • What is its mass? This can be calculated by applying the laws of Conservation of Energy and Momentum to the formation of the resonance. - p (-p)* T-, p-,m-mp T*,P*,m*,v*

40. C.of E. T- + m-c2 + mpc2 = m*c2 + T* (1) C.of M. p- = p* and hence p-2c2= p*2c2 (2) Relativistic kinematic relationships Total Energy E = m0c2/ (1-v2/c2) Momentum p = {m0/ (1-v2/c2)}v Eliminating v, E2 = p2c2 + m02c4. Also E= T + m0c2 and hence E2= T2 + 2m0c2T + m02c4 Thus p2c2 = T2 + 2m0c2T In (2) we then have T-2 + 2m-c2T- = T*2 + 2m*c2T* and substituting for T* from (1): m*c2 = {(m-+ mp)2c4 + 2T-mpc2} Substituting pion, proton masses and T- = 195 MeV we find: m* = 1235 MeV/c2

41. The Total centre-of-mass Energy is E =m*c2 = 1236 MeV. To see this impose a back velocity on the laboratory such that the resonant state is brought to rest ( i.e. - v*). Then the pion and the proton will have equal and opposite momenta. pCM pCM m*c2 To find the spin state of the resonancewe needscattering theory.

42. Scattering Theory -Spin-less particles, Elastic Scattering,CM reference frame r inc = eikz Solution of: -2/2m(2 inc/z2) = E inc. As E=pCM2/2m and pCM = k this reduces to 2 inc/z2 = -k2inc. Incident pion wave Proton  Unscattered pion wave z Scattered pion wave scatt

43. scatt is a solution of: -2/2m(2scatt) = (E -V(r))scatt • scatt= f() eikr/r • at large r where E>> V • Solution independent of  as there is cylindrical symmetry about the incident beam direction. • The total wavefunction after the p interaction consists of an unscattered part (eikz) as well as scatt. • final = eikz + scatt

44. How are inc, scatt related to observables? • We know * =  2 is the probability of a particle’s existence at a point in space. If there are: • n pions/unit area in the incident beam and • the CM beam velocity is v then the number of pions incident per second per unit area is: nveikze-ikz = nv. The number of pions elastically scattered at angle  per unit area per second is : nvf()f*()eikre-ikr/r2

45. Experimentally we measure the number of pions scattered into unit solid angle (unit area /r2) at angle  per second. This is: nv f() f*(). Then the differential cross-section for elastic scattering del/d() = number of pions scattered into unit solid angle at angle  per second / number of pions incident per unit area per second i.e. del/d() = f() f*() . el is the cross-section for elastic scattering. el/unit area is the probability for scattering into 4 solid angle.

46. el = 4 d el/d () d  = 4 f()f*() d  Now, we cannot specify the x- and y- positions of the incident pions, momentum p, relative to a target proton. p angular momentum state lp = p  d d 2 fm 1013 fm

47. The pions can interact, in principle, with the protons in ALL possible angular momentum states lp. This allows us to rectify the asymmetry in the scattering - initial plane wave, final spherical scattered and plane wave - by expressing the plane waves in terms of ingoingandoutgoing spherical waves representing a pion-proton state of angular momentum number l = lp. inc = eikz = eikrcos = 1/kr 0 (2l +1) Pl(cos){eikr - e-i(kr-l)}/2i where Pl(cos) is a Legendre polynomial. outgoing wave ingoing wave

48. When the p interaction occurs only outgoing waves will be affected. Assuming only elastic scattering occurs ( i.e. no inelastic scattering) then these will be advanced (if pinteraction attractive) or retarded (if pinteraction repulsive) compared to the situation if there were no pinteraction. i.e. there will be a phase shift in the outgoing waves. By convention this is 2(l). Thus the wavefunction describing the total situation after scattering: final =1/kr0(2l +1) Pl(cos){ei(kr+ 2(l)) - e-i(kr-l)}/2i = eikz +f() eikr/r (3)

49. As: eikz = 1/kr 0 (2l +1) Pl(cos){eikr - e-i(kr-l)}/2i we then find that, by rearrangement of (3): f() = 1/2ik 0 (e2i(l) -1)(2l +1) Pl(cos) =1/2ik 0 ei(l) (ei(l) - e-i(l) )(2l +1) Pl(cos) = 1/2ik 0 ei(l) 2isin (l)(2l +1) Pl(cos) = 1/k 0 ei(l) sin (l)(2l +1) Pl(cos) Now el = 4 d el/d () d  = 4 f()f*() d . As 4 Pl(cos) Pl’(cos) d  = 4/(2l+1) when l=l’’ = 0otherwiseel = 4/k2 0 sin2(l)(2l +1)

50. Applying scattering theory to the 1235 Mev/c2 resonance • Scattering theory applies to spinless particles only and predicts: el = 4/k2 0 sin2(l)(2l +1) • We measureT = el + R • When does T = el ??? • i.e.When is R = 0 ???