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Nucleon: T = ½, m t =  ½. For a nucleus , by extension: m t = ½ (Z - N).

16.451 Lecture 13: Isospin and symmetries 16/10/2003. 1. Nucleon: T = ½, m t =  ½. For a nucleus , by extension: m t = ½ (Z - N).

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Nucleon: T = ½, m t =  ½. For a nucleus , by extension: m t = ½ (Z - N).

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  1. 16.451 Lecture 13: Isospin and symmetries 16/10/2003 1 • Nucleon: T = ½, mt =  ½. For a nucleus, by extension: mt = ½ (Z - N). • If neutrons and protons are really “identical” as far as the strong interaction is concerned, then nuclei with the same mass number but (Z,N) interchanged ought to be very similar. These are called “mirror nuclei”, e.g. 11B (5,6) and 11C (6,5) • Energy spectra line up after correction for Coulomb energy difference in the ground state. 

  2. 2 Two nucleon (NN) system and isospin: • Pauli exclusion principle: wave function for identical fermions must be antisymmetric • if the particle labels are exchanged • How do we tell what symmetry the isospin configurations have? T = 0 or 1 for NN. • Use symbolic representation:  = ½ and  = - ½ • The 4 configurations (m1, m2) are: (), (), () , () • () and () are symmetric– exchanging the symbols (1,2) has no effect. • These correspond to total isospin (T, mT) = (1, 1) and (1, -1) • (), () states correspond to mT = 0, but they have mixed symmetry. • Solution: make symmetric and antisymmetric combinations of the mixed states: • symmetric: () + ()  () + () (T=1, mT = 0) • anti – : () - ()  () - () = - {() - () } (T=0, mT=0 Bottom line: T = 1 states are symmetric,T = 0 antisymmetric. (Same for spin, S)

  3. 3 implications of isospin symmetry: (Krane, ch. 11) • pp and nn systems are always T = 1 • np system is () , ie it can be partly T = 1 and partly T = 0 • for a nucleus, mT = ½ (Z-N) and T = |mT|, ie lowest energy has smallest T • Example: “isobaric triplet” 14C, 14N, 14O: 14O: Z = 8, N= 6 mT = + 1, T = 1 14C: Z = 6, N = 8 mT = -1, T = 1 14N mT = 0, T = 0 (g.s) and T = 1 (8 MeV)

  4. 4 Final example, evidence for isospin: Consider the deuteron, 2H = (np) bound state (d) Quantum numbers: mT = 0, T = 0 J = 1 + (S = 1, L = 0,  = (-1)L ) How do we know it has T = 0 ? “Isospin selection rules”: The reaction: 1) d + d   + 4He occurs, but isospin analysis:(T = 1 deuteron also works) 2) d + d  ° + 4He does not isospin analysis: (only T = 0 prevents this!) Bottom line: T is conserved by the strong interaction. Energy depends on T but not on mT

  5. Isospin and Quarks: 5 There are a total of 6 quarks in the Standard Model (u,d,s,c,t,b – more later!) but only two play a significant role in nuclear physics: u and d. Not surprisingly, isospin carries over into the quark description: the “up” quark has isospin T = ½ “up” and similarly for the “down” quark: Isospin addition for the proton: p = (uud), mt = ½ + ½ - ½ = ½  neutron: n = (udd), mt = ½ - ½ - ½ = - ½  What about the delta? Addition of 3 x isospin- ½ vectors: T = 1/2 or 3/2; T = 3/2 is the : ++ = (uuu), + = (uud), ° = (udd), - = (ddd)  What about antiquarks? same isospin but oppositemt  e.g. pion: (+, °, - ) 

  6. Lifetime: 885.7  0.8 sec (world average) Back to the neutron: 6 

  7. 7 Neutron Beta Decay: • a fundamental WeakInteraction process • lifetime  is relatively long: (lecture 6!) • large  implies small transition rate , therefore ‘weak’ interaction V(r) • compare to  resonance decay: +  p + °, a strong interaction process, • with  = 5.7 x 10-24 seconds!!! • precision studies of neutron decay are a very important testing ground for the • “Standard Model” of fundamental interactions, as we shall see.... • interaction is almost pointlike, that is, the neutron disappears and the decay • products appear almost instantaneously at the same place. (Fermi theory) • modern picture (Assignment 3): W- boson

  8. “after” “before” Closer look: electron energy spectrum 8 (energy cons.) (momentum) Define the “Q – value”: (in general, Q > 0 for a reaction to proceed) From Particle Data Group entries: Q = 0.78233  0.00006 MeV ( 60 eV!)

  9. Fit to expected spectrum shape including detector model. Endpoint smeared with detector resolution Electron Energy Spectrum from “PERKEO” expt. at ILL reactor, France: 9 Bopp et al., Phys. Rev. Lett. 56, 919 (1986) Note: presence of neutrino affects this shape dramatically – otherwise it would be a sharp peak at a value determined by momentum /energy conservation!

  10. 10 Case study: “state of the art” neutron lifetime measurement

  11. decay proton, to detector transmitted beam incoming n beam neutron detector decay volume, length L 11 Outline of the method: decay rate: measure rate by counting decay protons in a given time interval (dN/dt) and normalizing to the neutron beam flux (N) Ideally done with “cold neutrons”, e.g. from a reactor, moderated in liquid hydrogen... Issues: 1. precise decay volume ? 2. proton detection ? 3. beam normalization ? ...

  12. 12 Neutron beam distribution – definitely not monoenergetic: • ~ MeV neutrons from a reactor are “moderated” by scattering in a large tank of • water (“thermal”) or liquid hydrogen (“cold”) • after many scatterings, they come to thermal equilibrium with the moderator and • are extracted down a beamline to the experiment • velocity distribution is “Maxwellian”: energies in the meV range (kT = 26 meV @ 293K) Krane, Fig 12.4

  13. 10B + n   + 7Li + 2.79 MeV  = ovo/v with o = 3838 b (barns) kinetic energy of ionized fragments can be converted into an electrical signal  detector Neutron detection at low energy: (Krane, ch. 12) 13 • several light nuclei have enormous neutron capture cross sections at low energy: • (recall, cross sectional area of a nucleus, e.g. 10B is about 0.2 barns, lecture 4) • key feature: cross sections scale as 1/velocity at low energy

  14. decay proton, to detector transmitted beam incoming n beam (DC beam, thermal energy spread) 10B neutron detector decay volume, length L Decays and transmitted detector counts accumulated for time T Put this together for detector counting rates as shown: 14 neutrons must be captured to be detected!

  15. proton detector variable length Penning trap (16 electrodes) thin 10B foil to capture beam n’s  particle detectors for capture products protons spiral around field lines when let out of the trap Experimental details (all in vacuum, at ILL reactor, France): 15 • use Penning trap to confine decay protons • let them out of the trap after accumulation interval T • measure the ratio Ndet/Ndecay as a function of trap length L  slope gives 

  16. Rate versus trap length L proton energy from timing: about 34 keV (after kick – raw energy is only 0.75 keV) Result: 893.6  5.3 seconds (1990) PDG average: 885.7  0.8 (2003) Amazing results: 16

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