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Make up schedule. Monday Tuesday Wednesday Thursday Friday. 8:00am 9:00 10:00 11:00 12:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00. Apr 24???. Apr 25???. April 11 114 Ferguson. April 27 leave early morning by car. April 11 catch a 1:00pm flight.

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8:00am 9:00 10:00 11:00 12:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00

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  1. Make up schedule Monday Tuesday Wednesday Thursday Friday 8:00am 9:00 10:00 11:00 12:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00 Apr 24??? Apr 25??? April 11 114 Ferguson April 27 leave early morning by car April 11 catch a 1:00pm flight Apr 24???

  2. The Nuclear pp cycle producing energy in the sun 6 protons  4He + 6g+ 2e + 2p 26.7 MeV Begins with the reaction 0.26 MeV neutrinos

  3. 500 trillion solar neutrinos every second!

  4. ALL NEUTRINOS ARE LEFT-HANDED Helicity = ms/s = -1 ALL ANTI-NEUTRINOS ARE RIGHT-HANDED Helicity = ms/s = +1

  5. Dirac Equation(spin-½ particles) ( p my  0 0 sj -sj0 j  ( 0p0  •p y  my 0 s -s0 0 p •s - p •s0 p • ( ) = ( ) where p •s = px( ) + py( ) + pz( ) 0 1 1 0 0 -i i 0 1 0 0 -1 = ( ) pz px-ipy px+ipy-pz

  6. Our “Plane wave” solutions(for FREEDirac particles) y(r,t) = aexp[i/h(Et-p • r)]u(E,p) ae(i/h)xmpmu(E,p) which gave ( p mu = ( )( ) E/c-mc -p•s uA p•s -E/c-mcuB from which we note: uA= ( p • s )uB uB= ( p • s )uA c E-mc2 c E+mc2

  7. Dirac Equation(spin-½ particles) g 0 -pxg1-pyg 2-pzg 3= my E c recall g 5=ig 0g 1g 2g 3 multiply from left by (-ig 1g 2g 3) -ig 1g 2g 3g 0 = +ig 0g 1g 2g 3 = g 5 since g mg n=-g ng m -ig 1g 2g 3g 1 = -i(g 1)2g2g 3 = +ig2g 3 = +i(s2s3)( )( ) = +i(is1)( ) = s1I since (g i)2 = -I 0 1 -1 0 0 1 -1 0 -1 0 0 -1 so -px g 1 - px s 1I -ig 1g 2g 3g 2 = s2I -ig 1g 2g 3g 3 = s3I (g 5 -(p • s)I)y = -img 1g 2g 3y E c

  8. This gives an equation that looks MORE complicated! How can this form be useful? (g 5 -(p • s)I)y = -img 1g 2g 3y E c For a ~massless particle (like theor any a relativistic Dirac particle E >> moc2) E=|p|c as mo0 (or at least mo<<E) ( |p|g 5 -(p • s)I)y = 0 Which then gives: or: (g 5 -p • sI)y = 0 ^ s0 0s sI = What do you think this looks like? ^ p • sI is a HELICITY OPERATOR!

  9. In Problem Set #5 we saw that if the z-axis was chosen to be the direction of a particle’s momentum were all well-defined eigenspinors of Sz i.e. (p • sI)u(p)= u(p) ^ “helicity states” (g 5 -p • sI)y = 0 ^ So • 5 “measures” the helicity of y g 5y = (p • sI)y ^

  10. Looking specifically at g5u(p) = = uA uB uB uA For massless Dirac particles (or in the relativistic limit) (p • sI)u(p) ^ g5u(p)=

  11. We’ll find a useful definition in the “left-handed spinor” Think: “Helicity=-1” (1-5) 2 uL(p)=u(p) In general NOT an exact helicity state (if not massless!) Since 5u(p)= ±u(p) for massless or relativistic Dirac particles 0 if u(p) carries helicity +1 u(p) if u(p) carries helicity -1 if neither it still measures how close this state is to being pure left-handed separates out the “helicity-1 component” Think of it as a “projection operator” that picks out the helicity -1 component of u(p)

  12.  5v(p) = -(p· I)v(p) Similarly, since for ANTI-particles: again for m 0 (1+5) 2 we also define: vL(p)=v(p) with corresponding “RIGHT-HANDED” spinors: (1+5) 2 (1-5) 2 uR(p) =u(p) vR(p)=v(p) and adjoint spinors like since 5†=5 since 5 = -5

  13. Chiral Spinors Particles uL= ½(1- 5)u uR= ½(1+ 5)u uL= u½(1+ 5) uR= u½(1- 5) Anti-particles vL= ½(1+ 5)v vR= ½(1- 5)v vL= v½(1- 5) vR= v½(1+ 5) 1 - 5 2 1 + 5 2 Note:uL+ uR= ( )u + ( )u= u 1 - 25 + (5)2 4 and also: ( ) ( ) u = 1 - 5 2 1 - 5 2 ( )u 2 - 25 4 1 - 5 2 = ( )u = ( )u

  14. Chiral Spinors Particles uL= ½(1- 5)u uR= ½(1+ 5)u uL= u½(1+ 5) uR= u½(1- 5) Anti-particles vL= ½(1+ 5)v vR= ½(1- 5)v vL= v½(1- 5) vR= v½(1+ 5) 1 + 25 + (5)2 4 note also: ( ) ( ) u = 1 + 5 2 1 + 5 2 ( )u 2 + 25 4 1 + 5 2 = ( )u = ( )u 1 - (5)2 4 while: ( ) ( ) u = 1 + 5 2 1 - 5 2 ( )u = 0 Truly PROJECTION OPERATORS!

  15. Why do we always speak of beta decay as a process “governed by the WEAK FORCE”? What do DECAYS have to do with FORCE? Where’s the FORCE FIELD? What IS the FORCE FIELD? What VECTOR PARTICLE is exchanged? e- What’s been “seen” We’ve identified complicated 4-branch vertices, but only for the mediating BOSONS… Not the FERMIONS! _ n e p semi-leptonic decays + +  e _  e leptonic decay 

  16. We’ve also “seen” the inverse of some of these processes: _ e e+ p n e e  

  17. The semi-leptonic decays(with participating hadrons) must Internally involve the transmutation of individual quarks: e _ e d d u ?? u d d u u d d u u e+ _ ?? e + u d _ ?? 

  18. Protons, quarks, pions and muons are all electrically charged so do participate in: e- p  p e- Can we use QED as a prototype by comparing to or lepton baryon Charge-carrying currents imply a charged vector boson exchange! e- p ??? (we’ve already seen gluons carry color) n e

  19. Then +1 -1/3 de+ u e might explain -decay! _ e 0  +2/3 e ℓ-ℓ coupling quark-flavor coupling and explains  decays  To explain + decay: e e+ u d requires a +1 charge carrier but coupling only to the left-handed particle states _ What about  decays? coupling strength modulated by left-handed components

  20. We’ve seen the observed weak interactions: m-e- +e+m np+e- +e p+e n+e+ could all be explained in terms of the interaction picture of vector boson exchanges if we imagine a the existence of a W  me e- W - m- ue e- W - d d e+ W +? W -? u e

  21. We’ve identified two fundamental vertices to describe the observed “weak” interactions. mW m- eW e- uW d or Lepton coupling Changes electric charge! Changes mass! Quark coupling Flips isospin! Changes mass! Changes electric charge! Some new “weak charge” that couples to an energy/momentum carrying W ±

  22. Continuing the analogy to eme qJmleptonAm e g e qJmleptonAm In general for a Quantum Mechanical charge carrier, the expression for “current” is of the form but these newest currents would have to allow eOme coupling to a “weak-field” W m Which must carry electric charge (why?) but not couple to it (why?)

  23. If this interaction reflects a symmetry, how many weak fields must there be? U(1) (the photon) one field SU(2) 3 fields YANG MILLS: SU(3) 8 fields COLOR: (gluons) U(1)is clearly inadequate U(2)would mean 3 weak fields we know we need W+, W- Could there be a neutral W0 ? But YANG-MILLS assumes we have “ISO”DOUBLET states!

  24. We’ve discovered we do have: Left-handed weak iso-doublets (in a new weak “iso”-space) u d e e- Right-handed weak iso-singlets uR dR eR L L NOT part of a doublet… NOT linked by the weak force to neutrinos NOTE:there is NO(e)R

  25. Left-handed weak iso-doublets (in a new weak “iso”-space) u +½ d -½ e +½ e- -½ Right-handed weak iso-singlets uR0 dR0 eR0 L L With ISO-SPIN we identified a complimentary “hypercharge” representing another quantum value that could be simultaneously diagonalized with ISO-SPIN operators. We generalize that concept into a NEW HYPERCHARGE in this “weak” space.

  26. YL = 2Q – 2I3weak Left-handedI3weak uL+½ dL-½ (e)L+½ e-L-½ Right-handed uR0 dR0 eR0 YL = 2(2/3) – 2(1/2) = 1/3 YL = 2(-1/3) – 2(-1/2) = 1/3 = -1 YL = 2(0) – 2(+1/2) = -1 YL = 2(-1) – 2(-1/2) YR = 2(Q) – 2(0) = 2Q

  27. Not all weak participants have ELECTRIC CHARGE • Its NOT electric charge providing the coupling • All weak participants (by definition) carry weak iso-spin • u +½ • d -½ • e +½ • e- -½ • u +½ YL = 1/3 • d -½ YL = 1/3 • e +½ YL = -1 • e- -½ YL = -1 But interactions are only well-defined by the theory if the fermion legs to a vertex have equal coupling strengths uW d L eW e- L L YL = 2Q – 2I3weak L

  28. With DOUBLET STATES and an associated “charge” defined we can attempt a Yang-Mills gauge-field model to explain the weak force but with some warnings...

  29. The Yang-Mills theory requires introducing a 3rd field: Could this be the photon? How do we distinguish this process from  exchange? e-W 0 e- Maybe the noted U(1) symmetry is part of a much larger symmetry: U(1) SU(2) ? U(1)U(1)×SU(2) UEM(1)UY(1) ×SUL(2)

  30. Straight from U(1) and the SU(2) Yang-Mills extension, consider: some new Yang- Mills coupling charge-like coupling to a photon-like field Ti=i/2for left-handed doublets = 0for right-handed singlets This looks like it could beU(1) with =qand B  A Yg1 2 This all means we now work from a BIG comprehensive Lagrangian summed over all possible fermions f to include terms for u, d, c, s, t, b, e, , , e, , 

  31. which contains, for example: plus similar terms for ,  , c, s, ,  ,t, b,

  32. Straight from U(1) and the SU(2) Yang-Mills extension, consider: some new Yang- Mills coupling charge-like coupling to a photon-like field Ti=i/2for left-handed doublets = 0for right-handed singlets This looks like it could beU(1) with =qand B  A Yg1 2 This all means we now work from a BIG comprehensive Lagrangian summed over all possible fermions f to include terms for u, d, c, s, t, b, e, , , e, , 

  33. which contains, for example: plus similar terms for ,  , c, s, ,  ,t, b,

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