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Straight from U(1) and the SU(2) Yang-Mills extension, consider:

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. T i =  i /2 for left-handed doublets = 0 for right-handed singlets. This looks like it could be U(1) with = q and B   A .

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Straight from U(1) and the SU(2) Yang-Mills extension, consider:

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  1. 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, , 

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

  3. The expression above exclusively links left-handed states with onlyleft-handed states, right-handed with right! But  = R+L Which of course means for the most general electron states _ _ _ e = eR+ eL e = eR+eL and =0 =0 There are no cross terms!

  4. L and R-handed components are simply distinguished by different interactions. For now reproducing ONLY the (1st generation) lepton terms, we see the terms include:   g1 2 L eL g2 2 L eL {(L†eL†) B} {(L†eL†)W ·  }  0 YL  0 + ( ) ( ) ( ) 1 0 0 -1 0 1 1 0 0 -i i 0 W1 + W2 + W3 eLW+ L LW- eL these quantum states are the charged currents W+ and W- then writing: W±= (W1±iW2) 1 2

  5. plus completely NEUTRAL INTERACTIONS eL†eLW3 ~eL†eLB and which involve NO CHANGE in electric charge! The “primordial” -field, B is blurred by the existence of a chargeless W0! but B CANNOT be the -field we know (A) anyway since ~L†LB In fact neutrinos are simultaneously coupled to both neutral fields as well ~L†LW3 The exchange of both B and W3 contribute to all L†Linteractions.

  6. Looking more closely at the  coupling terms: † † This is the effectivefield: the quantum mechanical admixture of both states! † -1 this weak chargeless vector boson is the physically observable state! -g1Bm + g2W3m Let’s define:Z0 = g12 + g22 setting: and Z0 = -Bsin + W3cos

  7. Now looking at the coupling to the charged electrons: † † Why is there different coupling here? Both eL and eRmust couple to the photon as well! A neutral state NOT coupling to † would have to be orthogonal to Z0. That would have to be: A = Bcos + W3sin g2Bm + g1W3m = g12 + g22 Now we assume thisis the photon!

  8. -g1Bm + g2W3m Z0= = -Bsin + W3cos g12 + g22 g2Bm + g1W3m = Bcos + W3sin A = g12 + g22 We can invert these definitions to get with which we see

  9. Each handedness of the electron couples IDENTICALLY to A (the ) the same coupling strength -g1g2 g12+g12 -ee†eA eL†eLA compare to the measured electric charge!

  10. Each handedness of e- couples to Z, but differently! but, remember, only the left-handedness of e- coupled to neutrinos through W± fields

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