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Electroweak Mixing

  1. Jun 26, 2011 #1
    In the GWS electroweak model, there are two fundamental charges: weak hypercharge and the third component of weak isospin (henceforth referred to as hypercharge and isospin respectively). The gauge boson of hypercharge is the B0, and those of isospin are the W+, W0, and W-. The B0 and W0 mix to make the photon and Z0. What I don't understand is what determines the charges acted upon by the photon and Z0. The Z0 acts on isospin just like the Ws, but the photon acts on electric charge, which, in the context of electroweak theory, is defined to be the isospin plus half the hypercharge. How is this derived?
     
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  3. Jun 27, 2011 #2

    Bill_K

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    Z0, You can't derive it, of course, since it's a theory, but you can explain how we were led to it, and why it seems reasonable. W+ and W- always couple to particles in pairs: e and νe, μ and νμ, u quark and d quark. It's as if the particles formed doublets in a symmetry group, which was called weak isospin. The W+ and W- acted like stepping operators of an SU(2) group, but the third component W0 was missing.

    Neutral weak currents were discovered in 1973 in processes such as neutrino-electron scattering. The first idea was that they were mediated by the missing W0. However things did not fit. Although the charged weak currents are known to be purely left-handed, the neutral current was found to be predominantly left-handed with a smaller right-handed component. So it did not simply fit into a triplet. But, if you had a triplet W plus something else there would be a fourth degree of freedom.

    Weinberg's idea was that the fourth degree of freedom was electromagnetism. If so, it's clear how electromagnetism must fit in. Particles in each weak multiplet differ in charge by one. Also the average charges of the multiplets are displaced. By analogy with the strong interactions, we describe this by introducing a weak hypercharge Y and the formula Q = J3 + Y/2. Thus eL- and ve form a doublet with total charge -1 (hence Y = -1) while eR- forms a singlet with total charge -1, hence Y = -2.

    Weinberg said the B and the W0 were mixed into two orthogonal states Aμ = Bμ cos θW + Wμ3 sin θW and Zμ = -Bμ sin θW + Wμ3 cos θW where θW is a weak mixing angle.

    Ok, here's the key point. When you write out the electroweak neutral current interaction in terms of these rotated states,
    g Jμ3 Wμ3 + ½ g' JμY Bμ = (g sin θW Jμ3 + ½ g' cos θW JμY) Aμ + (...) Zμ
    the coefficient of Aμ must be the electromagnetic current JμEM = e(Jμ3 + 1/2 JμY). (This implies g sin θW = g' cos θW = e.)

    The field Zμ is simply whatever is orthogonal to that. It turns out to be JμNC = Jμ3 - sin2 JμEM
     
    Last edited: Jun 27, 2011
  4. Jun 27, 2011 #3

    jtbell

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    I thought GWS theory came before the discovery of weak neutral currents (c. 1968 versus 1973), and predicted their existence. That was one of the reasons the observation of neutral currents was such a big deal: it was a big step in confirming GWS theory.
     
  5. Jun 27, 2011 #4

    Bill_K

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    Thanks, I stand corrected.
     
  6. Jun 27, 2011 #5

    Vanadium 50

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    Bill is right, but you are starting from a bad place.

    1) The "fundamental charge" of weak isospin is not T3. It's T. This is where the SU(2) comes in - it means that the amount of weak charge held by a particle cannot be represented as a simple number (that would be a U(1) theory) but must be represented by a matrix.

    2) Before symmetry breaking, you have a w1, w2, w3. You don't know what electric charge is at that point in the derivation, so it's premature to label them by their charge. When going through the derivation, you will discover that one linear combination of w's has positive charge, one has negative charge, one combination of the B and the w3 has the same couplings as the photon (and is identified with it), and the orthogonal combination is the Z.
     
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