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Helicity interactions of the Z0

  1. Jun 10, 2013 #1
    The Z0 is a linear combination of W0 and B0 bosons, so unlike the charged current interaction it can interact with both handednesses (is that a word). In the search to quantify this mixing, people measured the cross section for muon (anti)neutrinos to scatter off electrons. This is a reaction that can only happen via neutral currents.

    Consider the neutrino case. It's LH, because all neutrinos are. The electrons are an equal mix of RH and LH states. If you look at the interaction, it's proportional to the following:

    [tex] \bar{\nu} \left( g_R \frac{(1-\gamma^5)}{2} + g_L \frac{(1+\gamma^5)}{2} \right) e [/tex]

    Now according to my book, in the final C.S. these pick up factors according to spin rotation matrices:

    [tex] \frac{\mathrm{d}\sigma}{\mathrm{d}y} \propto g_L^2 (1-y)^2 + g_R^2 [/tex]

    Fine.

    However, if the neutrino is LH (so appears in this vertex in the same way as a RH antineutrino) then why is there a RH antineutrino x RH electron term? In other words:

    [tex] \bar{\nu}_{RH} [ g_R P_R + g_L P_L ] (e_{RH} + e_{LH}) \\
    = \bar{\nu}_{RH} [ g_R P_R^2 + g_L P_L^2 ] (e_{RH} + e_{LH}) \\
    = \bar{\nu}_{RH} [ g_R P_L^\dagger P_R + g_L P_R^\dagger P_L ] (e_{RH} + e_{LH}) \\
    = g_R \bar{\nu}_{RH} P_L^\dagger e_{RH} + g_L \bar{\nu}_{RH} P_R^\dagger e_{LH} \\
    = 0 + g_L \bar{\nu}_{RH} e_{LH}
    [/tex]

    So the [itex] g_R \bar{\nu}_{RH} e_{RH} [/itex] term disappears to 0. How then is there a term proportional to [itex]g_R^2[/itex] in the differential C.S. equation?

    Thanks!
     
  2. jcsd
  3. Jun 10, 2013 #2

    Bill_K

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    Science Advisor

    Looks to me like you've written a charged current, not a neutral current. This would not couple to the Z0. Especially if you intend that to be a muon neutrino, there's a problem!
     
  4. Jun 10, 2013 #3
    OK, right I think I'm understanding this a little better. So what I should have written, for muon-electron scattering, is:

    [tex] \bar{\psi}_{\nu_\mu} \left[ g_R^{\nu_\mu} \frac{(1-\gamma^5)}{2} + g_L^{\nu_\mu} \frac{(1+\gamma^5)}{2} \right] \psi_{\nu_\mu} [/tex]
    multiplied by the term for the electron vertex:
    [tex] \bar{\psi}_{e^-} \left[ g_R^e \frac{(1-\gamma^5)}{2} + g_L^e \frac{(1+\gamma^5)}{2} \right] \psi_{e^-}
    [/tex]

    Is this correct?

    So this should give (for the real neutrino) that only LH e -> LH e or RH e -> RH e is ok and the other terms go.

    Compare with the neutrino-antineutrino -> electron anti-electron case for which the electron vertex is:
    [tex] \bar{\psi}_{e^+} \left[ g_R^e \frac{(1-\gamma^5)}{2} + g_L^e \frac{(1+\gamma^5)}{2} \right] \psi_{e^-}
    [/tex]
    in which we should find that only RH e+, LH e- or LH e+, RH e- is allowed.

    Firstly, have I got both of these right? Secondly, how is it that the first current gives LH,LH & RH,RH but the second give LH,RH & RH,LH? I assume this must be to do with the fact that it's a conjugate wavefunction for an antiparticle rather than a particle but what's the mathematical detail to this?

    Thanks again!
     
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