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.(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Helicity interactions of the Z0

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