Parity operator notation for neutrino oscillation probabilities

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Doofy
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Just a quick question about notation really here. In neutrino oscillation we can calculate a probability of an oscillation occurring between two flavour eigenstates - invariably denoted [itex]P(\nu_{\alpha} \rightarrow \nu_{\beta})[/itex]. I've got some confusion about what happens to this when we apply operators for P, CP and CPT.

Charge conjugation turns particles to anti-particles, so I'm thinking this transformation would be denoted [itex]P(\nu_{\alpha} \rightarrow \nu_{\beta}) \rightarrow P(\overline{\nu_{\alpha}} \rightarrow \overline{\nu_{\beta}})[/itex]

Time reversal would presumably be [itex]P(\nu_{\alpha} \rightarrow \nu_{\beta}) \rightarrow P(\nu_{\beta} \rightarrow \nu_{\alpha})[/itex]

The combination of C and T operations would therefore be [itex]P(\nu_{\alpha} \rightarrow \nu_{\beta}) \rightarrow P(\overline{\nu_{\beta}} \rightarrow \overline{\nu_{\alpha}})[/itex]

My question is how would the parity transformation be incorporated here?

ie. how would one write [itex]\hat{P} P(\nu_{\alpha} \rightarrow \nu_{\beta}), \hat{C}\hat{P} P(\nu_{\alpha} \rightarrow \nu_{\beta})[/itex] and [itex]\hat{C}\hat{P}\hat{T} P(\nu_{\alpha} \rightarrow \nu_{\beta})[/itex] ?
 
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The application of P will switch the chirality of a fermion. For neutrinos, this would be rather problematic, as, so far as we know, there are no right-handed neutrinos. (This is related to the fact that the weak force violates P maximally.) C, on the other hand, will reverse all charges, but will not reverse chirality; so, it will take you from a left-handed neutrino to a left-handed anti-neutrino. But, as you can check for yourself, the anti-neutrino state which must be included to allow spin-conservation in gauge interactions involving a left-handed neutrino is actually the right-handed anti-neutrino. (Equivalently, the existence of a left-handed anti-neutrino would require the existence of a right-handed neutrino.) So, again, we have a maximal violation, this time of C. It's only if we apply CP (turning a left-handed neutrino into a right-handed anti-neutrino) that we should expect to get a meaningful conjugation.

Now, as it happens, we already know that the weak force slightly violates CP, as well; and, it's expected that there should be some CP violation in the neutrino sector; but, this ought to be a small effect compared to the size of the oscillation effects.

CPT, on the other hand, should be an exact symmetry. There's a good deal of work showing that a violation of CPT necessarily entails a violation of local Lorentz symmetry.