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To account for parity violation the weak interaction matrix element is written in the form

$$ M \propto \sum_{i} C_i (\bar u_p O_i u_n) (\bar u_e O_i (1 + {{C'_i}\over{C_i}}\gamma^5)u_\nu) $$

Why does time reversal invariance imply that the $C_i$ coefficients must be real?

I was thinking that maybe it might be related to some kind of hermiticity condition, in order to have $|M_{if}|=|M_{fi}|$. Am I on the right path?

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    $\begingroup$ You shouldn't write it this way--- the operator you use should be (1+\gamma^5) to be a projection operator on helicity, since one of the helicities you are implicitly introducing for the neutrino when you write it as a Dirac spinor is not physically there. The time-reversal is not related to Hermiticity, it's a stronger condition. $\endgroup$
    – Ron Maimon
    Commented Jul 23, 2012 at 7:16
  • $\begingroup$ Yeah, at a later step one proves that $C_i = \pm C'_i$, but my question is "why does time reversal-invariance imply that the $C_i$ must be real"? I edited my question accordingly, since it seems I haven't been clear enough. $\endgroup$
    – Editortoise-Composerpent
    Commented Jul 23, 2012 at 8:50
  • $\begingroup$ It shouldn't be a "step". There are no steps. It's just true that it's a projection, and if you consider it a "step" you are imagining a completely different theory for all C'_i/C_i not equal to $\pm 1$. If you have a source that treats this as a step, throw it away, it is junk. The correct derivation of the nature of coefficients in a Lagrangian is not by "T invariance" (which is nonsense--- why should it be T invariant? It's not P invariant) but by requiring that the Lagrangian is real, for grassman fields, that there is an involution which makes L real. This is the important thing. $\endgroup$
    – Ron Maimon
    Commented Jul 23, 2012 at 19:36

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Time-reversal operator is anti-unitary, meaning, basically, that for any c-number $a$: $$T\,a\,T^{-1} = a^*$$ Now, If you have a T-invariant Lagrangian term ${\cal L}_{term}$: $$T{\cal L}_{term}T^{-1}={\cal L}_{term}$$ Then if multiply in by $a$: $$ Ta{\cal L}_{term}T^{-1}=TaT^{-1}T{\cal L}_{term}T^{-1}=a^*{\cal L}_{term}$$ So you need $a$ to be real if you want it to be T-invariant.

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