# In this example, how can we have CP conservation with C violation?

Consider a simple two-body decay process $$X\to Y+Z$$ where $$X$$ is a boson, and $$Y,Z$$ are fermions. If $$C$$ is violated, $$\Gamma(X\to Y+Z)\neq \bar{\Gamma}(\bar{X}\to\bar{Y}+\bar{Z}).\tag{1}$$

However, if we further assume that CP is conserved and since the CP conjugate of a left-handed fermion is its right-handed antifermion, we must have $$\Gamma_1(X \to Y_L + Z_L) = \bar{\Gamma}_1(\bar{X}\to \bar{Y}_R + \bar{Z}_R),\\ \Gamma_2(X\to Y_R + Z_R) =\bar{\Gamma}_2(\bar{X}\to \bar{Y}_L + \bar{Z}_L).\tag{2}$$

To see see whether $$(1)$$ and $$(2)$$ can be simultaneously allowed, we add up the equation in Eq.$$(2)$$. It follows that $${\scriptsize \Gamma_1(X \to Y_L + Z_L) +\Gamma_2(X\to Y_R + Z_R) = \bar{\Gamma}_1(\bar{X}\to \bar{Y}_R + \bar{Z}_R)+\bar{\Gamma}_2(\bar{X}\to \bar{Y}_L + \bar{Z}_L).}\tag{3}$$

Question The LHS (RHS) of Eq.$$(3)$$ cannot be identical to the LHS (RHS) of Eq.$$(1)$$. If it were so, then Eq.$$(3)$$ (which involves an equality) would contradict Eq.$$(1)$$ (which involves an inequality). That would mean, it is not possible for CP to be conserved if C is violated, which is incorrect.

But why can't we regard the LHSs (RHSs) of Eq.$$(2)$$ to be the two partial decay widths of $$X\to Y+Z$$ ($$\bar{X}\to\bar{Y}+\bar{Z}$$). Why is that unjustified?

Probably the easiest way to see this is to look at $$W^+$$ decay into a positron and a neutrino in the SM with only one generation, so CP is conserved, and C and P are violated maximally.

The relevant terms in the lagrangian are proportional to $$W^+_\mu \bar \nu _R \gamma^\mu e_L + W^-_\mu \bar e _R \gamma^\mu \nu_L ~,$$ which transform under C to $$W^-_\mu \bar e _L \gamma^\mu \nu_R + W^+_\mu \bar \nu _L \gamma^\mu e_R ~,$$ essentially zero (you may easily assume neutrino masses and the right handed neutrino are not there, say in the early 80s!). These terms are absent, and no decays of this type go!

Likewise, under CP, the original couplings transform to $$W^-_\mu \bar e _R \gamma^\mu \nu_L +W^+_\mu \bar \nu _R \gamma^\mu e_L ~,$$ that is, the original terms. So, you note the effective chirality flips between the C and CP (original!) cases--that's the role of P.

So, consider $$X= W^+$$, $$Y= e^+_R$$, $$Z=\nu_L$$, for the decay of a $$W^+$$ to a left-chiral neutrino and a right chiral positron; in the W rest frame, the positron will prefer the direction of the spin of the W.

The r.h.s. width in (1) will be zero, since, as we saw, $$W^-$$ cannot go to a right-chiral electron and a ("nonexistent") left-chiral antineutrino!

Now, in the CP conserving case, I correct your process to $$\Gamma_1(X \to Y_R + Z_L) = {\Gamma}_1(\bar{X}\to \bar{Y}_L + \bar{Z}_R),\\ \Gamma_2(X\to Y_L + Z_R)=0 = {\Gamma}_2(\bar{X}\to \bar{Y}_R + \bar{Z}_R).\tag{2}$$ It is the usual mnemonic of reversing the chiralities for antiparticles in the SM.

There is no inconsistency to be seen.