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CP violation

In quantum field theory (QFT), ${\rm CP}$ symmetry or ${\rm CP}$ violation is a property of the Lagrangian. For a ${\rm CP}$ violating QFT, in general, the absolute square of the Feynman amplitude of a processes ${\rm A+B\to C+D}$ and its conjugate process involving the antiparticles ${\rm \bar{A}+\bar{B}\to\bar{C}+\bar{D}}$ are unequal (see here): $$|\mathcal{M}_{AB\to CD}|^2\neq|\mathcal{M}_{\bar{A}\bar{B}\to\bar{C}\bar{D}}|^2.\tag{1}$$ Since the phase space factors must be identical in both cases, Eq.$(1)$ would also imply that the scattering cross-sections are unequal: $${\large\sigma}_{AB\to CD}\neq{\large \sigma}_{\bar{A}\bar{B}\to\bar{C}\bar{D}}.\tag{2}$$


T violation

Next, assuming the theory to be a CPT invariant QFT, ${\rm CP}$ violation requires time-reversal violation, (or ${\rm T}$-violation, for short) so that CPT is conserved. Now, a signature of T-violation is that the process ${\rm A+B\to C+D}$ and its inverse ${\rm C+D\to A+B}$ must be differentiable.


Question

At what level the processes ${\rm A+B\to C+D}$ and ${\rm C+D\to A+B}$ inherit a distinction?

Does T-violation come from the Lagrangian making $$\mathcal{M}_{\rm AB\to CD}\neq \mathcal{M}_{\rm CD\to AB}$$ or does it come from the difference in the phase space factors$^1$ making $${\large\sigma}_{AB\to CD}\neq{\large \sigma}_{CD\to AB}?$$


$^1$ For completeness, I recall that for a $2-2$ scattering, $A+B\to C+D$ is given by $$\sigma=\frac{1}{v_{\rm rel}}\frac{1}{4E_AE_B}\int\frac{d^3p_C}{(2\pi)^{3}2E_C}\int\frac{d^3p_D}{(2\pi)^{3}2E_D}(2\pi)^4\delta^{(4)}(p_A+p_B-p_C-p_D)|\mathcal{M}_{AB\to CD}|^2$$ and $C+D\to A+B$, in the above formula we have to make the interchange $A\leftrightarrow C$ and $B\leftrightarrow D$.

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