# How is a T-violation inherited in a QFT?

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 $$B\leftrightarrow D$$.