# C, P and CP in particle physics

We know that the charge conjugation (C) and parity (P), both are violated maximally in particle physics. Then what is the need for looking for CP symmetry in order to distinguish matter from antimatter? Why is CP symmetry important?

Imagine a process where particle $A$ decays to particle $B$. This decay is described by its decay amplitude $M_{A\to B}$ (basically some coefficient in a Lagrangian). For the antiparticles, $\overline A$ and $\overline B$, this decay amplitude is $M_{\overline A\to \overline B}$.
If CP is not violated, these two decay amplitudes are the same complex number: $$M_{A\to B} = M_{\overline A\to \overline B} =: |M|\text{e}^{\text i\varphi}$$
Now we introduce the CKM-matrix and its phase factor $\text e^{\text i \Omega}$. The two amplitudes pick up this factor and its conjugate: $$M_{A\to B} \to M_{A\to B}\,\text e^{\text i \Omega}\quad \text{and}\quad M_{\overline A\to \overline B}\to M_{\overline A\to \overline B}\, \text e^{-\text i \Omega}$$
Any physical observable like a cross section is proportional to $|M|^2$, so until now, these two would still behave the same. But what if there's an intermediate state, $X$: $A\to X\to B$ and $\overline A\to \overline X\to \overline B$. The decay amplitudes are now given by \begin{align*} M_{A\to B} &= M_{A\to X} + M_{X\to B} = |M_1|\text{e}^{\text i\varphi_1}\text e^{\text i \Omega_1} + |M_2|\text{e}^{\text i\varphi_2}\text e^{\text i \Omega_2},\\ M_{\overline A\to \overline B} &= M_{\overline A\to \overline X} + M_{\overline X\to \overline B} = |M_1|\text{e}^{\text i\varphi_1}\text e^{-\text i \Omega_1} + |M_2|\text{e}^{\text i\varphi_2}\text e^{-\text i \Omega_2}. \end{align*} When we calculate the difference of the absolute squares, we get: $$|M_{A\to B}|^2 - |M_{\overline A\to \overline B}|^2 = -4|M_1||M_2|\sin(\varphi_1-\varphi_2)\sin(\Omega_1-\Omega_2).$$ Now there's a difference in the decay of particles and antiparticles. If we eliminate the CKM-phase factor ($\Omega_{1,2}\to 0$), the above difference would vanish and particles would again behave the same as antiparticles.