# How does CPT symmetry hold while Charge, Parity and Time can all be violated?

To get it straight, Charge symmetry can be broken by U-236 atoms, Parity can be broken by K-one atoms and Time symmetry can be broken by quarks. So, how do they together hold true?

• CPT is different than C, P, or T taken by themselves. This answer might be useful: physics.stackexchange.com/questions/44628/… Commented Jul 26, 2021 at 13:39
• Also possibly helpful en.m.wikipedia.org/wiki/CPT_symmetry Commented Jul 26, 2021 at 13:47
• If the question is generally about how a combined symmetry can hold true when the individual symmetries are broken, consider the following example: take a square and split it evenly into four quadrants, coloring the upper-left and lower-right blue and the upper-right and lower-left red. This structure is not invariant under reflection about either the horizontal or vertical axis, but it is invariant under the successive applications of both reflections. In the same way, violations of some combination of C, P, and T symmetries don't imply the violation of CPT symmetry. Commented Jul 26, 2021 at 21:42

As a basic analogy consider the following shape in 3D. It is not symmetric under reflections of the $$x$$ axis. Neither it is symmetric under reflections of the $$y$$ axis or the $$z$$ axis. It is also not symmetric under any combinations of two of these reflections, say if you first reflect the $$x$$ axis and then reflect the $$y$$ axis. However if you reflect all three axis the shape transforms into itself.
The interaction that violates most discrete symmetries in the Standard model is the $$W^{\pm}$$ interaction (which is part of the weak interaction). The corresponding term in the Lagrangian looks like this (I write for quarks, for the leptons it looks similar), $$$$\int d^4x\Big((u_{iL})^\dagger V_{ij} W_\mu^{+}\bar{\sigma}^\mu d_{jL}+\mathrm{h.c.}\Big)$$$$ Under the spatial reflection ($$\mathcal{P}$$) the left-handed and right-handed components of the spinors are interchanged $$$$\mathcal{P}: \psi_L(t,\vec{x})\mapsto \psi_R(t,-\vec{x}), \quad \psi_R(t,\vec{x})\mapsto \psi_L(t,-\vec{x})$$$$ $$$$\mathcal{P}: W_0^{+}(t,\vec{x})\mapsto W_0^{+}(t,-\vec{x}), \quad W_k^{+}(t,\vec{x})\mapsto -W_k^{+}(t,-\vec{x}),\,k=1,2,3$$$$ Similarly the charge conjugation ($$\mathcal{C}$$) not only conjugates the fields but also exchanges the left and right components of the spinors, $$$$\mathcal{P}: \psi_L(t,\vec{x})\mapsto -i\sigma_2\psi_R(t,\vec{x})^\dagger, \quad \psi_R(t,\vec{x})\mapsto i\sigma_2\psi_L(t,\vec{x})^\dagger$$$$ $$$$\mathcal{P}: W_0^{+}(t,\vec{x})\mapsto -W_0^{-}(t,\vec{x}), \quad W_k^{+}(t,\vec{x})\mapsto -W_k^{-}(t,\vec{x}),\,k=1,2,3$$$$
Because only the left components participate in the weak interaction, we immediately see that $$\mathcal{P}$$ and $$\mathcal{C}$$ are both violated - they both transform the interaction term into the one containing only the right components of the spinor fields i.e. of the form $$u_R^\dagger W_\mu\sigma^\mu d_R$$. However the combination of these two transformations transforms the left-handed spinors to the left-handed spinors. So you may suspect that the combined transformation $$\mathcal{CP}$$ would leave this interaction term invariant. This happens to be true for the similar interaction with $$Z$$-bosons - it violates $$\mathcal{C}$$ and $$\mathcal{P}$$ separately but is invariant under the combination $$\mathcal{CP}$$. However the interaction with $$W$$ bosons contains the Cabbibo-Kobayashi-Maskawa (CKM) matrix $$V_{ij}$$. Under the combined $$\mathcal{CP}$$ transformation this interaction term transforms not into itself but into the one with complex conjugated CKM-matrix, $$$$V_{ij}\mapsto V_{ij}^\star$$$$ So if $$V$$ is complex (which it is) this term violates the combined $$\mathcal{CP}$$ symmetry. However the time reversal transformation is antilinear - i.e. it involves the complex conjugation of all parameters and the resulting transformation of the interaction term is again the complex conjugation of the CKM-matrix. So separately $$\mathcal{CP}$$ and $$\mathcal{T}$$ change this interaction term but their combination $$\mathcal{CPT}$$ leaves it invariant.