# Is CPT symmetry a direct consequence of Special Relativity?

I'm wondering about the origin of $$\mathcal{CPT}$$ symmetry in the Standard Model. The Wikipedia entrance makes me understand it is a direct consequence of Special Relativity. Is it right?

Yes. Here is a sketch of how it works, although it can be proven under general assumptions. A Lorentz transformation takes the form $$\begin{bmatrix} \cosh{y} & 0 & 0 & \sinh{y} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ \sinh{y} & 0 & 0 & \cosh{y}\\ \end{bmatrix}$$ with $$y$$ an arbitrary real number called the rapidity. Lorentz-invariant field theories have to be invariant with respect to such transformations. However in quantum scattering theory (and to each order of perturbation theory in QFT) all the amplitudes are analytic functions of $$y$$. You'll notice that if you plug in $$y = - \mathrm{i} \pi$$: $$\begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &-1\\ \end{bmatrix}$$ Which is in fact a $$PT$$ transformation. I believe I first saw this argument from Ron Maimon.
• And what about the $\mathcal{C}$? May 23 at 15:15