The CPT group of the Dirac field.

*(English)*Zbl 1073.81033Summary: Using the standard representation of the Dirac equation, we show that, up to signs there exist only two sets of consistent solutions for the matrices of charge conjugation \((C)\), parity \((P)\), and time reversal \((T)\), which give the transformation of fields \(\psi_C(x)=C \overline\psi^T(x)\), \(\psi_\Pi(x_\Pi)=P \psi(x)\) and \(\psi_\tau(x)=T \psi (x_\tau)^*\), where \(x_\Pi=(t,-\vec x)\) and \(x_\tau=(-t,\vec x)\). These sets are given by \(C=\pm\gamma^2\gamma_0\), \(P=\pm i\gamma_0\), \(T=\pm i\gamma^3\gamma^1\) and \(C=\pm i\gamma^2\gamma^0\), and \(C=\pm i\gamma^2\gamma_0\), \(P=\pm i\gamma_0\), \(T=\pm \gamma^3\gamma^1\). Then \(P^2=-1\), and two successive applications of the parity transformation to fermion fields necessarily amount to a \(2\pi\) rotation. Each of these sets generates a nonabelian group of 16 elements, respectively, \(G_\theta^{(1)}\) and \(G_\theta^{(2)}\) which are nonisomorphic subgroups of the Dirac algebra, which, being a Clifford algebra, gives a geometric nature to the generators, in particular to charge conjugation. It turns out that \(G_\theta^{(1)}\cong DH_8 \times \mathbb{Z}_2\subset S_6\) and \(G_\theta^{(2)}\cong 16E\subset S_8\), where \(DH_8\) is the dihedral group of eight elements, the group of symmetries of the square, and \(16E\) is a non trivial extension of \(DH_8\) by \(\mathbb{Z}_2\), isomorphic to a semidirect product of these groups; \(S_6\) and \(S_8\) are the symmetric groups of six and eight elements. The matrices are also given in the Weyl representation, suitable for taking the massless limit, and in the Majorana representation, describhing self-conjugate fields. Instead, the quantum operators \(C,P\) and \(T\), acting on the Hilbert space, generate a unique group \(G_\Theta\), which we call the CPT group of the Dirac field. This group, however, is compatible only with the second of the above two matrix solutions, namely with \(G_\theta^{(2)}\), which is then called the matrix CPT group. It turns out that \(G_\Theta\cong DC_8\times\mathbb{Z}_2\subset S_{10}\), where \(DC_8\) is the dicyclic group of 8 elements and \(S_{10}\) is the symmetric group of 10 elements. Since \(DC_8\cong Q\), the quaternion group, and \(\mathbb{Z}_2\cong S^0\), the 0-sphere, then \(G_\Theta\cong Q \times S^0\).