jheimann@bbncc5.UUCP (John Heimann) (08/29/85)
Doug Gwyn asks how one relates mass to CPT. The answer is this: suppose a system is invariant under CPT. Then the Hamiltonian H commutes with the CPT operator Ucpt: [Ucpt, H] =0. It is possible to show formally that the CPT operator and total angular momentum commute: [Ucpt, J^2]=0; whereas Ucpt and Q, the (electronic, barionic, leptonic) charge anticommute: {Ucpt, Q} = 0. Now suppose P(m) (I write P since Psi isn't an ASCII character) is an eigenstate of the Hamiltonian with mass eigenvalue m, representing some particle: H P(m) = m P(m). Then commutation of H with Ucpt implies that H (Ucpt P(m)) = Ucpt (H P(m)) = Ucpt (m P(m)) = m (Ucpt P(m)). Thus P'(m) = Ucpt P(m) is also an eigenstate of H with the same mass. It has the same total spin since Ucpt commutes with J^2 but opposite (electric, baryonic, leptonic) charge Q. That is to say, it is the anti-particle of the particle represented by P(m). John