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From: jheimann@bbncc5.UUCP (John Heimann)
Newsgroups: net.physics
Subject: Antiparticle Mass and CPT
Message-ID: <29@bbncc5.UUCP>
Date: Thu, 29-Aug-85 15:12:13 EDT
Article-I.D.: bbncc5.29
Posted: Thu Aug 29 15:12:13 1985
Date-Received: Sun, 1-Sep-85 09:12:08 EDT
Reply-To: jheimann@bbnccv.UUCP (John Heimann)
Organization: Bolt Beranek and Newman, Cambridge, MA
Lines: 18


	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