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In Quarks and Leptons by Halzen and Martin p. 105 it says:

The bonus embodied in the Dirac equation is the extra twofold degeneracy. This means that there must be another observable which commutes with $H$ and $\mathbf{P}$, whose eigenvalues can be taken to distinguish the states.

Why is this the case?

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  • $\begingroup$ What do you mean "why is this the case"? Do you want to know why there must be an observable commuting with them? That's a generic property - all sets of observables can be completed into a complete set of commuting observables where all degeneracies are lifted. $\endgroup$ – ACuriousMind Nov 20 '15 at 15:27
  • $\begingroup$ The bonus embodied in the Dirac equation is the extra twofold degeneracy. Why does this mean that there must be another observable which commutes with H and P, whose eigenvalues can be taken to distinguish the states? $\endgroup$ – Joseph Dewdney Nov 20 '15 at 15:40
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Let there be two observables $H,P$ such that there is a twofold degeneracy in all eigenstates, i.e. a full eigenbasis is given by states $$ \lvert h,p \rangle_1,\lvert h,p\rangle_2$$ with eigenvalues $h,p$ for $H,P$ respectively. The subscript is meant to denote that $\lvert h,p \rangle_1,\lvert h,p\rangle_2$ are linearly independent.

Then there is a third observable $S$ commuting with $H,P$ which can simply be defined as $$ S\lvert h,p \rangle_1 = \lvert h,p \rangle_1\quad\land\quad S\lvert h,p\rangle_2 = -\lvert h,p\rangle_2$$ since these states form a basis. Since we made all eigenvectors of $H$ and $P$ also be eigenvectors of $S$, $S$ is diagonal in this basis, and the three observables pairwise commute.

Note that the choice of eigenvalue here is completely arbitrary, no one forces us to pick $1,-1$ (but I chose to do so because this anticipates the physical meaning of the observable as helicity, which tells us whether the spin is algined with momentum or not, in this case), the only thing that matters is that the eigenvalues are different.

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  • $\begingroup$ But $S$ should correspond helicity not spin? $\endgroup$ – Sebastian Riese Nov 20 '15 at 16:42
  • $\begingroup$ @SebastianRiese: Oops, yes, indeed. $\endgroup$ – ACuriousMind Nov 20 '15 at 16:43

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