Constants of the motions are those dynamic variables that commute with the Hamiltonian. For a free particle ()
With the usual definition of the orbital angular momentum
we find that none of the components of commute with . Thus the orbital angular momentum is not a constant of the motion. On the other hand we find that
does commute with and thus is a constant of the motion.
We now prove this statement. To do this, we simply verify that this result is true for . Clearly,
For the first term in equation 5.51,
For the second term in equation 5.51, we make use of the relationship
where . Using we have
Thus, we have shown that
By symmetry it now follows that quite generally
As a consequence we have that and can be simultaneously diagonalized.
The bonus embodied in the Dirac equation is the extra twofold degeneracy. This means that there must be another observable which commutes with and , whose eigenvalues can be taken to distinguish the states. On inspection, we see that the operator commutes with and ; is the unit vector pointing in the direction of the momentum, . The ``spin'' component in the direction of the motion, , is therefore a ``good'' quantum number and can be used to label the solutions. We call this quantum number the helicity of the state.