1. P. L. Csonka, M. J. Moravcsik andM. D. Scadron:Phys. Lett.,15, 353 (1965).
2. P. L. Csonka, M. J. Moravcsik andM. D. Scadron:Phys. Rev. Lett.,14, 861. (1965).
3. As an illustration consider the example of a photon. If it is linearly polarized perpendicularly to the scattering plane (i.e. parallel to $$\hat m$$ ) its spintrinsic parity is +1, because the polarization vector ε is a pseudovector, and so is $$\hat m$$ . If it is linearly polarized in the scattering plane, then its spintrinsic parity is (−1), because the only nonzero components of ε are along $$\hat l$$ and $$\hat n$$ and both $$\hat l$$ and $$\hat n$$ switch sign when a parity transformation is performed. If the photon is circularly polarized, then it is a linear combination of the two previously mentioned states, and, therefore, it is not in an eigenstate of spintrinsic parity.
4. P. L. Csonka, M. J. Moravcsik andM. D. Scadron: UCRL 14222 (to be published).
5. It may happen that both ℒ and ℒ- vanish. That is a “dynamical accident” and then this particular experiment cannot be used to determine η4. measuring a different linear combination ℒ’ +,ℒ’ - may decide the question.