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The answers to a similar question Is a single photon always circularly polarized? suggest that the angular momentum of the elliptically polarized photon can be known depending on the probability amplitude of the Right and Left circularly polarized components. In that case, the angular momentum isn't an integer multiple of the reduced Plank's constant.

In this light, what are the Selection rules determining the interaction of elliptically polarized photons interact with matter?

Can elliptically polarized excitons emit photons upon conserving their non-integer angular momentum?

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The spin angular momentum of an elliptically-polarized photon is not well defined, where the term "not well defined" must be understood in the technical sense it has within quantum mechanics.

In that case, the angular momentum isn't an integer multiple of the reduced Plank's constant.

The angular momentum, indeed, isn't an integer multiple of $\hbar$, but it also isn't not an integer multiple of $\hbar$. It isn't anything. It does not have a pre-assigned value, and it does not take one until and unless you choose to perform a projective measurement of spin angular momentum.

That said:

In this light, what are the selection rules determining the interaction of elliptically polarized photons interact with matter?

when understanding the interaction of an elliptically-polarized single photon with matter, the simplest (though not necessarily the only possible) way is to decompose it as a superposition of circularly-polarized components, figure out what outcomes each component produces, and then form the total outcome as the corresponding linear superposition of the individual outcomes.

What that looks like will depend on the details.

  • It's possible that both circular polarizations have equal absorption probabilities leading to different states, as in e.g. a $|1s\rangle$ state of hydrogen absorbing $|R\rangle$ or $|L\rangle$ photons to go into the $|2p_+\rangle$ and $|2p_-\rangle$ electronic excited states. Then, a photon in the elliptically polarized state $$|\psi_\mathrm{ell}\rangle = \alpha |R\rangle + \beta |L\rangle$$ will be absorbed to produce the excited state $$|\psi_\mathrm{exc}\rangle = \alpha |2p_+\rangle + \beta |2p_-\rangle.$$

  • It's also possible that one of the circular polarizations will be absorbed and the other one won't, so $|g\rangle|L\rangle$ will go to $|e\rangle|0\rangle$ (excited atom, no photon) but $|g\rangle|R\rangle$ will not induce any change. In this situation, $|g\rangle|\psi_\mathrm{ell}\rangle = |g\rangle\left(\alpha |R\rangle + \beta |L\rangle\right)$ will go to $$\alpha |g\rangle|R\rangle + \beta |e\rangle|0\rangle.$$ This is an entangled state between the photon and the atom.

  • It's also possible that both circular polarizations will be absorbed, but they will do so with different transition amplitudes. In this case, the result will also be an entangled state between the photon and the atom.

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