Photons with half integer angular momentum - what's happening?

Question

I have just read this article - what is happening?

Analysing these beams within the theory of quantum mechanics they predicted that the angular momentum of the photon would be half-integer, and devised an experiment to test their prediction. Using a specially constructed device they were able to measure the flow of angular momentum in a beam of light. They were also able, for the first time, to measure the variations in this flow caused by quantum effects. The experiments revealed a tiny shift, one-half of Planck's constant, in the angular momentum of each photon.

Read more at: http://phys.org/news/2016-05-physicists.html#jCp

Answer 1

Nothing is happening. At least, nothing except that a new generalized quantity suggestively called "angular momentum" was defined and subsequently measured. But nothing we know about the usual angular momentum of photons is changed by this in any way.

Standard total angular momentum is $J = L + S$, where $L$ is the orbital and $S$ the spin angular momentum. In three dimensions and usual setups, $L$ and $S$ are not independently conserved quantities - it is only the total $J$ that is conserved. Since $S$ is integral for photons and $L$ is always integer, $J$ is always integral. Additionally, $S$ and $L$ do not separately correspond to actual transformations one can do on light as they do not preserve the transversality of the electromagnetic wave.

All that the paper "There are many ways to spin a photon: Half-quantization of a total optical angular momentum" by Ballantine, Donegan and Eastham does is consider a situation (a light beam) where there is at least one component of $L$ and $S$ that is independently conserved and generates a consistent transformation (one that preserves tranversality), so that a "generalized" angular momentum $J_\gamma = L+\gamma S$ can be defined in that direction. If you choose $\gamma=\frac{1}{2}$, it is obvious that you get half-integer values for $J_{1/2}$.

The significance of this paper (paraphrasing their own words) is firstly that they actually figured out an experimental measurement of $J_{1/2}$ and secondly that this hints at a possible "fermionization" of photons in situations where $J_{1/2}$ is a good operator, i.e. a description of the photonic system by an equivalent fermionic system. However, it must be stressed that this $J_{1/2}$ is not the usual total angular momentum, let alone spin, and hence does not contradict the usual statement of "photon angular momentum comes in integral multiples of $\hbar$". It's a generalization of the usual angular momentum $J_1$ that shows, in some situations, a half-integer quantization.

Answer 2

Just as a supplement to ACuriousMind's answer, it is worth noting that buried in the bottom of their paper they actually show what the "spin 1/2" eigenstates are in terms of the regular basis:

$|j=1/2\rangle=\frac{1}{\sqrt{2}}(|1, -1 \rangle + |0,1\rangle$)

$|j=-1/2\rangle=\frac{1}{\sqrt{2}}(|-1, 1 \rangle + |0,-1\rangle$)

where $|l, \sigma\rangle$ is the angular momentum in the normal $|l,s\rangle$ basis. Written out explicitly, it is clear:

1. That these are eigenstates of $L+S/2$, and
2. That this trick could only work for an integer or half-integer $\gamma$, and
3. That there's nothing too special going on here.

Still, it can still be interesting to frame an old system in a new way. I knew about the possibility of anyons in low dimensions, but I still would not have guessed that the very natural and common reduction in symmetry caused by picking a propagation axis might be sufficient for this effect. However, that might is an important qualifier: since the authors don't actually demonstrate fractional statistics or a procedure to measure them, this remains to be seen.

Edit: Emilio asks for a concrete demonstration of point two:

We want

$(L+\gamma S)(\alpha|l_1,1\rangle+\beta|l_2,-1\rangle)=j(\alpha|l_1,1\rangle+\beta|l_2,-1\rangle)$ .

This is the most general angular momentum superposition possible with a fixed total $j$, since there are only two spin possibilities. Furthermore, we know from beginning QM classes that an eigenstate of $j$ will have all these possible elements with some Clebsch-Gordon coefficients.

Applying the operators:

$((l_1+\gamma)\alpha|l_1,1\rangle+(l_2-\gamma)\beta|l_2,-1\rangle)=j(\alpha|l_1,1\rangle+\beta|l_2,-1\rangle)$

giving the conditions

$l_1+\gamma=j$

$l_2-\gamma=j$,

or $l_2-l_1=2\gamma$, $l_1\neq l_2$ .

Since $l$ are integers, this implies that $\gamma$ must be a half-integer.