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I think an inescapable consequence of the angular momentum algebra is that a particle with spin-$j$ must have $(2j+1)$ spin projections in any direction. However, photons seem to evade this conclusion. Why?

Solidification
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  • If the photon had spin 1 with respect to $SO(3)$ (as we sometimes colloquially say) there would indeed be a paradox. But you have to use the massless little group in this case as https://physics.stackexchange.com/q/578059/ describes. – Connor Behan Dec 05 '21 at 13:46

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The technical answer is that spin is property associated with representation of the Poincare group P. The representations are are induced from the "little group" i.e. the sub-group of P that leaves the particle's four momentum fixed. For a massive particle we can go to the rest frame in which the four momentum is $p=(m,0,0,0)$. This vector this is left fixed by rotations SO(3), so for a massive particle spin is a property of rotations and their spin $j$ representations.

For a massless particle there is no rest frame and the reference momentum must be a null vector $p =(|{\bf p}_0|, {\bf p}_0)$. The little group now consists of space rotations SO(2) about the three-vector ${\bf p}$, together with operations that are generated by infinitesimal Lorentz boosts in directions perpendicular to ${\bf p}_0$ combined with compensating infinitesimal rotations. Remarkably the combined operations mutually commute, possess all the algebraic properties of Euclidean translations, and the resulting little group is isomorphic to the symmetry group SE(2) of the two-dimensional Euclidean plane. Wigner argued that the translations must do nothing and so the spin of a massless particle is asociated with the one-dimensional representations of SO(2).

A more intuitive explanation is that for something moving at the speed of light, any vector originally have a component perpendicular to the drection of travel will be Lorentz trasformed to one pointing in the direction of travel

mike stone
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  • Does the angular momentum commutation relations ($[J_x.J_y]=i\hbar J_z$ and its cyclic permutations) not apply to photons? Is there a way to see this? – Solidification Dec 05 '21 at 14:02
  • There is a discussion of the algebra in our paper https://arxiv.org/pdf/1501.04586.pdf. – mike stone Dec 05 '21 at 14:13
  • @mikestone Does the algebra that you have discussed in your paper depend on how the position operators for photon are defined (like there are no position operators for photons satisfying Newton Wigner axioms, but there are other variants like Pryce and Hawton operators), or is it independent of that? – KP99 Dec 05 '21 at 16:12
  • @KP99 Although we explored the literature I doubt that I know all the possible mathematical definitions of position operators, but however one selects one, the sideways shifts in the beam seem to be physical.... – mike stone Dec 05 '21 at 17:19
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The electromagnetic field is transversal because of charge conservation. The state of polarisation or spin that is perpendicular to the direction of propagation is therefore never excited.

my2cts
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