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Photon is a spin-1 particle. Were it massive, its spin projected along some direction would be either 1, -1, or 0. But photons can only be in an eigenstate of $S_z$ with eigenvalue $\pm 1$ (z as the momentum direction). I know this results from the transverse nature of EM waves, but how to derive this from the internal symmetry of photons? I read that the internal spacetime symmetry of massive particles are $O(3)$, and massless particles $E(2)$. But I can't find any references describing how $E(2)$ precludes the existence of photons with helicity 0.

Qmechanic
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Siyuan Ren
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    related: http://physics.stackexchange.com/q/46643/ –  Aug 31 '13 at 20:37
  • A discussion about the essence of photon’s spin and differences to the massive case: http://physics.stackexchange.com/q/19229/ – Incnis Mrsi Aug 18 '14 at 12:34

1 Answers1

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It derives not from the internal symmetry itself but from the fact that it is a gauge symmetry.

Your symmetry group assignments are not those of the symmetry group but of the little group of the representation. If you assume in addition that the representation is irreducible, you end up in the massless case (with little group ISO(2)=E(2)) with a helicity representation, which picks up from a vector representation only the transversal part, corresponding to a gauge symmetry. Because of reflection symmetry (parity), there are two helicity degrees of freedom. Under the connected part of the Poincare group, this splits into two irreducible representations of fixed helicity, corresponding left and right circular polarization.

This is described in full detail in Section 5.9. of the quantum field theory book (Part I) by Weinberg. In particular, the 2-valuedness (rather than the 3-valuedness) of the helicity is discussed after (5.9.16).

Arnold Neumaier
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  • That book has a chapter on massless particles, but does not mention E(2)-like little group. – Siyuan Ren Jun 09 '12 at 13:25
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    @KarsusRen: It mentions it on p.70 under the name ISO(2), which is just an alternative tradition for writing E(2). – Arnold Neumaier Jun 10 '12 at 10:23
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    A freely available presentation by Nicolis that follow's Weinberg's is here: http://phys.columbia.edu/~nicolis/GR_from_LI.pdf –  Aug 31 '13 at 20:38
  • @Arnold Neumaier: do you know a simple explanation how the Poincaré sphere structure appears directly from representations? – Incnis Mrsi Aug 14 '14 at 16:27
  • @IncnisMrsi:There are two helicity degrees of freedom, and any 2-level system has a fundamental SU(2) representastion, described by a poincare sphere = bloch sphere. – Arnold Neumaier Aug 17 '14 at 12:50
  • Poincaré sphere is equivalent to Bloch sphere in the sense of quantum information. Physically equivalent they are not because of different groups. Actually I didn’t read postings completely, neglected to look at Nicolis’ paper, and had today to invent the concept of little group (including E(2)) myself, from the Lorentz group. While I discovered it myself, I learned that *E(2) action* on the Poincaré sphere *is not transitive*. It doesn’t make circular polarization to anything else (I just discovered that linear polarization isn’t Lorentz invariant – interesting). This is an answer. – Incnis Mrsi Aug 17 '14 at 15:47
  • @ArnoldNeumaier I'm very interested in weinberg book Vol.1 but I can't find exactly where he explain the operator J3 in equation (2.5.39) couldn't have eigenvalue 0. (I refered to this equation only because it is a central result that considered is as definition) I only find in page 90 he brings topological reasoning for helicity could be integer or half-integer. but where he says this half-integer couldn't be zero? – moshtaba May 16 '19 at 18:46
  • @moshtaba: The Lorentz gauge condition together with gauge invariance eliminates this possibility. – Arnold Neumaier May 17 '19 at 08:59
  • @ArnoldNeumaier Thanks, but could you please refer to a specific page of Weinberg's book (as you said in your answer he discussed this problem in Vol.1) where he explicitly (or implicitly but exactly) explains this elimination? – moshtaba May 17 '19 at 09:28
  • @moshtaba: Section 5.9, after (5.9.16). Note that there are only two polarization vectors $e^\mu$. – Arnold Neumaier May 17 '19 at 12:11
  • Page 73, the next paragraph under equation 2.5.47, actually. Lorentz transformations don't change helicity, so you could have possible said that each particle with different helicity is different, but electromagnetic and gravitational forces obey space inversion, therefore, both helicity $\pm \sigma$ are called a photon or just defined one particle. – physicsbootcamp Jul 06 '22 at 15:40