2012-06-08 17:11:06 8 Comments

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.

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## 1 comments

## @Arnold Neumaier 2012-06-08 17:17:23

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).

## @Siyuan Ren 2012-06-09 13:25:17

That book has a chapter on massless particles, but does not mention E(2)-like little group.

## @Arnold Neumaier 2012-06-10 10:23:02

@KarsusRen: It mentions it on p.70 under the name ISO(2), which is just an alternative tradition for writing E(2).

## @user4552 2013-08-31 20:38:40

A freely available presentation by Nicolis that follow's Weinberg's is here: phys.columbia.edu/~nicolis/GR_from_LI.pdf

## @Incnis Mrsi 2014-08-14 16:27:55

@Arnold Neumaier: do you know a simple explanation how the Poincaré sphere structure appears directly from representations?

## @Arnold Neumaier 2014-08-17 12:50:35

@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.

## @Incnis Mrsi 2014-08-17 15:47:12

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

on the Poincaré sphereE(2) action. It doesn’t make circular polarization to anything else (I just discovered that linear polarization isn’t Lorentz invariant – interesting). This is an answer.is not transitive## @moshtaba 2019-05-16 18:46:30

@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?

## @Arnold Neumaier 2019-05-17 08:59:14

@moshtaba: The Lorentz gauge condition together with gauge invariance eliminates this possibility.

## @moshtaba 2019-05-17 09:28:00

@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?

## @Arnold Neumaier 2019-05-17 12:11:28

@moshtaba: Section 5.9, after (5.9.16). Note that there are only two polarization vectors $e^\mu$.