S. Coleman: "Spin is a concept that applies only to particles with mass"

Question

Coleman ("Quantum Field Theory Lectures of Sidney Coleman", p. 400) states:

"Spin is a concept that applies only to particles with mass, because only for a particle of non-zero mass can we Lorentz transform to its rest frame and there compute its angular momentum, which is its spin. For a massless particle, there is no rest frame, so we can't talk about the spin. We can however talk about its helicity, the component of angular momentum along the direction of motion."

Is it not possible to define / discuss spin without referring to the rest frame? In some cases, an expression for total angular momentum $\textbf{L} = \textbf{J} + \textbf{S}$ has one piece which is coordinate invariant, which can therefore be identified as the inherent angular momentum, or spin.

Why then do we routinely refer to the spin of the photon as $\pm 1$ (or $\pm \hbar$)?

Answer 1

I want to extend and possibly explain the comment made by Prahar which is correct. Taking Wigner by his word, when we say particle we are really talking about an irreducible projective representation of the Poincare group. The crux of your question lies in what these representations are, which depends on whether you are looking at a massive or massless particle. Constructing them is usually done via the "little group method".

First, you can note that the projective representations of the Lorentz and Poincare group are real representations of their respective covering groups $SL(2,\mathbb{C})$ and $SL(2,\mathbb{C}) \times \mathbb{R}^4$.

In the massive case, for a momentum $p = (m, \vec{0})$, the little group is $H_p = SU(2)$, and you get the spin you know from quantum mechanics.

In the massless case, for $p = (p^0,0,0,p^0)$, the little group is a different one and the analysis is more complicated. See for a reference chapter 1, section 3 of Local quantum physics by Haag. In the end, you get the helicity, not spin.

Answer 2

For a massless spinning particle the decomposition ${\bf J}={\bf L}+ {\bf S}$ is frame dependent. See the last section of our paper arXiv:1501.04586 for a physical picture of why this is so,

Answer 3

The statement is incorrect. Photons have spin 1.

However in the standard theory there is no expression possible for electromagnetic spin. There is only one for the total field angular momentum, namely $${\bf J} = {\bf r} \times {\bf P}\,,$$ where ${\bf P}$ is the Poynting vector. Note that this expression fails for the angular momentum of a circularly polarised plane wave, for which it gives zero. Expressions exist that are not gauge, Lorentz or even rotationally covariant (as referred to in the answers of Mika and Mike Stone).

To have spin conservation you need a Lagrangian that has the correct symmetry, as Noether's theorem indicates. The standard Lagrangian lacks this symmetry. The so-called Fermi Lagrangian does have the correct symmetry$^*$. It gives the following expression for electromagnetic spin density $$S^{0\mu\nu} = \partial^0 A^\mu A^\nu$$ or $$ {\bf S} = \varepsilon_0 {\dot {\bf A}} \times {\bf A}\,.$$ This expression is valid in the Lorenz gauge and in units with $c=1$. It gives the density of spin for a general electromagnetic field, not just for a single plane wave as many other expressions do.

$^*$ The textbook by Mandl&Shaw (1991) bases field quantisation on this Lagrangian, following Fermi.

Edit after comments:

While every statement in it is verifiably correct, my answer has stirred deep emotions. This was not my intention at all. I was accused of 'bad faith' and even a higher being was invoked. This unfortunate lack of objectivity obstructs an open scientific discussion on the fundaments of field theory.

The critical discussion of the fundaments of main stream physics is an essential part of main stream physics.