# Photon spin and total angular momentun

I'm studying the photon spin and I assume that the angular momentum of spin is equal to 1 (from QED). In my book it's written, related to photons, that it doesn't make sense distinguishing between spin and angular momentum. Can you explain me why?

• Which book? Which page? – Qmechanic Jun 18 '18 at 8:37
• For the record: Phys.SE generally encourages users to cite references, even if they are in a different language, or not easily available. – Qmechanic Jun 18 '18 at 9:22
• It is currently impossible to provide a full and precise answer to this question without an exact reference to the text that you're using. The details of the language matter, and very much so, and by deciding to withhold it (because you decided that your answerers don't speak the language?) you're substantially crippling your question. Provide a full reference and let your interlocutors worry about the translation. – Emilio Pisanty Jun 19 '18 at 12:49
• – Emilio Pisanty Jun 19 '18 at 17:37
• Just to be a curmudgeon, the spin angular momentum is not $1$. The total spin quantum number is $j=1$, which means the spin angular momentum is $\hbar\sqrt{j(j+1)} = \sqrt 2\hbar$. – JEB Jun 19 '18 at 18:18

There's a number of things that your book could mean by that phrase, and it's impossible to tell exactly what the text entails without the precise wording. However, there's a number of salient points to keep in mind.

• Spin is a type of angular momentum. Angular momentum, in its very essence is the conserved quantity that corresponds, via Noether's theorem, to rotational invariance: in other words, if the hamiltonian of a system is rotationally invariant then angular momentum is conserved, and angular momentum acts as the generator of rotation transformations. For particles with spin, it is the spin that acts as the rotation generators, so that alone should seal the deal, but we also know that it can be exchanged into the more usual mechanical angular momentum (via the Einstein-de Haas effect).

The same goes for photons ─ their spin acts as the generator for the rotations of the internal degrees of freedom of the electromagnetic field, i.e. the vector aspects of the EM field's polarization, and it can equally well act mechanically (a tool known as an optical spanner) to transfer angular momentum to material particles.

• On the other hand, spin is not the only type of angular momentum that light can hold. Instead, just like matter, light can hold orbital angular momentum, which comes from how its linear momentum density is distributed in space and therefore from how its wavefronts and spatial dependence are laid out. And, as in the link above, optical spanners can also be used to translate it to mechanical angular motion.

• That said, there is a fundamental issue in trying to split the total angular momentum of light $\mathbf J$ into spin and orbital components $\mathbf J = \mathbf L + \mathbf S$. There's a ton of subtlety involved if you want to do the maths right, mostly to do with the gauge-freedom aspects of QED (with which you can start e.g. here), but the core idea is that you cannot rotate the polarization of light arbitrarily and keep a straight toe to the Maxwell equations: if you have a wave that's linearly polarized along $x$ propagating along $z$ and you do a 90° turn about the $y$ axis, then the wave will no longer be transverse and it will break the Gauss law.

This ultimately means that it is hard to give a fully bullet-proof definition of the spin angular momentum of a photon, but there's plenty of definitions that (while not bulletproof) are plenty good for an overwhelming majority of practical purposes.

Finally, if you want a comprehensive yet readable introduction to the subject of the angular momentum of light, I would recommend this PhD thesis:

R.P. Cameron. On the angular momentum of light. PhD thesis, University of Glasgow (2014).

• @ Emilio Pisanty while I understand what you are saying here is an excerpt from your link: "One sees immediately that, even if circularly polarized, a plane wave cannot carry an angular momentum of any type. This last statement has led to some debate, but the resolution of this seeming paradox is simply that the perfect plane wave is only ever found in textbooks. Real beams are limited in extent either by the beams themselves or by the measurement system built to observe" – Árpád Szendrei Jun 19 '18 at 17:12
• "The spin angular momentum (SAM) of light is connected to the polarization of the electric field. Light with linear polarization (left) carries no SAM, whereas right or left circularly polarized light (right) carries a SAM of ± ̄ h per photon. them, and this finite aperture always gives rise to an axial component of the electromagnetic field [ 4 ]. – Árpád Szendrei Jun 19 '18 at 17:13
• For the case of circular polarization, the axial component of the electromagnetic field is an unavoidable consequence of the radial gradient in intensity that occurs at the edge of the beam or the measurement system. A detailed treatment of these edge effects, for any arbitrary geometry, always returns a value of the angular momentum, when integrated over the whole beam, of ± ̄ h per photon [ 4 ] for right-handed and left-handed circular polarization, respectively." – Árpád Szendrei Jun 19 '18 at 17:13
• @ÁrpádSzendrei I'm not particularly interested in debating semantics; I've had exactly this debate before and, frankly, it is extremely boring. It is a complete fallacy to attempt to restrict the use of the word "orbital" for classical or Bohrian orbits, because that rules out OAM as a descriptor for the AM held by electrons in atoms just as much as it does photons. Grown-up physicists use QM, not the Bohr model, and the OAM I refer to is the one-size-fits-all version from the grown-up QM hydrogen atom. If you want to use the Bohr model, go to 1913. – Emilio Pisanty Jun 19 '18 at 17:31
• That's a misrepresentation of that claim. It's hard to find experiments that provably strictly require a quantized field (and thus photons) for their description, regardless of the degree of freedom in question. OAM has some gauge-freedom issues as to its definition and its unambiguous separation from spin (already explained here, and explained in depth in the cited literature), but the OAM of a single photon is as uncontroversial as its linear momentum. – Emilio Pisanty Jun 19 '18 at 18:15