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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?

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    $\begingroup$ Which book? Which page? $\endgroup$ – Qmechanic Jun 18 '18 at 8:37
  • $\begingroup$ It's an italian book. I wrote wrong, it's written that It doesn't make sense trying to distinguish beetween spin and angular momentum related to photons $\endgroup$ – user198587 Jun 18 '18 at 9:11
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    $\begingroup$ For the record: Phys.SE generally encourages users to cite references, even if they are in a different language, or not easily available. $\endgroup$ – Qmechanic Jun 18 '18 at 9:22
  • $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Jun 19 '18 at 12:49
  • $\begingroup$ Related: What is the orbital angular momentum (OAM) of individual photons?. $\endgroup$ – Emilio Pisanty Jun 19 '18 at 17:37
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I am going to give two ways to answer this question as per the comments:

  1. Classical way:

Spin is the intrinsic angular momentum of a particle.

The other type of angular momentum is orbital angular momentum.

Orbital angular momentum is what we are talking about when we talk about an electron orbiting a nucleus in classical way.

In case of a photon, your book is saying that there is no sense talking about difference between total angular momentum and spin because usually photons do not have orbital angular momentum, there is nothing they would orbit in a classical way (except black holes).

So in case of an electron (a bound electron), total angular momentum is spin and orbital angular momentum.

But in case of a photon, spin equals total angular momentum (again this is the classical view introduced in 1913).

This was the case until the 1990's, and this idea of orbital angular momentum was introduced by Niels Bohr for an electron revolving around a nucleus in a circular orbit.

  1. as per QM

Now after the 1990's they discovered that EM waves (so this is not for single photons) can have something else then spin. Though in a classical way the EM wave is not orbiting anything, still there is another type of angular momentum that it can possess and it can rotate micro objects.

Here is an excerpt from the link in the other answer:

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

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 ].

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.

Now I was talking about the classical view because it is much easier to understand. But the real world now is based on QM and this tells us that electrons are not orbiting in a circular way the nucleus, and that there is another type of angular momentum of EM waves (other then spin). As you can see from the excerpt this type of angular momentum is different from the classical view.

To further explain that please look at this answer here:

https://physics.stackexchange.com/a/381381/132371

Now, if you want a direct mechanical detection of the angular momentum carried by a single-photon excitation of an OAM mode, then that's unlikely to be feasible - in the same way that it's likely to be infeasible for the linear momentum of that state, because both are very small and very hard to measure. In that respect, atomic absorption experiments showing modified selection rules are likely to be conceptually sufficient, but I'm unsure whether the experiment has been done as yet.

So you see it is very hard to talk about OAM for a single photon, and that is what you are asking. OAM is usually defined for the EM wave itself, for a herd of photons.

Please also see this answer:

https://physics.stackexchange.com/a/99019/132371

If you read the wikipedia article on orbital angular momentum of light you will see that in the first place it is a classical electromagnetic concept, where the light has a vorticity, i.e. a helical motion around the axis of the vortex.

When one goes to the quantum detail of photons one can define an OAM against this classical axis for each photon in this specific classical electromagnetic beam. Thus OAM is not an intrinsic characteristic of photons, but only on photons in special beam distributions, as in the figure:

So you can see OAM is not an intrinsic characteristic of photons , but only of photons in special distributions.

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    $\begingroup$ This is utterly and completely wrong. The orbital angular momentum of a photon is a perfectly valid quantity - did you try a literature search before posting? $\endgroup$ – Emilio Pisanty Jun 19 '18 at 7:17
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    $\begingroup$ Photons can have orbital angular momentum eprints.gla.ac.uk/67185/1/Yao-Padgett-2011.pdf $\endgroup$ – alanf Jun 19 '18 at 15:11
  • $\begingroup$ Just to be clear, this answer is wrong regardless of any attempt to use semantics in a way that conflicts with the broader literature. This answer makes the explicit claim that "in case of a photon, spin equals total angular momentum", which is flatly incorrect; whether you want to call the additional term OAM or something else, the angular momentum of light does not consist solely of spin. $\endgroup$ – Emilio Pisanty Jun 19 '18 at 17:34
  • $\begingroup$ This is again incorrect. Light has orbital angular momentum in classical electromagnetism just as much as it does in QED, and you do not need to invoke photons or quantum mechanics to explain it. $\endgroup$ – Emilio Pisanty Jun 19 '18 at 18:03
  • $\begingroup$ Please see this answer: physics.stackexchange.com/a/99019/132371 "When one goes to the quantum detail of photons one can define an OAM against this classical axis for each photon in this specific classical electromagnetic beam. Thus OAM is not an intrinsic characteristic of photons, but only on photons in special beam distributions, as in the figure:" $\endgroup$ – Árpád Szendrei Jun 19 '18 at 18:08
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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).

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  • $\begingroup$ @ 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" $\endgroup$ – Árpád Szendrei Jun 19 '18 at 17:12
  • $\begingroup$ "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 ]. $\endgroup$ – Árpád Szendrei Jun 19 '18 at 17:13
  • $\begingroup$ 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." $\endgroup$ – Árpád Szendrei Jun 19 '18 at 17:13
  • $\begingroup$ I do understand that you call this orbital angular momentum, but still you have to please tell me if you think this is completely the same type of OAM as for other types of particles like an electron orbiting a nucleus (in a classical way). Of course as per QM we cannot talk about an electron orbiting but still in a classical view the original understanding of OAM is for a bound electron. It was proposed by Niels Bohr as revolving in a circular orbit. $\endgroup$ – Árpád Szendrei Jun 19 '18 at 17:20
  • $\begingroup$ @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. $\endgroup$ – Emilio Pisanty Jun 19 '18 at 17:31

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