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Update @ 21.01.2018

People investigate and talk about orbital angular momentum (OAM) of photons. For example, see this well-cited paper here and the PRL here. The latter starts with the remark

It is well known that photons can carry both spin and orbital angular momentum (OAM).

By definition, a photon is a one-particle state with definite momentum and helicity. In quantum mechanics, the OAM doesn't commute with momentum, and therefore a momentum eigenstate is not an OAM eigenstate. By that logic, individual photons cannot have definite OAM. Does it mean that the situation changes in quantum field theory and particles with definite momentum can also have definite OAM?

Maggiore's book on Quantum Field Theory talks about the spin (or helicity, to be more precise) of photons by working out the action of the spin operator on one-particle photon states (see my answer here) brushing aside the action of OAM operator.

  • I'm eager to understand the quantum field theory perspective of the OAM of individual photons. What is the OAM of individual photons?

  • Is there some physical example that cannot be explained without assuming that individual photons carry nonzero OAM in addition to spin (or helicity, to be more precise)?

Emilio Pisanty
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SRS
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2 Answers2

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This is an interesting mix of misconceptions:

By definition, a photon is a one-particle state with definite momentum and helicity. In quantum mechanics, the OAM doesn't commute with momentum, and therefore a momentum eigenstate is not an OAM eigenstate. By that logic, individual photons cannot have definite OAM.

No, a correct definition of a photon doesn't need to assign it a definite momentum or helicity, or even a well-defined frequency. The most convenient basis expansions tend to have these properties, but that's not inherent to the definition of a photon.

The short of it is that when you quantize electromagnetism, you start off by finding a suitable basis of vector-valued functions $\mathbf f_n(\mathbf r)$ in which to expand the vector potential as $$ \mathbf A(\mathbf r,t) = \sum_n\bigg[a_n(t)\mathbf f_n(\mathbf r)+a_n(t)^*\mathbf f_n(\mathbf r)^*\bigg], $$ where $a_n(t)$ is the generalized coordinate corresponding to the mode $\mathbf f_n(\mathbf r)$, setting things up so that its Poisson bracket with its conjugate is $\{a_n,a_m^*\}=\delta_{mn}$, and then you quantize by replacing $a_n(t)$ with the annihilation operator of that mode, so that a state with a single photon on that mode is $a_n^\dagger|0⟩$.

Now, here's the important thing: there is no requirement that the mode functions $\mathbf f_n(\mathbf r)$ be plane-wave states with circular polarization. It's a convenient choice, but it's not the only possible choice. Photons are excitations of the classical mode in question. Thus, if the classical mode is a plane wave, the photon will have a well-defined linear momentum, but if it is e.g. a Laguerre-Gauss or a Bessel mode, it will have a well-defined orbital angular momentum.

And, exactly as with the basis mode functions themselves, a photon with well-defined angular momentum can be understood as a superposition of photons with well-defined linear momentum (and vice versa), in the same way that you can expand a plane wave in terms of Bessel functions and vice versa. More importantly, this extends to linear combinations of modes with different frequencies: these give single-photon wavepackets, which evolve in time and are not eigenstates of the field hamiltonian, but are still $N=1$ eigenstates of the photon-number operator and therefore equally valid as single-photon states as the single excitations of a monochromatic plane wave.


OK, thus far for the standard description of how to deal with orbital angular momentum within the larger framework of quantum electrodynamics and quantum field theory, or within the more restrictive subsets of those that are often called quantum optics. However, just because you can describe something in a quantum way doesn't mean that you need to, but unfortunately ruling out alternative possible explanations, as you ask in your second question,

Is there some physical example that cannot be explained without assuming that individual photons carry nonzero OAM in addition to spin (or helicity, to be more precise)?

is a rather hard proposition.

However, OAM in this respect is no different to any other degree of freedom of light, and for any experiment that requires photons and a quantum-mechanical description on a given coordinate, you can produce an working experiment built on OAM, from Mandel dips to Bell-inequality violations to quantum cryptography, for which a good review is

G. Molina-Terriza, J.P. Torres and L. Torner. Twisted photons. Nature Phys. 3, 305 (2007).

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.


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

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

Emilio Pisanty
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  • "Is there some phenomena that cannot be explained without x" is pretty much always an unreasonable request. One can probably formulate all of physics without any reference to energy, for example, it just would be incredibly inconvenient. – Rococo Jan 21 '18 at 19:50
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    @Rococo Indeed it is, but it is still a very valuable question that needs to be asked repeatedly even if sometimes there isn't a clear answer. Large swathes of properties often ascribed to "photons" can be reproduced just as well by coherent states, and therefore by classical states of the field, possibly with some shot noise. (As a similar example, there's nothing quantum about gauge transformations, which show up already in lagrangian classical mechanics.) Keeping a close tab on what does and does not require quantum mechanics for its explanation is an important thing to do. – Emilio Pisanty Jan 21 '18 at 19:58
  • @EmilioPisanty Maggiore's book on quantum field theory defines a photon in the same way I did. Please see the fist first paragraph of page 100 here. https://books.google.co.in/books?id=yykTDAAAQBAJ&pg=PA100&dq=The+fact+that+massless+particles+are+helicity+eigenstates&hl=en&sa=X&ved=0ahUKEwjzg5nQ-OnYAhWLqY8KHbJaAb0Q6AEIKDAA#v=onepage&q=The%20fact%20that%20massless%20particles%20are%20helicity%20eigenstates&f=false – SRS Jan 21 '18 at 20:54
  • @SRS As I said in the answer, that is a valid choice, but it's not the only possible choice. There is nothing special about the single-photon states defined by Maggiore (beyond being eigenstates of $H$, $\mathbf p$, $\mathbf s$) - they're what comes out if you start with a plane-wave basis expansion, but if you start with (say) a cylindrical-harmonics expansion, you'll get single-photon states with different properties. – Emilio Pisanty Jan 21 '18 at 21:00
  • @EmilioPisanty This is a great answer! I struggled hard to get my head around this. The QFT books I'm familiar with do not give this description of photons, and I've barely seen a quantum optics book. – SRS Jan 22 '18 at 07:24
  • @EmilioPisanty I have a related question. Is this expansion in terms of modes other than plane waves also possible for other relativistic fields such as the electron field? Can we have one-electron wavepackets in QFT? – SRS Jan 22 '18 at 07:38
  • @EmilioPisanty I had one more question. Single-photon wavepackets do not have definite linear momentum. Does it not mean that they also do not have definite helicity? – SRS Jan 22 '18 at 12:38
  • @SRS Then get reading! Scully & Zubairy, Gerry & Knight, and Cohen-Tannoudji's Photons and atoms make great starting points. On your technical questions: (i) obviously yes, the basis expansion is used across the board in QFT. (ii) Single-photon wavepackets can have definite linear momentum, at least in one dimension, and even in situations where they have support over all possible momenta (with, say, a gaussian envelope) it is still perfectly possible to make them have well-defined helicyt by simply enforcing that for each plane-wave component. – Emilio Pisanty Jan 22 '18 at 18:35
  • +1 Excellent answer! I have a follow up question. In spectroscopy, non zero OAM photons have much lower cross sections when compared to ones with zero OAM (due to transition probability). Does this mean that most light that we observe have OAM of 0? – Superfast Jellyfish Sep 19 '20 at 07:38
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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:

helical em beam

Different columns show the beam helical structures, phase fronts, and corresponding intensity distributions.

Edit after edit of question:

As far as I understand, in case of a free non-relativistic massive particle moving in a straight line, the value of L=r×p can be made to vanish (for all times) by choosing the origin of coordinates on the trajectory. Therefore, a free particle in classical mechanics need not have a nonzero orbital angular momentum.

This is a misunderstanding. Angular momentum can be defined whenever an axis that can give an $r\times p$ can be defined. It is a mathematical equation. It is when conservation of angular momentum enters the picture that the specific axis defines a specific $L$

As photons do not bind in potential wells, (except black holes by gravitation which is another story), there is no orbital% angular momentum because no orbits.

edit after discussion in comments:

%Definition orbital angular momentum as in this link.

anna v
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    There is "external orbital angular momentum" and "internal orbital angular momentum." The latter is indeed intrinsic. – Eric Walker Oct 29 '15 at 04:41
  • @EricWalker internal orbital angular momentum for elementary particles is called spin. The photon is an elementary particle. – anna v Oct 29 '15 at 04:59
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    See https://en.wikipedia.org/wiki/Orbital_angular_momentum_of_light. – Eric Walker Oct 29 '15 at 05:02
  • @EricWalker Light is composed of photons, photons are not light. Buildings are composed out of bricks. Bricks are not buildings. Anyway, the link you quote is on the top of my answer from which I took the illustration. I am talking of internal here, and it is intrinsic to the beam of light, not to the photons. – anna v Oct 29 '15 at 05:07
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    You may be correct. But if you are, this leaves open another question that has been on my mind. How can a gamma photon emitted from an $I=2$ or $I=3$ nuclear transition carry away only one $1ℏ$ unit of angular momentum (i.e., its intrinsic spin)? – Eric Walker Oct 29 '15 at 05:11
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    @EricWalker angular momentum is conserved. The nucleus picks up the balance in the new state. – anna v Oct 29 '15 at 05:13
  • We're talking about conservation of angular momentum, not linear momentum. – Eric Walker Oct 29 '15 at 05:14
  • Spin is just intrinsic to a particle angular momentum. It is not conserved separately – anna v Oct 29 '15 at 05:15
  • Understood. But there are isomeric transitions from $2+$ to $0+$, to give one example, in which a single photon is produced. That photon has to carry 2 units of angular momentum in order for the system to conserve total angular momentum. – Eric Walker Oct 29 '15 at 05:16
  • If this is true, I am not up to date with nuclear physics, the photon would be carrying off angular momentum with respect to the nucleus similar to the intrinsic angular momentum described above in the beam of light. The angular distribution of the gamma would not be isotropic with respect to the nucleus orientation . – anna v Oct 29 '15 at 05:22
  • That's correct -- it's not. There is a whole field of angular correlation studies that look at things like the angular distribution of photons emitted from polarized nuclei. – Eric Walker Oct 29 '15 at 05:24
  • (I guess this angular momentum wouldn't be "intrinsic," in the sense that it's required for a photon to be a photon. So I have been using the term incorrectly.) – Eric Walker Oct 29 '15 at 05:36
  • "There is no orbital angular momentum because no orbits" is an extremely strange thing to say; as you have already noted the orbital angular momentum of light is a perfectly standard concept, it has extremely broad support in the literature, and there are perfectly valid reasons for the choice of the name. – Emilio Pisanty Jan 21 '18 at 17:16
  • @EmilioPisanty I cannot tire of saying : light is emergent from photons, photons are not light. Photons are not in orbits, to have orbital angular momentum ( except around black holes). They have angular momentum against any r vector possible to define to cross with their momentum p, but it is not an orbit .In the atoms there are no photons, only electrons and nuclei. Photons transiting may take angular momentum against a nucleus axis , but that is not an orbit either. – anna v Jan 21 '18 at 17:33
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    "It is not an orbit" - and neither is that of the motion of an electron around an atom; if you're OK with saying that a $2p$ electron in hydrogen "doesn't have orbital angular momentum" because it "doesn't have an orbit", then I guess that position is consistent. (There is no other way: QFT regards both the EM and the electron field as equivalent. You either take the equivalence or you reject QFT as a valid description of nature at that level.) – Emilio Pisanty Jan 21 '18 at 17:43
  • Also, a technical comment: thinking of OAM as "taking angular momentum against an axis" probably reflects a mistaken identification of the OAM of helical-phasefront beams as extrinsic (i.e. dependent on the coordinate origin) instead of intrinsic (i.e. independent of the origin). Light's OAM is intrinsic (reference), and it is every bit a property of the photons as their momentum is. (And again, if you're happy rejecting the latter, then sure, that's a consistent position.) – Emilio Pisanty Jan 21 '18 at 17:46
  • @EmilioPisanty again you confuse light with photons, I am answering about photons. Of course light does have intrinsic angular momentum , as seen in the link in the answer. Quantum mechanically electrons have orbitals, bound states with ,ml quantum numbers . Photons do not have orbitals. This shows the way photons with their intrinsic spin +/-1 build up the circular polarization of light https://en.wikipedia.org/wiki/Spin_angular_momentum_of_light – anna v Jan 21 '18 at 18:17
  • @EmilioPisanty I see where we miscommunicate. You are calling an angular momentum of the photons in a circularly polarized light,orbital angular momentum and I call it just "angular momentum "against the axis of light. Individual photons do not orbit around the vorticity axis, they are not bound, light is a superposition of photons not an interaction to generate bound states. The intensity of light (suerposition) orbits, but not the individual photons that make it up, imo. – anna v Jan 21 '18 at 18:26
  • No, at no point am I referring to the spin angular momentum of light as encoded in its polarization. The viewpoints you express in this answer and in several comments, in seeing light as somehow "separate" from photons, are simply not compatible with the QFT description of nature, and there is as yet absolutely no experimental evidence to doubt that description. Individual photons in QFT can and do have OAM (in the same way that electrons have OAM, while neither photons nor electrons 'orbit' any axis) and that's something one just has to accept (or reject QFT altogether). – Emilio Pisanty Jan 21 '18 at 18:41
  • @EmilioPisanty There exist bound states in field theory https://journals.aps.org/pr/abstract/10.1103/PhysRev.84.350 , to contrast with unbound ones. The photon cannot be in a bound state. – anna v Jan 21 '18 at 18:54
  • Indeed - but a state being bound has nothing to do with the appropriateness of the term 'orbital' in OAM. Coulomb waves (the continuum eigenstates of the hydrogen atom) are not bound but they have well-defined orbital angular momentum in exactly the same way as bound states, and the same is true for e.g. the Bessel basis for a free massive particle in 2D. – Emilio Pisanty Jan 21 '18 at 18:59
  • The QM literature treats the qualifier 'orbital' as referring to the spatial dependence of whatever the relevant (wave)function is, nothing more. Any indication that bound states or "orbits" are required is in direct contradiction with the usage in the literature, i.e. the content at the end of this answer is somewhere between misleading and advocating for extremely non-standard use of terminology without indicating it. – Emilio Pisanty Jan 21 '18 at 19:00
  • @EmilioPisanty I have made an edit with the definition of orbital angular momentum I am assuming. I searched and did find this https://www.nature.com/articles/srep29212 , for example, and I see that a redefinition has been going on in the specialized field. – anna v Jan 21 '18 at 19:23