I'm looking for a physical argument that can demonstrate if the OAM modes are unique or not. In your above answer you have an "If...." and it is that clause I am trying to dig deeper into.

Planck's black-body equation incorporates the fact that photons are mass-zero, spin 1 particles and thus have a two-dimensional space of internal states (i.e. for fixed momentum). By what formalism you describe the many-photon state space can have no influence on the black-body spectrum. As Curtis mentioned, for specificly designed micro cavities (not falling under the general notion of a (macroscopic) black body) the thermal equilibrium state may well depend on the modes of this cavity. For instance, a flat cavity which has macroscopic dimensions in two orthogonal directions and is ony a fraction of the considered wavelength wide in the third orthogonal direction can't emit thermal radiation with a linear polarization along the narrow direction. So Planck's formula will probably fail for this wavelength (if you take the full form which takes into account emissivity and absorbance it may well stay correct ???).

I have two points to offer here. First, from OAM mode generation perspective, it usually requires large aperture(lens, hologram) with highly ordered phase distribution, with spiral phase plate being the most trivial example. In other words if SAM is associated with the rotation of the particular plane wave or a light ray, while the OAM is collective phenomena associated with rotation around beam axis.

I do not see how you would obtain the energy radiated in the given solid angle as it will change across the wave front. Again the helical beam is quite prominent example. With plane waves there is no such a problem .

From purely mathematical standpoint I agree with Curtis, the basis provided by orthogonal linearly polarized mode is full. I see here the OAM as the alternative basis rather than the additional set.

As for your concern for the "if..." clause as far as I understand the angular spectrum distribution is a finite function, i.e. it can be expanded in Fourier series on a complete basis. Therefore linearly polarized modes are sufficient in this case. You may use alternative basis, but I do not see why for your particular problem you should prove that OAM modes can be represented as a superposition of linearly polarized modes.

The black body radiation depends on the density of states. The standard derivation assumes that the temperature and size of the black body are such that many states are excited - having characteristic wave lengths which are much smaller than the black body. In this case the density of states is not sensitive to the exact shape and properties of the cavity (the waves traverse many wavelengths before interacting with the boundary ).

If however you were to cool down the system so that only a few low levels are excited, then the continuum description of the density of states has to be modified, and you might be sensitive to specific modes.

Hi Derek,

The OAM modes derive from the Gauss-Laguerre basis, and as such they are a complete orthonormal basis for any spatial distribution of an optical field. From this point of view they are no different (at least conceptually) from any other basis or equivalently modes, e.g. the Hermite-Gauss basis. It does happen, of course, that the OAM modes have some rather nice properties and can be used, for example, to encode information [e.g. Wang et al. "Terabit free-space data transmission employing orbital angular momentum multiplexing", Nature Photonics 6, 488–496 (2012)], but they are an extrinsic property of the photon field, not an intrinsic property of the constituent photons as is the case the photon polarisation.

@Andy. I feel your reply is getting closer to satisfying the particular worry I had and getting closer to the nub to the problem. So in a nutshell, are you saying that a fundamental difference between polarization and OAM is that you can do single-photon experiments with polarization, but you cannot with OAM?

@Derek, I don't think that your rephrasing is quite correct as you certainly can do single photon experiments with OAM - there are several papers on this, for example some of the nonlocality work of Zeilinger and co-workers, and the very nice work by Molina-Terriza and co-workers.

But you are correct in that the reason why you can do single photon experiments with OAM is different from why you can do experiments with polarisation. Polarisation, as a representation of the intrinsic spin of the photon, is confined to two polarisation states (spin \pm 1). You can certainly do continuous variable spatial interference (e.g. Young's double slit), or alternatively by imposing a finite set of modes, as in the case of OAM, you can study a much larger Hilbert space. But the difference is that the accessible Hilbert space in the OAM case is a consequence of the system. For example the detectors you use to monitor the spatial distribution of the photon field, or the source that is used to generate the spatial part of the field. If we ignore certain obvious constraints (e.g. size of optical elements and wavelength), then the accessible spatial Hilbert space is infinite and I can describe this using any set of orthogonal modes I want, and with appropriate optical elements I have complete freedom to add or subtract angular momentum from my field (or indeed from any individual photon).

However because the polarisation represents the intrinsic spin of the photon, then I am far more constrained in what kind of transformations can be applied and in the total Hilbert space that available (for example for encoding information, either quantum or classical). In the bug vs feature category, however, of course this means that polarisation is almost always more robust against environmental noise than OAM, and it also implies that OAM is most useful (perhaps only useful?) for direct line of sight or multi-mode fibre applications.

Hope this helps

@Andy. Ok, so how do we now take your description to make a clear statement why N=2 (and no more) in Planck's BB equation?

@Derek. Isn't the point here that in the derivation of the BB spectrum you are looking at the infinitesimal radiation. In the canonical treatment one is viewing the output from the BB through a pinhole so as to minimise disturbance of the BB, but equally this is an argument of energy density per unit volume. In such a limit it is simply not possible to talk about OAM as these are modes that have larger than infinitesimal surface area. So to put it another way: if I perform the BB measurement, I am observing the BB through an aperture and am hence forcing the output radiation into a single spatial mode, and hence I cannot expect to get more spatial information out. Again, alternatively (from the point of view of the interior of the BB) I am considering the total energy within the BB, and hence attempting to determine the energy density per unit wavelength. So in this case I am already using the spatial degrees of freedom and there is not the possibility to double count them by also using OAM encoding.

@Derek When calculating the BlackBody radiation you need to take into account all the possible states of the light in FREE SPACE as there is no prefenratial state. All the possible states of the photon can be obtained from a linear superposition of the orthogonal polarisation. In terms of Mathematical properties it means your space is of dimension two and the 2 vectors that define a complete base are the 2 polarisation vectors. So the 2 is a dirrect consequence of how many vectors you need to define to map all the different states of your photon. For the case of Hermite-Gaussian modes you actually map the system which is not any more a photon on is own. Here the base is bigger and you can have N modes where N can be larger than 2. In this the base of you system is no more a 2 vector set but a N vector.

This can easyly be seen using a simple optical fibre. It is possible to excite different modes of the fibre with the same polarisation. It will depend only on how the excitation is done.

The transformation to convert from one base to the other will depend strongly on your system.

To go back to the black body radiation one of the first statements in the derivation is : "Consider a cube of side L with conducting walls filled with electromagnetic radiation in thermal equilibrium at temperature T". Here it implies a very specific system and also that L is taken large enough so it is larger than the wavelength so you can sum on the different modes 2L/m where m is an integer. If L is taken smaller than the wavelength this statement do not hold any more and the derivation needs to be done using the actual optical Density Of Modes of the system. In this case, it can be interesting to do the sumation according to the modes of the system.

So there is a fundamental difference between a photon on its own bouncing inside a box and a photon inside a system which limits its possible states. I hope it is an helpful answer.

That prefactor is related to the number of modes the photon can be emitted into. This includes two polarization modes for each spatial mode. The number of spatial modes is determined by the parameters of the system (e.g. its dimensionality and size), and on the wavelength of the photon. This number does not depend on the specific basis used for the counting, and it will be the same for OAM modes as for plane waves (which are the common choice of basis). The reason for this is the degeneracy of the spatial modes: all of them are related to a photon with the same energy (that depends only on its wavelength), and therefore, any set of orthonormal modes that span this degenerate subspace is applicable, and by definition contains the same number of modes. As correctly stated by Andrew, there is no need to double-count.

Conservation of linear momentum and conservation of angular momentum states in any classical field are not completely independent (only the spin states or polarization states has to be accounted for). This is why we only counted k-space instead of L-space but if you prefer to do so, counting the density of states based on all possible L values should lead to the same statistics of photons, such as Planck's law of radiation.

A can give you a very detailed reply... but can you be bothered enough to make my effort worthwhile!

OAM should not be compared to spin in this discussion, but to the plane wave modes that are usually invoked in derivations of Planck's formula. Polarization ("spin") is an "internal" degree of freedom, while OAM modes are -- as correctly pointed out by Andrew -- one of many possible ways to encode "orbital" modes -- just as e.g. plane wave modes. So whether you use plane waves or OAM's, you always have to multiply by 2 to take care of the 2 possible polarizations.

Planck's equation is concerned with black-body radiation primarily at the classical limit treating energy (E=hv) as integral multiples of a basic unit hv. Using spherical symmetry, 4pi, and 2 polarization directions, Planck's equation is applied as 8pi, confirmed by experiment and as the present scholars have indicated above.

There is no "physical justification for dropping them off" because angular momentum was not an experimental or theoretical characteristic or mode for deriving the discrete energy of emitted and absorbed photons, to wit: Planck's constant and equation. Following upon Planck's work, the Dirac constant, also known as the reduced or rationalized Planck's constant (hbar=h/2pi) was derived for angular momentum in treating other modes for elementary particles such as the electron.

"Compton's experiment verified...and thus demonstrated that the energy and momentum are quantized according to E=hv, P=hbark". (Bohm, 1951) Where P=momentum, k=direction, and hbar=Dirac's constant.

This information is taken from Bohm (1951) and Lipkin (1967).

I am very grateful for your questions as they sent me on a wonderful journey of discovery - the origins and historical foundations for quantum theory.

The orbital angular momentum (OAM) mode is a superposition of plane waves, so it's counted in different k mode. Any filed distribution could be seen as the superposition of plane waves(different k), Any polarization could be seeen as the superposition of two directions (Ex and Ey). But if we take the single photon into consideration, this does not quite explain well.

For a photon what matters is elicity and the two possible values are +1 and -1.

For a massless particle it is not possible to separate the spin from the angular momentum. Indeed there is no rest frame, that's why elicity is important. The state of a photon is uniquely defined by its momentun and elicity.

the angular momenta are related to two states of circular polarization, but each of them can be decomposed into two linear polarization states with a 90 deg phase change. If you superimpose the two couples of linear polarization along the same transverse axis, you still remain with two independent polarization states. You can decompose the polarization ellipse in either way (two circular or linear one, varying the phase shift) but you always remain with two possible independent states. There is no way to do it in another way. We live in a 3D space and light is a transverse wave!

You can construct rotating waves with linear combinations of Complex numbers. If you find a copy of Chandrasekhar's "Radiative Transfer" transfer. you will have the best explanation I have ever seen. It's about avoiding multiple counts of the same thing.