Two recent papers about photon orbital angular momentum have set my head spinning as well. I'll try to define "property" as something you can change in one arm of a Mach-Zehnder interferometer, and "distinct property" as a change in the interference pattern from one property change that can not be imitated by making a change in a different property.

My Mach-Zehnder interferometer is built in free space (not in waveguides or fibers) on an optical table.

I can change the direction of the momentum with a wedge prism in one arm, or the magnitude via doppler shift with some piezo-driven mirror. These result in various fringe patterns in 3D at the output which can be probed and mapped. Momentum is one property.

I can change the polarization state in one arm using a rotating half wave plate. This will reduce the amplitude of any fringe pattern (or asymmetry in intensity between the two outputs. Polarization is one property.

I can affect the population of the helicity states in one arm using a quarter wave plate. I'm guessing that the macroscopic property of circular polarization is related to this. This will be distinct from the changes caused by the half-wave-plate-induced polarization changes when looking at the difference between the patterns in the two outputs of the interferometer. Helicity is one property.

Now, reading that photons with non-zero orbital angular momentum, or vorticity are less exotic and more commonplace than has previously been thought, I'd like to understand better if it must now be considered as fundamental property as the others above.

Question: Is orbital angular momentum, or "vorticity" a fourth property of photons that would work in my definition of distinct, independent properties?

Are there more I haven't thought of?

below: One implementation of a Mach-Zehnder interferometer using an extended, collimated, coherent source. From here.

enter image description here

  • $\begingroup$ How about speed and Oscillating frequency $\endgroup$ Commented Jul 28, 2017 at 22:03
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    $\begingroup$ Four houses at most; after that it has to be a hotel. $\endgroup$ Commented Feb 22, 2020 at 2:38
  • $\begingroup$ What has been the reaction to these cited papers? It is surprising to me that no one had noticed the OAM of photons radiated from from cyclotron motion. $\endgroup$
    – aRockStr
    Commented Feb 22, 2020 at 3:28

2 Answers 2


Short answer: The orbital angular momentum (or vorticity) of photons is part of the spatial degrees of freedom. It would therefore replace the momentum degree of freedom and should not be consider as an additional degree of freedom.

Longer answer: Photons as associated with creation and annihilation operators that can in general be represented by:

$$ a_s^{\dagger}({\bf k}) , a_s({\bf k}) $$

where the subscript $s$ represent the spin, which manifests as the polarization of the the light, and ${\bf k}$ is the propagation vector. (Here I only consider propagating fields. For virtual fields there would also be the angular frequency $\omega$ as an independent degree of freedom.) The spin degree of freedom is reduced to a two-dimensional Hilbert space due to gauge invariance. The propagation vector is a three-dimensional vector, which gives three infinite dimensional Hilbert spaces. The latter represents the spatio-temporal degrees of freedom.

In addition to the spin and spatio-temporal degrees of freedom, one also has the particle number degree of freedom. This comes from the fact that one can have states with multiple photons that are mutually orthogonal.

This more or less exhausts the different degrees of freedom that one can have with photons. All the different properties of photons are in some way "encoded" in terms of these degrees of freedom.

For instance:

  • Helicity is part of the spin degree of freedom and it comes about because one cannot measure the spin of the photon in its rest frame (because it doesn't have one). As a result the only well defined quantity related to the spin degree of freedom is actually its helicity.

  • Time-bins come about from two steps. In the first step one converts the three components of the propagation vector to two (transverse) components plus the angular frequency via the vacuum dispersion relation $c|{\bf k}|=\omega$. (It also assumes that we pick a particular propagation direction, which in turn may imply a paraxial approximation.) The next step is then to go to the inverse Fourier domain of the angular frequency, which is time.

  • Orbital angular momentum (OAM) is based on a particular modal expansion for the two transverse spatial degrees of freedom. These two transverse spatial degrees of freedom can be seen as the inverse Fourier domain of the two remaining components of the propagation vector, after the conversion referred to in the previous point. One can represent these two spatial degrees of freedom in terms of any modal expansion. However, there are some special modes that are eigenstates of the angular momentum operator. These OAM modes include the Laguerre-Gauss modes and the Bessel modes, each forming a complete orthogonal basis. For some reason (which is perhaps beyond the scope of this question) it became popular to use such OAM modes in quantum optics.

So how would one measure the OAM of a photon? One can do it with a two step process. All OAM modes have an azimuthal phase dependence given by $\exp( i \ell \phi)$, where $\phi$ as the azimuthal coordinate and $\ell$ is an integer representing the azimuthal index, and which is proportional to the OAM of the mode. First, one would need to remove this helical phase with a hologram or spatial light modulator having the conjugate phase. The next step is to couple the result after the modulation into a single mode fibre. Since a mode with a nontrivial helical phase cannot couple into a fibre, the light that does couple into the single mode fibre had to have the right azimuthal index before the modulation step. Thus we can determine the AOM of the mode.

If one has a classical light beam with a vortex in it, one can observe the vortex by looking at the interference between this vortex beam and a plane wave (beam without a vortex). The vortex shows up as a forked fringe within the interference pattern. Higher order vortices would give forks with multiple prongs.

To address the actually number of degrees of freedom (which I apparently missed in the question), we consider them one by one:

  • Polarization: this is easy, there are only two degrees of freedom, which describes a two-dimensional space. This space is represented by the surface of a sphere, which is called the Poincare sphere. This is an convenient set of degrees of freedom to use in experiments because they are easily manipulated using various polarization components.

  • Particle number: here we have a countable infinite number, as indicated by the Fock basis. However it is a difficult set of degrees of freedom to use, because one needs to prepare photonic quantum states. Nevertheless, it is a vibrant field of research at the moment and new methods to prepare such state are invented constantly. Moreover, there is an added bonus, because each photon comes with a full complement of all the other degrees of freedom.

  • Spatio-temporal: here we also have a countable infinite number. To see this one needs to remember that a valid complete orthogonal basis for a set needs to have its elements taken from that set. The Fourier transform may give one the impression that the number of degrees of freedom is uncountably infinite, but the basis elements associated with the Fourier transform are plane waves, which are not normalizable and therefore not elements of the set of functions with well-defined Fourier transforms. Therefore, we need to select a basis with elements that are normalizable. In the context of an actual experiment with paraxial optical beams, and focussing only on the spatial degrees of freedom, the Laguerre-Gauss (LG) modes are a convenient set to use (but there are many others that one can use). There are a countable infinite number of LG modes. Hence a countable infinite number of degrees of freedom. A similar argument can be used for the temoral degrees of freedom. However, this is a theoretical quantity. In a practical experiment, one would always only have access to a finite number of these due to the limitations of the experiment. Any physical optical setup has a finite space-bandwidth product associated with it, which determines the number of degrees of freedom that can pass through the system. In the temporal domain, the time-bandwidth product plays a similar role.

In summary: the only properties that photons can have as those associated with particle number, spin and spatio-temporal degrees of freedom. While spin has only two degrees of freedom, the other degrees of freedom are (theoretically) countably infinite.

  • $\begingroup$ In the spirit of a good SE answer, can you also address the question at the level at which I've asked it? This answer is not helpful to me as I don't understand how I would apply this in the experiment I have described. Thanks! (hint: axicon) $\endgroup$
    – uhoh
    Commented Jul 23, 2017 at 6:34
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    $\begingroup$ Minor nitpick: It's not that we cannot measure the photon's spin in its rest frame, it's that it doesn't have a rest frame to begin with. $\endgroup$
    – ACuriousMind
    Commented Jul 23, 2017 at 9:50
  • $\begingroup$ @ACuriousMind: that's what I meant. I included a clarification. $\endgroup$ Commented Jul 28, 2017 at 14:26
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    $\begingroup$ @uhoh: I've added a couple of paragraphs about the experimental observation of OAM, both for the single photon and the classical case. Not sure what the axicon is supposed to do though. Axicon are usually used to produce Bessel beams. $\endgroup$ Commented Jul 28, 2017 at 14:28
  • $\begingroup$ @flippiefanus OK thanks, it will take me a bit of time to read through, I can see you've added quite a lot, thanks! That axicons are used for one thing does not mean they are not used for another. I don't know how it works (yet) but see for example nature.com/articles/srep21877 Both kinds of beams need a null in the center for instance, but I don't know how they get a spiral phase with the axicon and the lens - misalignment perhaps? $\endgroup$
    – uhoh
    Commented Jul 28, 2017 at 14:46

The photon wavefunction has an essentially infinite number of degrees of freedom. Proof: the shape of a wave front describing a photon wavefunction can be described by Zernike polynomials. The terms are orthogonal: change the amplitude of one, and it does not affect the amplitudes of the other terms (other than the necessity of normalizing the sum of all amplitudes). Further proof: every wavefunction has a non-infinitesimal spectral bandwidth. All spectra (an infinite number) that have that bandwidth are valid states of the photon. The number of terms of a Fourier transform used to describe such a spectrum is infinite and the terms are orthogonal.

Interference in a double-slit interferometer illustrates the fact that the "position-state" of the photon can exist in superposition. The relative likelihood of the photon passing through either slit can be measured by alternately blocking each slit.

Much the same thing can be done by putting an array of blockable pinholes all over a sheet of opaque material blocking a beam, and covering/uncovering the different pinholes while measuring what gets through. Each pinhole represents a different orthogonal spatial mode. By adjusting the timing of the covering/uncovering, the spatiotemporal statistics of a series of light pulses (in principle) can be measured.

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    $\begingroup$ I think it's fixed now, thanks! $\endgroup$
    – S. McGrew
    Commented Feb 22, 2020 at 3:08
  • $\begingroup$ Thanks! I see, these are not necessarily all simultaneously and independently measurable (which I didn't stipulate in the question), but we could at least say that we independently set an arbitrarily large number of Zernike coefficients by passing a photon through some complex apodizing screen. We could independently measure a photon's wavelength and linear polarization with a grating and polarizing beamsplitter, so I know that the answer to the question I want to ask (but don't yet know how to word correctly) is "at least two". $\endgroup$
    – uhoh
    Commented Feb 22, 2020 at 3:17
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    $\begingroup$ Vastly more than two, if I understand your question: an infinite number. I have edited my answer to broaden the explanation. $\endgroup$
    – S. McGrew
    Commented Feb 22, 2020 at 3:41

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