# Does the frequency of photons decrease at outer radii of a helical mode?

A helical mode is a mode of the electromagnetic field in which the wavefront is characterized by one or more helixes. Along the direction of travel, at the center, lies an "optical vortex." The number of helixes is an integer that gives the "topological charge," and it can be $\pm1, \pm2, \pm3,$ etc. As a wavefront passes through one wavelength $\lambda$, the helixes sweep through an angle of $2\pi$.

The accepted answer to a recent question about the path that photons follow in an such a mode was that they follow a straight line, but there is a delay in the phase that depends upon position. Assuming this answer is correct (which I don't), does it imply that photons at outer radii of the wavefront have lower frequencies, so that the phases of the photons at different radii will remain aligned along the line that a given helix sweeps out? Does this mean that a helical mode is not possible for monoenergetic photons?

The idea that photons only travel in straight lines at best oversimplifies things, and at worst (such as in this case) is just plain wrong.

Photons are the quanta or excitations of an electromagnetic field, so any concepts you may know from classical EM theory about field solutions also applies to the possible modes a single photon can be in. The only type of mode where a photon is truly going in a straight line is an infinite plane wave (which was a definite wavevector $\mathbf k$ and thus momentum $\mathbf{p = \hbar k}$). If a photon is in any other field configuration, then it is in a coherent superposition of momentum states and thus in general is in a coherent superposition of going in multiple directions at once. Even a typical (Gaussian) laser beam which seems to go in a straight line will have some spreading or diffraction which is due to not being an infinite plane-wave (e.g. the beam is partially localized and so obeys an uncertainty relation between the cross-sectional beam size and the transverse wavevector/momentum).

If you have a simple enough field (i.e. one that is smooth and doesn't change too much on the scale of the wavelength), then at every point in the field you can define a local wavevector or ray (note that this is an approximation similar to the ray optics of geometrical optics). Now the reason for the change in phase in a cross-section of the beam can be understood in a purely geometric way. As shown in the image below, if you have two points initially in phase but with local wavevectors or rays pointing in slightly different directions, then after some propagation, the transverse cross-section of the beam along the optical axis will have accumulated different phases at different points in the beam, which is what is shown schematically in the image below. So at the outer ring of an OAM beam the field does not contain light of a different wavelength (or frequency), but due to this obliquity factor of the local wavevectors, there is a difference in the wavelength projected on the average/propagation axis. Now as for what the mode looks like, if you have an OAM beam with a cross-section that looks like the image below e.g. an LG_30 mode at center of beam (brightness represents amplitude and color represents phase). After propagation the beam will grow in size, but also the phase will become twisted due to an accumulation of a quadratic phase e.g. for an LG beam this is given by $exp(-ikr^2/R(z))$. This looks like the image below • Interesting discussion. I was wondering whether the line of equal phase was really straight, which seems not to be the case (except, perhaps, for a brief moment of time). – Eric Walker Oct 29 '15 at 19:01
• You got it almost right. Photons don't travel, at all. It doesn't even make sense to talk about photons until a measurement has been made. So in that sense the straight lines in free wave solutions are also just a geometric and not a physical quantity. You can never measure them with photons. – CuriousOne Oct 29 '15 at 19:42
• @EricWalker You are correct, lines of equal phase will generally not be straight. For a Laguerre-Gaussian mode this only happens at plane that has the narrowest beam waist (z=0). – Punk_Physicist Oct 29 '15 at 20:40
• @CuriousOne, your argument does not seem to be general. Would you extend it to x-ray and gamma photons, for which our experiments make it clear that they travel along well-defined lines, from point a to point b? – Eric Walker Oct 29 '15 at 21:58
• @EricWalker: The reason that we see "tracks" in high energy physics is because individual location measurements on the "particles" are "weak" in the sense that their precision is not close to the limits set by the uncertainty relation. If we keep doing weak measurements that don't change the momentum of the "particle" much, then we can reconstruct classical tracks in the same way as we measure orbits of planets. See e.g. "The wave mechanics of alpha ray tracks" by Neville F. Mott, Proceedings of the Royal Society, London, A126, 79-84 (1929), which is a "pedestrian" analysis of the problem. – CuriousOne Oct 30 '15 at 4:58