The wave front of the optical vortex beam is helical. Does it mean that the photon travels in a helical path? When the optical vortex beam is focused on a screen, an annular ring with dark center is observed. If I take a small arc of this ring only, block rest of the ring and let it propagate, would I see the light beam actually following a helical path? How would this happen? Of course, Optical vortex beam has orbital angular momentum but still it's not making sense to me. How the photon should be travelling in a helical path?
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$\begingroup$ Photons don't travel, at all. A photon is simply a quantum number that belongs to an excitation of the quantum field. $\endgroup$– CuriousOneCommented Jun 8, 2015 at 0:13
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No, the photons do not travel in a helix, they travel in a straight line but with a phase delay that is dependent on position. Looking across the beam's wavefront there is a phase delay that is dependant on the polar angle $\theta$ around the beam axis.
If we take a simple helical mode's complex amplitude as $\zeta(r,\theta,z) = u(r,z) e^{-ikz} e^{il \theta}$ where $u(r,z)$ is the amplitude at radius $r$, $z$ is the distance traveled by the beam, $k$ is the wavenumber and $l$ is a fixed integer. The $e^{il \theta}$ part shows that the only thing special about the beam is this fixed phase delay that is dependent on $\theta$.
Of course when we propagate the whole beam the phase changes with $z$ as with any other beam. If we plot the line when the modulo phase shift equals zero, this would trace out a helix, hence the name.
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$\begingroup$ ok, got that. So does it mean that if i take a small arc of the annular ring and let it propagate, it will not follow the helical path but propagate as any ordinary beam? $\endgroup$– celosCommented Jun 8, 2015 at 18:51
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$\begingroup$ Although, the ring is an eigenfunction of the Fourier transform, but a small arc of it is not, so the small arc will not remain invariant during propagation... $\endgroup$– ptomatoCommented Jun 10, 2015 at 4:40
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1$\begingroup$ Why the centre of the beam is dark in vortex beam? I read somewhere that it is because at the center the light with all phase from 0 to 2pi superimpose to give resultant zero. But if it is true then for optical vortex beam with fractional topological charge should not have dark core because then at the core not all the phase will be present. please give your opinion. @docPhil $\endgroup$– celosCommented Jul 2, 2015 at 5:43
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1$\begingroup$ "photons travel in a straight line" is just as misleading as "photons travel in a helix". Momentum eigenstates (e.g. plane-waves) are the only modes with a definite momentum (direction), and any other field configuration is a superposition of such states (and thus doesn't have a definite direction). $\endgroup$ Commented Oct 29, 2015 at 18:52