Meaning of the Orbital Angular Momentum of Super Chiral Light

• I
Gold Member
Summary:
Super Chiral light has a higher angular momentum than normal light. What does that really mean?
In this article it discusses the generation of something called super chiral light and claims with metamaterials they can make it have very high angular momentum like l=100. What does that really mean? How does that relate in magnitude to the normally computed linear momentum of a photon p=h/λ on some equivalent basis such as per unit energy? Thanks.

http://www.sci-news.com/physics/metasurface-enhanced-laser-super-chiral-light-08370.html

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anuttarasammyak
Gold Member
I myself do not know well about angular momentum of light so OP causes more fundamental question to my mind.

In the system where total momentum is zero (COM system), we can discuss angular momentum without defining the position of the origin from where we calculate angular momentum. For the system of a photon where such COM system does not exist, how we should care about the origin for its angular momentum calculation ?

vanhees71
Gold Member
It's good to study in detail the multipole expansion of electromagnetic waves to begin with. It's the expansion in terms of energy, ##\vec{J}^2##, and ##J_z## eigenstates. See, e.g., Jackson classical electrodynamics. One should keep in mind that physically these are observable quantities (in the sense of a mode decomposition), while the distinction in "orbital" and "spin" angular momentum for the em. field must be taken with a grain of salt since it's in principle not possible to decompose ##\vec{J}## into ##\vec{L}## and ##\vec{S}## in a unique gauge-invariant way. So one has to carefully read each paper, in which sense this decomposition is to be understood in the context.

bob012345
Cthugha
Usually, when talking about the orbital angular momentum (OAM) of light one discusses paraxial beams as here it is possible to define OAM and spin angular momentum at least in such a manner that the decomposition problems rightfully mentioned above can be avoided well enough.

In that framework, OAM is the angular momentum light fields may have due to the spatial distribution of the phase pattern. Assuming a nicely-behaved medium that is lossless and homogeneous, light propagates in a direction perpendicular to phase fronts (planes of constant phase). For a plane wave propagating along the z-direction, the light field will have the same phase for all x- and y-positions at a given z-position. The only phase gradient is along the z-direction and the light field will propagate that way.

Now, there are many other possible phase patterns besides plane waves. Most of them are not stable when propagating, but some are. Most notably Laguerre-Gaussian modes. These have a ring-like shape and show an angular phase gradient. So, when going full circle along the x-y-plane, the local phase at each point will be given just by the angle inside the plane and will show some gradient, e.g. 2 pi. If this is the case, the planes of equal phase will look different and the light field will still propagate along the direction of phase gradients. So most of the movement will still be along the z-direction, but due to the circular phase gradient in the x-y-plane, there will be some corkscrew-like motion added on top to that. Essentially, you will get a light field that moves a tiny little bit slower in the z-directions and the "missing" velocity components are now redirected into the x-y-plane, but the exact direction differs at each point, so that you get a circula type of motion of the total light field. However, when doing the math, you will find that this motion is balanced when integrating over the full plane, so that this kind of motion is stable. The fact that these light fields move a tiny little bit slower along the z-direction even on the single photon level has already been observed:
https://science.sciencemag.org/content/347/6224/857 (There is also a free version of the paper available on the ArXiv).

Such a light field with a 2 pi circular phase gradient will carry one unit of OAM. If you want more OAM, you need more velocity inside the x-y-plane, so you need to make the gradients steeper. If you use a 4 pi gradient, you will get 2 units of OAM. If you use a 200 pi gradient, you will get 100 units of OAM. Now, if you want to calculate this in units comparable to the standard momentum of a photon, you can also do this in the "standard" way. The full momentum vector now has components along the z-direction and components along the x-y-plane. You just need to consider the momentum projection on the x-y-plane at each point and you need to know the intensity distribution (or probability density for single photons) inside the plane at each point and you can calculate the OAM (with respect to the center of the ring-shaped beam) carried by the light field in total or each individual photon.

strangerep, bob012345 and vanhees71
Gold Member
....
Such a light field with a 2 pi circular phase gradient will carry one unit of OAM. If you want more OAM, you need more velocity inside the x-y-plane, so you need to make the gradients steeper. If you use a 4 pi gradient, you will get 2 units of OAM. If you use a 200 pi gradient, you will get 100 units of OAM. Now, if you want to calculate this in units comparable to the standard momentum of a photon, you can also do this in the "standard" way. The full momentum vector now has components along the z-direction and components along the x-y-plane. You just need to consider the momentum projection on the x-y-plane at each point and you need to know the intensity distribution (or probability density for single photons) inside the plane at each point and you can calculate the OAM (with respect to the center of the ring-shaped beam) carried by the light field in total or each individual photon.
Thanks. What does a '200pi' gradient mean? Does it mean that this exists only inside an optically dense medium? In other words, can this high OAM light exist in free space?

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Cthugha
At this point a picture might be more helpful than lots of words. Miles Padgett is usually considered to be one of the researchers who did a lot of pioneering work on the OAM of light. Please have a look at one of his talks which he made public here: https://www.gla.ac.uk/media/Media_263851_smxx.pdf

Have a look at slide 6 first. On the left a standard plane wave is shown. On the right, a beam with OAM is shown for comparison. The planes shown there are planes of equal phase. Also the directions of E-fields, B-fields and the local direction of the momentum vector are shown.

Now have a look at slide 8. Here, the different columns show different properties of the light fields. From top to bottom the different rows show beams carrying a different amount of OAM. The column on the left hand side again shows the planes of equal phase for comparison. For example for the plane wave you see circles that denote these planes. Let me call this the x-y plane. The light field propagates perpendicular to these planes along the z-direction. The next column shows the intensity of the beam in the x-y-plane for the beams with different OAM. This is a Gaussian beam for the plane wave and a ring for beams carrying OAM. The important column is the next one. This is the phase profile of the beam in the x-y-plane. It is color-coded. Red corresponds a phase of zero. Violet is pi/2. light blue corresponds to pi and 3/2 pi is on the border between green and yellow.

Fpr the plane wave you find just one color. The phase is the same everywhere in the plane. For the beam carrying one unit of OAM (l=1) in the next row, you see 0 phase on the right side, pi/2 phase at the bottom, pi phase on the left and 3/2 pi at the top. You simply get a circular phase gradient, which amounts to 2 pi in total. In the next row you see the beam carrying two units of OAM. Again, you get 0 phase on the right, but you already get pi at the bottom and 2 pi again at the left side, which is obviously identical to phase 0. On the top you get pi again and on the right side you go back to 0 (or 2 pi). In other words: You get two 2 pi gradients, so that you get a steeper phase gradient of 4 pi when going full circle. In the next row, you see the l=3 beam. You already know the drill. Here you get three 2 pi gradients, so you get a gradient of 6 pi in total.

As the light field is periodic with 2 pi, you can do this up to (almost) arbitrary numbers of 2 pi gradients. You can have OAM of 100, 200 or 500. All of these beams can propagate in free space and are stable. They do not need to propagate inside a dense medium.

Finally have a look at slide 9. This shows the local Poynting vector at different positions in the beam. Here you get the idea, why this kind of propagation relates to OAM. Essentially, going up to larger values of OAM just means that you tilt these beams further away from the z-direction. As long as these momentum components are small compared to the momentum component along the z-direction, the beam will be stable.

DrClaude, vanhees71, strangerep and 1 other person
Gold Member
I was hoping this amounted to a way to increase the kick of light per unit energy on some equivalent basis but I think it's ultimately not. For example, comparing the total angular momentum of a beam for a given energy from it's linear momentum acting on some micro device vs. the maximum angular momentum created using this technique using the same energy.