Wouldn't any structured beam of light be expected to travel slower than a plane wave?

There aren't many new, actual bona-fide discoveries in classical optics these days.

I saw this news item in Phys.org: Observation of twisted optical beam traveling slower than the speed of light

The researchers first noticed the slow speed of twisted light when conducting experiments with Gaussian laser light and light with 10 twists. "We realized that the two beams didn't arrive at the detector at the same time," Karimi said. "The twisted light was slower, which was surprising until we realized that the twists make the beam tilt slightly as it propagates. This tilt means that the twisted light beam doesn't take the straightest, and thus fastest, path between two points."

In waveguides, for example few-mode optical fibers with very small index differences, each mode propagates at a different velocity. The differences can be big - a few percent. Roughly speaking, the more complex the mode looks, the more transverse k, the less longitudinal k.

The same is true with defined beams in free space. As soon as you start adding spatial structure, by focusing a wide beam to a narrow beam, or making more complex shaped beams, anything where you have more transverse k vector, the beam slows down, and it is 100% consistent and calculable from Maxwell's equations.

It may be quite difficult to verify for experimental reasons, I can understand that, and it's always good to verify something that hasn't been verified before for many reasons - student research experience, putting the expensive equipment to the test, increment publication list, etc. It's good.

My Question: Is it something that is absolutely expected to happen - any structured beam in space will be slower than an infinite plane wave in space? Wouldn't any structured beam of light be expected to travel slower than a plane wave?

The Phys.org article links to the paper Observation of subluminal twisted light in vacuum Frédéric Bouchard, Jérémie Harris, Harjaspreet Mand, Robert W. Boyd, and Ebrahim Karimi Optica Vol. 3, Issue 4, pp. 351-354 (2016)

Subluminal means slower than the speed of plane wave light (in the same medium), and my point is that any finite beam of light will always be subluminal - travel slower than a plane wave in the medium.

edit: I just noticed this at the bottom. "...their calculations have predicted may travel around 1 femtosecond faster than the speed of light in a vacuum" so apparently they don't think so!

If it's possible to slow the speed of light by altering its structure, it may also be possible to speed up light. The researchers are now planning to use FROG to measure other types of structured light that their calculations have predicted may travel around 1 femtosecond faster than the speed of light in a vacuum.

Read more at: http://phys.org/news/2016-03-optical-slower.html

• Without knowing anything about this experiment or the people responsible, I'll say that I always treat announcements in optics with caution for two reasons. First the usual science-press claim-magnification-and-distortion field, but the second is that I've seen the millennium announced in optics more than a few times in my lifetime: it is a field with considerable subtlety and people who are not expert or narrowly expert in the field can make mistakes. If you want to know what the paper really claims, read the abstract and conclusions yourself. – dmckee Jul 20 '16 at 14:49
• @dmckee Wow you have a very elegant way of saying something that I'd probably put in much cruder terms and then delete, and then maybe say anyway. I'm asking here in stack exchange for what other people familiar with the wave equation have to say, not what the paper says - basically for exactly the reason you so nicely put. – uhoh Jul 20 '16 at 14:56
• Phase and group velocity can be greater than $c$, but energy velocity must be $\leq c$. There's a discussion here around pg 55. – garyp Jul 20 '16 at 17:15
• By the way, the title is very good. I wouldn't change that. – David Z Jul 20 '16 at 17:44
• What you see as pressure to close questions is actually pressure to improve them. We have high standards, and questions are closed when they don't meet those standards, with the expectation (or at least the hope) that they will be edited, improved, and then reopened. This is why you get a lot of feedback in comments. (FWIW, questions which people think are irredeemably bad get very little feedback.) As for editing, it's best to read and possibly respond to comments and build up a list of improvements to make. Then edit once and make all the changes you've accumulated. – David Z Jul 20 '16 at 18:13

Yes, you are right. A structured light beam would necessarily travel slower than a plane wave - slower than the speed of light. However, there is room for much confusion here.

When one says that a plane wave travels at the speed of light, one tacitly assumes that this speed is measured along the direction of the propagation vector. If instead one measures it along some direction that makes a nonzero angle with the propagation vector one can distinguish between the speed of the energy/power in that direction or the speed of the wave fronts (the component of the phase velocity) in that direction.

Since a structured light beam is constructed out of a spectrum of plane waves, one would need to consider their combined effect along the general direction of propagation of the entire light beam. The pertinent speed in this case is the one related to the energy/power of the beam and not the phase velocity. Therefore, one needs to compute the weighted average of the projections of the speeds of the individual plane waves along the general direction of propagation.

To give a mathematical treatment, I'll assume the structured light beam is represented by a single photon state and I'll ignore polarization. Then one can express the light beam by $$|\psi\rangle = \int |\mathbf{k}\rangle G(\mathbf{k}) d^2 k ,$$ where I'm ignoring some unimportant details (you can add it if you wish). Here $|\mathbf{k}\rangle$ represents the plane waves and $G(\mathbf{k})$ is the angular spectrum, which is normalizes $$\int |G(\mathbf{k})|^2 d^2 k = 1 .$$

To compute the velocity we define an operator $\hat{V}$ and to project this onto the direction of propagation of the entire beam (which we assume to be the $z$-direction), we compute $\hat{z}\cdot\hat{V}$. (Please forgive the abuse of notation, the hat over the $\hat{V}$ denotes an operator while the hat over $\hat{z}$ indicates a unit vector). This operator is defined so that $$\langle\mathbf{k}|\hat{V}|\mathbf{k}'\rangle = c \hat{k}\delta(\mathbf{k}-\mathbf{k}') ,$$ where $\hat{k}$ is a unit vector in the direction of propagation.

So when we compute this speed we get \begin{align} \langle\psi|(\hat{z}\cdot\hat{V})|\psi\rangle &= \int \int \langle\mathbf{k}|(\hat{z}\cdot\hat{V})|\mathbf{k}'\rangle G(\mathbf{k}) G^*(\mathbf{k}') d^2 k d^2 k' \\ &= \int \int c (\hat{z}\cdot\hat{k})\delta(\mathbf{k}-\mathbf{k}') G(\mathbf{k}) G^*(\mathbf{k}') d^2 k d^2 k' \\ &= \int c (\hat{z}\cdot\hat{k}) |G(\mathbf{k})|^2 d^2 k . \end{align} Now, since $(\hat{z}\cdot\hat{k}) < 1$ and since the angular spectrum is normalized, one finds that the result of this calculation would always be smaller than the speed of light.

So how does the width of the beam affect this result? Which one would be faster, a wide beam or a narrow beam? Note that the wider a beam the closer it comes to be like a plane wave. So we expect that a wide beam would be faster than a narrow beam.

Due to the inverse relationship between the size of a function and that of its Fourier transform, a wide beam would have a narrow angular spectrum and visa versa. So, if my beam is expressed as $$g(\mathbf{x})=N \exp(-|\mathbf{x}|^2/w_0^2)$$ in some plane with a fixed value of $z$, then its angular spectrum would be $$G(\mathbf{k})=N' \exp(-|\mathbf{k}|^2 w_0^2/2) ,$$ with normalization constants $N$ and $N'$ and beam radius given by $w_0$.

Note that the spectrum is centered on the beam axis. So for a narrow spectrum (large value of $w_0$) the dot-products $(\hat{z}\cdot\hat{k})$ are close to 1. So the resulting speed would also be close to the speed of light. On the other hand, for a wide spectrum (small value of $w_0$) the dot-products would include values that are significantly smaller than 1, which will bring down the value of the integral and therefore give a speed that is significantly smaller than the speed of light.

So, to summary: a wide beam would propagate faster than a narrow beam.

• OK great! So if the angular spectrum $G(\mathbf{k})$ were calculated for a Gaussian beam, it might look something like $( \ k_x \mathbf{\hat{x}}, \ \ 0 \mathbf{\hat{y}}, \ \ \sqrt{k_0^2-k_x^2} \mathbf{\hat{z}}\ ) \ exp(- ((k_x/k_0)^2/\theta_{rms}^2)$, then as $\theta_{rms}$ went to zero the final integral would reduce to $c$? And if $\theta_{rms}$ were say 0.1 for a beam of order $10 \lambda$ wide, would it equal something like $\sqrt{1^2-0.1^2} \ c$ or $0.995 \ c$? I'm just trying to make sure I understand what's happening here. – uhoh Sep 20 '16 at 5:54
• @uhoh: See the addition in my answer. – flippiefanus Sep 20 '16 at 10:29
• I'll mark it as accepted and give a point just for the "Yes, you are right." part (that always makes my day), but the rest of it is also really helpful as well!! Thank you for taking the time to work through this mathematically to a nice conclusion. – uhoh Sep 20 '16 at 10:43