About spatial energy distribution of light emitted by a laser pointer

Question

This question is triggered by a section in the Veritasium video "Infinite Slit Paradox: Something Strange Happens When You Trust QM" about Feynman's path integral that came out on March 5, 2025.

At 25:02 into the video there is a section titled Proof that light takes every path That section is presented by Casper Mebius.

Casper Mebius is a team member of the Veritasium team, he is credited as director and co-writer.

Casper Mebius first presents reflection of lamp light against a mirror.

For the next demonstration most of the mirror is covered with a sheet of black cardboard, and vertically underneath the light source a reflective diffraction grating is placed. Then diffracted light reaches the camera.

Then Casper Mebius mentions that the light from the lamp propagates away from the lamp as a wide cone, so it will reach everywhere anyway.

What if a laser pointer is used, and the beam is pointed not at the diffraction grating, but at an angle to that direction. Casper Mebius proceeds to point the laser pointer such that if the black sheet would not be there the laser pointer would reflect in the mirror, towards the camera.

The screenshot from the video shows the result:
On the black sheet there is a bright spot from the laser pointer visible, and coming from the reflective diffraction grating there is a faint spot visible.

Caspar Mebius is convinced that this demonstration is proof that light takes every path, with the camera registering light when the phases line up.

However, it seems to me this demonstration is not sufficient.

For any beam forming there are two diffraction effects that affect the distribution of emitted energy.

• A well know diffraction effect is Beam divergence There is a minimum possible value. Of course, in this demonstration by Casper Mebius beam divergence is not a factor.

• Diffraction in general: at any discontinuity energy radiates out in all directions, including at a high angle to the direction of source of the light.

The (edge of the) collimating lens of the laser pointer gives rise to diffraction effect. By far most of the energy remains in the beam, but some energy goes out as a wide cone.

I expect that the latter is at work in that Casper Mebius demonstration. I think the bright/faint difference is a clue. I expect that they had to turn the camera to its highest sensitivity, and that the bright spot (on the black sheet) is far over-exposed.

For a laser pointer (and any beam-emitting laser source in general): of course the luminosity of the wide-cone-diffracted-light is so much lower that it is overwhelmingly unlikely to ever affect the course of any experiment. It's easy to be unaware of it.

It would appear that Casper Mebius' thinking is dependent on assumption that for a laser pointer there is zero luminosity outside of the collimated beam.

Question:
Is this particular scepticism valid?
Is there enough diffraction at the point where the laser light exits the laser pointer to account for the faint spot?

To avoid misunderstanding: I'm not objecting to the proposition that light takes all paths; I don't have an opinion on that interpretation.

The point I'm raising is: I think this Casper Mebius demonstration is not sufficient.

[Later edit]

I just visited the comment section of that video. In multiple comments, each with many likes, it is pointed out that the cheap laser pointer that is used has a lot of stray light. So it appears that Casper Mebius didn't even think to avoid stray light.

So that demonstration is refuted with much simpler means than I suggested in this question.

Answer 1

This is a self-answer. I submitted the question, and soon after that I gained a better insight.

In my opinion the decision to include that diffraction demonstration by Casper Mebius, and the way it was included, was a serious error. The presentation fails on two levels: implementation error, and conceptual misunderstanding.

First: the implementation error.
The simple laser pointer used by Casper Mebius emits a lot of stray light; light is spilling out in all directions. The faint dot that Casper Mebius thinks is corroboration of light-takes-all-paths is that stray light. The demonstration is inconclusive.

Moreover: even if you construct a laser source optimized to produce the best beam collimation possible, stray light cannot be eliminated entirely. Whenever a propagating wavefront interacts with an edge - any edge - there is diffraction. Even with a perfect lens: a lens has a perimeter, that's an edge. (Recommendation: youtube video on the 'Huygens Optics' channel about diffraction)

Summary:
While you can optimize to minimize diffraction, diffraction cannot be eliminated entirely. Any experimental setup designed to illustrate the light-takes-all-paths notion can also be acounted for in terms of diffraction effect.

Conceptual misunderstanding: The light-takes-all-paths notion is an interpretation of quantum mechanics. Just as in the case of, say, the Copenhagen interpretation, there is no way to either confirm or disconfirm the interpretation.

Related to that: it is essential to be very cautious about attribution of physical reality to the details of a calculation strategy.

Example from electric engineering:
Calculations of currents and voltages in electric oscillations circuits. Keeping track of sines and cosines is tedious, a more effecient calculation strategy is to move the calculation to complex number space. Representing current and voltage in exponential notation requires introduction of an imaginary current and and imaginary voltage. When the equations are solved the solution is converted back representation in physical current and physical voltage.

So that is an example where an efficient calculation strategy exists, but there is no reason to attribute physical reality to the details of the calculation. It's just faster to do the mathematics that way.

Summary: I submit that intrinsically the light-takes-all-paths notion can neither be confirmed nor disconfirmed. It's not accessible to measurement.

Answer 2

I am not sure I agree with your interpretation. Wave optics and path integral are equivalent, so observing diffraction (including beam divergence) is rather a direct confirmation of the path integral formalism. Granted, the latter is pedantic in the context of optics, but the experiment is a useful illustration before applying it in more abstract contexts like QFT.

What is true is that it does not illustrate the quantum nature of light. Everything can be explained classically. In this sense, yes the video is misleading by advertising the path integral as a purely quantum phenomenon. Actually, they left out a key part of the Feynman's thought experiment. The original setup was considered with a single photon source, in which case you are truly experimenting with the quantum behaviour of light. It is therefore less of a stretch to explain the resulting interference pattern with path integral rather than classical wave optics.

So in short, it is a good illustration of the path integral (which I think was the main objective), but not of the quantum nature of light (which was not really the topic).

Hope this helps.

Answer to comments

• Stray light is not incompatible with the path integral formalism. As already mentioned, the stray light is due to the diffraction which causes the laser beam to widen and causes it to have a gaussian distribution ("fuzziness" with no sharp edge). But diffraction (and wave optics in general) is mathematically equivalent to the path integral formalism, so it is not a failure of the experiment, it is a direct confirmation.

• I was responding to both your question and self-answer. Once again, it is another illustration of the wave nature of light, just like the historical Arago spot.

• A clear separation of the Lagrangian into kinetic and potential, $L=K-V$ is pretty specific to classical mechanics, so there is nothing deep in relating the principle of least action to these objects.

• Minimisation for the least action is important, you cannot just get away with merely stationarity. Mathematically, it is highly relevant as it allows to construct solutions by compactness arguments. Physically, you can derive it from more fundamental models. For example, in mechanics/optics, it emerges from the Schrödinger equation using WKB. Again, to make it single valued, you need to invoke minimisation.

Answer 3

At 28:10 in the video, he shows that a light that shines in all directions. Without a grating, it lights up everything. With a grating, it produces many spots where you would not classically expect.

He repeats the same experiment with a laser that lights up just one spot. With a grating, he only gets one spot.

If light explores all paths, and all paths are the same for the two cases, why is there any difference, with or without the grating? Plainly, the two lights don't explore all paths in the same way.

Light is described by a quantum mechanical wave function. The amplitude and phase of the wave function determine the probability of finding a photon at a particular spot on the mirror. The screen or mirror absorbs the photon and emits a reflected photon.

The path integral tells you how to calculate the wave function that arrives at each point of the screen. The biggest contribution is from the paths near a straight line. As the video says, these are the paths near a minimum in the action. Phases from nearby paths reinforce.

The video says that all paths have an equal weight in the sum. Since all points on the screen have near straight paths leading to them, this leads to the conclusion that all points on the screen should be light. And for the first light, they are.

But for the laser, the screen is dark except at the bright spot. So the paths leading to that spot are brightly lit. Straight path leading to other spots are dark. So light isn't exploring those paths.

But the diffraction grating shows a spot. So light is exploring paths at least to that spot.

So what gives?

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This isn't an answer. It is an extension of the question. It isn't what an answer is for. But if I ask my own question, it would be a duplicate of this.

Answer 4

I watched the video and came to the same conclusion. I suspected upon watching that if an aperture were introduced to block off stray light and higher order spatial frequencies from the beam that you would not observe such an effect.

Answer 5

The video as you describe it indeed relies on flare and it does not demonstrate what it claims to.

Light does travel by all paths but it also interferes. The end result, actually the far field result, in this case is a fairly sharply defined laser beam. Outside the beam destructive interference occurs.

Answer 6

The experiment might be demonstrating the laser diffraction in the air. The laser beam creates a light cone. If the experiment was done in highest vacuum possible and we would see delta in the intensity of the reflection that would prove this explanation.

Note: The intensity would be a function of vacuum level, it is impossible to get an intergalactic vacuum in our experiments so there would be some diffraction in vacuum chamber as well.