Sigh. Approaching retirement age and still deeply confused about something I first encountered in highschool 40 years ago.

Consider the usual double slit experiment. Make the light source be a laser with a beam 1 mm wide. And put it 5 meters away from the slits.

On the other side of the slits the photon shows the well known diffraction pattern, alternating dark and light bands. Good, very tidy. But consider that single photons can diffract.

So a single photon comes down the beam. The beam is 1mm wide with very little scatter, and 5 m long. So the momentum of a single photon is very tightly bounded. And moving objects near but not in the beam don't change things on the other side of the slits. Objects such as students doing the experiment in highschool, for example. If you don't get any red on you, you won't change the pattern or the brightness.

On ther other side of the screen the photon can turn quite a corner, for example 30 degrees. The energy does not change very much, since it is still the same pretty red color from the laser.

How does it manage to turn this corner and conserve momentum?


3 Answers 3


The slits themselves receive a tiny impulse from each photon. If a photon is diffracted to the left, the slits get nudged to the right. Every time a photon changes direction, it requires something else to gain momentum in the opposite direction, whether a solar sail or a star bending light by gravity. Since the slits are usually anchored to the ground and the impulse is so small, the effect is not observable.

Your question actually came up in a series of debates between Albert Einstein and Niels Bohr on whether quantum mechanics made any sense. Einstein argued that the impulse of a photon on the slits would allow the measurement of the photon's position and momentum at the same time, contrary to quantum theory. Bohr replied that the necessary precision of the slit momentum measurement would--through Heisenberg's Uncertainty Principle--make the slit's position uncertain enough to destroy the interference pattern, negating any measurement of the photon's position.

  • $\begingroup$ "Every time a photon changes direction, it requires something else to gain momentum in the opposite direction, whether a solar sail or a star bending light by gravity.", can you please elaborate on the case of gravitational lensing? Why would the star gain momentum in the opposite direction, when light travels along a geodesic along curved space around a star? $\endgroup$ Commented Dec 7, 2021 at 7:08
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    $\begingroup$ @ÁrpádSzendrei a photon mutually changes momentum with a star a similar way as ʻOumuamua mutually changed momentum with our Sun. $\endgroup$ Commented Dec 7, 2021 at 7:52
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    $\begingroup$ @ÁrpádSzendrei : I think you mean to say the star and the photon travel along geodesics in the space-time consistent with the stress-energy concentration of both of them... $\endgroup$ Commented Dec 7, 2021 at 15:47
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    $\begingroup$ Both photon and star travel along a geodesic. Not just one of them. (Although the star is immensely more massive). So in an idealised 2 body situation, both gain momentum towards the other as they pass by - the stars motion changes a totally undetectable amount, the photons motion changes a detectable amount (lensing effect). Overall momentum remains conserved though. $\endgroup$
    – Stilez
    Commented Dec 8, 2021 at 3:35
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    $\begingroup$ @Peter-ReinstateMonica There is no rest frame for a photon. $\endgroup$
    – Mark H
    Commented Dec 8, 2021 at 10:41

It actually imparts momentum on the slit. Same as with the typical solar sail, although the slit is much smaller and firmly affixed so it does not move.

(Also it might be better to think of photons as very small and a laser beam sending many pulses into the EM field like many pebbles dropping into a pond. But considering a photon as long as its path is also wise when you consider Feynman path integral implications. My thinking is an excited electron/atom is disturbing the EM field with virtual photons, this can take milli or micro seconds or longer or shorter. When a probable path is preferred the real photons goes ... See Feynman path integral.)

  • $\begingroup$ "firmly affixed" As Galileo appears to have said: "and yet it moves". en.m.wikipedia.org/wiki/And_yet_it_moves $\endgroup$
    – my2cts
    Commented Dec 6, 2021 at 8:34
  • $\begingroup$ @my2cts Consider "it does not move" to be shorthand for "the movement is so small, due to the Earth's immense inertia, that it is not currently practical to measure." (Back-of-the-envelope calculation: reflected radiation pressure applies a force on the order of 6 Newtons per gigawatt, reduced by the square of the cosine of the angle of incidence. Applying a continuous force of 6 N to an object with the mass of the Earth, produces a change in its momentum on the order of $10^{-24}\,\text{m}/\text{s}^2$.) $\endgroup$
    – zwol
    Commented Dec 6, 2021 at 16:44

Other answers have suggested that the diffracted photons are receiving momentum through interaction with the slit filter, but I find this hypothesis untenable, as the portion of light that interacts with the wall of the slits bounces back and decoheres, not contributing to the diffraction pattern, so we can fully ignore their contribution here.

Quantum amplitudes are linear, hence removing or filtering a portion of a wavefront does not count as an interaction, hence it cannot transfer momentum to the diffracted amplitude.

The correct answer lies elsewhere: as you pointed the transversal section of the beam wavepacket is assumed to have net momentum drift of zero, and a small Gaussian spread. We need to think about the evolving 2D cross-section of the beam as the system of interest.

As you might know, Gaussian beams are the optimal cross-sectional shape for light to keep itself from spreading. We can think of it as a semi-rigid phase of light.

After the diffraction, it loses the Gaussian transversal section that keeps the beam transversally "packed", and it "dissolves" as if it was transitioning from our conceptual semi-rigid phase to a liquid-like phase which a much higher beam divergence (we certainly can measure beam divergence in rate of increase of beam spread over longitudinal distance)

The Gaussian shape is special because it "saturates" the uncertainty principle inequality such that:

$$ \Delta x \Delta p \propto \hbar $$

This gives its special property of optimal shape conservation over time. Just that, in this case, the "time" variable of the 2D packet is the longitudinal axis of propagation, and instead of actual spatial coordinates, the spread happens in angular coordinates.

When our Gaussian 2D beam crosses the diffraction slit, the Gaussian pattern is "cut" in two slices. What happens to these individual slices of our beam is that their transversal position are resolved better than before, so their transversal momentum must spread out (otherwise they would violate Heisenberg's uncertainty inequality)

An increase in the uncertainty of a variable after a measurement seems to contradict the conservation of the variable, but this comes from the idea that photons existed along the trajectory, and everything suggests this picture of a photon like a small dot carrying momentum along a trajectory is misleading and incorrect

  • $\begingroup$ What does it mean "loses the gaussian" etc.? What is the nature of that step? Is it not an interaction with the slits? Single photons can turn this corner. The average over many photons is not going to save momentum conservation for a single photon. I can let photons go through quite perceptably separated. $\endgroup$
    – Dan
    Commented Dec 7, 2021 at 3:09
  • $\begingroup$ But there have been ways to do (essentially) single photon experiments for quite a while, and these days you can even make photon pairs and measure one's wavelength and direction so that you know almost exactly the momentum of the other. So "it averages out in the end" isn't really helpful in my opinion. $\endgroup$
    – uhoh
    Commented Dec 7, 2021 at 13:31
  • $\begingroup$ Maybe I misunderstand what you have written but you seem to imply that momentum is not conserved individually for each diffracting particle ("photon") but for their statistical ensemble as a collective; otherwise I cannot interpret this "their distribution respect that symmetry and in average their transversal momentum components should average to zero". I am told that this interpretation is quite out of fashion, so to speak, although this amazon.com/Quantum-Mechanics-Development-Leslie-Ballentine/dp/… is an excellent book. $\endgroup$
    – hyportnex
    Commented Dec 7, 2021 at 18:38
  • $\begingroup$ I've added some clarifications. Please review $\endgroup$
    – lurscher
    Commented Dec 7, 2021 at 20:21
  • $\begingroup$ Does your reasoning hold for double slit experiments with single electrons? $\endgroup$
    – Mark H
    Commented Dec 7, 2021 at 22:10

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