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Sometimes, the slits are in the range of nanometers, but I often bump into comments saying you can try this at home with lasers and polarized glass. Recently, I even found clearly macroscopic pictures (eg. a bedromm wall) showing interferences of sunlight with itself.

So I wonder if macroscopic interferences are just analogous to the particle-level ones, or if, on the contrary, they have the very same causes and nature.

As I recently learnt (thanks to @anna v), light rays are emerging from the electromagnetic field, which itself comes from the superposition of a lot of photons' wavefunctions. Therefore, one could suppose that the answer to the question is : what you see at the macroscopic level (including sun light and lasers) comes from the electromagnetic field, not from the photon's wave function.

The problem is that you may observe a wave collapse (interferences disappearing) with the laser by neutralizing polarisation with a third polarized glass, which seems to indicate a typically quantum behaviour...??? I'm puzzled.

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    $\begingroup$ Anna V has a point of view, but it's vaguely presented, and thus difficult to draw any conclusions from. I believe she's talking about coherent states, and that what she says is correct, but it's ... vague. So that makes statements such as your " comes from the electromagnetic field, not from the photon's wave function" vague and difficult to address. $\endgroup$ – garyp Sep 19 '19 at 12:03
  • $\begingroup$ @garyp : 1°/ do you have any better answer to the question ? 2°/ do you have read the original discussion with anna v ? here : physics.stackexchange.com/questions/503491/… and she references : motls.blogspot.com/2011/11/… $\endgroup$ – Michelange Baudoux Sep 19 '19 at 12:24
  • $\begingroup$ 1.) I have a point of view that is satisfying to me (and others). I'll see if I can provide an answer, but I'm not confident I will. 2.) I haven't, but I will. $\endgroup$ – garyp Sep 19 '19 at 14:05
  • $\begingroup$ Just remember a superposition of "a lot of" single-photon states is still a single-photon state. $\endgroup$ – flippiefanus Sep 22 '19 at 3:55
  • $\begingroup$ @flippiefanus : did I not write "the superposition of a lot of photons' wavefunctions", with "wavefunctions" being a plural, and each wavefunction being, per definition, a superposition of all potential states ? $\endgroup$ – Michelange Baudoux Sep 22 '19 at 4:02
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The short answer is it depends.

Questions you have to ask yourself:

  • What is your detector system's resolution?

  • What wavelength are you going to use?

  • How far can you project?

All of these things affect the slit separation that you can tolerate.


As an experimentor I find that there is a basic skill that is badly under-taught: letting the theory tell you how to design your experiment.

Taking the case of normal incidence on the slit plane for simplicity, the angle at which with $n$th bright fringe appears is given by $$ \sin \theta_n = \frac{n \lambda}{d} \;,$$ and if we project over a distant $l$, the linear distance from the central maximum is \begin{align} \Delta x_n &= l \tan \theta_n \\ &\approx l \frac{n \lambda}{d} \;. \tag{small angle approx.} \end{align}

To design a working experiment or demonstration you have to ensure that $\Delta x$ is reasonable for the detector system you are using.

Case study: classroom demonstration.

You're going to have students peering at the pattern with the Mark I Eyeball from meters away. A separation of a few centimeters is mandatory and tens of centimeters is better. Say, $\Delta x_1 = 0.1\,\mathrm{m}$

You are also going to be using a visible wavelength to support those eyeballs. Use $\lambda = 500\,\mathrm{nm}$ until you know what laser you actually have.

And finally you're going to be projecting over a few meters at most. Choose $l = 4\,\mathrm{m}$ as easibly achievable in most classrooms (ten meters is not unreasonable in some spaces, but there aren't many rooms where you could use thirty meters).

And now the math gives you the answer. For the parameters of my classroom demo, you end up with $$ d = l \frac{\lambda}{\Delta x} = 2 \times 10^{-5}\,\mathrm{m} $$ or a line density of 50 lines per mm. That a bit more than is achievable with a laser printer, but much lower density than even cheap plastic diffraction gratings.

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  • $\begingroup$ Neat, now I know how to proceed concretely. But would that setup (with polarized glass added) demonstrate quantum wavefunction collapse, EM field interferences, or light ray interferences? $\endgroup$ – Michelange Baudoux Sep 19 '19 at 21:35
  • $\begingroup$ And, most importantly, will neutralizing the polarization with a third piece of glass work in the first place ? $\endgroup$ – Michelange Baudoux Sep 19 '19 at 21:36
  • $\begingroup$ @MichelangeBaudoux please note that interference is not interaction, it is superposition of the wavefunctions for the problem at the quantum level. See this experiment youtube.com/watch?v=RRi4dv9KgCg $\endgroup$ – anna v Sep 20 '19 at 3:37
  • $\begingroup$ That experiment is puzzling, especially given the MIT does not provide the textbook of it, and given the experimental setup is not completely clear to me in the video. I'll have to dig destructive light interference later. It seems that the "destructed" light is in fact still there, just "folded", not completely destroyed. This would require "invisible light" to come back from the interferometer, but nature supplies plenty of that with radio waves & infrared, doesn't it ? $\endgroup$ – Michelange Baudoux Sep 20 '19 at 12:27
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The answer of dmckee is excellent for working out how big your slits need to be.

On you second part, by "quantic" I assume you mean quantum. The existence of double-slit interference does not in any way prove that anything quantum is going on at all. Classical waves undergo double-slit interference. (It is common to do a classroom demonstration with waves on a tub of water.)

The double-slit interference experiment by Young was actually originally done with sunlight. (Young did not have access to a laser as it would not be invented for hundreds more years).

Including orthogonal polarisers over the two slits would kill the interference, but the connection between the polarisers and quantum wavefuntion collapse is not perfect. Ideally you would place a non-demolition single photon detector at either slit to detect the photons and destroy the interference this way. (Although if you manage this you should send a paper summarizing it to a good journal - it is exactly the kind of experiment that is currently very cutting edge.)

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  • $\begingroup$ "cutting edge" %^) $\endgroup$ – Michelange Baudoux Sep 19 '19 at 21:58
  • $\begingroup$ more seriously, what do you mean by "the connection between the polarisers and quantum wavefuntion collapse is not perfect" ? Would it be demonstrating quantic decoherence just a bit or not at all ? $\endgroup$ – Michelange Baudoux Sep 19 '19 at 22:00
  • $\begingroup$ "Quantic" was an error (abuse of English to mimic french "Quantique"). sorry about that. $\endgroup$ – Michelange Baudoux Sep 19 '19 at 22:07
  • $\begingroup$ More simply put, can one tell me (as this site claims users.csbsju.edu/~frioux/polarize/POLAR-sup.pdf) whether the three polarizer experiment is a quantum effect or not? Thanks ! $\endgroup$ – Michelange Baudoux Sep 19 '19 at 22:10
  • $\begingroup$ The problem is that "Is it quantum or not" doesn't really make sense as a physics question. Quantum physics describes the world we live in, so in a physics sense everything is quantum (ice skating is quantum). If you want to find something that could not have possibly been anticipated without quantum physics then the polariser/double-slit experiment is a bad bet, because it was discovered (and explainable) using the wave physics of light long before quantum physics was discovered. $\endgroup$ – Dast Sep 20 '19 at 11:43

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