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Consider the following alterations to the double slit experiment:

  • turn the observer on and off rapidly
  • move the observer further and further away, until the point at which it can no longer effectively observe

What happens to the wave function when the observer turns off and then back on rapidly, to the electron in flight? When moving the observer further away, at what point does the wave function no longer treat it as an observer?

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  • $\begingroup$ @CuriousOne So this video is wrong? youtube.com/watch?v=7u_UQG1La1o $\endgroup$
    – J.Todd
    Jun 1 '16 at 2:45
  • $\begingroup$ If it says that physics depends on observation, then it's wrong. $\endgroup$
    – CuriousOne
    Jun 1 '16 at 2:46
  • $\begingroup$ @CuriousOne so placing an observation instrument next to a double slit experiment doesn't change the outcome? Why not? I thought that was the whole point of the double slit experiment. $\endgroup$
    – J.Todd
    Jun 1 '16 at 2:48
  • $\begingroup$ @CuriousOne So you're claiming the double slit experiment is invalid? Im trying to figure out if you're a physicist or a troll. Please explain. $\endgroup$
    – J.Todd
    Jun 1 '16 at 2:55
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    $\begingroup$ @Viziionary - please be sure that modern physics (based on quantum mechanics) absolutely requires to be formulated relatively to an observer and his or her or its observations. These facts were realized around 1925, incorporated inseparably to the mathematical formalism, rewarded by numerous Nobel prizes, and every single insight in the following 90 years reinforced the point. Observers are absolutely fundamental for the application of the laws of physics. Everyone who tells you something else is a crackpot. But the dependence works differently than you think, too. $\endgroup$ Jun 1 '16 at 3:37
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I wrote my Master Thesis on partially coherent classical wave fields applied to gratings, so I will try to give some insight on what exactly about the double slit makes it a quantum problem and what is really just classical wave mechanics. This should simplify the discussion a bit by separating the two issues (at least hopefully).

General comment on the double slit

In my opinion most of the properties of the double slit that are usually attributed to quantum mechanics can actually be perfectly reproduced with classical wave fields from statistical sources. By that I mean e.g. an electromagnetic field that has some statistical phase-fluctuations due to the source process. Most light can actually be represented in that way in particular from astronomical sources. Even some laser processes can, since the notion of coherence can easily be formalized as statistical correlations of the wave field (1). The only regime where this breaks down (to my knowledge) is the few photon limit and some non-ergodic processes/pulses.

With this approach you can get everything you want when thinking of the double slit experiment:

  • Interference pattern on a screen for sources with small angular and wavelength spread
  • Coherence can be modelled (2)
  • You can couple to classical waveguide detectors ("observers" that are very well modeled classically).

Relation to the Question

The situation the OP described can also be created this way, with a purely classical "observer". Just put a fully incoherent re-emitter (something that absorbs the power at one point and re-emits it incoherently as a half-spherical wavefront. Not sure something like that exists, but a simple dipole antenna probably comes pretty close) instead of this "quantum observer". If you put it directly into one of the slits you will make the interference pattern disappear due to statistical phase fluctuation. As the observer is moved further away the pattern would change obviously, in the far distance limit it would go back to the original pattern. I can try a simulation of that since I have a program that does these kinds of things, but I think this qualitative insight should be sufficient for the question.

Where is quantum mechanics?

So why do people even talk about quantum mechanics when looking at the double slit experiment? Historically it was used to show that other particles (e.g. electrons) have wave-character. In the many-particle limit we could even describe that with the formalism described above. The only point where we get a problem is the few-particle limit. What people start discussing then is how the particle has a certain position when detected on the screen, which then leads to arguments about the measurement problem. I will not go into detail about this here, other people like @LubošMotl know a lot more about it than me and I suggest listening to their advice. What I wanted to emphasize though is that this is a completely separate issue from propagation through the double slit and the interference caused by it.


(1) see e.g.

  • M Born and E Wolf. Principles of Optics. 7th ed. Cambridge University Press, 1999.
  • S. Withington, D.J. Goldie, C.N. Thomas. "Partially Coherent Optical Modelling of the Far-Infrared Imaging Arrays on the Cooled-Aperture Space Telescope SPICA". In: Annalen der Physik (2013).

(2) see e.g. this answer of mine

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To me, an 'observer' in this sense means something that can take a 'measurement' of the photon (or electron) as it travels through the slits (like an observation instrument), and thus altering its wave-function. If the observer can't take measurements, then he can't alter the events, and the same thing will happen whether or not he is there.

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  • $\begingroup$ could you please tell us the reason why this is not the case, so we can argue? instead of just down voting? $\endgroup$ Jun 1 '16 at 7:24
  • $\begingroup$ I downvoted it because you're not answering the question at all. The OP is asking how the behavior of the experiment changes when the observer/observation is gradually being removed. You didn't discuss it, instead, you assumed "If the observer can't take measurements..." - The last sentence is an oxymoron by itself. The observer is defined as an agent that does make observations, so if he doesn't, he's not an observer. Again, why don't you read e.g. this RPF text which I would imagine to be the answer? feynmanlectures.caltech.edu/III_01.html#Ch1-S6 $\endgroup$ Jun 1 '16 at 7:37
  • $\begingroup$ @Luboš Motl I totally agree with you, I'm just trying to make it clear by that last sentence. $\endgroup$ Jun 1 '16 at 7:42
  • $\begingroup$ It's great that you agree but you still haven't answered the question at all, so it's still downvoted. $\endgroup$ Jun 1 '16 at 7:44
  • $\begingroup$ @Luboš Motl According to physics.stackexchange.com/help/how-to-answer "Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer". Am I not helping somehow to the right direction? $\endgroup$ Jun 1 '16 at 7:51
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It does not depend on an observer or being observed at all. Interference requires things to be arranged in a perfect way. If you introduce anything into the experiment in an attempt to observe you will disturb the pattern. If you shine light on the experiment you mess it up. If you place devices somewhere in the experiment you will mess up the pattern by either blocking photons or diverting electrons depending on the type of experiment. There are many ways to destroy the pattern but looking has nothing to do with it.

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  • $\begingroup$ The statement that "interference requires things arranged in perfect way" is just self-evidently wrong. In quantum mechanics, all probability amplitudes for all possible outcomes always interfere with each other, whether something is declared "perfect" or not, not to mention that the word "perfect" clearly can't be given any general definition. ... Inserting new components changes the experiment but it doesn't change the fact that all the evolution may be undone in principle. $\endgroup$ Jun 1 '16 at 7:11
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    $\begingroup$ @Luboš Motl according to my Modern Physics book (Beiser), wave-functions add and not 'probabilities', and thus the wave-function itself causes interferences, and not 'probabilities', as in the case of double slit. And until now, it is not so well established that a photon can interfere (in the sense of changing the wavefunction) with another as they travel through space, since they are bosons. Otherwise, we can't just assume that a photon from a known source will have the same wavelength and frequency, if there are other photons passing through its path, $\endgroup$ Jun 1 '16 at 7:38
  • $\begingroup$ because those photons will alter the wavefunction. $\endgroup$ Jun 1 '16 at 7:38
  • $\begingroup$ Every not-completely-stupid textbook of QM agrees that the wave functions add. But not-completely-stupid textbooks also inform their readers that the wave functions are composed of probability amplitudes, templates to calculate subjective probabilities, they are not objectively real waves. ... Your claims about the absence of knowledge whether photons interfere are just plain ludicrous. The interference of photons is what is observed in the simplest Young double slit experiment that was already done centuries ago. Because we know that light is a stream of photons, those interfere, too. $\endgroup$ Jun 1 '16 at 7:41
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    $\begingroup$ @Luboš Motl Ok. But can you confirm from the experiments that it's the photons that are actually interferring, and not that a single photon creates an interference with its own wave? $\endgroup$ Jun 1 '16 at 8:47

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