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My question is clear, that I ask: What do we mean by "observation" in 2-slit experiment for electrons (or any other wave-particle)?

You know, we say that :"if we observe the electron, it shows a particle-like behavior; and if we do not observe it, it shows a wave like behavior"

So, if the observation is through sending and getting photons, is it just through our eyes or through any other interruption by light beams?
I'm confused! Or if my sentences are silly, what's the correct approach to understand the term observation?

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    $\begingroup$ possible duplicate of What is an observer in quantum mechanics? $\endgroup$ Commented Dec 14, 2014 at 16:45
  • $\begingroup$ Who said that "if we observe the electron, it shows a particle-like behavior; and if we do not observe it, it shows a wave like behavior"? Interference fringes can be observed, and they are one of the reasons for us to attribute wave-like behavior to particle in this experiment. $\endgroup$
    – Sofia
    Commented Dec 14, 2014 at 17:06
  • $\begingroup$ You mean that, we always see the diffraction pattern? I don't think so and as far as I know, if we observe the particle-wave entity, it shows a particle-behavior...!! $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 17:15
  • $\begingroup$ You are quite confused. Let's make order. Each electron and electron behaves according to the wave-function, i.e. as a wave. 1) First the electron passes through the two slits. A particle doesn't pass simultaneously through two slits. 2) Beyond slits, the wave exiting one slit produces interference with the wave exiting the other slit. That, for each electron and electron. So, beyond the wall with two slits, for each electron and electron we have an interference tableau. Please note, so far we didn't measure, didn't observe, didn't disturbed the tableau in any way. But the tableau is there. $\endgroup$
    – Sofia
    Commented Dec 14, 2014 at 17:39
  • $\begingroup$ (continuation) Now we proceed to the measurement. We don't do that by eye, but by apparatuses. They measure, and if thy are clever enough they can even produce a summary report. But, let's return to what they record. Assume beyond the slits a photographic plate. The electron has an extraordinary property: although the tableau of interference that it produces, occupies some volume, the electron delivers its energy just to one single molecule. THIS is the particle-behavior. So, an electron impresses the plate here, another electron there, and in the end we get the whole interference tableau. $\endgroup$
    – Sofia
    Commented Dec 14, 2014 at 17:56

4 Answers 4

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Your question is not silly. In the two slit expariment you measure the position if the photon, that is a particle-like quality. What happens is that a photodetector records when it absorbs a photon. That gets recorded in the memory. Some schools of physics interpret the interaction of the photon with the detector as a measurement. Others interpret that the measurement only happens when a conscious being "reads" the measurement from the detector.

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  • $\begingroup$ where should we put the detector? if between the slit and the curtain, then we force the electron not to reach the curtain and if on the curtain it's the same... if we do not count, the nature has no problem with herself? i mean the electrons reach the curtain and they are somehow detected... then why shouldn't they show a particle like behavior then?! $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 16:45
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    $\begingroup$ the courtain is the detector. Because the particle like behaviour only shows when they hit the detector. Before that, the equation that describes its possible behaviour is a wave equation. This is not an easy topic, you should read a lot about it to start to understand it. $\endgroup$
    – user65081
    Commented Dec 14, 2014 at 16:49
  • $\begingroup$ i know the curtain is a detector... i mentioned it in the comment. i want to know the different between the detector as a curtain and the detector as an observer which makes observation and ruin the diffraction pattern... What makes one detector as an"observer" or "non-observer"?!! $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 16:59
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    $\begingroup$ what makes a detector an observer is a matter of a lot of discussion. You can write an entire book about it. What matters is that regardless of who is the observer, the results will be the same (a diffraction pattern). $\endgroup$
    – user65081
    Commented Dec 14, 2014 at 17:10
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    $\begingroup$ @MojtabaAkbarzadeh > What makes one detector as an"observer" or "non-observer"? The observer is the part which you describe as a classical object. Once your description in terms of wavefunction limits to some subsystem, then other objects are basically observers if they can be well enough described by classical mechanics. If instead you took the detector onto account and described the whole system of detector+electron via wavefunction, then you'd have another place where observation occured (e.g. a computer reading the detector). To have meaningful predictions, you need some observer anyway. $\endgroup$
    – Ruslan
    Commented Dec 14, 2014 at 18:27
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When you send the photons through the double-slited wall, they form a diffraction pattern on the other side, which is a wave-like phenomenon. It makes no sense to think of the photons as "particles" anymore, since they would need to cross both slits simoultaneously in order to create the diffraction pattern.

So a way to find out what's happening is to place photodetectors in the slits. But what happens when you place the photodetectors is that these absorb the photons which hit them, not allowing them to cross the wall. This means that we are able to find where the photon was going to cross, but we didn't let it cross the wall at all. The rest of the photons will cross the wall across the other slit, but they will not form the expected diffraction pattern.

But if we remove the photodetectors and send the photons trough the wall, even in tiny packets small enough to consider that we are sending them one by one, we get the diffraction pattern. This would mean a single photon is crossing both slits simoultaneously, which is definitely a wave property and not a particle one.

I reccomend Feynman's lecture on the subject or the first chapter of Quantum Mechanics by Claude Cohen-Tannoudji.

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  • $\begingroup$ I'm afraid if the one photon experiment is done or it's just a theory... $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 17:11
  • $\begingroup$ More strange is to interpret the fringes behind an edge. They occur in single particle experiments too. $\endgroup$ Commented Dec 14, 2014 at 17:16
  • $\begingroup$ Can you give me a reference which has experimentally done the single particle phenomena, please?! $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 17:29
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    $\begingroup$ @MojtabaAkbarzadeh From Wikipedia en.wikipedia.org/wiki/Double-slit_experiment "A low-intensity double-slit experiment was first performed by G. Taylor in 1909, by reducing the level of incident light until photon emission/absorption events were mostly nonoverlapping. A double-slit experiment was not performed with anything other than light until 1961, when Claus Jönsson of the University of Tübingen performed it with electrons. In 1974 the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi repeated the experiment using single electrons ..." $\endgroup$ Commented Dec 14, 2014 at 18:55
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A free moving electron does not emit photons. If one want to see an electron you have to illuminate it. Sending photons, the electron reflect (absorb and re-emit) photons which one can detect then. By doing this the electron get disturbed and changes his direction. The fringes on the observation screen will be destroyed. Since one cannot synchronize the electron's state with the photon's state the interaction between them always has an uncertainty and the result gets blurry. The uncertainty of each interaction (between the electron and the photon and between the resulting photon and the measurement instrument) is the reason for the impossibility to conclude about the position of the electron during the transition of an edge (or a slit or a double slit).

On the over side, the wave characteristic of the electron in the interaction with an edge (slit, double slit, diffraction foil) is an interpretation of some thing what is not observable. To be more precise when ever one observe the electron it is a particle with some uncertainty in the electron's state (velocity, position, momentum, ...) which one describe with wave function. That is very useful for calculations. Duality means that both features - wave characteristic and particle characteristic - coexist in every moment.

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    $\begingroup$ wave and particle behaviors co exist, but we can not see the effect of both features simultaneously...(And, all i mean is that, what if we do not observe...Isn't there any other photons to disturb the motion of the electron??) $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 17:23
  • $\begingroup$ @MojtabaAkbarzadeh Yes, there is an interaction between the electron and the edges of the slit(s) or an edge. Every material has electrons on it surface and they are interacting with the electron. This interaction takes place by the exchange of photons. $\endgroup$ Commented Dec 14, 2014 at 19:07
  • $\begingroup$ @MojtabaAkbarzadeh There I collected some pictures from electron diffraction. Sorry, it is in German. Can give translation to you if you need. Pay attention to the fact that light is always bended to behind the geometrical shadow, but electrons are always bended away from the shadow. $\endgroup$ Commented Dec 14, 2014 at 19:17
  • $\begingroup$ I would be grateful if you give me the translation. Because i get nothing of the German version. $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 19:33
  • $\begingroup$ @MojtabaAkbarzadeh That will be take some time. Meanwhile tip the pictures and there will be an English text too. $\endgroup$ Commented Dec 14, 2014 at 19:41
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Double slit experiment with electrons is a demonstration of a quantum behavior. When we say observe we mean expose to some kind of interaction. So when electron travels from its source towards the double slit and then passes through, and hits the detector, we see that it is a particle. But, when many electrons pass through the slit we observe a interference pattern so we try to resolve this. Some people said its because electrons behave like waves but that would imply that every individual electron behaves like a wave and that was too strange. So people tried to resolve this by assumption that electrons interact in such a way to form this pattern...of course, new double slit exp was done, this time with only one electron at a time traversing to the slits and to detector and same pattern was observed. So, we have to conclude that free electrons away from interactions behave like waves and more strangely, that they pass through both slits at the same time. So, their position is not determined. There is some kind of probability that they are at some place and this probability wave travels to the detector and only there electron, because of interaction, becomes localized that is to say, nature assigns definite position to the electron. Observation therefore means some kind of interaction.

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    $\begingroup$ of course, there should be some interaction, but the electron has so many interaction before we enter an interaction (through the measurement), like electron-electron interaction, electron and photons interactions, electron and slit interaction, and non of these interactions ruin the diffraction pattern and just our measurement of knowing the place of electron makes it blur... $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 18:36
  • $\begingroup$ Not we knowing, but special kind of interaction which makes electron localized. We dont need to know there is no need for the conscious observer to make the state collapse. $\endgroup$ Commented Dec 14, 2014 at 18:38
  • $\begingroup$ yes, conscious or unconscious doesn't make a difference. But the act of observing elude the wave pattern... they are not waves anymore...!!! $\endgroup$
    – P.A.M
    Commented Dec 14, 2014 at 19:36
  • $\begingroup$ Agreed on that.. $\endgroup$ Commented Dec 15, 2014 at 14:07

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