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If we shine light on an electron wave passing through a double slit then the interference pattern is destroyed. This is now understood as decoherence coming from interaction of the electrons with light.

What I do not understand is why decoherence doesn't occur from the material making up the barrier (in which we have the slits). I mean why do we need the light to cause decoherence and why in the absence of the same does the electron behave like an isolated system?

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    $\begingroup$ I think part of the answer is obvious. The slit is such that its internal degrees of freedom do not interact with the electron. My question is realistically, what material is the slit made of and what should it be made of if we would want it to decohere the electron? $\endgroup$ May 30, 2017 at 8:43
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    $\begingroup$ My similar question was closed and referred here. Vadim's answer is helpful on the experimental setup to reduce decoherence. However, can anyone explain why decoherence from the electron wave function interacting with the macroscopic system of atoms doesn't always make the system become classical and spoil the experiment? $\endgroup$
    – Peter A
    Dec 24, 2020 at 12:04
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    $\begingroup$ In short: this is because when quantum fluctuation of certain object's observable is "small", one can treat such observable classically so this classical parameter can't decohere the system of interest. A longer and more mathematical version of my answer is shown below, hiding among other answers. $\endgroup$
    – Bohan Xu
    Mar 8, 2023 at 17:16

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What I do not understand is why decoherence doesn't occur from the material making up the barrier (in which we have the slits).

I bet that under the influence of modulated EM radiation, which leads to plasmonic or phononic excitations in the material, the pattern of the deflected electrons changes.

Justification.
The electron - as well as a photon - that passes through the slit, couples to the surface electrons of the slit edge(s). In principle, this is one of the technologies of spin electronics. The incoming radiation couples to a solid. At the end of the solid, the coupling reverts to energy of the outgoing radiation. This manipulates the direction of the outgoing radiation (electrons, photons). The main thing is to focus the radiation. enter image description here
Source: Yang Longkun, Li Pan, Wang Hancong, Li Zhipeng. Surface plasmon polariton waveguides with subwavelength confinement

If the deflection of the electrons (leading to an intensity pattern on a screen) has to do with the coupling to the surface of the slit, then the manipulation of the edge electrons by plasmons or phonons changes the deflection. Thermal radiation up to a certain intensity, on the other hand, has no effect on the interactions at the edge. Thermal radiation would induce neither plasmons nor phonons. This is the reason why the fringe pattern remains undisturbed by the impact of electrons.

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Below is the answer that I posted to a similar question.

One should distinguish the idealized model discussed in textbooks and real interferometers. Decoherence is indeed an issue in many experiments, which is why realizing such interferometers in practice has been challenging.

What is more surprizing, is that some degree of decoherence is present even in the simplest discussions of the two-slit experiment, as not all the particles arrive at the screen - some of them escape in space, while others land on the non-transparent part of the wall with the slits. Thus, the first attempts of literally realizing an Aharonov-Bohm experiment in solid state devices, with the two particle beams confined within two waveguides (arms of a ring) resulted in phase rigidity - AB oscillations with phase either $0$ or $\pi$. To make the phase change continuously, one had to introduce artificially particle losses, as in this paper. Since then the decoherence in AB interferometers was studied extensively, both experimentally and theoretically. There have been even proposals of using controlled decoherence for measurements, as here.

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  • $\begingroup$ Dear Vadim. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) $\endgroup$
    – Qmechanic
    Dec 24, 2020 at 10:12
  • $\begingroup$ @Qmechanic I totally agree. I wanted to link to the other answer, but coudln't do it for some reason. $\endgroup$
    – Roger V.
    Dec 24, 2020 at 10:15
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The essential thing for making the interference pattern is that the electron can reach a point on the screen in two ways (in this case) where each way picks up a phase that destructively interferes with the other, crucially neither of the paths can cause any other asymmetrical contribution to the overall phase - this means the two paths can't interact permanently with the environment.

Now, the non-intuitive thing about quantum particles is that they don't necessarily go in straight lines - in this case the electron paths that do meet up to interfere, could have passed through both slits without interacting with any slit material on the sides. It doesn't matter for the dark peaks that other electron paths collide with the slits or external environment far away (in fact some will, and in those cases you lose the high-peak visibility on the screen as you lose some electrons you should have detected otherwise).

If you add a "detector" in one of the slits (or, as in your question, you let photons interact with the paths), the electron path's interaction with the detector will to some degree change the phase of that path permanently, which means the dark peaks will lose some of the possible destructive interference (the ref. Roger Vadim linked shows a paper where they did just this). Incidentally an interesting aspect of these detectors-in-the-slits setups is that if the detectors are not perfect, i.e. they will misfire, you get back some of the lost destructive interference, as the overall phase is restored in the cases where one detector fires correctly and the other misfires at the same time!

This discussion is idealized - of course there will be paths where the electron do interact with the sides of the slits and reflect and add a momentum that could (theoretically) be measured, etc... all these other paths contribute to reducing the fidelity of the interference.

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In short: this is because when quantum fluctuation of certain object's observable is "small", one can treat such observable classically so this classical parameter can't decohere the system of interest.

In details: suppose you have system 1 (system of interest) and system 2 (environment) with Hamiltonian $H=H_1 \otimes I_2+ I_1 \otimes H_2 +A_1 \otimes A_2$. The last term can create decoherence. However, suppose $\langle AnyRelevantState|A_2|AnyRelevantState \rangle$ has fluactuation that is smaller than any relevant quantity here, then we can assume this observable of the environment to be classical. Mathematically, we can look at the time evolution of the reduced density matrix (its time derivative) of system 1 and substitute $\hat{A_2}=A_{2,0}+(\hat{A_2}-A_{2,0})$ where $A_{2,0}$ is the classical value and the stuff inside the parentheses, the quantum fluctuation, can be ignored when it is sandwiched by any relevant state. The time derivative of the reduced density matrix of system 1 would be a self-contained equation that only depends on $A_{2,0}$ and state of system 1 instead of the state of system 2, and there would be no decoherence within this approximation.

The photon used to detect where did the electron go, will decohere the electron because the $A_2$ relevant here is the position operator of the photons (or the position dependent field operater), roughly speaking. Since those photon's position has significant fluctuation (because they are colliding with the electron), the classical approximation for $A_2$ is invalid, so decoherence can occur.

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The electron is always has its wave properties. If the electron is undisturbed by photons then it has found a path thru the slits based on "interfering with itself" and there will be a pattern. "Interfering with itself" means the EM field is very dynamic, even before the electron has begun its journey the chosen path is predetermined by virtual photons/forces. For example Feynman theory says the particle considers all paths but chooses the one that is a multiple of wavelength, this restriction is why we have no photons in the dark areas and all in the light in the bright areas for example, same for electrons.

For the electron DSE when a light is shone, the interaction causes the EM field to interact/recalculate the electrons (and photons) path once again, the interact involves all the forces present in the EM field ..... the double slit no longer matters at this point. The interference is absent with no double slit.

When you use the word decoherence .... yes many electrons hit the slit directly and decohered .... many did not passing thru.

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