If we fire photons at the double slit, we see an interference pattern on the screen. We know that if a quantum system interacts with any outside object (such as a detector, etc), its wavefunction collapses. Is there any chance that the photon could interact with the slits? They are large classical objects that can potentially collapse a wavefunction. Why does the w.f. only collapse when a quantum system interacts with a detector?
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1$\begingroup$ Are you familiar with Feynman's Lectures on Physics? W.f. can interact with the slit and it is true irrespective of whether we are seeing electrons or not. The difference is created by a variable intensity light source which allows us to see the electrons: the elcetrons can interact with these photons leading to w.f. collapse. See chapters 1 and 3.2 from the above link. $\endgroup$– KP99Commented Dec 25, 2022 at 15:24
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$\begingroup$ look at my answers here physics.stackexchange.com/questions/285142/… and here physics.stackexchange.com/questions/506916/… $\endgroup$– anna vCommented Dec 25, 2022 at 19:29
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2 Answers
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To speak of an interference pattern does not do justice to the phenomenon. The characteristic intensity distributions of a light source can be seen not only behind slits but also behind individual edges. Even from photons that are emitted individually and one after the other, the intensity distribution will appear on the observation screen after some time of exposure. The interference for the intensity distribution for a corner and single photons is truly unexplainable.
Is there any chance that the photon could interact with the slits?
This actually leads to a better description of the phenomenon. The surface electrons of the slit interact with the photon or electron beam. Phononic effects quantise the direction of deflection of the particles and the characteristic intensity distribution is the result of this quantised interaction.
To reduce the phenomenon of deflection, one would have to influence the phononic processes. The photolithographic industry, with its current limitations of achievable structure sizes, should have the greatest interest in pursuing this way. Possibly, the theory would then also move.
answered Dec 27, 2022 at 5:38
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$\begingroup$ In diffraction process, how to describe the edge in the sense of particle-wave duality? and Does a particle interact with walls of a slit? $\endgroup$ Commented Dec 27, 2022 at 5:42
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If we fire photons at a black screen that has two narrow holes, then we see an interference pattern on a screen that is placed behind the double hole screen.
We know that if a quantum system interacts in an irreversible way with any outside object (such as a detector, etc), its wavefunction collapses. We call this kind of thing a measurement.
Now, the black screen with the holes tries to detect a photon. When that succeeds, then the wavefunction disappears. And when that fails, then the wavefunction disappears from where there existed photon detectors, and inside the two areas without any photon detectors a wavefunction with large amplitude appears. I mean the holes are the areas without any photon detectors.
When we do the double slit experiment, we use a detector with two holes in it, and all those results where the detector, despite of the holes, manages to collapse the wavefunction, those results we ignore.
So then we are left with such cases in which wavefunction was reversibly squeezed, but not irreversibly collapsed.
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$\begingroup$ Many photons are reflected by the double slit .... the anti-pattern can be visible on the opposite wall in a dark room ..... like you say they go elsewhere to collapse. $\endgroup$ Commented Dec 27, 2022 at 2:39