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How does one detect which slit a photon has passed through in a double slit experiment?

Would the double slit experiment change if the photons were to pass through vacuum tubes instead of slits?

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  • $\begingroup$ What do you mean "pass through vacuum tubes"? Do you mean two tubes? $\endgroup$ – rghome Dec 10 '19 at 17:21
  • $\begingroup$ Yes. Instead of 2 slits the photons have to pass through a tube. $\endgroup$ – Oliver Van Der Togt Dec 10 '19 at 17:50
  • $\begingroup$ You detect photons in the double slit experiment the same way you detect them in any other context. Some kind of photographic equipment, or a photo-multiplier tube if the light level is low enough. $\endgroup$ – puppetsock Dec 10 '19 at 18:39
  • $\begingroup$ @Oliver Van Der Togt Your response in the comments is even more confusing. Do you mean two tubes instead of two slits or one tube instead of two slits? $\endgroup$ – rghome Dec 11 '19 at 8:11
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Now you are specifically asking about detection, which is not well defined in QM. You mean measuring, and QM measurements are always measurements of specific observables.

It is very important to differentiate measurement, and interaction.

There is no holistic act of "observing all properties of a system at once" like there is in classical mechanics - a measurement is always specific to the one observable it measures, and the measurement irrevocably alters the state of the system being measured. People often use "detect a particle" as shorthand for "perform a position measurement of a particle".

I actually asked a question about this:

If a photon truly goes through both slits (at the same time), then why can't we detect it at both slits (at the same time)?

Now there are different types of interactions, and yes it is possible to interact with a photon at a slit without destroying the interference pattern.

The three main types of interaction are:

  1. elastic scattering, in this case the photon keeps its energy, phase and the interference pattern is still there

  2. inelastic scattering, in this case the photon gives part of its energy to the atom, changes phase and the interference pattern disappears

  3. absorption, in this case the photon gives all its energy to the absorbing atom/electron, and the photon ceases to exist. This is what happens at the actual screen, when you see a dot.

Most detectors usually do 1. or 2., and in case of elastic scattering, the pattern is still visible.

Although the electrons (which were shot one by one) could still pass through the filtered slit, the filter caused more of the electrons to undergo inelastic scattering rather than elastic scattering. As the physicists explained, an electron undergoing inelastic scattering is localized at the covered slit, and acts like a spherical wave after passing through the slit. In contrast, an electron passing through the unfiltered slit is more likely to undergo elastic scattering, and act like a cylindrical wave after passing through that slit. The spherical wave and cylindrical wave do not have any phase correlation, and so even if an electron has a probability to pass through both slits, the two different waves that come out cannot create an interference pattern on the wall behind them.

What constitutes measuring in the double slit experiment?

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    $\begingroup$ This contains irrelevant copy/paste material and I don't mean the very large quotes. You keep posting the same text. $\endgroup$ – my2cts Dec 10 '19 at 20:01
  • $\begingroup$ @my2cts he is asking about detection, measurement. $\endgroup$ – Árpád Szendrei Dec 10 '19 at 20:40
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In order to contribute to an interference pattern, the wave associated with a photon must pass through both slits. That's all that can be said.

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  • $\begingroup$ What about single slits and what about edges. Also behind these obstacles fringes occur. So a statement, that a photon has to go through 2 slits, couldn’t be the full answer. $\endgroup$ – HolgerFiedler Dec 11 '19 at 5:03
  • $\begingroup$ The question was about a double slit. I was not suggesting that other effects could not produce an interference pattern. $\endgroup$ – R.W. Bird Dec 11 '19 at 14:29
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You just put a detector in front of every slit. This implies that there is no interference pattern. I mention this as usually questions on the double slit experiment are about interference.

Your second question needs clarification.

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