# Single slit experiment

Suppose we are doing a double slit electron diffraction experiment and we place a detector behind the slits to detect which slit the electron passes through. If we do this the diffraction pattern will be destroyed and instead we get just two peaks - one behind each slit.

Suppose we use just one slit, then we would expect to get a diffraction pattern described by the $$\mathrm{sinc}(x)$$ function, as shown in the figure below.

Now suppose we place a detector behind the slit just as we could do in the double slit experiment. Will this also destroy the diffraction pattern leaving just a diffuse band?

• There is no wave function collapse. – my2cts Jan 1 '20 at 1:50
• Hi. I've edited your question to try and make it clearer what you are asking. If you don't like my change please feel free to roll it back. – John Rennie Jan 1 '20 at 7:29
• The "diffuse band" is just what people talk about to simplify the thought experiments whereas in reality you will always get either a single-slit pattern or a double-slit pattern (extra fringes in the middle spot). The "diffuse band" is just people ignoring the peripheral bands of the single-slit pattern. All experiments are centered around whether or not the middle band of the single-slit pattern gains extra fringes. – Maximal Ideal Jan 1 '20 at 7:45
• @MaximalIdeal So with detector present or not a single slit will always produce interference pattern with small lateral waves? – jbradvi9 Jan 2 '20 at 11:58
• @jbradvi9 There's one thing I forgot. In most discussions, they might presuppose that the detectors are infinitely precise (or maybe the slits are infinitely small to pinpoint the exact location of the electron). In that case, there really won't be any single-slit pattern even in principle. In a realistic scenario, we don't have infinitely precise detectors. Of course, in the thought experiment we may assume we do have infinitely precise detectors. – Maximal Ideal Jan 2 '20 at 20:22

If the detector is capable of revealing which part of the slit the electron passes through, even the single-slit diffraction pattern will disappear. Any diffraction pattern depends on uncertainty about the paths of the detected particles. Remove that uncertainty, and the diffraction pattern disappears.

• To clarify, when you say "capable of revealing which part of the slit the electron passes through," I assume you mean in the limit that the detector is infinitely precise. (Sorry for being nitpicky.) – Maximal Ideal Jan 2 '20 at 20:00
• It has nothing to do with uncertainty because electrons and photons don’t know any better. The interference goes away because either you turned on a light to see or you intercepted photons emitted from the electron. Either one will effect the trajectory of the electron and therefore effect the pattern. – Bill Alsept Jan 2 '20 at 20:31
• That is an interesting, forcefully stated, misleading, and incorrect statement. – S. McGrew Jan 2 '20 at 20:54
• @MaximalIdeal , the more precise the locstion-measuring detector, the less uncertain the electron's path through the slit, and the less diffraction will be detectable (that is, the side fringes will be more blurred). – S. McGrew Jan 2 '20 at 21:11
• @BillAlsept The interference goes away when the wavefunction is entangled with that of a detector that distinguishes both paths. – my2cts Jan 2 '20 at 21:13

The "diffuse band" that you mentioned is just what people talk about to simplify the thought experiments. Either they do this by ignoring the peripheral bands, or they assume the detectors are infinitely precise. This is done because the discussion is about what happens to the middle band of light.

In reality you will always get either a single-slit pattern or a double-slit pattern, and really all experiments are centered around whether or not the middle band of the single-slit pattern gains extra fringes within it.

If there are no detectors, we get a double-slit pattern (the two single-slit patterns interfere with each other). If there is a detector, we get single-slit patterns (the two single-slit patterns don't interact).

So with detector present or not a single slit will always produce interference pattern with small lateral waves?

If the slits openings are not infinitely small and the detectors are not infinitely precise then yes, but see S. McGrew's answer for a caveat.

For a detector you would need either real photons or virtual ones as part of the detection process. Any interaction with the electron after is passes the slit allows the electron to "recalculate" its path, this new path is now not involving the slit at all, therefore NO pattern. It is the same for the double slit.

It is the whole setup (source + slit(s) + target) that influences the possible paths of a photon or electron, in dark areas there are no photons/e, bright areas have them all. The tight geometry of the the DSE or SE is what limits the paths and shows the pattern (p.s. the word interference is very historical and somewhat misleading). High probability paths are ones where the path length is related to the wavelength.