# Most basic double slit experiment question

• rasp

#### rasp

Mechanically speaking, where is the photon emitter aimed when there are 2 slits? If the experimenter was using an accurate emitter, wouldn't he/she expect the photons to hit only where it was aimed, e.g. either one slit or neither slit? Is it that the emitter oscillates, so it can aim at each of the two slits?

Mechanically speaking, where is the photon emitter aimed when there are 2 slits? If the experimenter was using an accurate emitter, wouldn't he/she expect the photons to hit only where it was aimed, e.g. either one slit or neither slit? Is it that the emitter oscillates, so it can aim at each of the two slits?

The emitter is aimed in the direction of both slits. While it may be possible to build a device that aims more or less precisely, this would defeat the purpose of the double slit experiment. If you know that the photon has to be going through the left slit and that there is zero probability of it going through the right slit, then you wouldn't get any interference.

No, it doesn't oscillate. Think of it like a light bulb rather than a gun. A light bulb fires photons in every direction. If you had a very dim light bulb that only fired one photon at a time that would be the type of emitter you would use (if you were interested in what happens when one photon goes through at a time). It doesn't need to alternate as one in every 100 photons (for example) goes through the left slit and one in every 100 goes through the right slit and the other 98 don't go through either. You can know this probability for a given setup but you wouldn't know which slit any given photon would go through, kind of like you know a coin has a 50 percent chance of coming up heads, 50 of coming up tails, but no idea which the next flip will bring. The difference between the coin and a photon, however, is that unless we measure which slit the photon went through (or force it to only go through one) it behaves as though it went through both (kind of).

If fact, you don't even need a sophisticated emitter to perform a basic version of the double slit experiment. You can get a cheap laser pointer. Then put something opaque with a single tiny slit in it in front of the pointer (maybe you don't need to do this step?). Then put another screen behind it with two tiny slits in it very close together. Then another screen behind that one with no slits. I would use black for the two with the slits and white for the slitless one. You won't see one photon at a time but you will see the probability distribution of where a single photon would go. I.e. the bright spots on the white paper indicate a higher probability that the dark spots.

Actually, here's a thread from this forum where someone else did this:

even if it went through one slit we would still get an interference pattern.
they do this experiment with electrons and when they fire them one at a time they still get an interference pattern , and i think it is impossible or very unlikely to be able to fire 2 electrons with the same exact path.

even if it went through one slit we would still get an interference pattern.

No, this isn't true. If you cover one of the slits, the interference pattern disappears.

they do this experiment with electrons and when they fire them one at a time they still get an interference pattern

Exactly. This is the whole reason the experiment is so mind blowing. Even though they fire one at a time it appears as though the electron is going through both slits. The reason why it looks like it's going through both slits is that you get an interference pattern when both slits are open and none when you cover (or measure) one of the slits, regardless of how many electrons you fire at a time.

The reason why the interference pattern disappears when you cover a slit is because you drop the probability that the electron will go through that slit to zero. The interference pattern is a result of the probabilities of all possible paths (kind of, it's a tiny bit more complicated - check out Richard Feynman's book QED) for the electron being added up. I.e. The interference pattern isn't a result of individual electrons interfering with each other but each electron interfering with itself, thus providing evidence for the wave/particle duality of matter.

If you knew, with 100% certainty which path the electron was taking then you would no longer see the quantum strangeness, as the effect relies on uncertainty.

No, this isn't true. If you cover one of the slits, the interference pattern disappears.

Then what i learned in school about single slit interference is wrong

If i understand it right, there is diffraction from 1 slit, but no interference, cause there are no alternate paths for the photon.
But what me always puzzled is: what does it mean, "the photon is measured at one slit" ?
How can you measure a photon without absorbing it ?

If i understand it right, there is diffraction from 1 slit, but no interference, cause there are no alternate paths for the photon.
But what me always puzzled is: what does it mean, "the photon is measured at one slit" ?
How can you measure a photon without absorbing it ?

If I remember right, it's absorbed and re-emitted. Also, as a matter of interest, these detectors fail to detect the photon a certain percentage of the time. If you put detectors at both slits, the times where you successfully detect the photon going through a specific slit will have a pattern that shows no interference and the times where you fail to detect it will behave as if the detectors weren't even there at all, i.e. the interference pattern.

Ok, if measurement means, it is absorbed and re-emitted, its clear for me why there is no interference. Thank You.

can't a photon interfere with itself .

yes it can.
As Soup pointed out:
"The interference pattern is a result of the probabilities of all possible paths (kind of, it's a tiny bit more complicated - check out Richard Feynman's book QED) for the electron being added up."

Actually its in principle a result of ONE photon/electron interfering with itself and not a result of multiple photons/electrons interfering with each other. There the analogy with waterwaves false short.

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so the analogy with water waves is incorrect.

Sorry, cragar. I think we got our words twisted up a little. A single slit can exhibit diffraction, which is a type of interference. But when I was talking about "the interference pattern" I was referring to the double slit interference, usually referred to as such and I wasn't thinking about diffraction as a type of interference, sometimes called "the diffraction pattern". I was a little unclear.

So, yes, if you cover one slit you'll still get an "interference" pattern (but I believe this only happens when the slit is certain widths) but this isn't the same pattern that we get when both slits are open.

I believe the analogy might be a bit missleading yes.
btw. Richard Feynmans QED was one of the best books i ever read.

i see , thanks for making it clear .

so the analogy with water waves is incorrect.

That's right. It's close and it's a good demonstration. But, like many analogies, it is not a perfect description of what's actually going on at the quantum level.

But what me always puzzled is: what does it mean, "the photon is measured at one slit" ?
How can you measure a photon without absorbing it ?

By running it through a polarizer. That has the effect of providing which-slit information. This is sometimes done in double slit experiments.

By running it through a polarizer. That has the effect of providing which-slit information. This is sometimes done in double slit experiments.

Thank you very much for destroying my belief i understand something ;)
Another question: is the speed of light thru a polarizer = c ?
If it is slower, could i not argue its not the "same" photon ?

Thank you very much for destroying my belief i understand something ;)
Another question: is the speed of light thru a polarizer = c ?
If it is slower, could i not argue its not the "same" photon ?

In fact, maybe it isn't the same one! Anyway its not me that you would be arguing against, it would be the photon. They have a union.