The pattern of one photon at a time passing through large-distance setups?

[tade]

Let's say that we're sending a collimated beam of EM radiation through a slit that's 3 cm wide, and it diffracts and forms a clear interference pattern on a screen, with the bands of brightness and darkness, or of high and low intensities.

And next, we reduce the intensity of the beam until it emits only one photon at a time. Let's say one photon every two seconds.
If we obtain the collective results of multiple photons' landing positions on the screen, what will the resulting pattern look like?

And let's say we make it a double-slit experimental set-up, and the distance between the two slits is 5 cm, they are spaced 5 cm apart, I'm also wondering what the resulting pattern will look like.

 

[PeterDonis]

we obtain the collective results of multiple photons' landing positions on the screen, what will the resulting pattern look like?

The same as the pattern you saw at high intensity.

let's say we make it a double-slit experimental set-up, and the distance between the two slits is 5 cm, I'm also wondering what the resulting pattern will look like.

You can find that out from many, many references online that discuss double slit experiments.

 

[tade]

The same as the pattern you saw at high intensity.You can find that out from many, many references online that discuss double slit experiments.

because i was really wondering, i was really unsure as to whether the actual results would be of the sameso let's say the first example, with a slit that's 3 cm wide, what do you think is occurring when a single photon is moving through the slit, in order for the collective pattern to be identical

 

[PeterDonis]

what do you think is occurring when a single photon is moving through the slit in order for the collective pattern to be identical

The phase of the wave function is the same regardless of the intensity, and the relative phase of the wave functions coming from the two slits is what predicts the pattern on the detector.

 

[tade]

The phase of the wave function is the same regardless of the intensity, and the relative phase of the wave functions coming from the two slits is what predicts the pattern on the detector.

though sorry, i was referring to my first example, a single slit which is 3 cm wide

 

[PeterDonis]

i was referring to my first example, a single slit which is 3 cm wide

Basically the same answer: the only difference is that in this case it's the phase of the wave function coming from the single slit that predicts the pattern on the detector, regardless of intensity.

 

[tade]

Basically the same answer: the only difference is that in this case it's the phase of the wave function coming from the single slit that predicts the pattern on the detector, regardless of intensity.

do you mean like the summing up of all the phases forming a "circular arc"

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c3
"Divided into segments, each of which can be regarded as a point source, the amplitudes of the segments will have a constant phase displacement from each other, and will form segments of a circular arc when added as vectors."

 

[PeterDonis]

do you mean like the summing up of all the phases forming a "circular arc"

For single slit diffraction, yes. This is because you have to take into account the finite width of the slit in order to get an interference pattern with a single slit.

For the double slit case, the usual analysis does not get this complicated, because you don't need to take into account the finite width of the slits. All you need is the distance between the slits. The phase at each point on the detector is just the sum of two phases, one from each slit, each one depending on the path length from that slit to the given point on the detector. That is enough to get the double slit interference pattern.

A more complicated analysis of the double slit case would involve adding together two single slit patterns of the kind shown in the article you reference. This is not usually done because diffraction effects from each slit individually are usually negligible in a double slit experiment.

 

[tade]

For single slit diffraction, yes. This is because you have to take into account the finite width of the slit in order to get an interference pattern with a single slit.

For the double slit case, the usual analysis does not get this complicated, because you don't need to take into account the finite width of the slits. All you need is the distance between the slits. The phase at each point on the detector is just the sum of two phases, one from each slit, each one depending on the path length from that slit to the given point on the detector. That is enough to get the double slit interference pattern.

A more complicated analysis of the double slit case would involve adding together two single slit patterns of the kind shown in the article you reference. This is not usually done because diffraction effects from each slit individually are usually negligible in a double slit experiment.

so its one photon at a time (one every two seconds), and they are traveling in a direction perpendicular to the plane of the slit(s)

and so from those kinds of experimental results, would it be correct to state that just a single photon traveling perpendicularly is interacting with the slit across its entire width of 3cm?

or for the double-slit case, just a single photon traveling perpendicularly interacting with both slits which are spaced 5 cm apart

 

[PeterDonis]

would it be correct to state that just a single photon traveling perpendicularly is interacting with the slit across its entire width of 3cm?

No, because in the slit itself, there is nothing to interact with!

The only interaction is at each edge of the slit, where the photon wave function is cut off--beyond each edge of the slit you have solid material, not empty space, and in this simplified analysis, solid material completely stops the photons. This simplified model captures that interaction by treating each point along the slit itself (where there is empty space) as a point source and adding together the resulting phases. But that does not mean there is any interaction happening in the empty space of the slit itself.

As just noted, this is a simplified model. We use it not because it is exactly correct, but because it gives answers that are good enough in practice and because any more complicated (and therefore more correct) model would be computationally much more expensive.

or for the double-slit case, just a single photon traveling perpendicularly interacting with both slits which are spaced 5 cm apart

The photon can be said to interact with both slits in this case, but only in the limited manner described above.

 

[tade]

The photon can be said to interact with both slits in this case, but only in the limited manner described above.

so as the photons are traveling perpendicularly, it seems like just a single photon can have a "window of interaction" which is at least 5 cm wide across, i guess that'd be really really large for a single photon

 

[jtbell]

A more complicated analysis of the double slit case would involve adding together two single slit patterns of the kind shown in the article you reference. This is not usually done because diffraction effects from each slit individually are usually negligible in a double slit experiment.

This is true if the spacing between the slits is much larger than the width of the slits. If they are comparable in size then you have both "interference" and "diffraction" aspects in the intensity pattern.

https://openstax.org/books/university-physics-volume-3/pages/4-3-double-slit-diffraction

 

[PeterDonis]

as the photons are traveling perpendicularly, it seems like just a single photon can have a "window of interaction" which is at least 5 cm wide across, i guess that'd be really really large for a single photon

Heuristically, since the photon's wave function is a plane wave, its momentum is highly certain (in your images, it is pointing from left to right), so its position is highly uncertain. This high position uncertainty is what you are referring to (not really correctly) as a "window of interaction". In practice, this is described as the photon source being highly collimated (i.e., it needs to be outputting photons whose momentum is highly certain).

 

[tade]

Heuristically, since the photon's wave function is a plane wave, its momentum is highly certain (in your images, it is pointing from left to right), so its position is highly uncertain. This high position uncertainty is what you are referring to (not really correctly) as a "window of interaction". In practice, this is described as the photon source being highly collimated (i.e., it needs to be outputting photons whose momentum is highly certain).

and if the photons have a low uncertainty in position, but if the beam is wide and intense, the experiment for an interference pattern would still have a "wide window" right?

 

[Nugatory]

if the photons have a low uncertainty in position, but if the beam is wide

Those two conditions are mutually exclusive. The beam width is (sort of, within the limitations of a B-level thread, heuristically) a proxy for the uncertainty in position.

 

[vanhees71]

Let's start from scratch.

First of all one has to unlearn the wrong idea that a photon were some localized (or localizable) massless point particle. That picture of very early quantum theory (Einstein 1905) is outdated since 1925, where one learned that the quantum phenomenology can be described by relativistic quantum field theory (and until today that's the only theory we have for that phenomenology and it's among the best theories ever in being consistent with all observations at very high precision).

It's way better to think about a single photon, which is a very specific state of the electromagnetic field (a socalled one-"particle" Fock state), as an electromagnetic wave with the lowest possible intensity possible for a given frequency. It behaves in all aspects like such an electromagnetic wave except when it is detected, because then it can make one excitation of the material in a detector. This means that whenever you detect a single photon electromagnetic field state you can detect it only once. It's not possible to detect "half a photon" at one place and the other half at another place. This implies that if you use a photoplate (or more modern a CCD camera) a single-photon state of the em. field will be detected at one place, i.e., it leaves one spot on the photoplate or CCD cam.

If you want to demonstrate the wave nature of light, i.e., an electromagnetic wave, you must use a coherent source, i.e., it should be light with a pretty well defined wave vector (and thus also pretty well defined frequency). In an idealized limit it's something that's close to a plane wave of a given frequency and wave vector. If you use such light to illuminate a double slit, you'll see a nice intereference pattern behind the slit.

Now if you use a single-photon source and observe the single photons one by one hitting the screen, you'll see that each photon will indeed leave only one spot on the screen, i.e., the single photon is not observed as somehow "smeared" as a classical em. wave (quantum-field theoretical it's a socalled coherent state, which is a state of the em. field that has not a specified photon number). For each single photon you cannot predict, where it will hit the screen behind the double slit but if you wait long enough to collect very many photons, you'll find the same interference pattern as with a classical em. wave. This means that the intensity of the classical em. wave translates into the detection probability distribution of the single photon on the screen, where each single photon is "prepared" to have the same properties (i.e., pretty well defined wave vector or using the Einstein-de Broglie relation, momentum $\vec{p}=\hbar \vec{k}$).

In this way modern QFT delivers a consistent description what in the old quantum theory was called "wave-particle dualism": On the one hand you have "particle properties" of single photons: It's detected always only at one spot on the screen. On the other hand you have "wave properties" of an ensemble of equally prepared single photons: The pattern found after collecting very many such photons on the screen is given by the interference pattern of the classical em. wave, but the meaning of the corresponding intensity is a probability distribution for the detection of the single photon at any place of the screen.

What's also important to note is the "contextuality", i.e., it also depends on the measurement you make on the em. field or on the single photon state of the em. field: If you place the screen very close to the slits, you won't observe an interference pattern at all. For the classical em. wave you'll simply see the two slits on the screen. That's easy to understand using Huygens's principle: From each point in the two slits there comes a spherical wave, and the pattern on the screen is due to the superposition of all these partial waves. If you put the screen very close to the slits, the spherical waves originating from one slit won't have much overlap with those coming from the other slit, and thus there's no interference between different partial waves, having a phase difference wrt. each other when hitting a given point on the screen, which explains why then there's no interference pattern. For a single photon you have to use the same wave picture but interpret the intensity of the classical wave as a probability distribution for hitting a given point on the screen. Putting the screen pretty close to the slits, given our argument with the classical wave, you can always with pretty much certainty say, through which slit each photon came, but you won't see the interference pattern when collecting many photons. You also won't be able to predict through which screen each photon will come but only when you observe it you know pretty well through which screen it came. This gedanken experiment shows, how modern QFT resolves the other mind-boggling feature of the old wave-particle dualism, i.e., that it depends on the measurement you make, whether you observe particle- or wave-like features of the same situation, and why it's impossible to observe both aspects with one experiment, i.e., if you put the screen very close to the screen, you gain "which-way information" for each single photon and observe "particle properties", but then you don't observe wave properties. If you put the screen very far away from the slits it's the other way: You'll find interference patterns when sending very many single photons through the slits but for each single photon it's impossible to say, through which of the slit it came. I.e., if you want an accurate observation of particle properties, you don't find any wave properties and vice versa. Of course you can make a compromise and put the screen not too close and also not too far from the slits. Then you get an interference pattern with "lower contrast", i.e., you'll neither be able to say with certainty through which slit a single photon came nor will you see a well developed interference pattern, but what you see can be predicted using the modern QFT.

 

[tade]

Those two conditions are mutually exclusive. The beam width is (sort of, within the limitations of a B-level thread, heuristically) a proxy for the uncertainty in position.

hmm, what about considering the positional uncertainty of each individual photon, and imagining that you only have emitters which are "a single photon wide", and to get a wider beam, you put many of these emitters next to each other, kinda like imagining that you're tying together a bundle of sticks or pencils

 

[PeterDonis]

what about considering the positional uncertainty of each individual photon, and imagining that you only have emitters which are "a single photon wide"

That's what the sources you have been describing are: emitters of highly collimated photons where each photon is significantly "wider" than the spacing of the two slits (5 cm in your example).

to get a wider beam, you put many of these emitters next to each other

Why would you bother when a single emitter already gives a "width" that is more than wide enough for your experiment?

If you want each emitter to have a narrow beam, meaning much smaller uncertainty in (transverse) position than the spacing of your slits, then that emitter will be very poorly collimated and photons emitted from it will have a large uncertainty in their direction. That's not what you want for a double slit experiment. Adding together a bunch of these poorly collimated emitters won't fix that; you'll just have multiple photons with a large uncertainty in direction instead of one.

 

[tade]

If you want each emitter to have a narrow beam, meaning much smaller uncertainty in (transverse) position than the spacing of your slits, then that emitter will be very poorly collimated and photons emitted from it will have a large uncertainty in their direction. That's not what you want for a double slit experiment.

hmm, oh yeah, if you try to make the beam narrower, meaning a smaller positional uncertainty, then it'll have a larger momentumal uncertainty and the directions will be more spread out in a wider cone, kinda paradoxical

 

[tade]

hmm, oh yeah, if you try to make the beam narrower, meaning a smaller positional uncertainty, then it'll have a larger momentumal uncertainty and the directions will be more spread out in a wider cone, kinda paradoxical

@PeterDonis ok let's say that its one photon every two seconds for an emitter and its "effective width" is much shorter than 5 cm

and if the photons have a low uncertainty in position, but if the beam is wide and intense, the experiment for an interference pattern would still have a "wide window" right?

and if we put a bunch of these beams together, kinda like tying together a bundle of pencils, such that the resulting beam is wider, and we limit the whole thing such that its still one photon every two seconds, will the result be as though the setup has gained a larger "effective width"?

and next, if we turn up the intensity, no longer a dripping of a tap, but a full waterfall, will there be no doubt that it'll be as though it has gained a larger "effective width"?

 

[tade]

Let's start from scratch.

First of all one has to unlearn the wrong idea that a photon were some localized (or localizable) massless point particle. That picture of very early quantum theory (Einstein 1905) is outdated since 1925, where one learned that the quantum phenomenology can be described by relativistic quantum field theory (and until today that's the only theory we have for that phenomenology and it's among the best theories ever in being consistent with all observations at very high precision).

It's way better to think about a single photon, which is a very specific state of the electromagnetic field (a socalled one-"particle" Fock state), as an electromagnetic wave with the lowest possible intensity possible for a given frequency. It behaves in all aspects like such an electromagnetic wave except when it is detected, because then it can make one excitation of the material in a detector. This means that whenever you detect a single photon electromagnetic field state you can detect it only once. It's not possible to detect "half a photon" at one place and the other half at another place. This implies that if you use a photoplate (or more modern a CCD camera) a single-photon state of the em. field will be detected at one place, i.e., it leaves one spot on the photoplate or CCD cam.

If you want to demonstrate the wave nature of light, i.e., an electromagnetic wave, you must use a coherent source, i.e., it should be light with a pretty well defined wave vector (and thus also pretty well defined frequency). In an idealized limit it's something that's close to a plane wave of a given frequency and wave vector. If you use such light to illuminate a double slit, you'll see a nice intereference pattern behind the slit.

Now if you use a single-photon source and observe the single photons one by one hitting the screen, you'll see that each photon will indeed leave only one spot on the screen, i.e., the single photon is not observed as somehow "smeared" as a classical em. wave (quantum-field theoretical it's a socalled coherent state, which is a state of the em. field that has not a specified photon number). For each single photon you cannot predict, where it will hit the screen behind the double slit but if you wait long enough to collect very many photons, you'll find the same interference pattern as with a classical em. wave. This means that the intensity of the classical em. wave translates into the detection probability distribution of the single photon on the screen, where each single photon is "prepared" to have the same properties (i.e., pretty well defined wave vector or using the Einstein-de Broglie relation, momentum $\vec{p}=\hbar \vec{k}$).

In this way modern QFT delivers a consistent description what in the old quantum theory was called "wave-particle dualism": On the one hand you have "particle properties" of single photons: It's detected always only at one spot on the screen. On the other hand you have "wave properties" of an ensemble of equally prepared single photons: The pattern found after collecting very many such photons on the screen is given by the interference pattern of the classical em. wave, but the meaning of the corresponding intensity is a probability distribution for the detection of the single photon at any place of the screen.

What's also important to note is the "contextuality", i.e., it also depends on the measurement you make on the em. field or on the single photon state of the em. field: If you place the screen very close to the slits, you won't observe an interference pattern at all. For the classical em. wave you'll simply see the two slits on the screen. That's easy to understand using Huygens's principle: From each point in the two slits there comes a spherical wave, and the pattern on the screen is due to the superposition of all these partial waves. If you put the screen very close to the slits, the spherical waves originating from one slit won't have much overlap with those coming from the other slit, and thus there's no interference between different partial waves, having a phase difference wrt. each other when hitting a given point on the screen, which explains why then there's no interference pattern. For a single photon you have to use the same wave picture but interpret the intensity of the classical wave as a probability distribution for hitting a given point on the screen. Putting the screen pretty close to the slits, given our argument with the classical wave, you can always with pretty much certainty say, through which slit each photon came, but you won't see the interference pattern when collecting many photons. You also won't be able to predict through which screen each photon will come but only when you observe it you know pretty well through which screen it came. This gedanken experiment shows, how modern QFT resolves the other mind-boggling feature of the old wave-particle dualism, i.e., that it depends on the measurement you make, whether you observe particle- or wave-like features of the same situation, and why it's impossible to observe both aspects with one experiment, i.e., if you put the screen very close to the screen, you gain "which-way information" for each single photon and observe "particle properties", but then you don't observe wave properties. If you put the screen very far away from the slits it's the other way: You'll find interference patterns when sending very many single photons through the slits but for each single photon it's impossible to say, through which of the slit it came. I.e., if you want an accurate observation of particle properties, you don't find any wave properties and vice versa. Of course you can make a compromise and put the screen not too close and also not too far from the slits. Then you get an interference pattern with "lower contrast", i.e., you'll neither be able to say with certainty through which slit a single photon came nor will you see a well developed interference pattern, but what you see can be predicted using the modern QFT.

So Pete said that "This high position uncertainty is what you are referring to (not really correctly) as a "window of interaction"." (#13), and I was thinking that the "window" appears to be pretty "solid", as it is able to interact with both slits in the perpendicular/transverse direction when they are 5 cm apart, and that's for just a single photon

and i mentioned distances like 3 cm and 5 cm, and quantum weirdness can be quite hard to believe, and so i was also wondering what do you think are the largest distances which have been demonstrated so far for the "window" of a single photon

 

[PeterDonis]

let's say that its one photon every two seconds for an emitter and its "effective width" is much shorter than 5 cm

Then you won't see an interference pattern on the detector.

and if we put a bunch of these beams together, kinda like tying together a bundle of pencils, such that the resulting beam is wider, and we limit the whole thing such that its still one photon every two seconds, will the result be as though the setup has gained a larger "effective width"?

No. You still won't see an interference pattern on the detector.

and next, if we turn up the intensity, no longer a dripping of a tap, but a full waterfall, will there be no doubt that it'll be as though it has gained a larger "effective width"?

See above.

 

[vanhees71]

That's what the sources you have been describing are: emitters of highly collimated photons where each photon is significantly "wider" than the spacing of the two slits (5 cm in your example).

Sure, this you must have, if you want to observe two-slit diffraction patterns to begin with. It's part of the coherence, i.e., the photon momenta must be sharply defined enough to illuminate both slits (almost) homogeneously to get a full-contrast interference pattern.

Why would you bother when a single emitter already gives a "width" that is more than wide enough for your experiment?

If you want each emitter to have a narrow beam, meaning much smaller uncertainty in (transverse) position than the spacing of your slits, then that emitter will be very poorly collimated and photons emitted from it will have a large uncertainty in their direction. That's not what you want for a double slit experiment. Adding together a bunch of these poorly collimated emitters won't fix that; you'll just have multiple photons with a large uncertainty in direction instead of one.

Indeed, then you'll not see nicely developed interference patterns, because rather you can aim at a single slit with your narrow "beam" and there's nothing coming through the other slit.

 

[tade]

Sure, this you must have, if you want to observe two-slit diffraction patterns to begin with. It's part of the coherence, i.e., the photon momenta must be sharply defined enough to illuminate both slits (almost) homogeneously to get a full-contrast interference pattern.

Indeed, then you'll not see nicely developed interference patterns, because rather you can aim at a single slit with your narrow "beam" and there's nothing coming through the other slit.

hmm sorry, a bit confused as to to whom or to what you're replying to, no worries

 

[tade]

Then you won't see an interference pattern on the detector.No. You still won't see an interference pattern on the detector.See above.

ok, thanks so much for the conceptual clarification

and i mentioned distances like 3 cm and 5 cm, and quantum weirdness can be quite hard to believe, and so i was also wondering what do you think are the largest distances which have been experimentally demonstrated so far for the "window" of a single photon

 

[PeterDonis]

and i mentioned distances like 3 cm and 5 cm, and quantum weirdness can be quite hard to believe, and so i was also wondering what do you think are the largest distances which have been experimentally demonstrated so far for the "window" of a single photon

If you mean what are the largest slit separations used in a double slit experiment with photons, I don't know. I doubt it is very much larger than 5 cm since making photon sources with collimation that good is very difficult.

For a single slit, 3 cm might already be too large to get significant diffraction effects with visible light, since it is about a hundred thousand times larger than the wavelength.

 

[tade]

If you mean what are the largest slit separations used in a double slit experiment with photons, I don't know. I doubt it is very much larger than 5 cm since making photon sources with collimation that good is very difficult.

For a single slit, 3 cm might already be too large to get significant diffraction effects with visible light, since it is about a hundred thousand times larger than the wavelength.

so double slit experiments with only one photon at a time

and for photons of any wavelength

 

[PeterDonis]

so double slit experiments with only one photon at a time

This adds the problem of making a usable source that faint to the problem of collimation that I already mentioned.

and for photons of any wavelength

This adds the problem of making usable sources and detectors to the other problems. The fact is that we only really have usable photon sources for some particular narrow ranges of wavelengths that happen to be easier to handle: visible light, infrared, UV (to some extent although making usable detectors becomes problematic since UV can ionize most atoms), and radio waves (but here the long wavelengths create other problems for double slit experiments--interference effects with radio waves are much better handled by other techniques).

 

[PeterDonis]

so double slit experiments with only one photon at a time

and for photons of any wavelength

As far as theoretical predictions go, they are the same regardless of the intensity of the source or the wavelength, provided that the slit sizes and spacing are in the appropriate range for the wavelength and collimation of the source.

 

[StevieTNZ]

Dear @tade
Perhaps you might find this book useful - https://www.routledge.com/Quantum-Reality-Theory-and-Philosophy/Allday/p/book/9781032122380

 

[jtbell]

For a single slit, 3 cm might already be too large to get significant diffraction effects with visible light, since it is about a hundred thousand times larger than the wavelength.

A very wide single slit is simply two instances of edge diffraction, facing each other. If you move the edges closer together, the two edge-diffraction patterns merge in a complicated way. The result has to be calculated using the methods of Fresnel diffraction. When the slit becomes narrow enough, you get the familiar simple single-slit pattern of Fraunhofer diffraction.

 

[tade]

Then you won't see an interference pattern on the detector.No. You still won't see an interference pattern on the detector.See above.

and also, regarding the "bundle of pencils", I'd just like to ask, what are your thoughts on the simplified notion (if I am hopefully allowed to mention this notion, for the purposes of trying to cover the scenario more comprehensively) that making it a "full gushing waterfall" will enable the wave(function)s of the multiple "pencils" to blend together, and thereby resulting in an overall larger "effective width"

So I understand that you've said on this "See above.", relating to the prior answers, though from my perspective, what I'm adding might be a new different line of reasoning, or approaching the scenario from a different viewpoint

 

[PeterDonis]

what are your thoughts on the simplified notion (if I am hopefully allowed to mention this notion, for the purposes of trying to cover the scenario more comprehensively) that making it a "full gushing waterfall" will enable the wave(function)s of the multiple "pencils" to blend together, and thereby resulting in an overall larger "effective width"

I've already told you: that's not how it works.

from my perspective, what I'm adding might be a new different line of reasoning, or approaching the scenario from a different viewpoint

Your perspective is flawed.

You should be spending your time and effort learning the math of QM and how it is used to make predictions. You should not be spending your time trying to dream up ever more elaborate scenarios that are just going to be flawed because you don't understand how QM works. That is not a good use of your time, and it's not a good use of the time of the posters here who are responding to you.