I Young's slit experiment with single photons

[Marilyn67]

Hello,

I have a little problem understanding Young's slit experiment with single photons :

I have understood for a long time that each photon impact on the screen corresponds to a photon sent by the source, and that, if we don't try to find out by which path the photon has passed, of course, each impact of photon contributes to gradually draw an interference pattern, ok.

My problem comes from the description that is often made in popular science articles :
We are told : "The photon interferes with itself", which amounts to saying that the wave function of a photon has passed through the 2 slits.

The problem is this :

If we move away from the median line between the 2 slits, the distance traveled by the photon from one slit is different from the distance traveled by the photon from the other slit.

I conclude that this (single) photon can't interfere "with itself" since when the complementary part of the photon arrives on the screen, the first part has already produced an impact since it arrives ahead of the complementary part (a single photon can't produce 2 impacts, it's impossible, since there is reduction of the wave function at the first impact).

Necessarily, I tell myself that to have an interference, at different distances, (all over the screeen) it's "different" photons that interfere with each other, because the sending rate of single photons is sufficiently fast. That's right ?

If yes, then I deduce that with a much slower rhythm (we wait for the impact of a photon on the screen to send the next one), the pattern is no longer an interference pattern but a pattern of classical diffraction. That's right ?

This seems consistent with the idea that in this very specific case, by knowing at what "moment" a photon arrives, we can deduce what path it has actually traveled by deducing the distance it has traveled from time (potentially measurable, even if it is not measured) and C. (d=C.t).

How does the mathematical formalism of quantum mechanics manage to predict this phenomenon, predictable but a priori surprising since there is no longer a single ψ wave function, but as many ψ wave functions as photons ?

I guess this process (changing pattern with a very slow rate) doesn't work with electrons because the speed of an electron is fundamentally undetermined before measurement, whereas the speed of a photon in vacuum is always fundamentally determined (C), if I'm not talking nonsense. (?)

Is there a reasoning error somewhere ?
Above all, tell me if I wrote one or more nonsense in my message, it's very important for me

In advance, thank you for your answers.

Cordially,
Marilyn

 

[Dale]

I conclude that this (single) photon can't interfere "with itself"

Nature simply doesn’t care about your conclusion. As beautiful and certain as it seems to you, it isn’t the way she works. A single photon does in fact interfere with itself, in the usual meaning of the phrase.

it's "different" photons that interfere with each other

Under your own justification for why a photon cannot interfere with itself, it would be even more impossible for other photons to interfere. If the picoseconds you are talking about in your concern make a difference, then the much longer time between photons will be a huge issue.

 

[Marilyn67]

Hello Dale,

Thank you very much for your very quick response, but I don't understand.

How is interference possible if the 2 complementary parts of the photon don't coincide at a specific place and time ?

If I take an extreme case and only send a photon after one second ore one minute ?

There must be a limit, right ?

I may have misunderstood your answer.

Cordially,
Marilyn

 

[vanhees71]

This is another example for the harm done by still writing bad introductions to quantum-mechanics textbooks by using the outdated picture of photons as some point-particle like entities representing the electromagnetic field. That's utterly wrong. A photon is just a one-quantum Fock state of the quantized electromagnetic field. You should think of it as a wave rather than a point-particle like object.

The phenomenolgy simply is that if you do the double-slit experiment with a single photon the interference pattern of the corresponding electromagnetic wave (i.e., the energy density of this em. wave at the observation screen) provides the probability distribution for detecting a photon at a given place on this screen. To see this probability distribution in your experiment you have repeat it very often, i.e., accumulating many single photons going through the double slits.

With this simple picture all the apparent difficulties in understanding, why single photons show interference phenomena as do classical electromagnetic wave in this specific sense of probabilities for detecting a photon at the screen, are gone.

 

[PeroK]

How is interference possible if the 2 complementary parts of the photon don't coincide at a specific place and time ?

The photon is not a classical point particle and does not follow a classical trajectory. Or, in this case, does not follow the superposition of two classical trajectories.

Instead, there is a certain probability amplitude for detection of a photon at a given point at a given time. Moreover, there is no classical certainty about when a photon is emitted from source. There is, therefore, a nonzero probability of the photon detection event at points that are not equidistant from the slits.

 

[PeroK]

PS in basic terms the uncertainty in time means there is an overlap between the times for each component of the superposition when the probability amplitude is nonzero for a given point on the detection screen.

 

[Nugatory]

In advance, thank you for your answers.

Although no substitute for a serious introduction to quantum electrodynamics, you will find Feynman’s layman-friendly and almost math-free book “QED: The strange theory of light and matter” helpful.

 

[Dale]

How is interference possible if the 2 complementary parts of the photon don't coincide at a specific place and time ?

There are no two complementary parts.

A photon is a specific state of the electromagnetic field. Such states are spread out, and more specifically the position operator cannot even be defined for them. Saying that it has two parts and that those two parts are in different places is simply not related to how these states actually behave.

A single photon is the entire state, not just a part of the state. The state itself is a wave, and it behaves as a wave does, including exhibiting interference. The math is solid, the theory is solid, and the experimental evidence is solid.

 

[Marilyn67]

Thank you all for your very concise answers.

Yesterday evening again, I would have certified that I was right, especially since the idea was consistent with the fact that the knowledge of the path always coincides with a classic pattern and vice versa.
So this unavoidable condition is always respected here, except that the opposite is happening and that we cannot therefore know the path traveled in this way.

This is another example for the harm done by still writing bad introductions to quantum-mechanics textbooks by using the outdated picture of photons as some point-particle like entities representing the electromagnetic field. That's utterly wrong. A photon is just a one-quantum Fock state of the quantized electromagnetic field. You should think of it as a wave rather than a point-particle like object.

The phenomenolgy simply is that if you do the double-slit experiment with a single photon the interference pattern of the corresponding electromagnetic wave (i.e., the energy density of this em. wave at the observation screen) provides the probability distribution for detecting a photon at a given place on this screen. To see this probability distribution in your experiment you have repeat it very often, i.e., accumulating many single photons going through the double slits.

With this simple picture all the apparent difficulties in understanding, why single photons show interference phenomena as do classical electromagnetic wave in this specific sense of probabilities for detecting a photon at the screen, are gone.

You are absolutely right, and when I ask myself a quantum question, I always think first in terms of waves, those that we use in the classical field to have the best possible intuition at the start, and then, I think about what quantum mechanics could bring as a change.
I realize that I was wrong with quantum mechanics.

The photon is not a classical point particle and does not follow a classical trajectory. Or, in this case, does not follow the superposition of two classical trajectories.

Instead, there is a certain probability amplitude for detection of a photon at a given point at a given time. Moreover, there is no classical certainty about when a photon is emitted from source. There is, therefore, a nonzero probability of the photon detection event at points that are not equidistant from the slits.

The explanation is there...!
I really never thought of that.
This explains why we can't know what path has been traveled, because we don't know "t".

PS in basic terms the uncertainty in time means there is an overlap between the times for each component of the superposition when the probability amplitude is nonzero for a given point on the detection screen.

This is the clearest and most convincing explanation !
This "overlap" makes me understand why doing the same experiment on an astronomical scale, such as the Earth-Moon distance for example, and decreasing the photon firing rate would not change anything on another scale of time and space ! (it was a question I had in mind and I don't need to ask it anymore).

Although no substitute for a serious introduction to quantum electrodynamics, you will find Feynman’s layman-friendly and almost math-free book “QED: The strange theory of light and matter” helpful.

Thank you, I took good note of it !

I would like to tell you all, that if I asked you this question with the experience of Young's slits, it's because there is an idea that I have had in mind for some time and I still have to think about it before to submit it to you (it's much more complex and it involves things that are apparently well established in quantum mechanics).

As soon as I put it on paper, I'll open a new topic in this section, and I'll put the link in this topic to make sure you don't miss it, because your knowledge is really precious to me !

I'm really curious to know what you will think of it because I can't find the answer anywhere and it seems that the experiment has never been tried in this way.

See you soon,
Marilyn

 

[Marilyn67]

Sorry Dale, our messages crossed.

The math is solid, the theory is solid, and the experimental evidence is solid.

I don't doubt it for a second, and I thank you for all the knowledge you bring to me here.

I have an experimental protocol in mind, precisely exploiting our knowledge of quantum mechanics, and I don't know what to think of it for the moment.

As said before, I still have to think about it before submitting it to you, and I really don't know what to think of it for the moment with my modest knowledge.

Have a good evening,
Marilyn

 

[Marilyn67]

Hello,

As promised, I'm coming back to you
I made a schoolgirl mistake in my protocol, and this one is in the trash...
Sorry for my last off-topic post, that was stupid of me.

@PeroK: Thanks for the very clear explanations and I understood this problem of indeterminacy on when the photon was emitted and the resulting non-zero probabilities on non-equidistant points, but a bigger problem has been bothering me ever since yesterday.

Instead, there is a certain probability amplitude for detection of a photon at a given point at a given time. Moreover, there is no classical certainty about when a photon is emitted from source. There is, therefore, a nonzero probability of the photon detection event at points that are not equidistant from the slits.

PS in basic terms the uncertainty in time means there is an overlap between the times for each component of the superposition when the probability amplitude is nonzero for a given point on the detection screen.

Ok, but my problem is this, now :

We can easily do the equivalent of Young's slit experiment by replacing it with a Mach-Zehnder interferometer.

The advantage is to be able to have a very short path on one side, and a very very long one on the other (detour with mirrors, fiber optics, and even also (why not) materials that slow down light to a snail's speed).

Let's go crazy and imagine a path of 1 meter on one side and (for example), the Earth-Moon distance on the other side (I'm taking an extreme example).
The time gap is huge !

My "classic" intuition with a continuous laser tells me that I would have interference. Ok.

But if now I take single photons again (example, one photon per second), I agree that I still have an uncertainty on the emission time as with Young's slits, but I have a large trust interval beyond which I am certain that photon a was not emitted, with the counter and my clock in front of me, right ?

If I take up the concept of a wave, I have a single wave which passes through the short path and a single wave which passes through the long path.

The uncertainty as to when the emission is still there, but that's ridiculous compared to the large trust interval where I'm certain the photon hasn't been sent, so the probability of there being an overlap is zero, no ?

How to explain that the two waves interfere with such a degree of trust over such a large difference in duration, greater than 1 second here ?

I confess that I am lost !

Thank you in advance for your always very concise answers.

Cordially,
Marilyn

 

[PeterDonis]

How to explain that the two waves interfere with such a degree of trust over such a large difference in duration

You're assuming that they do--that is, that if we did the Mach-Zehnder interferometer experiment as you describe, with path lengths differing by many orders of magnitude, that you would still get the same result. But actually that is not what QM predicts! See below.

My "classic" intuition with a continuous laser tells me that I would have interference. Ok.

Only if the laser is on for the requisite amount of time--heuristically, long enough for the light travel time over both path lengths to elapse.

But in the MZI scenario you describe, that is not the case--at least, that seems to be the point of your formulating the scenario. And if it is not the case, then the MZI result will not be the same as the usual result: there will be a nonzero probability for the photon detector on the output arm that normally never detects a photon in the standard MZI experiment, to detect a photon.

 

[Marilyn67]

Hello @PeterDonis,

Thank you for your quick reply !

You're assuming that they do--that is, that if we did the Mach-Zehnder interferometer experiment as you describe, with path lengths differing by many orders of magnitude, that you would still get the same result. But actually that is not what QM predicts! See below.Only if the laser is on for the requisite amount of time--heuristically, long enough for the light travel time over both path lengths to elapse.

But in the MZI scenario you describe, that is not the case--at least, that seems to be the point of your formulating the scenario. And if it is not the case, then the MZI result will not be the same as the usual result: there will be a nonzero probability for the photon detector on the output arm that normally never detects a photon in the standard MZI experiment, to detect a photon.

So, if I understand correctly, in this case, with these orders of magnitude, we will have a classic diffraction pattern ?

My first intuition was therefore quantitatively wrong, (Young) but qualitatively, if we use a Mach-Zehnder interferometer, with two arms whose length can be modified at will, can present either an interference pattern or a classical pattern of diffraction, depending of the case...

Without going into big calculations, do you know from what approximate ratio a difference is obtained ?

Cordially,
Marilyn

 

[Dale]

Without going into big calculations, do you know from what approximate ratio a difference is obtained ?

This will happen when the path difference exceeds the coherence length. So it will depend on the source used and the setup of the optics.

 

[PeterDonis]

if I understand correctly, in this case, with these orders of magnitude, we will have a classic diffraction pattern ?

What experiment are you talking about? A Mach-Zehnder interferometer does not give a diffraction pattern. You have two photon detectors in the two output arms of the second beam splitter; in the normal MZI setup, where the beams in both arms coming into the second beam splitter are exactly in phase, only one detector will fire. Putting something that shifts the phase into just one arm will give a nonzero probability for the other detector to fire; but drastically increasing the length of one arm will also do that.

 

[Marilyn67]

Thanks @PeterDonis for your response,

What experiment are you talking about? A Mach-Zehnder interferometer does not give a diffraction pattern. You have two photon detectors in the two output arms of the second beam splitter; in the normal MZI setup, where the beams in both arms coming into the second beam splitter are exactly in phase, only one detector will fire. Putting something that shifts the phase into just one arm will give a nonzero probability for the other detector to fire; but drastically increasing the length of one arm will also do that.

I misspoke.
I speak of course from experience with an MZI.
Yes, when I was talking about diffraction pattern, I made a misstep.
I should have said simply that there was no interference pattern.

So you are confirming to me that in this extreme scenario (#11), we no longer have an interference pattern but a diffuse pattern ?

Cordially,
Marilyn

 

[PeterDonis]

I speak of course from experience with an MZI.

What kind of detectors does the MZI that you have experience with have? The typical description I see has photon counters at each output arm, which don't show patterns at all, they just give a count of photons. That is what I have been describing up to now.

If you have some sort of "pattern" detector at each output arm, in a standard MZI with nothing in each arm and each arm of equal length, I would expect to see a bright spot on the detector at one output arm (call this detector A) and nothing at all on the other detector (call this one detector B). There won't be any sort of "pattern" because there is nothing to form one: you just have light all in phase coming out one arm only.

In the MZI you describe, what you would expect to see will depend on how long you leave the laser on. If you leave the laser on for a long time (heuristically, significantly longer than the difference in light travel times between the arms), you would expect to see the same output as for the normal case above, except that at the very beginning and the very end of the experiment you might see the spot at detector A dimmer than the above and a dim spot (instead of nothing) at detector B. Heuristically, these differences at the beginning and the end are the times when there is only laser light in one arm (only the shorter arm at the beginning, only the longer arm at the end).

If you leave the laser on only for a short time (heuristically, significantly shorter than the difference in light travel times between the arms), you would expect to see a pair of dim spots, one each on detectors A and B for a short time, then a short period of darkness on both, then another pair of dim spots, one on each detector, for a short time. Heuristically, these correspond to the time when there is laser light only in the shorter arm, and only in the longer arm. The shortness of the laser on time means there is never a time when there is laser light in both arms.

Note that all of the above also assumes that the difference in path lengths between the arms is an exact multiple of the laser light wavelength, so that when there is light in both arms, the light in both arms is exactly in phase when it reaches the second beam splitter.

 

[Marilyn67]

Hello Dale,

This will happen when the path difference exceeds the coherence length. So it will depend on the source used and the setup of the optics.

I understood, but I was hoping for an example with numbers.
Example :
With a 1 meter arm and a 2 meter arm, a wavelength of 532 Nm, is it possible ?

At what rate of sending photons should we go down to have a diffuse pattern instead of an interference pattern in this case ?

Cordially,
Marilyn

 

[WernerQH]

There is an interesting video, discussed in the thread
https://www.physicsforums.com/threads/how-big-is-a-photon-interesting-video.1005059/
To achieve a coherence length of one meter is quite a challenge, requiring a very stable laser.
The demonstration is interesting because the intensity is reduced so much that there is only one photon in the apparatus at any time. (Extending this to the moon appears to be completely out of the question!)

 

[Marilyn67]

Thanks @PeterDonis,

I understand what you are saying and I agree with you.

What kind of detectors does the MZI that you have experience with have?

You ask the right question, I lack rigor !

I find out that I don't even know myself, but I'm sure I've seen it somewhere on arXiv !

When my PC crashed, I didn't save these documents !

It seems to me (?) that to form a visible pattern (spread in space), some experimenters used detectors composed of a lens (or a prism?) with a kind of CCD screen behind, like a camera, but I'm not sure.

Do you know how they do it ?

Is it possible to have a spatially distributed pattern on a screen on either side of the second semi-reflecting blade this way ?

Your help is invaluable to me !

Cordially,
Marilyn

 

[Dale]

I was hoping for an example with numbers

Whatever single photon source you have, the manufacturer’s documentation should have that information. I don’t. If it is a custom-built source then you would need to measure it yourself.

However, this ties back into @PeroK 's comments above. To get a longer coherence length means a smaller linewidth. A smaller linewidth means that the time uncertainty is larger. The two are unavoidably paired. You will find no conflict. If you make the path length very different then the only photons that will be able to self-interfere will be exactly those where the time uncertainty is large enough.

 

[Marilyn67]

Hello @WernerQH ,

There is an interesting video, discussed in the thread
https://www.physicsforums.com/threads/how-big-is-a-photon-interesting-video.1005059/
To achieve a coherence length of one meter is quite a challenge, requiring a very stable laser.
The demonstration is interesting because the intensity is reduced so much that there is only one photon in the apparatus at any time. (Extending this to the moon appears to be completely out of the question!)

Thank you for this very interesting link !
I just watched this video, and I had to take breaks !

Great video, great editing explained in detail and great explanations from the author !

This video brings me answers but asks me others !

I will have to read all the comments to try to see more clearly.

In particular, if I understand correctly, the laser + the filters are not a real source of single photons and there are always interferences.

Conversely, a fluorescent lamp is not coherent and there is never any interference.

So the error that was diagnosed by the Oxford Physicist seems to be the use of filters to produce single photons.

On the other hand, the video doesn't say what would happen with "real" single photons (from very expensive sources in laboratories such as those of Alain Aspect) by increasing the distance between the first experiment and the second and, the case where applicable (here), at what distance an interference pattern is changed into a pattern without interference.

Anyone have the answer ?

Cordially,
Marilyn

 

[Dale]

at what distance an interference pattern is changed into a pattern without interference.

Anyone have the answer ?

The answer is that there is no "the" answer. Each source has its own coherence length. You will have to find that either experimentally or in the manufacturer's documentation.

 

[PeterDonis]

The demonstration is interesting because the intensity is reduced so much that there is only one photon in the apparatus at any time.

This illustrates one of the dangers of using pop science videos. As it is stated, the quoted sentence is false. What the actual math says is significantly different: "the intensity is reduced so much that the expectation value of photon number is less than one". Lasers emit coherent states, which are not eigenstates of photon number, so you cannot say "how many photons" are in the apparatus at a given time.

 

[PeterDonis]

if I understand correctly, the laser + the filters are not a real source of single photons

In the sense that this kind of source emits coherent states, which are not eigenstates of photon number, yes.

and there are always interferences.

No. With a laser + filters source, there are still conditions that have to be met to get an interference pattern (as opposed to just a diffuse spot on the detector). It doesn't always happen.

Conversely, a fluorescent lamp is not coherent

Not quite. A fluorescent lamp emits a coherent state of the EM field; it just has a much shorter coherence length than a laser. In fact, pretty much any common light source emits a coherent state of the EM field, since that kind of state is the closest thing in quantum field theory to a classical EM field state. The main difference with lasers is having a coherence length long enough to enable meaningful interference phenomena to be easily detected.

and there is never any interference.

More precisely, for most light sources it is very difficult to set up an experiment that will detect meaningful interference phenomena.

 

[WernerQH]

Conversely, a fluorescent lamp is not coherent and there is never any interference.

There is always interference, but the phase relations between the waves may not be stable enough to lead to visible interference fringes. They tend to be washed out. You can produce interference patterns also with a fluorescent lamp. If you pass its light through a Fabry-Perot filter, for example, the coherence length can be as long as $10^4 \lambda$ when the filter's bandwidth is $\Delta \lambda / \lambda = 10^{-4}$.

On the other hand, the video doesn't say what would happen with "real" single photons (from very expensive sources in laboratories such as those of Alain Aspect) by increasing the distance between the first experiment and the second and, the case where applicable (here), at what distance an interference pattern is changed into a pattern without interference.

I don't know what "real" single photons are. You always have to wait some time for a photon to be produced, although there are of course experiments where they are created very quickly. The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy $E = h\nu$. The propagation of light, however, is well described as a wave phenomenon (Maxwell's equations).

 

[PeterDonis]

There is always interference, but the phase relations between the waves may not be stable enough to lead to visible interference fringes.

If you don't observe any interference, you can't say there was interference, at least as far as experimental results are concerned.

If you are basing your claim that "there is always interference" on some theoretical model, please show your work.

 

[PeterDonis]

The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy $E = h\nu$.

This is only correct for certain kinds of emission and absorption processes, not all of them. It is not correct for emission of light a laser, for example, since, as has already been noted, the EM field state emitted by a laser is a coherent state, and a coherent state is not an eigenstate of photon number.

 

[DrClaude]

This is only correct for certain kinds of emission and absorption processes, not all of them. It is not correct for emission of light a laser, for example, since, as has already been noted, the EM field state emitted by a laser is a coherent state, and a coherent state is not an eigenstate of photon number.

Even though laser light does not have a defined number of photons, it doesn't change the fact that

The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy $E = h\nu$.

 

[PeterDonis]

Even though laser light does not have a defined number of photons, it doesn't change the fact that

Since a laser emits a coherent state, how is the quoted statement a correct description of the laser emission process?

 

[DrClaude]

Since a laser emits a coherent state, how is the quoted statement a correct description of the laser emission process?

Whatever emitted the light (atoms, dye molecules, quantum wells, ...) emitted it one photon at a time. The quantum state of the emitters is of course compatible with the fact that the photon number in the light is indeterminate.

 

[PeterDonis]

Whatever emitted the light (atoms, dye molecules, quantum wells, ...) emitted it one photon at a time. The quantum state of the emitters is of course compatible with the fact that the photon number in the light is indeterminate.

I'm not sure how these two statements are consistent, although of course any attempt to describe what the actual math says in ordinary language is going to be imperfect. But the quantum state of the emitters, if the EM field is in a coherent state, is a superposition of the two energy levels involved. That means we cannot say that one photon has been emitted, because whether or not an emission has occurred from any particular emitter is indeterminate.

What we can say is that the frequency of the laser light is well-defined (at least in the idealized case where we ignore things like finite line width), since (at least as I understand it) a coherent state (at least in the idealized case) is an eigenstate of frequency. We just can't say that that frequency is a frequency of "photons" since the frequency and photon number operators do not commute.

As above, at least part of this might just be unavoidable imperfections in translating the actual physics and its mathematical description into ordinary language. I don't think we're in disagreement about the main points of the actual physics.

 

[Dale]

Even though a coherent state is not an eigenstate of the number operator it is still correct that

The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy E=hν.

If you prepare a thousand identical coherent states and measure the energy in each then you will find that it varies, but always by an integer multiple (lumps) of $h\nu$. So I agree that it is not an eigenstate, but I don’t think that fact invalidates the statement by @WernerQH

 

[PeterDonis]

Even though a coherent state is not an eigenstate of the number operator it is still correct that If you prepare a thousand identical coherent states and measure the energy in each then you will find that it varies, but always by an integer multiple (lumps) of $h\nu$.

That's because measuring the energy is equivalent to measuring the photon number; the energy operator is just $h \nu$ (more commonly written $\hbar \omega$ in the literature) times the photon number operator. So of course you're always going to get an eigenvalue of photon number times $h \nu$ as the measurement result.

As I said in response to @DrClaude, I think at least part of the issue here is that we are trying to describe physics in vague ordinary language instead of math, and there is no translation from the latter to the former that is fully accurate. My preference is to avoid having to do that as much as possible, and when it is unavoidable, to make the translation as direct as possible, by talking specifically about the EM field state (a coherent state) or the operator being measured (energy, aka $h \nu$ times photon number) instead of using vaguer terms like "occurs in lumps".

 

[PeterDonis]

That's because measuring the energy is equivalent to measuring the photon number; the energy operator is just $h \nu$ (more commonly written $\hbar \omega$ in the literature) times the photon number operator.

Actually, even this is not quite right, since the Hamiltonian also includes the zero point energy of $h \nu / 2$. @Dale might have been indirectly referring to this when he said the result of the energy measurement always "varies" by multiples of $h \nu$. But the fact that the zero point energy is included in the total energy means that the total energy is not a multiple of $h \nu$, and that makes it even less appropriate to say that the energy in the EM field "always occurs in lumps" of size $h \nu$, since that ignores the zero point energy. Giving the specific state and the specific operator being measured avoids issues like this.

 

[Marilyn67]

Thank you all very much for your very informative answers !
I can't find these answers elsewhere.

I know this is all complicated, and as PeterDonis points out, everyday language can cause confusion.

It is not easy to develop an experiment like this because it requires a lot of theoretical knowledge, and to put it into practice, I imagine the work of precision and adjustment that it represents, and also its cost.

I don't know who can do this alone in his cellar !

Then, apparently, you have to know how to interpret the results which can be positive biases or negative biases !

It's really not easy, and all the people who work and think in this area have merit !

I need to reread all your answers, to review this video again, to take notes and to make a synthesis to see more clearly.

Cordially,
Marilyn

 

[vanhees71]

I don't know what "real" single photons are. You always have to wait some time for a photon to be produced, although there are of course experiments where they are created very quickly. The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy $E = h\nu$. The propagation of light, however, is well described as a wave phenomenon (Maxwell's equations).

A real single photon is a normalizable Fock state of exact photon number 1, i.e.,
$$|\Psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 k \sum_{\lambda=-1,1} A_{\lambda}(\vec{k}) \hat{a}^{\dagger}(\vec{k},\lambda) |\Omega \rangle,$$
where $\lambda \in \{-1,1 \}$ are helicities (polarizations) and $|\Omega \rangle$ is the photon vacuum state. It describes an (asymptotic) free photon, and $A_{\lambda}(\vec{k})$ are square-integrable functions such that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 k |A_{\lambda}(\vec{k})|^2=1.$$
Photons never behave like massless point particles but as a quantum state of the electromagnetic field, i.e., as a wave. Particularly all the discussion about coherence, coherence lengths and times, etc. also apply to these single-photon states, and the interference patterns that can be measured by using very many single photons are those of the corresponding classical waves, reflecting the probability distributions for registering a photon at a given time and position of the detector (or a photoplate).

It hasn't been easy to produce true single-photon states. Unfortunately that's reflected by the inaccurate statement that you just have to use a very dim light source like a laser, but that doesn't produce true single-photon states but coherent states that are not eigenstates of the photon number. A very "dim" coherent state consists mostly of the vacuum, and the distribution of photon numbers is Poissonian. Although you can make the average photon number as small as you like (also below 1!), you never have a true single-photon state but a superposition of all number states, and for such a "dim coherent state" it's most probable to find no photon at all but some small probability to find 1, and even smaller probability to find 2, etc. photons.

 

[Marilyn67]

Hello,

I just have one more question for you.

More precisely, for most light sources it is very difficult to set up an experiment that will detect meaningful interference phenomena.

I know that commercially available small cheap laser pointers have much lower beam quality than a laser like helium neon laser or argon laser for example.

I myself own a small green laser pointer, class IIIb, (<20mW), 532 Nm.

If nothing other than the source is changed in the experience previously described on the link,

https://www.physicsforums.com/threads/how-big-is-a-photon-interesting-video.1005059/

but by replacing the neon helium laser with this type of laser pointer, will interference be observed as easily ?
Is the coherence length sufficient to observe interferences in the experiment setup depicted in the video ?

In advance, thank you for your response.

Cordially,
Marilyn

 

[Marilyn67]

by replacing the neon helium laser with this type of laser pointer, will interference be observed as easily ?
Is the coherence length sufficient to observe interferences in the experiment setup depicted in the video ?

For me, yes

 

[PeterDonis]

by replacing the neon helium laser with this type of laser pointer, will interference be observed as easily ?

Probably not because the cheap laser pointer probably has a much shorter coherence length. It probably also has a wider frequency width in its beam (no laser is perfect so they all have some finite frequency width, but the narrower the width the easier it is to observe interference).

 

[Marilyn67]

Thank you @PeterDonis ,

Probably not because the cheap laser pointer probably has a much shorter coherence length. It probably also has a wider frequency width in its beam (no laser is perfect so they all have some finite frequency width, but the narrower the width the easier it is to observe interference).

Your answer is interesting.

I didn't expect laser pointers with such poor quality..!

It would be interesting to repeat the experiment by modulating the distance of the second path to have an order of magnitude of the distance from which the results change... (a few decimetres, a few centimeters perhaps...)

See you soon,
Marilyn

 

[PeterDonis]

I didn't expect laser pointers with such poor quality..!

"Poor quality" is relative. Cheap laser pointers aren't designed to help you run interference experiments.

 

[Marilyn67]

"Poor quality" is relative. Cheap laser pointers aren't designed to help you run interference experiments.

Yes, it's true
It's all the more interesting to try
(I've done this before with Young's slits, my laser pointer and a lens)

Cordially,
Marilyn

 

[tech99]

At school I did the experiment using a sodium flame, before the invention of the laser. We passed the light through a slit to create a degree of coherence.

 

[Marilyn67]

Hello @tech99 ,

At school I did the experiment using a sodium flame, before the invention of the laser. We passed the light through a slit to create a degree of coherence.

Tell you about an experience with Young's slots ?
From what I understood at the beginning of the discussion, it's very easy to get visible interference with Young's slits because the path differences are very small at the point of impact.

I guess in the video setup a sodium flame would have given a diffuse pattern, right ?

Cordially,
Marilyn

 

[tech99]

There is an experiment shown on this video using sunlight as the source.

The light is passed first through a slit to provide spatial coherence. It is just a question of making the geometry so that the path length differences from each edge of the first slit to the double slits are much less than a wavelength. The pattern is very similar to the laser result, but not quite as sharp. The frequency coherence issue causes the sunlight to give coloured fringes. We used sodium light, which is sufficiently monochromatic to give clear fringes.

 

[Marilyn67]

Thanks @tech99 for your response,

I ask myself 2 questions that you may find silly :

It is just a question of making the geometry so that the path length differences from each edge of the first slit to the double slits are much less than a wavelength.

1/ How to obtain such precision when we know that the dilation due to temperature changes and that the vibrations present in the environment are of an order of magnitude well above 1/2 micron ?

2/ Why such precision for Young's slit experiment ?
If the difference in path length from each edge of the first slit to the double slits is even a few percent, isn't the interference pattern just shifted across the screen to the right or left to compensate this difference ?

Cordially,
Marilyn

 

[tech99]

The long triangle formed by the rays from each edge ensures that close tolerances are unnecessary. For instance, the (height - hypotenuse) is easily made much less than a micron. We do not require great precision.

 

[Marilyn67]

Ok, I understand the "how" but not the "why".

Indeed, to have interfringes "i" visible to the naked eye on the screen, a large length "D" and a small distance "a" between the two slits (i = λ.D/a) are required and here, "D" >> "a".

But we can imagine a much shorter triangle, and observe smaller interfringes i thanks to the equivalent of a microscope (ok it's more complicated, but a path difference less than the wavelength is not "fundamentally necessary", or is there another reason ?)

Is there a fundamental reason, or is it just a convenience to have wide interfringes ?

 

[tech99]

Ok, I understand the "how" but not the "why".

Indeed, to have interfringes "i" visible to the naked eye on the screen, a large length "D" and a small distance "a" between the two slits (i = λ.D/a) are required and here, "D" >> "a".

But we can imagine a much shorter triangle, and observe smaller interfringes i thanks to the equivalent of a microscope (ok it's more complicated, but a path difference less than the wavelength is not "fundamentally necessary", or is there another reason ?)

Is there a fundamental reason, or is it just a convenience to have wide interfringes ?

When we used sodium light we used a traveling microscope (which we used to call a cathetometer) to observe the fringes. There is a trade-off between fringe spacing and brightness, and a laser is much brighter than a sodium source.