Young's slits with incandescent light source

[PeterDonis]

I imagine that, as electrons are fermions and photons are bosons, the effects of any real aperture could be different.

Electron diffraction experiments, and AFAIK electron double slit experiments, have been done, and AFAIK they show the same general interference phenomena. There might be differences in the fine details, yes.

 

[gentzen]

Unless the source is emitting Fock states (which it isn't in any double slit experiment that has been done to date), there are not photons passing through the experiment. There is light (the quantum electromagnetic field) passing through the experiment, but you can't usefully describe it as photons. It will get detected as individual impacts on the detector screen, but that does not mean it is photons before then. It's not.

Would everyone's answers change if I said 'electron' instead of 'photon'?

Yeah, good question. I didn't give an answer, but I guess my answer would change. But why? I think it is easy to produce single electron states in practice. Or more precisely, states which can be described by single electron states with excellent accuracy. Do I mean the same thing by "single electron state" as PeterDonis means by "Fock state"? Not exactly, because a "Fock state" could also be a state with two electrons, or three electrons. What is not allowed (for a Fock state) is a superposition of a two electron state and a three electon state, i.e. superpositions between states with different numbers of photons or electrons.

So in 'double slit' experiments using electrons, what happens can be described with excellent accuracy by single electron states interfering with themselves.


However, there is also a different perspective: Weak light can often be described to excellent accuracy by a superposition of a zero photon state with a one photon state. And similar, there seems nothing wrong with seeing the single electron states mentioned above as superpositions between a zero electron state and a one electron state.
I guess the point is that I use non-relativistic QM for my mental picture of electrons, but a QFT description for my mental picture of photons. The crucial difference is that the QFT picture is a variable number of particles picture, while the non-relativistic QM picture is a fixed number of particles picture.

 

[Lord Jestocost]

Throwing another pebble in the pond...

Would everyone's answers change if I said 'electron' instead of 'photon'?
As I understand it, we get just the same interference effects in 'double slit' experiments using electrons. Is that just a coincidence? Maybe the effects aren't the same - has anyone done other optics-like experiments with electrons? There must be lots of diffraction patterns out there, for example.

There are also double-slit experiments with neutrons (“Single- and double-slit diffraction of neutrons” by A. Zeilinger, R. Gähler, C. G. Shull, W. Treimer, and W. Mampe, Rev. Mod. Phys. 60, 1067, 1988).

The outcomes of all these experiments confirm the predictions of standard quantum theory.

 

[sophiecentaur]

From this link: (QSNP)
"Fock state is a quantum state that contains a precise number of non-interacting, identical particles,"

Does this imply that a beam of electrons, which will all interact with each other, will not have a Foch state? It confuses me (not a bit).

Also, diffraction of electrons in electron optic equipment is hard to understand because of the need for phase information - phase of what 'wave'`? We've all seen electron diffraction patterns in school demos but what actually is going on there?

 

[sophiecentaur]

I also have a problem with treating all photons arriving from a (series of?) emissions as if they have perfect coherence. What sort of oscillating equipment could have zero bandwidth? There has to be some degree of decoherence so, apart from what classical wave theory tells us, what would be the effect of finite bandwidth on the resulting pattern? (filling in of nulls etc. which is also affected by slit widths)

 

[PeterDonis]

there seems nothing wrong with seeing the single electron states mentioned above as superpositions between a zero electron state and a one electron state.

It is possible to construct what are called coherent states (which is where we get the idea of a very weak light source emitting a superposition of states with zero and nonzero photons) for fermions, but I don't know if they have all of the same properties. But more importantly, I don't think those are what electron sources like a cathode ray tube emit. I don't think there are meaningful superpositions of different electron numbers in those states; at any point in space, at any given time, I think the electron state will be an eigenstate of electron number. (Btw, I was using "Fock state" in general to mean any such state, not just the ones with eigenvalue 1.)

Talking about "states" in the context of relativity creates problems because, heuristically, trying to adopt such a viewpoint in QFT requires choosing a particular frame, and the "states" you get then become artifacts of that particular choice of frame. For experiments where relativistic effects are negligible, this works all right, you can just pick the rest frame of the lab and pretend you're just doing NRQM in that frame (this even works for many experiments involving photons, such as the double slit--you just formally write down the same sorts of states you'd write down for any non-relativistic uncharged massless spin 1 particle, and ignore the fact that there is no rigorous mathematical derivation of any such thing from the underlying QFT). That seems to be more or less what we're doing in this thread.

 

[gentzen]

Does this imply that a beam of electrons, which will all interact with each other, will not have a Foch state? It confuses me (not a bit).

For the typical electron beam in an electron microscope, the direct interaction between the electrons is negligible. And even in electron beam writers, which can use much higher currents, the direct (Coulomb) interaction is rarely ever a problem.

There is indirect interaction in the sample, because the damage, charging, or other changes to the sample induced by earlier electrons do influences electrons which hit the sample later. But I wouldn't call this "all interact with each other". And it doesn't really impact the quantum description, at least not conceptually.

Also, diffraction of electrons in electron optic equipment is hard to understand because of the need for phase information - phase of what 'wave'`?

I guess I see your conceptual problem: The single electron state gets described by a time-harmonic wave, so in a certain sense an infinitely spatially extended and timeless description. It is a solution of a 3D Schrödinger (or Dirac) equation. One way to get used to this description is to study the scattering of a plane wave at a planar potential barrier/step. Some confusing aspects of this description already occur in this simple scenario, and can also be resolved. Sometimes these aspects still confuse me, while trying to understand more complicated multislice or Bloch-wave computations. Sometimes I have to go back to this simple scenario to clear my conceptual confusions.

Also confusing is the interaction of such an "incoming electron wave" with the phonons, plasmons, inner shells, ... in the sample. However, I once decided to not reply to [...] so I guess I should stick to that. So now I saved that part of my reply locally. It contained references to
https://github.com/elena-pascal/Thesis
https://github.com/EMsoft-org/EMsoft
Budhika G. Mendis
but was not easy to understand. It did describe my "rationalizations" of what they are doing. You could argue that it was kind of "original research".

 

[PeterDonis]

From this link: (QSNP)
"Fock state is a quantum state that contains a precise number of non-interacting, identical particles,"

I don't think it's necessary for the particles to be non-interacting for there to be eigenstates of particle number. But it is true that there are plenty of traps for the unwary lurking here, such as:

https://en.wikipedia.org/wiki/Haag's_theorem

Does this imply that a beam of electrons, which will all interact with each other, will not have a Foch state?

I think that, as I said in #36, at any point in the space occupied by the beam, at any given time, the state will be an eigenstate of electron number. Whether the term "Fock state" is strictly correct, given that yes, electrons interact because they're charged, I'm not sure. But I think that in any case the state is very different from a coherent state of light, or for that matter an incoherent state like that emitted by an incandescent source. The differences just don't happen to matter for the double slit experiment.

phase of what 'wave'`?

The electron's wave function.

 

[flippiefanus]

You seem to be implying that classical field theory cannot be used for any phenomena.

No, I said it cannot explain all phenomena. There are scenarios where classical electromagnetism is perfectly adequate. There is also scalar and vector diffraction theory.

 

[flippiefanus]

Throwing another pebble in the pond...

Would everyone's answers change if I said 'electron' instead of 'photon'?
As I understand it, we get just the same interference effects in 'double slit' experiments using electrons. Is that just a coincidence? Maybe the effects aren't the same - has anyone done other optics-like experiments with electrons? There must be lots of diffraction patterns out there, for example.

For a while people amused themselves by producing electron beams (in electron microscopes) with vortices in their fields, similar to how it is done in optical beams. The results are very similar for electrons and photons.

 

[PeterDonis]

For a while people amused themselves by producing electron beams (in electron microscopes) with vortices in their fields, similar to how it is done in optical beams.

Can you give a reference?

 

[flippiefanus]

Can you give a reference?

There are lots, see for example:
Lloyd, S. M., Babiker, M., Thirunavukkarasu, G., & Yuan, J. (2017). Electron vortices: Beams with orbital angular momentum. Reviews of Modern Physics, 89(3), 035004.

 

[Cthugha]

The stabilisation of the two source makes this scenario much more complicated and thus cannot be compared with the single source scenario. This strictly speaking becomes a multiple-photon effect.

No, why should it? In a standard laser, you also cannot distinguish between photons coupled out of the resonator at some moment and photons coupled out exactly one round trip later. One would also not consider this a multi-photon effect. It just matters whether all the excitations are within a single mode (and therefore indistinguishable). The light field and the emitter are still entangled during the coherence time of the light field, which means that the excitation of the entire system is in a superposition of being in the field or in the emitter. It does not change anything if you spread a single excitation over a single light field and two emitters instead.

Yes, fields indeed, but unless one introduces the notion of quantised fields (photons are not classical point particles), one would just be doing classical field theory that cannot explain all phenomena. It is not sufficient to just use probability amplitudes without considering quanta. There are experiments (violation of Bell's inequality) where the quantised nature of the EM field plays an important role.

Replacing the intensity with a probability amplitude that is normalised under $L^2$ (modulus square integrates to 1), one implies a quantum of the field, i.e., a photon. So one can just as well call it by that name. Reading through the description of the calculation procedure, I get the feeling that it is still based on photons, just without calling it that.

This is a thread about the double slit, which means first order-coherence, which mathematically just involves a single creation and annihilation operator at different positions (or times for first-order temporal coherence). It is not sensitive to operator ordering and therefore not "quantum".
Of course there are effects that require quanta, but they are completely irrelevant to this thread and the double slit.
Also, assuming that a probability amplitude that normalizes to 1 implies a quantum of the field is fundamentally misguided. This only applies when you put the quantum in already, so you calculate the probability amplitude for detection of a single photon using a single-photon sensitive detector assuming that a single photon was emitted. Probability amplitudes are completely general and could be used for, e.g., quadrature detection in homodyne detection or detection of currents in detectors that are not singe-photon sensitive. One can introduce photons, but there is no need to.

 

[flippiefanus]

In a standard laser, you also cannot distinguish between photons coupled out of the resonator at some moment and photons coupled out exactly one round trip later. One would also not consider this a multi-photon effect.

A laser produces a multi-photon state. Not sure what you were trying to say here. The state of two lasers that are coupled cannot be represented by a tensor product state. If you had a tensor product state you would not see interference. So clearly it becomes a multi-photon process. There is no way around it.

It is not sensitive to operator ordering and therefore not "quantum".

That is not what it means to be quantum. One can have a quantum formalism that does not even use operators. You can also have non-commuting operators in classical theories.

Also, assuming that a probability amplitude that normalizes to 1 implies a quantum of the field is fundamentally misguided

They why normalise? The whole point of the normalisation is to maintain the probability of measuring a single quantum.

Probability amplitudes are completely general and could be used for, e.g., quadrature detection in homodyne detection or detection of currents in detectors that are not singe-photon sensitive.

In homodyne detection you still measure photons. The probability distribution is normalised only because the probability amplitudes are normalised. In the end it some down to the same thing.

In general, I get the impression that you are missing the point. It is not that one cannot use classical theory to understand the double slit experiment. The point is that one can also use a quantum theory to study what happens in the double-slit experiment. Quantum theory is the more general one. Sometimes it is more complicated to use, but that is not always the case. The statement that there are some forms of light that cannot be modelled in terms of photons is nonsense. All forms of light can be represented as quantum states and those quantum states can be expressed as superpositions of Fock states.

 

[tech99]

From this link: (QSNP)
"Fock state is a quantum state that contains a precise number of non-interacting, identical particles,"

Does this imply that a beam of electrons, which will all interact with each other, will not have a Foch state? It confuses me (not a bit).

Also, diffraction of electrons in electron optic equipment is hard to understand because of the need for phase information - phase of what 'wave'`? We've all seen electron diffraction patterns in school demos but what actually is going on there?

I can remember at school doing the double slit experiemnt using sodium light from a sodium salt placed in a Bunsen flame together with an additional slit. This arrangement gave sufficient spacial and temporal coherence. The resulting fringes were observed with a travelling microscope, or cathetometer, and I think we observed them as virtual images, ie directly on the retina, but I am not sure now about this.

 

[TheHutch]

Perhaps I can summarise my conceptualisation of all this:

• Light sources, especially thermal, are messy, and usually should be modelled as generating a superposition of non-interacting identical particles (photons). Fundamental quantum uncertainty gives rise to this superposition and cannot be removed.
• Photons with different wavelengths are not identical, but still do not interact, so the whole can be modelled as a combination of single wavelength fields (sorry if that is the wrong term). Black body radiation, as per Plank, gives a probability distribution over these single wavelength fields.
• The superpositions, however, are ultimately superpositions of individual photon wave functions. It is these individual, non-interacting wave functions that give rise to interference patterns, i.e. the varying probabilities that photons get detected at particular points in space. (Loosely, a photon interferes with itself only - the Dirac quotation).
• Phases of photons do not come into it, it is the relative phases of the different paths making up the particle wave functions that count.
• Certain experimental set-ups can give rise to situations where photons cannot be regarded as non-interacting (e.g. entangled photon pairs?), requiring a more involved treatment.

Is that a fair summary?

 

[gentzen]

• The superpositions, however, are ultimately superpositions of individual photon wave functions. It is these individual, non-interacting wave functions that give rise to interference patterns, i.e. the varying probabilities that photons get detected at particular points in space. (Loosely, a photon interferes with itself only - the Dirac quotation).
• Phases of photons do not come into it, it is the relative phases of the different paths making up the particle wave functions that count.
• Certain experimental set-ups can give rise to situations where photons cannot be regarded as non-interacting (e.g. entangled photon pairs?), requiring a more involved treatment.

To this part I can mostly agree. Still, I can see other members here disagree with me, because they would use the word photon differently. But I don‘t think that such a fight about words is productive, as long as the underlying intuition is basically fine.

• Light sources, especially thermal, are messy, and usually should be modelled as generating a superposition of non-interacting identical particles (photons). Fundamental quantum uncertainty gives rise to this superposition and cannot be removed.
• Photons with different wavelengths are not identical, but still do not interact, so the whole can be modelled as a combination of single wavelength fields (sorry if that is the wrong term). Black body radiation, as per Plank, gives a probability distribution over these single wavelength fields.

I don‘t like this part, especially the phrasing and the concrete words.

„superposition of non-interacting identical particles (photons)“: An incoherent sum is a better description, in case that is what you wanted to say.
„modelled as a combination of single wavelength fields (sorry if that is the wrong term“: And also here, incoherent sum might be better, in case that is what you mean.

Additionally, saying that thermal light sources would be messy, confuses different levels: Their (instrumentalistic) modelling is not messy, only if you insist on having some „ontological“ image in your head, you get a messy looking image. But the word „messy“ should be reserved for stuff which actually makes real trouble experimentally or for the theoretical/computational treatment.

 

[PeterDonis]

A laser produces a multi-photon state.

A laser produces a coherent state. The expectation value of photon number in the state depends on the intensity; for low enough intensity it can be less than one.

The whole point of the normalisation is to maintain the probability of measuring a single quantum.

No, the point of normalization is to make sure that all probabilities sum to one. There can be others besides the probability of measuring exactly one quantum.

 

[PeterDonis]

• Light sources, especially thermal, are messy

Ok there.

• and usually should be modelled as generating a superposition of non-interacting identical particles (photons)

No, this is what we have not been doing. I don't think, for example, that this is a good description of a coherent state. Most of the rest of your bullets just build on this error.

• Phases of photons do not come into it, it is the relative phases of the different paths making up the particle wave functions that count.

This I think is true for the double slit experiment, but it contradicts many of your other bullets.

 

[Cthugha]

A laser produces a multi-photon state. Not sure what you were trying to say here. The state of two lasers that are coupled cannot be represented by a tensor product state. If you had a tensor product state you would not see interference. So clearly it becomes a multi-photon process. There is no way around it.

This statement is unfortunately incorrect. Already if you put a single coherent beam into a beam splitter, the output state considering both outpuut ports is a tensor product state. You trivially see interference between these two output beams. You need, e.g., a mixed state, where the off-diagonal matrix elements of the density matrix in Fock state are erased to remove interference.

That is not what it means to be quantum. One can have a quantum formalism that does not even use operators. You can also have non-commuting operators in classical theories.

You can make up your own definition of quantum, but then you disagree with the state of the art in the field. Identifying non-classicality in light fields is a relevant and still timely topic and it is absolutely clear that one must go beyond first-order coherence to identify non-classicality. See, e.g., this peer-reviewed paper (Phys. Rev. A 109, 022216 (2024)) that clearly states the common consensus:
"Consequently, any photonic test whose measurements constitute only first-order coherence can be simulated with the classical theory of coherence". This is so important that I typically started my own quantum optics lectures with this point in the last few years.

They why normalise? The whole point of the normalisation is to maintain the probability of measuring a single quantum.

No. Of course not. You normalize to get a probability of one when summing over all possible outcomes. This is by no means limited to single photons. When having a laser light field of 10 photons and 20 detectors, you would sum over a lot of probability amplitudes for many different numbers of detection events and their distribution among the detectors.

In homodyne detection you still measure photons. The probability distribution is normalised only because the probability amplitudes are normalised. In the end it some down to the same thing.

I am somewhat puzzled. Your statement makes no sense from a formal point of view. I have no idea what "the probability distribution" even refers to in this case and which probability amplitude you assume to be normalised. Fields? Photon numbers? Spatial distributions of individual photons?
Homodyne detection yields one of the field quadratures, which loosely correpsond to either the sine or cosine component of the field. These two are connected by an uncertainty relation and cannot be measured precisely at the same time. More precisely speaking, homodyne detection yields the projection of the Wigner function of the state of the light field on a projection axis given by the relative phase between the signal and the local oscillaator (if it is well defined). You can get a POVM for getting probability amplitudes for certain quadratures, but this is as always non-trivial in continuous variable quantum optics.

In general, I get the impression that you are missing the point. It is not that one cannot use classical theory to understand the double slit experiment. The point is that one can also use a quantum theory to study what happens in the double-slit experiment. Quantum theory is the more general one. Sometimes it is more complicated to use, but that is not always the case. The statement that there are some forms of light that cannot be modelled in terms of photons is nonsense. All forms of light can be represented as quantum states and those quantum states can be expressed as superpositions of Fock states.

I disagree. Let us revisit the very first post in this thread:

As far as I can see, real light (from an incandescent source) consists of a stream of uncorrelated photons, with entirely random phase relationships, so these short wave trains can only be the individual photons, and the only way interference patterns can emerge is if these photons interfere only with themselves, and not with each other. The purpose of the single slit is simply to constrain the spatial origin of photons making the interference pattern crisper. In other words, the interference is a quantum phenomenon of individual photons, and not some aggregate wave nature of light.

You are putting up a strawman. Nobody here ever stated that " there are some forms of light that cannot be modelled in terms of photons". You just made that up.
The very question raised in the initial question of this thread is whether one MUST consider double slit interference as a quantum phenomenon. The answer is a clear no.

You seem to intend to switch to other topics, but this would be more or less hijacking of this thread, so I strongly suggest to open a separate discussion for other questions not related to this thread.

 

[Cthugha]

Light sources, especially thermal, are messy, and usually should be modelled as generating a superposition of non-interacting identical particles (photons). Fundamental quantum uncertainty gives rise to this superposition and cannot be removed.

Mostly alright (although thermal sources may feel treated unfairly ;) ).
I tend to disagree about the uncertainty part. Especially for thermal light, this is a consequence of thermal equilibrium between the photon gas and the walls of the emitter. For thermal light, this is not really due to an uncertainty relation. Probably this is more of an issue of what you mean exactly by uncertainty.

Photons with different wavelengths are not identical, but still do not interact, so the whole can be modelled as a combination of single wavelength fields (sorry if that is the wrong term). Black body radiation, as per Plank, gives a probability distribution over these single wavelength fields.

There are some non-trivial points here because in the quantum-optics sense the term "photon" is usually understood to be a single excitation of the light field, while it is simultaneously also common to refer to a decomposition of the light field modes into monochromatic frequencies as"photon modes". If one talks about photon number distribution, one considers the first meaning. If one talks about "photons with different wavelengths" one means the latter.
Indeed, black body radiation may give a distribution among these frequencies. However, even when considering just a single of these monochromatic modes, one will encounter a typical photon number distributions within this mode (Bose-Einstein distribution) in the sense of the first meaning. Also, a single photon in the first meaning can be polychromatic and does not necessarily have a single wavelength. In practice, it never has because one would get an infinitely long single photon-wavetrain this way due to Fourier uncertainty.

The superpositions, however, are ultimately superpositions of individual photon wave functions. It is these individual, non-interacting wave functions that give rise to interference patterns, i.e. the varying probabilities that photons get detected at particular points in space. (Loosely, a photon interferes with itself only - the Dirac quotation).

This is problematic in terms of terminology because in the strict sense, there are no photon wavefunctions. The detection of a photon destroys it, so there are no eigenstates in the sense of ordinary QM without quantum field theory. I would rather say that for each photon mode (e.g. in the spectral decomposition you mentioned above) you will get a field (which will roughly be the QFT equivalent of the wave function you think about) that depends on the occupation number of that mode. In some limits these may look like individual waves (the individual photons you consider), but in most cases they do not.

Phases of photons do not come into it, it is the relative phases of the different paths making up the particle wave functions that count.

Okay.
In quantum optics, single photons do not have any well defined phase due to uncertainty relations, but it is a correct statement to state that in a well-prepared double slit experiment, the arising phases are of geometrical origin. In fact, in the very first double slit experiment, Young used a single pinhole to make sure that only the different paths matter.

Certain experimental set-ups can give rise to situations where photons cannot be regarded as non-interacting (e.g. entangled photon pairs?), requiring a more involved treatment.

True in terms of what you intend to express, but a bit loose in terms of terminology. However, for a single double slit more involved treatments are never needed. These only become relevant when one has more than one detector and non-trivial fields, e.g. in the double slit quantum eraser or things like that. One would still consider these photons as non-interacting, but entangled (which implies "not independent", which has a slightly different meaning than non-interacting - if you considered them as interacting you would expect some kind of interaction potential or something like that).

 

[sophiecentaur]

but I am not sure now about this.

You can use the eye to see an image in this way but it takes a bit of skill because the arrangement is pretty unstable. I guess a demo could have involved what is basically a projection directly on he retina; perhaps the image would be bright and it could be very much magnified. Would a class of students all be able to work it, though?

A laser produces a coherent state.

I have a problem with the use of the word 'coherent' as if it's absolute. As far as I can see, the output of a laser can be treated just the same as any oscillator by saying it is equivalent to a continuous oscillation which is modulated by noise. The bandwidth is limited by the geometry and uniformity of the etalon and the 'lasing ' medium and the following optics. Why would a Foch state be absolute? There must be an uncertainty / finite bandwidth. "One photon at a time' is a very artificial concept because the two slits will only perform when there's some form of reference to establish the pattern. If the one photon hits just one sensor then that experiment shows nothing.

 

[PeterDonis]

I have a problem with the use of the word 'coherent' as if it's absolute.

"Coherent state" is a specific term in QFT for a specific kind state:

https://en.wikipedia.org/wiki/Coherent_state

Of course a real laser will not emit an exact coherent state, but for most purposes one can ignore the imperfections and model the state it emits as a coherent state.

Why would a Foch state be absolute?

A Fock state is a very different kind of QFT state from a coherent state:

https://en.wikipedia.org/wiki/Fock_state

The differences are not really observable in a double slit experiment, but they certainly are in other experiments.

 

[sophiecentaur]

Nobody here ever stated that " there are some forms of light that cannot be modelled in terms of photons"

There are some interpretations of a phenomenon that are harder to use than others. It's true that most people (everybody, even?) cannot always reconcile the two approaches at all times. This thread seems to be demonstrating that.

Using one approach as a check on the other is probably the best way to go. It's as well to remember that any experiment will always involve equipment that uses classical equipment which has noise, finite bandwidth and physical size. Plus it's necessary to have a proper appreciation of the properties of photons; they just do not behave in an intuitive way.

 

[sophiecentaur]

The differences are not really observable in a double slit experiment, but they certainly are in other experiments.

I guess it's necessary to have actually used the idea of a Foch state in a real study so you can get familiar with it. That comment helps a bit - cheers.

 

[Cthugha]

I have a problem with the use of the word 'coherent' as if it's absolute. As far as I can see, the output of a laser can be treated just the same as any oscillator by saying it is equivalent to a continuous oscillation which is modulated by noise. The bandwidth is limited by the geometry and uniformity of the etalon and the 'lasing ' medium and the following optics. Why would a Foch state be absolute? There must be an uncertainty / finite bandwidth. "One photon at a time' is a very artificial concept because the two slits will only perform when there's some form of reference to establish the pattern. If the one photon hits just one sensor then that experiment shows nothing.

It was the great achievement of Roy Glauber (and also George Sudarshan) to come up with a framework that gives an absolute meaning to the word coherence. First-order coherence (as seen in the double slit or a Michelson interferometer) is indeed bandwidth-limited. First-order temporal coherence (g1(tau)) is just the Fourier transform of the power spectral density of your light field. No matter what the emitter is: if you do the same kind of spectral and spatial filtering to it, it will have the same first-order coherence properties - it may just become very faint along the way.

Glauber made the term coherence absolute by introducing coincidence count statistics involving two detectors - akin to the famous Hanbury Brown Twiss experiment, where just split a light field using a beam splitter and place two detectors at the exit ports. You just check how often one gets simultaneous detections. Here, you have two effects. On the one hand, the detection of a photon reduces the photon number of the light field, so the probability to detect another one should go down. On the other hand, the light source may be noisy and the photon number may be fluctuating. Then, the detection a photon flags the the instantaneous intensity will be above average and the probability to detect another one should go up.

For Fock states, the first effect wins: You just have a single photon, so that if one detector clicks you know that there is no other photon inside the light field, which could go to the other detector and the detectors will never show simultaneous clicks. For thermal light, the second effect wins. For coherent light both cancel exactly, so the probability to detect another photon just stays as it is. Obviously, this means that the light field is extremely robust to losses. This is an absolute definition of coherence. However, it depends only on the photon number distribution and is not related to phase at all (there are other coherence quantifiers such as quantum coherence that can do this). Therefore, I would like to emphasize that this absolute definition of coherence is not at all related to what one measures in the double slit or a standard Michelson interferometer - although it is of course not a coincidence that these coherent light fields usually show long coherence times and lengths.

Along similar lines, I also do not see a problem with "one photon at a time". Of course any real Fock state emitted will be some kind of pulse. You can send the emission into an HBT setup and check for how long the detection of a second photon will be suppressed. If you repeatedly send out these pulses with a waiting time that exceeds this time scale, where you know that only one photon can be emitted, you can repeatedly send these pulses through the double slit and wait for the interference pattern to emerge. It will just take incredibly long.