2-Slit: How do we know photon guns only produce single photons?

[julianb]

TL;DR Summary

In the "single photon" 2-slit experiment, one of the basic assumptions is that a photon gun only produces whole photons. But if a wave is emitted and the photon detector only detects whole-number photons, how do we know there's no energy unmeasured and thus unaccounted for?

In the "single photon" 2-slit experiment, I understand one of the basic assumptions is that a photon gun only produces whole photons, or, at least, it's assumed that only whole photons (or waves with whole photon level of energy) make it through the 2 slits.

But since a photon detector only detects whole-number photons, how do we know there's no energy unmeasured and thus unaccounted for?
How do we know that there's more light coming through the 2 slits than just discreet whole numbers of photons?

Subsequently, is it possible that interference in the "single photon" 2-slit experiment is coming from wave(s), even if the threshold of light energy that we can observe through experiment is in whole photons only?

I'm hoping for an answer more than, "because quantum physics tells us that light only comes in photon packets". (Setting aside that this is circular logic, please see this as an attempt to educate, not merely rebut.)
Me: Longtime hobbyist, first-time public poster. I've taken undergrad-level E&M Physics and on-and-off amateur research with online materials.

 

[Dale]

TL;DR Summary: In the "single photon" 2-slit experiment, one of the basic assumptions is that a photon gun only produces whole photons. But if a wave is emitted and the photon detector only detects whole-number photons, how do we know there's no energy unmeasured and thus unaccounted for?

I understand one of the basic assumptions is that a photon gun only produces whole photons,

It isn't quite correct as stated. The assumption is that every light source only produces whole photons. This was discovered with experiments using the photoelectric effect. We know that there is no unaccounted for energy by measuring the KE of the photo electrons.

 

[julianb]

It isn't quite correct as stated. The assumption is that every light source only produces whole photons. This was discovered with experiments using the photoelectric effect. We know that there is no unaccounted for energy by measuring the KE of the photo electrons.

Thanks, Dale.

So, the photoelectric effect is what's used to detect photons in the 2-slit experiment, correct?
Doesn't the original question apply to that, as well? (I don't want to sound rude, but, this is exactly what I was not hoping for - restating as fact the assumption I'm questioning. But in good faith, I'll continue. It's possible I may not have communicated my original point well, or that there was a bit of a misreading, so, please bear with me as we resolve any miscommunication.)

Given: the photoelectric effect requires a threshold amount of energy to excite an electron to be emitted.
Assumed: we only detect one electron being emitted in a hypothetical photoelectric effect (for sake of example, let's assume the basic setup of EM radiation hitting a metal plate).
Then, how do we know the EM radiation hitting the metal plate is only 1 photon's worth of energy?
How do we know that there isn't, say, a wave of 1.5 photons of energy hitting the place, but only 1 photon worth can be detected, since it takes a full photon to emit an electron from the metal plate?

tl;dr Yes, I'm absolutely asking if the single-photon conclusion of the PE is wrong.
And the 2-slit "single photon" experiment resulting in interference patterns is the evidence why I'm asking.

 

[PeterDonis]

the "single photon" 2-slit experiment

Can you give a specific reference?

The term "photon" is often misused, and the term "single photon 2-slit experiment" could mean two very different things:

(1) You have a typical light source, such as a laser, which emits a coherent state, but the intensity of the state is so low that the expectation value of photon number inside the apparatus is (approximately) one. Note that this is an expectation value only--the coherent state is not an eigenstate of photon number and is not properly described as containing "one photon".

(2) You have a very extraordinary light source, which could be made but it's very difficult, that emits a one photon Fock state, i.e., a state which is an eigenstate of photon number, with eigenvalue one. (There are also technical details about momentum and polarization, which we can gloss over here by assuming that they're set to predetermined values.)

The answers to the questions you are asking will be different for those two cases. That's why we need a specific reference, to know which case you are referring to. I strongly suspect it's (1), because I don't think any actual 2-slit experiment has been done under the conditions of (2), but I could be wrong.

 

[PeterDonis]

The assumption is that every light source only produces whole photons.

That assumption is only valid for case (2) in my post #4. It is not valid for case (1) since the state emitted from the source is not an eigenstate of photon number. It doesn't even have to have an expectation value of photon number that is an integer. In case (1), the only valid statement you can make about "photons" is that, with an appropriate detector, you will detect only whole photons. But that does not justify any such claim about the state prior to detection.

 

[PeterDonis]

I'm hoping for an answer more than, "because quantum physics tells us that light only comes in photon packets".

Quantum physics doesn't actually say that. It says that if we turn down the intensity of any light source, the light we detect will eventually resolve into discrete packets called "photons". But it does not say that the state of the light prior to detection is always discrete packets called "photons". That is only true for Fock states, which are very rare because they're so hard to produce.

 

[julianb]

Can you give a specific reference?

The term "photon" is often misused, and the term "single photon 2-slit experiment" could mean two very different things:

(1) You have a typical light source, such as a laser, which emits a coherent state, but the intensity of the state is so low that the expectation value of photon number inside the apparatus is (approximately) one. Note that this is an expectation value only--the coherent state is not an eigenstate of photon number and is not properly described as containing "one photon".

(2) You have a very extraordinary light source, which could be made but it's very difficult, that emits a one photon Fock state, i.e., a state which is an eigenstate of photon number, with eigenvalue one. (There are also technical details about momentum and polarization, which we can gloss over here by assuming that they're set to predetermined values.)

The answers to the questions you are asking will be very different for those two cases. That's why we need a specific reference, to know which case you are referring to. I strongly suspect it's (1), because I don't think any actual 2-slit experiment has been done under the conditions of (2), but I could be wrong.

Thanks, PeterDonis,

Either. And this definitely gets to the heart of the question and why I ask.
Your last sentence:
"... I don't think any actual 2-slit experiment has been done under the conditions of (2) ..."

If (2) doesn't exist, and if we're relying on (1) to show that "single photons" have interference patterns detected, then is it possible we're not accounting for light energy that's in between the divider with the 2 slits and the photon detector at the far-end?

(But if (2) does exist, then I'd want to know how we would know that said extraordinary light source actually emits light at eigenvalue one.)

 

[PeterDonis]

If (2) doesn't exist

I assume you mean, if a 2-slit experiment under case (2) has not been done. Photon sources that fall under case (2) certainly do exist and experiments have been done with them that confirm the predicted properties of Fock states under quantum field theory. I just don't know if any of those experiments were 2-slit experiments.

if we're relying on (1) to show that "single photons" have interference patterns detected

We aren't. In case (1), the term "single photons" is meaningless except in reference to the discrete detection events. You cannot use case (1) to justify a claim like "single photons have interference patterns", because the state prior to detection is not properly described as "single photons" to begin with. And as far as I know, no actual physicist in an actual textbook or peer-reviewed paper makes such a claim as you are implying in what I just quoted above. If you think one has, then you need to give a specific reference.

is it possible we're not accounting for light energy that's in between the divider with the 2 slits and the photon detector at the far-end?

We don't measure the energy at the slits, so it's meaningless to ask whether energy is conserved between the slits and the detector. In QM you can't make claims about conservation laws unless you measure the conserved quantities at the appropriate points.

We can do experiments where we measure the energy emitted by the source and the energy detected by the detector, and in those cases, we find that energy is conserved.

 

[julianb]

Quantum physics doesn't actually say that. It says that if we turn down the intensity of any light source, the light we detect will eventually resolve into discrete packets called "photons". But it does not say that the state of the light prior to detection is always discrete packets called "photons". That is only true for Fock states, which are very rare because they're so hard to produce.

Again, yes, you're understanding me at what I'm trying to prod at.

 

[Dale]

That assumption is only valid for case (2) in my post #4. It is not valid for case (1) since the state emitted from the source is not an eigenstate of photon number.

I don't think that is a requirement. The photoelectric effect, which demonstrates photons, is produced with a light source that also is not in an eigenstate of photon number.

The number of photons in such states is uncertain, but it is certain that it is a whole number (integer).

 

[julianb]

I assume you mean, if a 2-slit experiment under case (2) has not been done. Photon sources that fall under case (2) certainly do exist and experiments have been done with them that confirm the predicted properties of Fock states under quantum field theory. I just don't know if any of those experiments were 2-slit experiments.

Thanks again. This is really helpful.
I'd be interested to learn more about these experiments.

We aren't. In case (1), the term "single photons" is meaningless except in reference to the discrete detection events. You cannot use case (1) to justify a claim like "single photons have interference patterns", because the state prior to detection is not properly described as "single photons" to begin with. And as far as I know, no actual physicist in an actual textbook or peer-reviewed paper makes such a claim as you are implying in what I just quoted above. If you think one has, then you need to give a specific reference.

I mean ... it's the boiled down explanation, isn't it? Over and over again, explanations lead to "the quantum nature of light".
A cursory google search, just now, to demonstrate how easy and common this is to find - brings up a paper from 2023:
https://www.nature.com/articles/s41598-023-28264-1
The conclusion was Self-Interference was true.
But they still relied on single-photon detectors and laser input, beam-splitters, etc.

So this is where I get stuck in a state of confusion.
It would seem to me that, again, conclusions of self-interference keep being connected to experiments where it's acknowledged that there's a minimum detection threshold.

Crude analogy: If waves are analog, and we can only measure digital, how do we know we're not committing rounding errors?

We don't measure the energy at the slits, so it's meaningless to ask whether energy is conserved between the slits and the detector. In QM you can't make claims about conservation laws unless you measure the conserved quantities at the appropriate points.

We can do experiments where we measure the energy emitted by the source and the energy detected by the detector, and in those cases, we find that energy is conserved.

That may be true, but if the measuring at both source and detector are both subject to that confines of only being able to detect whole photons / whole electrons, then I don't get how that isn't subject to the same potential inability to detect energy in waves that's below 1 photon worth of energy.

(Or, potentially, as a tangent - that multiple waves are identical wavelength & frequency, but out of phase with destructive interference. If you had 2 identical waves of huge amounts of energy, but they were out of phase 50% with each other, we'd never detect them. Let alone if we had 2 waves with less than one photon of energy.)

 

[PeterDonis]

The photoelectric effect, which demonstrates photons

The photoelectric effect demonstrates that detection of light is in discrete units called "photons". It does not demonstrate that the light before detection must have been in discrete units called "photons". It can't, because, as you say, can produce the photoelectric effect using light sources that emit states like coherent states that aren't eigenstates of photon number.

 

[Dale]

The photoelectric effect demonstrates that detection of light is in discrete units called "photons". It does not demonstrate that the light before detection must have been in discrete units called "photons". It can't, because, as you say, can produce the photoelectric effect using light sources that emit states like coherent states that aren't eigenstates of photon number.

I don't think not being in an eigenstate is a relevant objection.

The question is if it is a whole number (integer), not if that whole number is fixed. Not being in an eigenstate means that the number of photons is uncertain. It does not mean that the number of photons could be a non-integer.

 

[julianb]

I don't think not being in an eigenstate is a relevant objection.

The question is if it is a whole number (integer), not if that whole number is fixed. Not being in an eigenstate means that the number of photons is uncertain. It does not mean that the number of photons could be a non-integer.

This is exactly what I'm asking about. Ultimately, how do we know that the amount of energy in a (EM) wave must be equivalent to an integer amount of photons? The PE tells us what we can measure is in multiples of photons, but still can't prove that there is not sub-photon levels of energy which is being overlooked. (In a way, what you're writing is still accurate - if we regard photons as "detectable" light energy, and only detectable in integers, then it's true, by definition.)

 

[PeterDonis]

The PE tells us what we can measure, but still can't prove that there's sub-photon levels of energy that we can't detect.

We know energy is conserved because whenever we measure the total energy of an isolated system, we find it has the same value. We aren't relying on the discrete nature of photoelectric detections to know that energy is conserved.

 

[Nugatory]

I don't get how that isn't subject to the same potential inability to detect energy in waves that's below 1 photon worth of energy.

The relationship between photons and waves is not what you are probably thinking. Photons appear in the theory when we try writing the electromagnetic field as a sum of various frequencies, a process somewhat reminescent of how we use fourier analysis to describe a sound wave as a sum of pure sinusoidal frequencies: https://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

Thus there's no such thing as "less than 1 photon worth of energy". Any non-zero electromagnetic field is always made up of photons and nothing else, the same way that there's nothing in a sound signal that is not represented the Fourier transform.

 

[julianb]

We know energy is conserved because whenever we measure the total energy of an isolated system, we find it has the same value. We aren't relying on the discrete nature of photoelectric detections to know that energy is conserved.

What are we relying on, then?

Also, one could agree that energy is conserved, while also ask if we cannot detect all of the (conserved) energy.

Like, hypothetically if a wave has 1.5 photons of energy in it, we could detect 1 photon at the start, 1 photon at the end, and that's conserved. And, it could still be true that 1.5 remains the whole time. But that doesn't negate the question of "what if we can't detect that 0.5?"

 

[julianb]

The relationship between photons and waves is not what you are probably thinking. Photons appear in the theory when we try writing the electromagnetic field as a sum of various frequencies, a process somewhat reminescent of how we use fourier analysis to describe a sound wave as a sum of pure sinusoidal frequencies: https://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

Thus there's no such thing as "less than 1 photon worth of energy". Any non-zero electromagnetic field is always made up of photons and nothing else, the same way that there's nothing in a sound signal that is not represented the Fourier transform.

Apologies, I don't follow.

So, the paper you linked starts off with this assumption: "Consider the well-known processes of emission and absorption of photons by atoms" ... which in turn, I understand, assumes that photons only exist in quantums. I acknowledge that it's been decades since I've had to do Fourier transforms and the level of calculus as the paper. But, it seems that the paper is exploring the nature of photons, which, by definition, assumes it's just looking at them as in discreet packets.

So, when you then wrote 'Thus there's no such thing as "less than 1 photon worth of energy".' -- I'm unsure how that follows? Does the paper demonstrate this? Or merely presume it?

Isn't that just still claiming (basically) "we can only detect photons ergo that must be the smallest amount of light"? I don't understand how that is any different from "the atom is the smallest possible particle."

 

[PeroK]

What are we relying on, then?

Also, one could agree that energy is conserved, while also ask if we cannot detect all of the (conserved) energy.

Like, hypothetically if a wave has 1.5 photons of energy in it, we could detect 1 photon at the start, 1 photon at the end, and that's conserved. And, it could still be true that 1.5 remains the whole time. But that doesn't negate the question of "what if we can't detect that 0.5?"

Waves don't have photons. That's not part of the theory of electromagnetism. Instead, the electromagnetic field interacts with matter in discrete quanta of energy. The quanta depend on the frequency of the light. This is what the photo-electric effect established. It doesn't matter how intense the incident light is, if it has too low a frequency, then it cannot interact with matter above the quantum related to its frequency. It makes little sense to talk about the total energy of an EM source in this context.

Likewise, if you carry out an experiment using a source that produces a very low intensity of light, then you detect the discrete interactions. This is essentially what a photon is.

 

[PeroK]

Isn't that just still claiming (basically) "we can only detect photons ergo that must be the smallest amount of light"? I don't understand how that is any different from "the atom is the smallest possible particle."

Your idea that there is a minimum quantum of light per se is a misunderstanding on your part. In principle, a photon can have any energy, no matter how small. The detection of the CMBR detects photons of a very low energy. For a normal light source, you can reduce the detection energy by moving the detector away from the source at high speed. In principle, the energy of any light could be attenuated to any value, no matter how small.

You are confusing the idea that light of a given frequency interacts in fixed quanta with there being a universal minumum quantum of light. Which there is not.

 

[julianb]

Waves don't have photons. That's not part of the theory of electromagnetism.

Thanks, PeroK.

To be clear, when I was writing "1.5 photons of energy", I mean that in terms of a hypothetical amount of energy, not believing or thinking that "waves have photons".

Instead, the electromagnetic field interacts with matter in discrete quanta of energy. The quanta depend on the frequency of the light. This is what the photo-electric effect established.

I'd like to learn more specifically about how the PE Effect established this.

As I understand the PE Effect, it still is limited by us being able to measure, at minimum, a single electron, correct?

It doesn't matter how intense the incident light is, if it has too low a frequency, then it cannot interact with matter above the quantum related to its frequency.

What do you mean by "intense" in this case?

If something has low enough frequency, then there's clearly energy in a wave, we agree, yeah?
And that energy is there regardless of whether or not it can, say, bump an electron from a metal plate, correct?

It makes little sense to talk about the total energy of an EM source in this context.

I agree, but maybe for not the reason you have.
You set the context as "[light that is too low a frequency] to interact with matter" but, if I read you correctly, only in the scope of "able to knock electrons off metal plates", essentially.

In essence, going back to my original question - and why I posted this about the 2-slit experiment and not the PE Effect, is that in the 2-slit experiment, the question is whether a wave can interfere with a wave.

Likewise, if you carry out an experiment using a source that produces a very low intensity of light, then you detect the discrete interactions. This is essentially what a photon is.

Again, I'm unsure what you mean by "intensity", and what you'd mean as a "very low intensity". Could you clarify these?

 

[julianb]

Your idea that there is a minimum quantum of light per se is a misunderstanding on your part. In principle, a photon can have any energy, no matter how small. The detection of the CMBR detects photons of a very low energy. For a normal light source, you can reduce the detection energy by moving the detector away from the source at high speed. In principle, the energy of any light could be attenuated to any value, no matter how small.

You are confusing the idea that light of a given frequency interacts in fixed quanta with there being a universal minumum quantum of light. Which there is not.

Okay, assume everything I've written for "energy" has to do with "amplitude and frequency of a wave". Essentially... the stuff that would make up photons in packets upon waveform collapse. (Do you get where the word "energy" used in a specific sense can be confusing with multiple contexts here, to someone who isn't a professional physicist?)

 

[PeroK]

I'd like to learn more specifically about how the PE Effect established this.

That shouldn't be hard to find.

As I understand the PE Effect, it still is limited by us being able to measure, at minimum, a single electric, correct?

I'm not sure what you mean by that. It detects electrons as a result of atomic ionisation. In the introduction to Ballentine's QM, he quotes an experiment of Franck and Hertz from 1914 which demonstrates "discreteness". These experiments are what kicked off QM. Discreteness was an experimental observation before it was part of a coherent theory of QM.

What do you mean by "intense" in this case?

Intensity means that amount of light. Essentially the total power of the EM field. The point about the photo-electric effect is that no amount of low-frequency light can achieve what a very weak source of higher-frequency light can do.

If something has low enough frequency, then there's clearly energy in a wave, we agree, yeah?
And that energy is there regardless of whether or not it can, say, bump an electron from a metal plate, correct?

Yes, see above. You can detect the energy of low-frequency light directly.

I agree, but maybe for not the reason you have.
You set the context as "[light that is too low a frequency] to interact with matter" but, if I read you correctly, only in the scope of "able to knock electrons off metal plates", essentially.

The photo-electric effect is not the only experiment. If you want to remove discreteness from modern physics, you'll have a hard job. The experimental evidence by now is overwhelming.

In essence, going back to my original question - and why I posted this about the 2-slit experiment and not the PE Effect, is that in the 2-slit experiment, the question is whether a wave can interfere with a wave.

That's simply mumbo jumbo. Quantum interference is ubiquitous. It's not just some weird effect in the double-slit experiment. Quantum interference appears as soon as you start working with the mathematics of QM.

Again, I'm unsure what you mean by "intensity", and what you'd mean as a "very low intensity". Could you clarify these?

In this context, low power.

 

[PeterDonis]

assume everything I've written for "energy" has to do with "amplitude and frequency of a wave".

Those aren't compatible observables in QM, so there is no state that has definite values for both. Nor does either have a definite minimum value.

the stuff that would make up photons in packets upon waveform collapse.

I don't know what this means. I strongly suggest that you take some time to look at the actual math involved, and try to state what you are trying to say in terms of the math instead of vague ordinary language.

 

[julianb]

That shouldn't be hard to find.

I'm not sure what you mean by that. It detects electrons as a result of atomic ionisation. In the introduction to Ballentine's QM, he quotes an experiment of Franck and Hertz from 1914 which demonstrates "discreteness". These experiments are what kicked off QM. Discreteness was an experimental observation before it was part of a coherent theory of QM.

Intensity means that amount of light. Essentially the total power of the EM field. The point about the photo-electric effect is that no amount of low-frequency light can achieve what a very weak source of higher-frequency light can do.

Yes, see above. You can detect the energy of low-frequency light directly.

The photo-electric effect is not the only experiment. If you want to remove discreteness from modern physics, you'll have a hard job. The experimental evidence by now is overwhelming.

That's simply mumbo jumbo. Quantum interference is ubiquitous. It's not just some weird effect in the double-slit experiment. Quantum interference appears as soon as you start working with the mathematics of QM.

In this context, low power.

Well, it _has_ been hard to find.

That's why I'm asking here.

And reframing "energy" as "power" isn't super helpful.

I don't appreciate your rudeness. I'm trying to understand, and being brushed off as "mumbo jumbo".

I'm sorry if you're not up to answering clarifying questions to people who don't have the extensive background in the subject. I understood this forum is entirely supposed to be about outreach to laypeople. If that's not what you're interested, please don't reply to me anymore.

 

[PeterDonis]

it _has_ been hard to find.

Where have you looked? So far you have given no specific references, even though I've asked you for them more than once.

I don't appreciate your rudeness. I'm trying to understand, and being brushed off as "mumbo jumbo".

I'm sorry if you're not up to answering clarifying questions to people who don't have the extensive background in the subject. I understood this forum is entirely supposed to be about outreach to laypeople. If that's not what you're interested, please don't reply to me anymore.

This is uncalled for. @PeroK is basically saying the same thing I said at the end of post #24: you need to be grounding your statements in math, not vague ordinary language, because physics is done in math. If you are not familiar with the math, my advice would be to take the time to learn it. I have found Ballentine to be a good textbook for learning the basic math of QM. Section 19.6 of Ballentine discusses photoelectric detectors.

 

[julianb]

To reframe and refresh:

Yes, I'm questioning the discreetness as an explanation for how things are, rather than how things can be measured.
And yes, I realize it's an underlying assumption in physics in the last 100 years.
And yes, that's absolutely why I'm asking about it, because I think it's worth considering if the wrong conclusion was made, and a lot of far-fetched metaphors are used to describe the universe rather than a much simpler answer: "we can't measure that small and we don't know".

The thing with the 2-slit experiment and why it's so compelling for this is that it gave us whole fields of physics where we legitimately wondered "what if there's infinite universes?" or "what if something interferes with itself?" or "can things just spontaneously appear?".

Rather than a simpler explanation of "there's more down there that we can describe mathematically but cannot actually explain."

So, yes, I'm looking for evidence that discreetness doesn't have a big problem with assuming "what we measure" is "all there is, measurable or not".

 

[renormalize]

This is exactly what I'm asking about. Ultimately, how do we know that the amount of energy in a (EM) wave must be equivalent to an integer amount of photons?

Because for the quantized electromagnetic radiation field, energy eigenstates (i.e., those carrying exactly known energy $E$ and having an angular-frequency $\omega$) are always simultaneously photon-number eigenstates (i.e., they consist of exactly $N = E/(\hbar\omega)$ photons). In contrast, energy/photon non-eigenstates, like coherent states, by the uncertainty principle yield varying values of energy and photon-number when repeated measurements are made on identically-prepared states.

 

[julianb]

Where have you looked? So far you have given no specific references, even though I've asked you for them more than once.


This is uncalled for. @PeroK is basically saying the same thing I said at the end of post #24: you need to be grounding your statements in math, not vague ordinary language, because physics is done in math. If you are not familiar with the math, my advice would be to take the time to learn it. I have found Ballentine to be a good textbook for learning the basic math of QM. Section 19.6 of Ballentine discusses photoelectric detectors.

Not everyone can dedicate their lives to math, PeterDavis.
The whole point of human civilization is specialization so we can do more as a whole.
It's also a superbly privileged economic take. :/

If you're not interested with interacting with laypeople, I'm not forcing you to reply to me. I'll find other people who are.

 

[PeterDonis]

I'm questioning the discreetness as an explanation for how things are

Why do you think any physicist has actually claimed that "discreteness is an explanation for how things are"?

Again, you need to give a specific reference that makes the claim you are asking about.

Without such a reference, further discussion in this thread is pointless.

 

[PeterDonis]

Not everyone can dedicate their lives to math

You don't have to dedicate your life to math to learn enough of the math for a specific area of physics you're interested in to be able to have a reasonably informed understanding.

PeterDavis

Please spell my last name correctly.

The whole point of human civilization is specialization so we can do more as a whole.
It's also a superbly privileged economic take. :/

I have no idea why you think this has anything to do with the discussion in this thread.

If you're not interested with interacting with laypeople, I'm not forcing you to reply to me. I'll find other people who are.

This is uncalled for. I think you need to reconsider your attitude.

 

[PeterDonis]

Thread closed for moderation.

 

[Nugatory]

Apologies, I don't follow.
So, the paper you linked starts off with this assumption: "Consider the well-known processes of emission and absorption of photons by atoms" ... which in turn, I understand, assumes that photons only exist in quantums.

That's not an assumption, it's the evidence-based starting point: All interactions between electromagnetic fields and matter, if examined at a sufficiently low intensity, involve the transfer of discrete amounts of energy and momentum. (We do make an assumption beyond that, namely that at high intensities it's the same physics but scaled up - we know what a drop of water is, we know what a rainstorm is, and we can model a rainstorm as a very large number of drops. There's no observational evidence to make us distrust this common-sense assumption so we'll run with it). That motivates us to seek a theory consistent with these observations - and, yes, such a theory might conceivably allow for other so far unobserved interactions as you suggest in your initial post.

So we proceed to develop such a theory, and eventually we find that the Hamiltonian can be written in the form at the bottom of page 7, as a superposition of quantum oscillators. At this point we decide that "quantized excitation of the electromagnetic field" is an unwieldy phrase and we decide to use the word "photon" instead to name the terms that appear in this Hamiltonian. So we aren't assuming "that photons only exist in quantums", we've constructed a viable theory that describes electromagnetic fields in terms of quanta, and now that we have a viable theory we've decided to name the quanta "photons".

So, when you then wrote 'Thus there's no such thing as "less than 1 photon worth of energy".' -- I'm unsure how that follows? Does the paper demonstrate this? Or merely presume it?

The Hamiltonian at the bottom of page 7, by construction from its starting point at Maxwell's equations, includes all the energy in any arbitrary configuration of an electromagnetic field and there's nothing but photons in that Hamiltonian - that's a demonstration.

Isn't that just still claiming (basically) "we can only detect photons ergo that must be the smallest amount of light"? I don't understand how that is any different from "the atom is the smallest possible particle."

The crucial difference is that there is no "smallest photon". Thus there is no amount of energy so small that it cannot be described as a superposition of photon states. It's analogous to how there's no volume of water that I cannot measure in "spoonfuls" if I have an infinite collection of measuring spoons, one for each real number.

To be fair, there is a LOT of corner-cutting and oversimplifying in that paper and that may reasonably leave you with the suspicion that we're trying to sneak some circular reasoning past you. However, I don't know of any simpler presentation that doesn't totally misrepresent what photons are. For more complete presentation, Lancaster and Blundell "Quantum Field Theory for the Gifted Amateur" is (barely) accessible to someone with a bachelor's degree background. Or if you want free online, there is Srednicki https://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf which is way more demanding but available free on the internet.

 

[PeterDonis]

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