Why is Photon Energy Quantized in Terms of Sine Wave Frequency?

[PeterDonis]

In the equation E=h f, could you please tell me what is E and what is f really?

Go read my post #28. I already addressed this question there.

 

[hutchphd]

In the original Planck formulation, I believe this was a statement about energy exchange with the walls in a cavity in a solid which was the model for a black body radiator at temperature T. Absent this ansatz, there were some nasty infinities in the thermodynamics. E and f were the usual classical objects for atoms and light.

 

[PeroK]

At a simpler level, without delving too deep into advanced theory, I'm trying to understand why, in the equation E=hf (for an individual photon's energy), the energy is dependent on the harmonic frequency of the wave (I don't think my question is ambigous?)

There is no EM wave associated with an individual photon. You're still mixing up two different theories of light. Consider diffraction:

There is a classical theory, describing light as an EM wave, where the diffraction pattern is explained by Huygens principle and an analysis using the classical wavelength of the light.

There is a quantum theory, where light is described probabilistically, which results in the same diffraction pattern. Moreover, in this theory, light interacts with matter in discrete quanta - called photons. And if we do diffraction with very low intensity light, we can see the diffraction pattern building up photon by photon. Note that each photon appears on the detection screen probabilistically. So, although each photon has an associated frequency they do not all diffract by the same angle.

However, when the pattern has built up we see that the photons collectively can be associated with a classical frequency $f$.

And, if we also measure the energy of each photon, we find that $E = hf$.

This is one example of how we see that the quantum theory is the fundamental theory, with the classical theory emerging as an approximation.

The classical EM wave is a similar case. The wave only appears as a result of the probabilistic behaviour of a sufficiently large number of photons. The individual photons are not themselves waves - and don't inherently have a wavelength and frequency. However, when the resulting phenomenon of light is studied, the energy of the photons corresponds to classical wavelengths and frequencies related to the energy.

Understanding this fully requires a study of the mathematics that underpins both theories. As, ultimately, the equivalence of the two theories where they overlap is a mathematical one.

I suggest you study Feynmans book fully, as this describes how classical wavelike phenomena emerge from a probabilistic quantum theory where light has no inherent wavelength or frequency at the fundamental level.

Maybe I should start more basic. In the equation E=h f, could you please tell me what is E and what is f really?

As above, E is the energy associated with an individual photon, and $f$ is the emergent frequency when a sufficient number of photons are involved for classical wavelike behaviour to be observed.

That equation is itself, therefore, something of a mixture of two theories of light.

 

[PeroK]

PS to understand your question you must understand the difference between the quantum mechanical and classical theories of light, and the relationship between them.

 

[PeroK]

PPS a related question is how can an electron have a wavelength and frequency? It's the same answer: when you apply the probabilistic quantum theory to a particle with mass, such as an electron, you get behaviour such as diffraction. Again, however, only when you do an experiment with a large number of electrons. And, the resulting diffraction pattern can be associated with that of a classical wave of a certain wavelength and frequency.

The only difference is that classically we associate light as a wave and an electron as a particle. When light exhibits particle-like behaviour or an electron exhibits wavelike behaviour we are surprised. But, ultimately, both behaviours are just two sides of the quantum coin.

 

[Fra]

My burning question is: why is this the case? Is there something fundamentally ingrained in nature that dictates the energy to be quantized in this manner, specifically in terms of a sine wave frequency? Why not in terms of a square wave, or any other waveform for that matter?

I am well aware that sine waves possess unique properties such as orthogonality and smoothness, and they are prevalent in numerous physical phenomena. However, my intellectual curiosity yearns for a deeper understanding — is there a more profound or fundamental reason behind the photon’s energy being quantized in this specific way?

I have spent years searching for answers, sifting through articles online, and reaching out to professors, but the answers I found were either too surface-level or they just skirted around the question. I’m at my wits' end here, and I am earnestly hoping that this community might offer a new perspective or point me towards resources that can finally put this longstanding query to rest.

Any thoughts, references, or guidance would be immensely appreciated.

Sometimes different people ask similar questions, but seeks an answer of different types.
I will just note that used to ask myself the same question, and its a good one, I don't think it's confused at all, on the contrary, I think it's related to the paradigm we use in physics. But this might fit better in the interpretational forum IMO, but perhaps someone can move it there.

I can just add a bit of how I reasoned if it may help. For me, I asked this question in a more general context of the foundations of QM, and why certain mathematics are more "fit" to describe nature? (ie it has nothing todo specifically with "photons")

It is perfectly fine to decompose functions in various bases, but from the perspective of information processing and retention of information under constraints of limited memeory, it seems all the mathematically possible ways, still does not do the job equally good, in particular when you want to describe "periodic phenomena". Note that "periodic phenomena" just means "repeating", a phenomena can be periodic without beeing a harmonic etc. The problem is how do you represent the transform? In limited memory you need to truncate? (ie discard data). Then you want to of course not discard data at random, but discard data that is LEAST important (which we often call "noise"). Ie. we should not occupy scarce resources such as memory with noise, so the task is - which transforms gives us the best outlook to "compress data", so we can choose to dismiss the noise and keep the rest.

Periodic phenomena are almost I would say something we actively look for, when repeating experiments with a given duration. One could even say that our method, "truncates" our observations so that we simply register short periods. We can not see slow phenomena (because it never reaches sufficient statistical confidence levels), we can also not see extremly fast pheonmena as it requires extremely fast sample rates. And for a given mesaurement devices, we have limts to all this. (a general note, without getting into any details).

In QM the answer to your question is supeficially I think related to how we DEFINE say, momentum or energy. Ie. as conjugate variables. And these conjugate variables indeed singles out the fourier transform as the natural partner, in the way that invariance with respect to one varibla, means that the other variable is constant. and this is exactlty the properties you want from a stable retention of compressed information. IF the retained part is "constant" then we have a stability. (And stability of say atoms, was one of the original mysteries of atomic physics: WHY is it stable, well it's because of quantization, but WHY is it quantized? Then we throw in the conjugate relation. But why? is there a deeper reeason?

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

This also relates to the HUP, and to actions, but I think to elaborate on this it becomes more interpretational and possibly speculative so I will pass. But this was just a short encouragement to you to say that I don't think your question is the least confused. But the answer lies in the foundational area I think. It's not specific to electromagnetism (at least not for me, the quyestion is bigger).

/Fredrik

 

[Fra]

Maybe I should start more basic. In the equation E=h f, could you please tell me what is E and what is f really?

See conjugate varibels, it relates to how E is "defined". But the question is of course, why do we "choose" to define it that way?

/Fredrik

 

[vanhees71]

At a simpler level, without delving too deep into advanced theory, I'm trying to understand why, in the equation E=hf (for an individual photon's energy), the energy is dependent on the harmonic frequency of the wave (I don't think my question is ambigous?) That is precisely Planck's relation, and I've struggled to find a satisfactory explanation online.

Any free electromagnetic field can be decomposed into plane-wave modes, and so you can for the field operators of the quantized theory.

The corresponding mode functions are all $u(t,\vec{x})=\propto \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x})$. Also they must fulfill the wave equation,
$$\frac{1}{c^2} \partial_t^2 u - \Delta u=0,$$
from which you get
$$\omega=c |\vec{k}|.$$
The free electromagnetic field is then represented by an infinite set of harmonic oscillators, labelled by $\vec{k}$ and the helicity $\lambda \in \{1,-1\}$ to describe the polarization states too. $\lambda=1$ corresponds to right-circular and $\lambda=-1$ to left-circular polarized em. waves.

Each harmonic oscillator then is described by creation and annihilation operators $\hat{a}^{\dagger}(\vec{k},\lambda)$ and $\hat{a}(\vec{k},\lambda)$. They obey bosonic commutation relations,
$$[\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=(2 \pi)^3 \delta^{(3)}(\vec{k}-\vec{k}') \delta_{\lambda \lambda'}.$$
A basis is then formed by the occupation-number (Fock) basis, starting from the ground state, the "vacuum", $|\Omega \rangle$, for which
$$\hat{a}(\vec{k},\lambda) |\Omega \rangle=0$$
for all $(\vec{k},\lambda)$ and the states $|N(\vec{k},\lambda) \rangle$, which are simultaneous eigenvectors of the number-density operators
$$\hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The energy and momentum can then be derived from the operators given as in classical electrodynamics by the corresponding energy-momentum tensor of the electromagnetic field. In terms of the number operators it results
$$\hat{H}=\sum_{\lambda} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \omega \hat{N}(\vec{k},\lambda),$$
$$\hat{\vec{P}}=\sum_{\lambda} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{(2 \pi)^3} \vec{k} \hat{N}(\vec{k},\lambda).$$
This can be interpreted that the mode $(\vec{k},\lambda)$ comes in discrete "quanta" of energy and momentum, $E=\omega$ and $\vec{p}=\vec{k}$ (note that I use natural units, where $\hbar=1$.

 

[vanhees71]

There is no EM wave associated with an individual photon. You're still mixing up two different theories of light. Consider diffraction:

Of course there is an em wave associated with an "individual photon" (where "individual" has to be taken as grain of salt since photons are of course indistinguishable in the usual sense of QT). A single-photon Fock state represents a corresponding mode of the electromagnetic field. In the usual momentum-helicity basis it's a plane em. wave with sharp wave vector $\vec{k}$ and helicity $\lambda$. One can interpret single-photon states only in this way. A naive particle picture doesn't work for photons, which are massless spin-1 quanta, which don't admit a full-fledged position observable, i.e., a photon cannot be prepared in a "localized state".

Of course you can try to do this by considering the em. field in a finite cavity. Then again you get field modes, and here only certain $\vec{k}$ are allowed due to the boundary conditions, but each field mode is a standing wave. The "intensity", i.e., the energy density of the em. field, spreads over the entire cavity. You cannot localize the "photon" any better.

 

[gentzen]

I will just note that used to ask myself the same question, and its a good one, I don't think it's confused at all, on the contrary, I think it's related to the paradigm we use in physics. But this might fit better in the interpretational forum IMO, but perhaps someone can move it there.

please don't

Maybe I should start more basic. In the equation E=h f, could you please tell me what is E and what is f really?

See conjugate varibels, it relates to how E is "defined". But the question is of course, why do we "choose" to define it that way?

This was a simple question. I don't get your answer. I guess an answer appropriate for the level of QuantumCuriosity42 could start like

In the original Planck formulation, I believe this was a statement about energy exchange with the walls in a cavity in a solid which was the model for a black body radiator at temperature T.

(i.e. giving a context in which E=h f has meaning). And then explain the meaning of E and f in that context, i.e. f is the frequency of the classical electromagnatic field. The energy exchange with the walls happens only in discrete energy packets, and E is the amount of energy in such a packet.

 

[Fra]

This was a simple question.

Was it?

I was wondering if it was, as it seemed like the answers given was not satisfactory the OP. And as I recognize the question from myself, so I thought I'll add another perspective, and just leave it there. And the OP can pick the sort of answer that was appropriate.

/Fredrik

 

[PeroK]

Of course there is an em wave associated with an "individual photon"

But, not in the sense of a classical EM wave, satisfying Maxwell's equations. Otherwise, a single photon would exhibit classical EM behaviour.

 

[hutchphd]

At the risk of exposing my profound lack of understanding of quantum field theory, I will attempt to answer your questions. If you are truly curious you must do the hard work to learn advanced quantum field theory from, say, Bogoliubov and Shirkov

TL;DR Summary: I've been on a multi-year quest, diving into internet resources and consulting professors, trying to grasp why photon energy is quantized in terms of sine wave frequency (E=h⋅ν), and not any other waveform. Despite understanding the unique properties of sine waves, I’m still in search of a deeper, more fundamental explanation. Any insights or resources to finally put this question to rest would be immensely appreciated!

Is there something fundamentally ingrained in nature that dictates the energy to be quantized in this manner

Yes of course there is.

Why not in terms of a square wave, or any other waveform for that matter

They are not eigensolutions for the free fields.

I am earnestly hoping that this community might offer a new perspective or point me towards resources that can finally put this longstanding query to rest.

Study enough to do field theory. This does not mean watch more videos

Could you expand on why these trigonometric functions are particularly suited for describing systems along a line? Is it tied to their properties, or is there a deeper physical reasoning?

They are the eigenstates of the quantized EM field in free space. One starts with the appropriate Lagrangian for the EM field (usually using the vector potentials). This produces equations that look much like many simple harmonic oscilllators and therefore proceeds using that usual formalism. The "number of photons n" corresponds to energy level of the appropriate oscillation mode $$E_{k,n}=\hbar \omega_k(n+\frac 1 2)$$

Would you be able to provide some resources where I can delve deeper into these topics?

Books about advanced quantum mechanics, preceeded by books about elementary quantum mechanics. See top

So, is the photon's energy as described by E=h*f just an approximation

no these are exact eigensolutions for the quantized free field

You are correct, but my original doubt remains. Why energy increases in relation to harmonic frequency.

Study tbooks. These are eigenmodes of harmonic oscillators
Your questions are all good ones but the answers are not simple. To really understand requires much concerted effort ( frankly more than I have been willing to invest ..... I presume any blunders in this colloquy on my part will be corrected by wiser hands)

 

[vanhees71]

But, not in the sense of a classical EM wave, satisfying Maxwell's equations. Otherwise, a single photon would exhibit classical EM behaviour.

Of course, it's a generic quantum state of the em. field, which cannot be in any way approximated by a classical theory (neither by a classical point-particle theory nor by classical electrodynamics).

 

[PeterDonis]

However, when the pattern has built up we see that the photons collectively can be associated with a classical frequency $f$.

Careful. The "photons" you describe here are a coherent state of the quantum EM field, which is not an eigenstate of photon number or energy. It has a definite frequency $f$, but there is no definite energy $E$ corresponding to $f$. So the Planck relation $E = hf$ has no meaning for this case since $E$ is not well-defined.

The term "photon" here really refers to the fact that the detections are quantized--at low enough intensity we see discrete dots appear on the detector screen. But that sense of the term "photon" has nothing to do with the Planck relation.

 

[hutchphd]

But that sense of the term "photon" has nothing to do with the Planck relation.

Quibble: Because photons do not interact directly, this interaction ( "detection"?) is central to the counting problem associated with light in a cavity "black body" at a defined temperature. So I disagree or we are drowning in the semantic sea?

 

[Vanadium 50]

Wish you hadn't said "quibble", because now I can't complain about quibbling without it sounding like I am complaining about you specifically. I'm not, but I think we (as a group) are,

Trying to summarize, our answer is;

• Sines and cosines are the solutions to the relevant differential equations.
• Sums of sins and cosines are also solutions to the relevant differential equations.
• However, sums of sines and cosines do not have a single well-defined energy. The only states that have a single energy are sines of one frequency (I drop the cosines here - a cosine is just an offset sine). For both sides of E = hf to be well-defined, E and f need to be single-valued.

The OP seems not to accept this. However, his objection is far from clear. He writes a lot...A LOT..and closes with restarting his original question. That leaves us only two options: a) repeat what we wrote, or b) quibble about each others answers.

If the OP can succinctly - succinctly - state his objection to the answers, it may be worth another circle around. Otherwise, this is probably hopeless.

 

[hutchphd]

It seems to me his objection devolves to "I don't understand anough physics to see where this comes from" and my intended answer was "there is a well-defined route but you have to walk it yourself and it is not available on youtube as a travelog".
And I love youtube!!

 

[Vanadium 50]

I don't want to put words in his mouth - so many are coming out so quickly that doesn't seem like a good place to stand - so I'll let him clarify. But I will say that this would go faster if he were to work out some of the mathematics himself.

 

[PeterDonis]

Quibble: Because photons do not interact directly, this interaction ( "detection"?) is central to the counting problem associated with light in a cavity "black body" at a defined temperature.

The case of light in a cavity is a different physical scenario from the case of light incident on a detector and being detected as discrete dots ("photons"). In a cavity we do not detect any photons, we just measure the black body temperature and the intensity of radiation at various frequencies. And of course the Planck relation was originally introduced by Planck to explain the curve of intensity vs. frequency, since the classical prediction was known to be egregiously wrong (the "ultraviolet catastrophe"). But the Planck relation in that case has no relationship to any detections of individual photons.

That said, even in the cavity case the radiation is a coherent state, not a Fock state, and so it is not in an eigenstate of energy and the $E$ in the Planck relation still does not describe "the energy of a photon" in any sense that corresponds to an actual observation. It is an abstract "energy" that appears in the distribution function.

 

[QuantumCuriosity42]

There is no EM wave associated with an individual photon. You're still mixing up two different theories of light. Consider diffraction:

There is a classical theory, describing light as an EM wave, where the diffraction pattern is explained by Huygens principle and an analysis using the classical wavelength of the light.

There is a quantum theory, where light is described probabilistically, which results in the same diffraction pattern. Moreover, in this theory, light interacts with matter in discrete quanta - called photons. And if we do diffraction with very low intensity light, we can see the diffraction pattern building up photon by photon. Note that each photon appears on the detection screen probabilistically. So, although each photon has an associated frequency they do not all diffract by the same angle.

However, when the pattern has built up we see that the photons collectively can be associated with a classical frequency $f$.

And, if we also measure the energy of each photon, we find that $E = hf$.

This is one example of how we see that the quantum theory is the fundamental theory, with the classical theory emerging as an approximation.

The classical EM wave is a similar case. The wave only appears as a result of the probabilistic behaviour of a sufficiently large number of photons. The individual photons are not themselves waves - and don't inherently have a wavelength and frequency. However, when the resulting phenomenon of light is studied, the energy of the photons corresponds to classical wavelengths and frequencies related to the energy.

Understanding this fully requires a study of the mathematics that underpins both theories. As, ultimately, the equivalence of the two theories where they overlap is a mathematical one.

I suggest you study Feynmans book fully, as this describes how classical wavelike phenomena emerge from a probabilistic quantum theory where light has no inherent wavelength or frequency at the fundamental level.As above, E is the energy associated with an individual photon, and $f$ is the emergent frequency when a sufficient number of photons are involved for classical wavelike behaviour to be observed.

That equation is itself, therefore, something of a mixture of two theories of light.

I understand the distinction you're making between the classical EM wave and the quantum view of light. I also grasp that when we talk about frequency, we are referring to the classical EM wave.

However, my core question remains: Why is this emergent frequency specifically related to sinusoidal (harmonic) waves? Why not square waves, triangular waves, or any other waveform for that matter? I'm curious about the inherent nature of light that leads us to describe its frequency using sinusoidal waves as opposed to any other shape.

Also, in what Feynman book can I see what you say about how classical wavelike phenomena emerge from a probabilistic quantum theory?

 

[QuantumCuriosity42]

PPS a related question is how can an electron have a wavelength and frequency? It's the same answer: when you apply the probabilistic quantum theory to a particle with mass, such as an electron, you get behaviour such as diffraction. Again, however, only when you do an experiment with a large number of electrons. And, the resulting diffraction pattern can be associated with that of a classical wave of a certain wavelength and frequency.

The only difference is that classically we associate light as a wave and an electron as a particle. When light exhibits particle-like behaviour or an electron exhibits wavelike behaviour we are surprised. But, ultimately, both behaviours are just two sides of the quantum coin.

Could you point me to a demonstration or a source where it is shown that the diffraction pattern can be specifically associated with a classical sinusoidal wave of a certain frequency?

 

[QuantumCuriosity42]

Wish you hadn't said "quibble", because now I can't complain about quibbling without it sounding like I am complaining about you specifically. I'm not, but I think we (as a group) are,

Trying to summarize, our answer is;

• Sines and cosines are the solutions to the relevant differential equations.
• Sums of sins and cosines are also solutions to the relevant differential equations.
• However, sums of sines and cosines do not have a single well-defined energy. The only states that have a single energy are sines of one frequency (I drop the cosines here - a cosine is just an offset sine). For both sides of E = hf to be well-defined, E and f need to be single-valued.

The OP seems not to accept this. However, his objection is far from clear. He writes a lot...A LOT..and closes with restarting his original question. That leaves us only two options: a) repeat what we wrote, or b) quibble about each others answers.

If the OP can succinctly - succinctly - state his objection to the answers, it may be worth another circle around. Otherwise, this is probably hopeless.

Firstly, I apologize for any misunderstandings or lack of clarity in my previous communications. I understand the foundational role of sines and cosines in solving the wave equations. My inquiry stems from a philosophical angle more than a purely mathematical one: Even if sines and cosines are solutions, I think we could decompose an electromagnetic wave using a basis other than sines and cosines. Why is the frequency in Planck's relation specifically tied to the frequency of a sine wave and not, for instance, a square or triangular wave? Is the sine/cosine (of infinite frequencies) basis the only orthogonal and periodic one? Perhaps my question borders on the philosophical, and the answer might not be rooted strictly within the current bounds of physics.

 

[PeterDonis]

Why is the frequency in Planck's relation specifically tied to the frequency of a sine wave

To the extent this question is well-defined (I have already posted multiple times about the issues with it), the answer has already been given by @Vanadium 50: sinusoidal waves are the eigenstates of energy, i.e., they are the states that have a definite energy. Any state for which the Planck relation can be well-defined must have a definite energy, otherwise the $E$ in the relation has no meaning. So any state for which the Planck relation can be well-defined must be a sinusoidal wave.

 

[PeterDonis]

I think we could decompose an electromagnetic wave using a basis other than sines and cosines.

Of course you can. But you won't be decomposing it in a basis of states with definite energy.

 

[Vanadium 50]

philosophical,

We don't discuss philosophy here.

But "why are the single energy solutions sines?" is no different than "why are graphs of quadratic equations parabolas?" They are the logical consequences of the equations, and even philosophers don't doubt these consequences.

 

[PeroK]

However, my core question remains: Why is this emergent frequency specifically related to sinusoidal (harmonic) waves? Why not square waves, triangular waves, or any other waveform for that matter? I'm curious about the inherent nature of light that leads us to describe its frequency using sinusoidal waves as opposed to any other shape.

That question has been answered already and is covered in any undergraduate textbook on EM. The wavelike solutions to Maxwell's equations are (sinusoidal) oscillating electric and magnetic fields. This led to Maxwell concluding that light was EM radiation in the first place.

Once you have recovered Maxwell's equation from quantum theory- which is mathematically hard - the derivation is the same.

Also, in what Feynman book can I see what you say about how classical wavelike phenomena emerge from a probabilistic quantum theory?

The Strange Theory of Light and Matter, which you said you'd already glanced at.

 

[PeroK]

Could you point me to a demonstration or a source where it is shown that the diffraction pattern can be specifically associated with a classical sinusoidal wave of a certain frequency?

The diffraction of light depends only on the wavelength. This is covered in numerous online physics material, such as the Khan Academy.

That the light waves are oscillating sinusoidal EM fields is irrelevant to diffraction. The diffraction of light was studied long before Maxwell. All that was known was that light was a wave of a measurable speed, frequency and wavelength. Maxwell showed us what sort of wave it was.

 

[Fra]

They are the logical consequences of the equations, and even philosophers don't doubt these consequences.

Yes, so provided OP understands the answers so far (?) it seems the residual question is:

"why is energy is defined by introducing the operator "ihd/dt" on the wavefunction?"

In CM we have concepts such as momentum and energy. In QM we in a sligthly ad hoc manner redefine all generalized momenta (from hamiltons formulation of CM) to a conjugate variable.

So strictly speaking "energy" in CM is not the same thing as "energy" in QM.

But as we want a classical correspondence we keep the name.

one mystery if QM is, how can we get away with just redefine things and still make it work? And if its not he only way of doing it, what is so special about conjugate variables? This was what i tried to comment on in my other post.

Once you get past this you realise the QM terms, unlike CM terms, are sort of defined in terms of information. This is also more natural as we are in QM describing things we can not direcetly see or visualize. Therefore we have to describe instead the information about the subatomic world in terms of data we collect in the macroscopic lab. So new abstract definitions seemed required.

Some of us wonder, how come certain mathematics describes nature in simpler ways than others?.

We decide for ourselves when we think there is something more to gain by asking why.

/Fredrik

 

[vanhees71]

Quibble: Because photons do not interact directly, this interaction ( "detection"?) is central to the counting problem associated with light in a cavity "black body" at a defined temperature. So I disagree or we are drowning in the semantic sea?

@PeterDonis is right with giving the caveat that a "very dim laser light" is not a single-photon state but still a coherent state. It's with good approximation a single-mode coherent state with a frequency $\omega$, but it's not an eigenstate of the electromagnetic field energy nor a photon-number eigenstate. The photon number is Poisson distributed. If the intensity, i.e., the mean photon number is (much) smaller than one, it's "mostly a vacuum state", and you see randomly single points at a time on the screen is just due to the fluctuations of this field, but it's not a single-photon source.

Of course, nowadays, it's no problem to do the double-slit experiment with true single-photon sources. Usually you just use spontaneous parametric-down conversion to produce entangled photon pairs and use one as the photons as the "trigger", i.e., to "announce" that there's also another photon in a single-photon state which you let go through the slits. Also then you get a random point on the screen, and collecting many such single-photons on the screen the diffraction (double-slit interference) pattern builds up as long as you do the experiment in such a way that it is impossible to observe through which slit each photon went. So here you observe "wave properties" of the single-photon state. If you somehow mark the photons, e.g., with polarizers in the slits, such that you can infer which-way information (in principle, it doesn't even matter if you really measure the polarization of each photon behind the slit) the double-slit interference pattern goes away, and you get the incoherent superposition of two single-slit interference patterns. In this sense you demonstrate "particle properties" of the single photons with this variant of the experiment.

 

[hutchphd]

That said, even in the cavity case the radiation is a coherent state, not a Fock state, and so it is not in an eigenstate of energy and the E in the Planck relation still does not describe "the energy of a photon" in any sense that corresponds to an actual observation. It is an abstract "energy" that appears in the distribution function.

I fear we are in fact adrift on the semantic sea.
When the thermodynamic counting is done, the associated parameter E involves projection onto the energy eigenbasis. This is a "measurement" only on paper, so I apologize for being unclear. But that is the "energy of a photon" in my mind.

 

[vanhees71]

Just to clarify the terminology:

In a cavity (which is easier because of the somewhat delicate "infinite-volume limit" in QFT) you have
$$\hat{H}=\sum_{\lambda} \sum_{\vec{k}} |\vec{k}| \hat{N}(\vec{k},\lambda).$$
A complete set of energy eigenstates of the free em. field thus is given by the corresponding Fock basis,
$$|\{N(\vec{k},\lambda) \}_{\vec{k,\lambda}} \rangle=\prod_{\vec{k},\lambda} \frac{1}{\sqrt{N(\vec{k},\lambda)!}} \hat{a}^{\dagger N(\vec{k},\lambda)} |\Omega \rangle.$$
The energy eigenvalues are
$$E[\{N \}]=\sum_{\lambda,\vec{p}} |\vec{k}| N(\vec{k},\lambda),$$
i.e., each occupied single-photon state contributes $E_{\vec{k}}=|\vec{k}|=\omega$.

The thermal state is given by the statistical operator,
$$\hat{\rho}=\frac{1}{Z} \exp(-\hat{H}/T)$$
with the partition sum
$$Z=\prod_{\vec{k}} \left [\frac{1}{1-\exp(-|\vec{k}|/T)} \right]^2,$$
where each factor comes from summing over $N(\vec{k},\lambda) \in \{0,1,2,\ldots\}$. With the two cases $\lambda=\pm 1$ giving the same factor (that's why each factor is squared in the product).

A single-mode coherent state is given by
$$|\alpha,\vec{k},\lambda \rangle=\sum_{n=0}^{\infty} \exp(-|\alpha|^2/2) \frac{\alpha^n}{n!} \hat{a}^{n}(\vec{k},\lambda) |\Omega \rangle.$$
The photon numbers are Poisson distributed,
$$P[N(\vec{k},\lambda)=n]=\exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!}$$
with
$$\langle n \rangle =\langle n^2 \rangle-\langle n \rangle^2=|\alpha|^2$$
The mean energy is
$$\langle E \rangle=|\alpha|^2 \omega, \Delta E=|\alpha| \omega.$$

What is now the precise question, we try to answer in this thread, concerning either of these states, loosely mentions in this thread?

 

[PeterDonis]

When the thermodynamic counting is done, the associated parameter E involves projection onto the energy eigenbasis.

But if you do that with a state that is not an energy eigenstate, such as a coherent state, you get either a superposition of multiple energy eigenstates with different amplitudes, or an expectation value (depending on what you mean by "projection"). Neither of these are "the energy of a photon".

But that is the "energy of a photon" in my mind.

I disagree. See above.

 

[vanhees71]

The energy of a single photon is of course well defined, because "a photon" is described by the single-photon Fock state, where $N(\vec{k},\lambda)=1$ and $N(\vec{k}',\lambda')=0$ for all $\vec{k}' \neq \vec{k}$, $\lambda \neq \lambda'$. This is an eigenstate of the em.-field energy with the said eigenvalue, $\omega=|\vec{k}|$. In the case of the now discussed "cavity QED" this is a proper state (in contradistinction to the infinite-volume case, where a plane wave never can represent a proper state).

 

[PeterDonis]

The energy of a single photon is of course well defined

For a Fock state, yes, as you say. But that does not mean the term "the energy of a photon" can be used in a meaningful sense for states that are not Fock states, which is what other posters in this thread are trying to do.