Photon interference and beamforming

[antonantal]

TL;DR Summary

If different photons don't interfere with each other, how can we explain beamforming in wireless communications?

In a double slit experiment with one photon sent at a time, the wave function Ψ of the photon reaching the screen is a superposition of Ψ₁ (photon going through first slit) and Ψ₂ (photon going through second slit). At the screen, Ψ₁ and Ψ₂ interfere (i.e. the photon interferes with itself), resulting in a probability density |Ψ₁ + Ψ₂|². If the amplitudes of Ψ₁ and Ψ₂ are both equal to A, then the peak probability density (and light intensity) is (2A)² = 4A².

Now suppose we send a second photon at the same time with the first one. First photon has wave function Ψ = Ψ₁ + Ψ₂, and second photon Φ = Φ₁ + Φ₂. Again, suppose Ψ₁, Ψ₂, Φ₁, Φ₂ have amplitudes A. Then, I imagine we can have three cases:

1. Each photon interferes with itself and with the other photon, resulting in a probability density |Ψ₁ + Ψ₂ + Φ₁ + Φ₂|², so peak light intensity (4A)²=16A²
2. Each photon only interferes with itself, resulting in a probability density |Ψ₁ + Ψ₂|² + |Φ₁ + Φ₂|², so peak light intensity 4A² + 4A² = 8A²
3. Each photon only interferes with the other, also resulting in peak light intensity = 8A²

Case 1 is disproven by experiment, so we are left with 2 or 3. And because single photon interference is observed experimentally, I conclude that case 2 is what happens. I guess this is why Dirac stated in his QM book that "each photon then interferes only with itself. Interference between two different photons never occurs".

Finally, suppose we replace the 2 slits with 2 antennas, and light with microwaves (lower frequency but still photons), a setup commonly used in wireless communications for beamforming. If we send the same signal on the 2 antennas, the resulting radiation pattern will have highs and lows depending on the angle, and we can control the angle by adding a phase shift between the antennas. In this case we clearly have interference but we cannot consider that each photon goes through both antennas and interferes only with itself. So, are different photons actually interfering with each other?

Clearly something is wrong in my reasoning, but I can't figure out what. Any help would be appreciated.

 

[PeroK]

I can give you part of the answer, assuming I understand the idea of beamforming.

The photon is not a point particle that follows a specific trajectory. In the double-slit experiment it may appear that a photon moves along an almost classical trajectory with a modicum of quantum uncertainty, but that is a simplified heuristic explanation.

You have a more complex setup and each photon, at least heuristically, is a superposition of all possible paths through the equipment. This results in a focused beam when the probabilistic QM calculations are done.

The other point to note is that the QM theory of light must produce classical EM. But, it's not always simple to map QM concepts from first principles through to the EM theory.

 

[vanhees71]

A classical electromagnetic wave is represented in quantum (field) theory as a socalled coherent state:
https://en.wikipedia.org/wiki/Coherent_state

 

[Cthugha]

Allow me to cite Nobel prize winner Roy Glauber (The whole article I quote from can be found here: https://arxiv.org/abs/nucl-th/0604021 ):

"When you read the first chapter of Dirac’s famous textbook in quantum mechanics [8], however, you are confronted with a very clear statement that rings in everyone’s memory. Dirac is talking about the intensity fringes in the Michelson interferometer, and he says, Every photon then interferes only with itself. Interference between two different photons never occurs.
Now that simple statement, which has been treated as scripture, is absolute nonsense."

Dirac made this statement way before the advent of quantum optics at a time when light was still poorly understood. All fields of course always interfere in the sense that you can always add two fields in a phase-sensitive manner. However, what we observe in typical interference patterns is usually a time or ensemble average over long times or many repetitions. In order to see an interference pattern, the relative phase between the two interfering beams is the relevant quantity that determines the pattern. If you take a single light field (which is what Dirac actually seems to imply by the term single photon) and split it into two, the relative phase between those fields is just given by geometric factors as both partial beams originate from the same field. Therefore, the relative phase is stable. If you take two different light fields originating from different sources, their phase is given by the emission process and varies randomly on short time scales. In the ensemble or time average you will then just average over all relative phases which washes out any interference pattern.

However, if you take two different light sources and ensure their relative phase is stable long enough to record an interference pattern, you will see that they interfere. This has been demonstrated experimentally time and again, e.g.:
Old publication from the 90ies - probably behind a paywall
Newer publication - should be open access

 

[PeroK]

Allow me to cite Nobel prize winner Roy Glauber

"Now that simple statement, which has been treated as scripture, is absolute nonsense."

One wonders what Glauber thought of Newton's Principia, let alone his corpusclar theory of light?

Dirac could do more than advance our understanding of quantum mechanics. It's impossible to future-proof theoretical physics.

 

[Lord Jestocost]

One should never read a statement without its context.

“Some time before the discovery of quantum mechanics people realized that the connection between light waves and photons must be of a statistical character. What they did not clearly realize, however, was that the wave function gives information about the probability of one photon being in a particular place and not the probable number of photons in that place. The importance of the distinction can be made clear in the following way. Suppose we have a beam of light consisting of a large number of photons split up into two components of equal intensity. On the assumption that the beam is connected with the probable number of photons in it, we should have half the total number going into each component. If the two components are now made to interfere, we should require a photon in one component to be able to interfere with one in the other. Sometimes these two photons would have to annihilate one another and other times they would have to produce four photons. This would contradict the conservation of energy. The new theory, which connects the wave function with probabilities for one photon gets over the difficulty by making each photon go partly into each of the two components. Each photon then interferes only with itself. Interference between two different photons never occurs.”

—Paul Dirac, The Principles of Quantum Mechanics, Fourth Edition, Chapter 1

 

[Cthugha]

One should never read a statement without its context.
[...]

Indeed. However, if one reads Glauber's article, it is exactly this full statement which he focuses on. When Hanbury Brown and Twiss performed their famous experiment which is considered as the birth of quantum optics by some, it was met with heavy criticism because many people thought that the two-photon interference seen there contradicts conservation of energy due to the line of reasoning presented by Dirac.

 

[antonantal]

If Dirac's statement is false, and two different photons do interfere with each other (assuming their respective sources are in perfect phase synchronization), and they also interfere each with itself, then don't we end up with a peak intensity that is 2 times greater than observed (case 1 from my initial post)?

 

[Cthugha]

If Dirac's statement is false, and two different photons do interfere with each other (assuming their respective sources are in perfect phase synchronization), and they also interfere each with itself, then don't we end up with a peak intensity that is 2 times greater than observed (case 1 from my initial post)?

I do not get what you mean. In case 1, the observed intensity will be 16 \(\displaystyle A^2\). The peak intensity in a double slit for constructive interference is 4 times the peak intensity that you observe when only one slit is open. For example for the first slit, the field present is \(\displaystyle \psi_1+\Phi_1\), so the field there is 2A for constructive interference which yields an intensity of $4 A^2$ which is exactly one quarter of what you get in the double slit case. This is what you observe in experiments.

 

[PeroK]

If Dirac's statement is false, and two different photons do interfere with each other (assuming their respective sources are in perfect phase synchronization), and they also interfere each with itself, then don't we end up with a peak intensity that is 2 times greater than observed (case 1 from my initial post)?

What does "two different photons" actually mean? You may have a two-photon state, in which case the photons are indistinguishable. That system may have superposition that constructively or destructively interferes with itself.

The analogy for two electrons would be exchange forces, which is a QM implication of their indistinguishability and the anti-symmetric constraint on the two-electron wave function.

 

[sophiecentaur]

The peak intensity in a double slit for constructive interference is 4 times the peak intensity that you observe when only one slit is open.

I need a bit of help with understanding this topic. In the world of RF, this is no surprise. If you disconnect one of a pair of antennae then the total Power will reduce by 3dB and the relative amplitude of the maximum will also reduce by 3dB, resulting in 1/4 maximum received power on axis.
Reducing transmitter power will still produce the same antenna pattern for the two antennae working together. I imagine that's much easier than doing the same thing with lasers but to drop the transmitter power to individual RF photons would be problematical.

I see that, once we are dealing with two 'independent' photon sources, Dirac's view could be reasonable but when the position of any photon in time and space cannot be known, why should the two photon sources not interact in the same way as just two normal level beams? i.e. why no interaction between photons when the conditions are appropriate?

This puts me in mind of how a pair of highly accurate oscillators will interact and 'pull each other', even when operating some distance between.

Please put me out of my misery.

 

[vanhees71]

You have to use the correct description of photons, which is QED, not some vague picture that it outdated since 1926 when QED has been first formulated by Jordan ;-).

 

[WernerQH]

Now suppose we send a second photon at the same time with the first one.

How do you do that?

The experimental situations that you describe are adequately described by classical physics. And Maxwell's equation are linear, you can just scale up the electric field. The interesting (didactic?) thing happens when you reduce the intensity (proportional to the square of the field). We know that at the lowest intensities emission and absorption of radiation can no longer be considered continuous, which is why we introduce the concept of photons. But you should not think of these photons as distinguishable particles, each carrying its own share of electromagnetic field. You have realized yourself that such a description becomes awkward. Dirac is very careful to distinguish between a "probable number of photons" and "probabilities for one photon".

An antenna can be described in QED as a coordinated (coherent!) pattern of microscopic currents (emission events), and it permits the calculation of the induced currents (absorption events) in the detector (another "antenna"). But you always have to add up all contributions, all possible paths the photons could have taken. It doesn't make sense to think of an "individual" photon having a particular path.

 

[antonantal]

Shouldn't basic QM (instead of QED) be enough for describing what happens in an idealized double slit experiment?
I am adding a mathematical description of my reasoning further below, maybe it is easier to spot if/where I go wrong.

What does "two different photons" actually mean? You may have a two-photon state, in which case the photons are indistinguishable. That system may have superposition that constructively or destructively interferes with itself.

Are you thinking about entangled photons? I am thinking two photons that are not entangled and are emitted at exactly the same time from two sources that are in perfect phase sync (an ideal case).

I do not get what you mean. In case 1, the observed intensity will be 16 \(\displaystyle A^2\). The peak intensity in a double slit for constructive interference is 4 times the peak intensity that you observe when only one slit is open. For example for the first slit, the field present is \(\displaystyle \psi_1+\Phi_1\), so the field there is 2A for constructive interference which yields an intensity of $4 A^2$ which is exactly one quarter of what you get in the double slit case. This is what you observe in experiments.

Let's consider the (position, momentum) coordinates $(x, p)$. The x-axis is along the plate (that has the slits), and the slits are two points on this axis: $x_1$ and $x_2$.
$p$ is the momentum component along the x-axis only, which determines the angle at which the photon will be detected on the screen. This means that in order to calculate the peak light intensity ratios between different cases (single slit, double slit, etc.), we just need to calculate the ratios between peak probability densities of $p$.
1) Single slit, single photon
The position-representation wave function is $\psi(x) = \delta(x-x_1)$
$\Rightarrow$the momentum-representation wave function is its Fourier transform: $$\tilde \psi(p) = \frac 1 {\sqrt {2\pi}} \int \delta(x-x_1) e^\frac {-ipx} {\hbar} = \frac 1 {\sqrt {2\pi}} e^\frac {-ipx_1} {\hbar}$$
$\Rightarrow$the probability density of p is: $$P(p) = |\tilde \psi(p)|^2 = \frac 1 {2\pi}$$
2) Double slit, single photon
$$\psi(x) = \frac 1 {\sqrt 2}[ \delta(x-x_1) + \delta(x-x_2)]$$
$$\Rightarrow \tilde \psi(p) = \frac 1 {2\sqrt {\pi}} \int [ \delta(x-x_1) + \delta(x-x_2)] e^\frac {-ipx} {\hbar} = \frac 1 {2\sqrt {\pi}} [e^\frac {-ipx_1} {\hbar} + e^\frac {-ipx_2} {\hbar}]$$
$$\Rightarrow P(p) = |\tilde \psi(p)|^2 = \frac 1 {2\pi} \left[1 + cos\left(\frac {p(x_1-x_2)} {\hbar}\right)\right]$$
The peak value of this is 2 times the one from single slit (since cos has max value 1).
The usual factor of 4 (known from optics) is because 2 slits also mean 2 times more photons in general, but now we are still using a single photon (as in the single slit case).

3) Double slit, two photons
Now I would expect to get the usual factor of 4 (2 times previous case), but I actually get a factor of 8 (4 times previous case). So, my question is what am I doing wrong here:
The two photons have equal wave functions (calculated in previous case):
$$\tilde \phi(p) = \tilde \psi(p) = \frac 1 {2\sqrt {\pi}} [e^\frac {-ipx_1} {\hbar} + e^\frac {-ipx_2} {\hbar}]$$
Each photon interferes with itself and with the other photon $\Rightarrow$
$$P(p) = |\tilde \phi(p) + \tilde \psi(p)|^2 = |2\tilde \psi(p)|^2 = 4 \frac 1 {2\pi} \left[1 + cos\left(\frac {p(x_1-x_2)} {\hbar}\right)\right] $$
The peak value of this is 8 times the one from single slit (instead of 4 as observed).

 

[PeroK]

Shouldn't basic QM (instead of QED) be enough for describing what happens in an idealized double slit experiment?

Basic QM is non-relativistic so shouldn't really be used to study photons, which are fundamentally relativistic.

That said, you can construct a heuristic argument based on the HUP (Heisenberg Uncertainty Principle), but you can't push that too far.

Moreover, even basic QM demands that you treat a system of two indistinguishable particles as precisely that.

In your final case, two photons, you have taken too literally the heuristic that "different photons can interfere with each other". They do not in the simplistic way you imagine.

Interference, it seems, is a broad term. And the interference observed in beamforming is not simple particle on particle interference - which is definitely not a thing in any QM.

As I mentioned above the nearest analogy i can think of is exchange forces for two electrons. You ought to read up on that. No one calls that interference, although the point is that a system of two or more electrons cannot be treated as independent particles. The indistinguishabilty of the electrons constrains the wavefunction and produces a different result than for two distinguishable particles. It would be wrong to call that interference and say that two electrons interfere with each other.

Instead, you need a proper treatment of the quantized EM field.

Your happy-go-lucky approach assuming basic QM for photons, with your own rules for distinguishability and interference is not theoretically valid at all.

Note that if you know the answer, then you can usually construct a naive model to give you that answer. But, even if you succeed in getting the right answer eventually, there is not necessarily any validity in the method.

 

[Cthugha]

Now I would expect to get the usual factor of 4 (2 times previous case), but I actually get a factor of 8 (4 times previous case). So, my question is what am I doing wrong here:

You just compare to the wrong results. In the double slit you need to get 4 times the intensity of a comparable result with no interference happening, which would be:
$$|\tilde \psi(p)|^2 + |\tilde \phi(p)|^2=\frac{1}{\pi}$$

So you would expect 4 times this value for the peak intensity in the interference case which is exactly what you calculated. There is no magic at work here by introducing photons. The peak intensity you get in the interference case is always 4 times the intensity you would observe for having the same total intensity present. You increased the total intensity present by having two photons present and did not consider this in your bookkeeping.

 

[vanhees71]

Shouldn't basic QM (instead of QED) be enough for describing what happens in an idealized double slit experiment?

Not if you discuss photons. Photons cannot be described by "basic QM" at all! You necessarily need quantum field theory. Particularly photons cannot be described by a wave function in the sense of "1st quantization".

The double-slit experiment with massive particles at small enough energies can be described with non-relativistic QM in the 1st-quantization formalism, of course. The math is, however, flawed also within this approach. For a correct treatment of the double-slit experiment, see, e.g., the Feynman Lectures vol. III:

https://www.feynmanlectures.caltech.edu/III_01.html

 

[antonantal]

You just compare to the wrong results. In the double slit you need to get 4 times the intensity of a comparable result with no interference happening, which would be:
$$|\tilde \psi(p)|^2 + |\tilde \phi(p)|^2=\frac{1}{\pi}$$

But this would mean to compare double-slit/two-photons to a case of single-slit/two-photons. In this comparison, the average number of photons (per emission interval) going through each slit of the double-slit is 1 while the number of photons (in same interval) going through the single-slit is 2. The experiment (from optics at least) assumes that each slit passes the same amount of light.

 

[Cthugha]

But this would mean to compare double-slit/two-photons to a case of single-slit/two-photons. In this comparison, the average number of photons (per emission interval) going through each slit of the double-slit is 1 while the number of photons (in same interval) going through the single-slit is 2. The experiment (from optics at least) assumes that each slit passes the same amount of light.

No, that is not correct. You always need to compare the same total intensity arriving. Typically, you also do not compare a double slit to an individual slit, but a double slit to the sum of intensity patterns of two individual slits, where each slit gets the same intensity as one of the double slits. You can only assume that "each slit passes the same amount of light" if you also consider the same amount of slits.

You can do a different comparison, but then you need to consider how the scaling works correctly as we see here.

 

[Delta2]

BTW I think the group of engineers that invented beamforming did so by considering the classical wave model of electromagnetic field, I don't think they tried to explain it using QFT/QED/QM.

 

[antonantal]

No, that is not correct. You always need to compare the same total intensity arriving. Typically, you also do not compare a double slit to an individual slit, but a double slit to the sum of intensity patterns of two individual slits, where each slit gets the same intensity as one of the double slits. You can only assume that "each slit passes the same amount of light" if you also consider the same amount of slits.

You can do a different comparison, but then you need to consider how the scaling works correctly as we see here.

In classical physics, the reason why double-slit has 4 times peak intensity of single-slit is this:
- at the single-slit, the electric field vector has magnitude $E_0$ and the wave has intensity $I_0 \sim E_0^2$
- at each slit of the double-slit, the electric field vector has the same magnitude $E_0$, and if they are in phase at the screen, the vector sum will be $2E_0$ and intensity $I \sim (2E_0)^2 = 4I_0$
The fact that the field is $E_0$ at each of the 2 slits in the double-slit and in the single-slit tells me that the intensity ($\sim E_0^2$) is the same at each of the 3 slits, which means that each slit passes the same amount of light. In other words double-slit uses 2 times more light than single-slit, so it is correct to compare double-slit/two-photons to single-slit/single-photon and expect to get the factor of 4.
Is this not true?

 

[antonantal]

In your final case, two photons, you have taken too literally the heuristic that "different photons can interfere with each other". They do not in the simplistic way you imagine.

Well, I just considered that interference means adding together the wave functions.
And this resulted in probability density $P(p) = |\tilde \phi(p) + \tilde \psi(p)|^2$
The only way we can get the expected factor of 4 is if $|\tilde \phi(p) + \tilde \psi(p)|^2 = |\tilde \phi(p)|^2 + |\tilde \psi(p)|^2$
That would mean the wave functions of the two photons are "orthogonal", so they do not interfere. This takes us back to Dirac's statement

 

[PeroK]

Well, I just considered that interference means adding together the wave functions.
And this resulted in probability density $P(p) = |\tilde \phi(p) + \tilde \psi(p)|^2$
The only way we can get the expected factor of 4 is if $|\tilde \phi(p) + \tilde \psi(p)|^2 = |\tilde \phi(p)|^2 + |\tilde \psi(p)|^2$
That would mean the wave functions of the two photons are "orthogonal", so they do not interfere. This takes us back to Dirac's statement

Sure, if you have interfering, distinguishable photons. The proper calculation will involve the relevant multi-photon states of the EM field.

I believe Dirac was correct and this is not photon on photon interference. Perhaps that's semantics, but distinguishable photons is not semantics; that's theoretically wrong.

 

[vanhees71]

BTW I think the group of engineers that invented beamforming did so by considering the classical wave model of electromagnetic field, I don't think they tried to explain it using QFT/QED/QM.

Well, the math for the single photon is almost the same since it's a linear theory. There it doesn't matter that the fields in QFT are operator valued. Nevertheless, there's no way to describe photons in the 1st-quantization formalism in a consistent way.

 

[PeterDonis]

I believe Dirac was correct and this is not photon on photon interference.

The key point to look at is whether your experiment has one light source or multiple light sources. In the double slit experiment, there is only one light source, which gets "split into two beams" because of the presence of two slits. In that case, which is the case Dirac was considering in the quote given, it is correct (at least heuristically) to say that photons do not interfere with each other, each photon only "interferes with itself".

However, in an experiment with multiple light sources, such as the Hanbury Brown Twiss experiment, if the relative phase of the light sources is controlled accurately enough, then photons from the two sources can interfere with each other. There is no way to account for that as "a photon interfering with itself", because a "single photon", whatever else you might say about it, must come from a single light source. If you have multiple light sources, you must have "multiple photons". This language is of course highly heuristic and conceals many technical issues, but it at least describes the key difference between the two types of experiments. Failing to observe that key difference can lead to confusion; I suspect it is the basic confusion the OP has.

 

[PeroK]

The key point to look at is whether your experiment has one light source or multiple light sources. In the double slit experiment, there is only one light source, which gets "split into two beams" because of the presence of two slits. In that case, which is the case Dirac was considering in the quote given, it is correct (at least heuristically) to say that photons do not interfere with each other, each photon only "interferes with itself".

However, in an experiment with multiple light sources, such as the Hanbury Brown Twiss experiment, if the relative phase of the light sources is controlled accurately enough, then photons from the two sources can interfere with each other. There is no way to account for that as "a photon interfering with itself", because a "single photon", whatever else you might say about it, must come from a single light source. If you have multiple light sources, you must have "multiple photons". This language is of course highly heuristic and conceals many technical issues, but it at least describes the key difference between the two types of experiments. Failing to observe that key difference can lead to confusion; I suspect it is the basic confusion the OP has.

My question is whether that is then an extension of the concept of interference?

I can see how the calculation works, but it feels wrong quantum mechanically. Like using the HUP for the double-slit for photons. It must be coincidence rather than theoretically robust. Like Bohr's hydrogen atom.

 

[vanhees71]

The key point to look at is whether your experiment has one light source or multiple light sources. In the double slit experiment, there is only one light source, which gets "split into two beams" because of the presence of two slits. In that case, which is the case Dirac was considering in the quote given, it is correct (at least heuristically) to say that photons do not interfere with each other, each photon only "interferes with itself".

However, in an experiment with multiple light sources, such as the Hanbury Brown Twiss experiment, if the relative phase of the light sources is controlled accurately enough, then photons from the two sources can interfere with each other. There is no way to account for that as "a photon interfering with itself", because a "single photon", whatever else you might say about it, must come from a single light source. If you have multiple light sources, you must have "multiple photons". This language is of course highly heuristic and conceals many technical issues, but it at least describes the key difference between the two types of experiments. Failing to observe that key difference can lead to confusion; I suspect it is the basic confusion the OP has.

The point of the HBT effect is that the phases of the "two photons" are uncorrelated. For a nice explanation of the HBT effect, see, e.g.,

https://arxiv.org/abs/nucl-th/9804026

 

[PeterDonis]

My question is whether that is then an extension of the concept of interference?

Why would it be? The basic idea is the same: you add amplitudes, not probabilities. That's ultimately where all quantum interference effects come from.

I can see how the calculation works

Which calculation are you referring to? Note that everything I said in my post was, as I stated, heuristic, and as I said, there are many technical issues that I didn't go into. I certainly was not talking about the workings of any particular calculation.

 

[vanhees71]

My question is whether that is then an extension of the concept of interference?

I can see how the calculation works, but it feels wrong quantum mechanically. Like using the HUP for the double-slit for photons. It must be coincidence rather than theoretically robust. Like Bohr's hydrogen atom.

The intensity interferometry was an extension of the concept of interference at the time when Hanbury Brown and Twiss published their seminal paper

https://doi.org/10.1080/14786440708520475

It was met with quite some skepticism first. Nowadays it's seen as the beginning of quantum optics. See the nice review by Gordon Baym, already quoted above:

https://arxiv.org/abs/nucl-th/9804026

 

[Charles Link]

To comment on the discussion between @antonantal and @Cthugha , for multiple slits, as the number of slits $N$ increases, the spread $\Delta \theta$ of each of the principal maxima narrows, so that the peaks get intensity proportional to $N^2$, but the $\Delta \theta$ is inversely proportional to $N$, thereby conserving energy.

One thing that has puzzled me somewhat is what happens when you have $N$ photons in the same mode of a cavity? The amplitude of the resultant E field must be proportional to the square root of $N$, but how that comes about at the individual photon level is something that I don't know if it has been completely determined. Photons each with random phases would achieve this result, at least for large $N$, but whether this result might come out of the QED is beyond me.

 

[hutchphd]

The amplitude of the resultant E field must be proportional to the square root of N,

In this case what are you considering the "resultant" E field? I am not seeing the argument here. The internal stimulated emission goes like N.

 

[Charles Link]

In this case what are you considering the "resultant" E field? I am not seeing the argument here. The internal stimulated emission goes like N.

To conserve energy, for $N$ photons, (energy=$N \hbar \omega$ ), with energy density proportional to $E^2$, we can't possibly have the photons or individual amplitudes in phase with each other, or the electric field amplitude $E$ would be proportional to $N$, and the energy density would be proportional to $N^2$.

 

[PeterDonis]

To conserve energy, for $N$ photons, (energy=$N \hbar \omega$ ), with energy density proportional to $E^2$

Remember that this is quantum physics, not classical physics, and you cannot have simultaneous eigenstates of non-commuting observables. To the best of my understanding, photon number and electric field are non-commuting observables, so there is no such thing as an $N$ photon state with energy density proportional to $E^2$ because no state can have definite values for both $N$ and $E$.

Most cavity states will not be eigenstates of $N$ (they will be more like coherent states, not Fock states), and I suspect they will not be eigenstates of $E$ either. So any relationship will involve expectation values, which can behave very differently from eigenvalues. (Note that a state that is not an eigenstate of $N$ is also not an eigenstate of the Hamiltonian, so it won't have a definite energy either, you have to work with the expectation value of energy.)

 

[vanhees71]

To comment on the discussion between @antonantal and @Cthugha , for multiple slits, as the number of slits $N$ increases, the spread $\Delta \theta$ of each of the principal maxima narrows, so that the peaks get intensity proportional to $N^2$, but the $\Delta \theta$ is inversely proportional to $N$, thereby conserving energy.

One thing that has puzzled me somewhat is what happens when you have $N$ photons in the same mode of a cavity? The amplitude of the resultant E field must be proportional to the square root of $N$, but how that comes about at the individual photon level is something that I don't know if it has been completely determined. Photons each with random phases would achieve this result, at least for large $N$, but whether this result might come out of the QED is beyond me.

That simply follows from the Hamiltonian for the cavity photons,
$$\hat{H}=\sum_k \hat{N}_k \omega_k,$$
where the sum runs over all single-photon cavity states. There's nothing mysterious in this. It's just one of the first exercises in QFT to deal with free particles/photons.

 

[Lord Jestocost]

Indeed. However, if one reads Glauber's article, it is exactly this full statement which he focuses on. When Hanbury Brown and Twiss performed their famous experiment which is considered as the birth of quantum optics by some, it was met with heavy criticism because many people thought that the two-photon interference seen there contradicts conservation of energy due to the line of reasoning presented by Dirac.

One should never mix up 'classical' interference effects with – so to speak – 'non-classical' interference effects. Dirac refers to classical interference effects which must in fact be understood as ensemble outcomes of individual entities where each of which is in a superposed state, i.e., Dirac refers to expectation values of single-photon observables. Collective many-photon interference effects should be understood in terms of many-photon observables.

Kirk T. McDonald remarks, for example, in “Bunching of Photons When Two Beams Pass Through a Beam Splitter”:

“Dirac has written [1] Each photon then interferes only with itself. Interference between two different photons never occurs. Indeed, a practical definition is that “classical” optics consists of phenomena due to the interference of photons only with themselves. However, photons obey Bose statistics which implies a “nonclassical” tendency for them to ‘bunch’.”