# What is the relation between electromagnetic wave and photon?

At the end of this nice video (https://youtu.be/XiHVe8U5PhU?t=10m27s), she says that electromagnetic wave is a chain reaction of electric and magnetic fields creating each other so the chain of wave moves forward.

I wonder where the photon is in this explanation. What is the relation between electromagnetic wave and photon?

• Please see my answer here. You can understand Willis Lamb's frustration and the waves and normal modes describe the electromagnetic field. Photons are then the changes of number state of each normal mode - they are like the discrete "communications" the whole EM field has with the other quantum fields of the World that make up "empty space". One can reinterpret this statement as Maxwell's equations being the propagation equation for a lone "photon", but only in terms of propagation equations for the mean of electric and magnetic field .... Dec 19, 2013 at 0:48
• ...observables when the EM field is in a superposition of $n=1$ Fock states (so it is "one photon propagating"). Dec 19, 2013 at 0:49
• Jul 29 at 19:45

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field.

QED is a complicated theory, so while it is possible to do calculations directly in QED we often find it simpler to use an approximation. The wave theory of light is often a good approximation when we are looking at how light propagates, and the particle theory of light is often a good approximation when we are looking at how light interacts i.e. exchanges energy with something else.

So it isn't really possible to answer the question where the photon is in this explanation. In general if you're looking at a system, like the one in the video, where the wave theory is a good description of light you'll find the photon theory to be a poor description of light, and vice versa. The two ways of looking at light are complementary.

For example if you look at the experiment described in Anna's answer (which is one of the seminal experiments in understanding diffraction!) the wave theory gives us a good description of how the light travels through the Young's slits and creates the interference pattern, but it cannot describe how the light interacts with the photomultiplier used to record the image. By contrast the photon theory gives us a good explanation of how the light interacts with the photomultiplier but cannot describe how it travelled through the slits and formed the diffraction pattern.

• This is news because all QM teachers told me that photons abstractions, proposed by QED, which is more exact than wave discription. However, this should not stop us from figuring out how two are related. Actually quanta = particles.
– Val
Dec 20, 2013 at 18:09
• @Val The way we actually calculate things in QED is with a perturbative expansion that involves photons. The underlying exact theory is one of several completely quantum fields. Dec 20, 2013 at 19:26
• There is a sense in which the classical description of light is retrieved as the classical limit of a coherent state of photons. I would say that this would be an appropriate answer to "where is the photon in the classical wave theory of light?" May 4, 2016 at 18:01
• @Prahar Yes, but you just said it yourself - that's not the reality. That's just "how it fits in the models"- it doesn't help you outside of the constraints of the models, and that's exactly what the OP is asking here. In the classical wave theory of light... there's no photons. Not one per wave, not "infinite amounts" per wave, just no photons, period. May 5, 2016 at 11:32
• I think that "excitation of a field instead of waves and particles" is one interpretation, and probably not the most popular one. Many people view fields only as a handy mathematical tool. Sep 30, 2018 at 7:56

In this link there exists a mathematical explanation of how an ensemble of photons of frequency $$\nu$$ and energy $$E=h\nu$$ end up building coherently the classical electromagnetic wave of frequency $$\nu$$.

It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons in a two slit experiment might give you an intuition of how even though light is composed of individual elementary particles, photons, the classical wave pattern emerges when the ensemble becomes large.

Figure 1. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

• the link to the blogpost you have provided above (the first link) is open to invited readers only. can you please provide any alternative document/post on how an ensemble of photons build classical electromagnetic wave. Jul 10 at 6:47
• Yes, unfortunately the user discontinued the open blog at the time I gave the link. Try arxiv.org/abs/1201.5536 Jul 10 at 8:31
• I was given this projecteuclid.org/journals/… and this projecteuclid.org/journals/… at some point. Jul 10 at 8:40

In 1995 Willis Lamb published a provocative article with the title "Anti-photon", Appl. Phys. B 60, 77-84 (1995). As Lamb was one of the great pioneers of 20th century physics it is not easy to dismiss him as an old crank.

He writes in the introductory paragraph:

The photon concepts as used by a high percentage of the laser community have no scientific justification. It is now about thirty-five years after the making of the first laser. The sooner an appropriate reformulation of our educational processes can be made, the better.

There is a lot to talk about the wave-particle duality in discussion of quantum mechanics. This may be necessary for those who are unwilling or unable to acquire an understanding of the theory. However, this concept is even more pointlessly introduced in discussions of problems in the quantum theory or radiation. Here the normal mode waves of a purely classical electrodynamics appear, and for each normal mode there is an equivalent pseudosimple harmonic-oscillator particle which may then have a wave function whose argument is the corresponding normal-mode amplitude. Note that the particle is not a photon. One might rather think of a multiplicity of two distinct wave concepts and a particle concept for each normal mode of the radiation field. However, such concepts are really not useful or appropriate. The "Complementarity Principle" and the notion of wave-particle duality were introduced by N. Bohr in 1927. They reflect the fact that he mostly dealt with theoretical and philosophical concepts, and left the detailed work to postdoctoral assistants. It is very likely that Bohr never, by himself, made a significant quantum-mechanical calculation after the formulation of quantum mechanics in 1925-1926. It is high time to give up the use of the word "photon", and of a bad concept which will shortly be a century old. Radiation does not consist of particles, and the classical, i.e., non-quantum, limit of QTR is described by Maxwell's equations for the electromagnetic fields, which do not involve particles. Talking about radiation in terms of particles is like using such ubiquitous phrases as "You know" or "I mean" which are very much to be heard in some cultures. For a friend of Charlie Brown, it might serve as a kind of security blanket.

• Wow, Lamb is actually making my rethink me admittedly amateur perspective on the matter. This quote blew my mind: " It is very likely that Bohr never, by himself, made a significant quantum-mechanical calculation after the formulation of quantum mechanics in 1925-1926." Mar 23, 2017 at 7:54
• This is not within the mainstream physics models at present, but a peculiar proposal not validated or supported by model calculations and predictions. Aug 3, 2018 at 3:54
• @anna_v to the limited extent I understand it, I believe that if you read the whole paper and not just the snippet I quoted here you would agree that Lamb's is mainstream physics with mainstream interpretation. Aug 3, 2018 at 19:54
• @annav, then again, the chosen answer interpreting everything as fields is not necessarily mainstream physics for many physicists (or, more importantly, not necessarily correct). I think this reference deserves a reading. Sep 30, 2018 at 8:01
• @Helen Inmy opinion quantum field theory has very many calculational successes in describing particle physics, where it is mainstream . One could argue about its region of validity, as with many mathematical models. For example QCD has more success with lattice QCD as the expansions of perturbative field theory do not work. I do not think that there is a problem with photons in the standard model, and photons are their own antiparticle. So I will not go to the trouble of reading the paper ( no link provided so it means a library or a paywall) where a prominent physicist discusses new theory Sep 30, 2018 at 8:49

In order to understand the wave particle dualism you have simply to understand what time is:

In 1905, the Newtonian unique time concept was replaced by a twofold time concept of observed coordinate time and proper time - the observed time is relative and observer-dependent, and it is derived from the intrinsic proper time of the observed particle ("The time measured by a clock following a given object"). Proper time is the more fundamental time concept.

You can understand the wave particle dualism if you consider the simplest case of a photon, that is a photon moving at light speed c. The spacetime interval of such photons (which corresponds to their proper time) is zero. That means that the event of emission and the event of absorption are adjacent in spacetime, the emitting mass particle is transmitting the momentum which is called photon directly to the absorbing mass particle, without any spacetime between them. That means that the particle characteristics are transmitted directly without need for any intermediate massless particle.

However, for observers the zero spacetime interval is not observable, e.g. between Sun and Earth are observed to be eight light minutes, even if the spacetime interval of the path of the photon is zero. In spite of the direct transmission of a momentum between two mass particles, observers observe an electromagnetic wave which is filling the gap of eight light minutes.

In summary, particle characteristics are transmitted directly according to the principles of spacetime intervals and proper time, whereas the wave is transmitted according the principles of the observed spacetime manifold.

Now you will ask: What about photons which are moving slower than c (through gravity fields and through transparent media)? The answer is that here quantum effects such as nonlocality are implied. But it is important to notice that the limit case of photons in vacuum moving at c may be explained and understood classically, without need for any quantum theory.

• This is perhaps the best explanation, as space and time collapse from the perspective of the photon, making the photon to be more like portals than a particle traveling. From its perspective emission and absorption happen simultaneously as space is effectively wrinkled to bring the two points together. One correction to make is that individual photons never move slower than c because they are mass-less. Rather, wave-fronts move slower because it takes time between absorption of one photon and re-emission of a different photon during interaction with matter. Jan 17 at 21:56

The photon dilemma

It is postulated by Planck that energy is quantized. Owing to classical electromagnetic theory light is an electromagnetic field. This field satisfies a wave equation traveling at the speed of light. Hence, light is an electromagnetic wave. Light consists of photons; and thus each photon carries a unit of energy. This behavior is demonstrated by the photoelectric and Compton effects. Since light is an electromagnetic energy, photons must also carry electromagnetic field and a unit of it. While photons are quantum objects, light still governed by Maxwell's classical theory. The photon model is not critically consistent with Maxwell equations, since it has a dual nature. In fact light as a wave is well described by Maxwell. Recall that Maxwell's equations don't involve the Planck's constant, and thus can not describe the particle nature of the photon. A complete Maxwell's equations should involve this missing element. In quantum electrodynamic paradigm, the photon is brought to interact with the electrons by invoking the idea of minimal coupling where electrons and photons exchange momentum. The photon appears as a mediator between charged particles.

At the same time while a moving charged particle has its self electric field, and magnetic field that depend on the particle velocity, the photon, the carrier of the electromagnetic energy is void of these self-fields because it has no charge and mass. Thus, a charge-less photon can't have electric and magnetic fields accompanying its motion.

The appropriate Maxwell's equations should then incorporate the photon linear momentum as well as its angular momentum. In such a case the new Maxwell's equations can then describe the dual nature of the photon. Like electric charge, the angular momentum is generally a conserved quantity. The question is how one can correct for these photon proprieties? One way to achieve that is to employ quaternions that generically allow many physical properties to be joined in a single equation. This is so because the quaternion algebra is so rich, unlike the ordinary real numbers. To this end we employ the position-momentum commutator bracket, and invoked a photon wavefunction. This wavefunction is constructed from the linear complex combination of the electric and magnetic fields.

The outcome of the bracket yields three equations defining the photon electric and magnetic fields in terms of its angular momentum. These equations turn out to be very similar to those fields created by a moving charge. Thus, the electric and magnetic fields of the photon doesn't' require a charge for the photon. It is intriguing that the photon has no charge and mass but has electric and magnetic fields as well as energy. These fields should also satisfy Maxwell's equations. Doing so, yields additional electric and magnetic charge and current densities for the photon. The emergent Maxwell's equations are now appropriate to describe the photon as a quantum particle. These additional terms in Maxwell's equations are the source in describing the photon quantum electrodynamics behavior. Some emergent phenomena associated with topological insulator, Faraday's rotation effect, Hall effect and Kerr's effect could be examples of this contribution terms to Maxwell's equations.

Here are the quantized Maxwell's equations incorporating the photon linear and angular momentum. These are the electric and magnetic fields due to the photon as a particle: $$$$\vec{L}\cdot\vec{E}=-\frac{3\hbar c}{2}\,\Lambda\,, \qquad\qquad \vec{L}\cdot\vec{B}=0\,,$$$$ and $$$$\vec{B}=-\frac{2}{3\hbar c}\,(\vec{L}\times\vec{E})\,,\qquad\qquad\vec{E}=\frac{2 c}{3\hbar}(-\Lambda\,\vec{L}+\vec{L}\times\vec{B})\,.$$$$ And these are the new Maxwell's equations: $$$$\vec{\nabla}\cdot\vec{E}=-\frac{4c}{3\hbar}\,\,(\vec{B}-\frac{1}{2}\,\mu_0\vec{r}\times\vec{J})\cdot\vec{p}+\frac{2}{3\hbar c}\,\vec{E}\cdot\vec{\tau}+\frac{\partial \Lambda}{\partial t}\,,\qquad \vec{\nabla}\cdot\vec{B}=\frac{4}{3\hbar c}\,\,\vec{E}\cdot\vec{p}+\frac{2}{3\hbar c}\,\vec{B}\cdot\vec{\tau}\,,$$$$ and $$$$\vec{\nabla}\times\vec{B}=\frac{1}{c^2}\,\frac{\partial\vec{E}}{\partial t}+\frac{2}{3\hbar c}\left(\Lambda\vec{\tau}+\vec{B}\times\vec{\tau }-\frac{\vec{P}}{\varepsilon_0}\times\vec{p}\right)-\vec{\nabla}\Lambda\,,$$$$

$$$$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}-\frac{2c}{3\hbar}\left(\mu_0\vec{J}\times\vec{L}+\frac{\vec{\tau}}{c^2}\times\vec{E}+2\Lambda\,\vec{p}\right)\,,$$$$ where $$$$-\Lambda=\frac{1}{c^2}\,\frac{\partial\varphi}{\partial t}+\vec{\nabla}\cdot\vec{A}=\partial_\mu A^\mu\,.$$$$ In the standard electrodynamics $$\Lambda=0$$ represents the Lorenz gauge condition.

• This answer is very confused. The photon does not "carry" the electric or magnetic fields, it is the Standard Model mediator of the EM interaction. It looks like you've confused classical and quantum concepts. Maxwell's equations do not need to incorporate anything quantum mechanical -- they are purely classical equations. I also don't know where those equations have come from. Nov 20, 2019 at 17:29

You report that in the video it is stated that an electromagnetic wave is "a chain reaction of electric and magnetic fields creating each other so the chain of wave moves forward." I disagree with this view. There is just one wave, that of the vector potential or more generally of the four potential. The electric and magnetic fields are just derivatives of the vector potential and do not "create each other".

Rejecting this explanation we then arrive at your deeper question: "What is the relation between electromagnetic wave and photon?"

Until a few years ago I shared the opinion of Willis Lamb, that the photon is a fictive particle. I finally changed my mind because such an explanation cannot account for low intensity diffraction experiments. Indeed, how can a single atom or molecule absorb a wave that is much larger that it? Note that I don't intend to fork off a discussion on this here but want to give my interpretation. This is that the vector potential describes the probability of a photon being absorbed, just like the Schrödinger and Dirac wave functions do for an electron. Indeed the Maxwell equations in vacuum can be written as a wave equation that closely resembles the Klein-Gordon equation. This interpretation implies that the photon indeed exists as a particle, much smaller than an atom and at least as small as a nucleon.

• "how can a single atom or molecule absorb a wave that is much larger that it? ", the same question can be asked how can an electrically small antenna ($"dimension"<< \lambda$), say, a Hertz dipole absorb an essentially infinite plane wave. It can, I have seen it; all waves all the way down, no photons needed... Nov 8, 2018 at 15:55
• @hyportnex your argument can easily be used to support the photon concept. Nov 8, 2018 at 19:49
• I have not seen any attempt neither do I believe that, say, a 5cm long ferrite loaded loop antenna's operation at around 550kHz can be usefully explained via photons and quantum physics but, please, go ahead. Nov 8, 2018 at 20:53
• @hyportnex your example pertains to the limit of many photons. That is why no QM is needed. Nov 8, 2018 at 23:30

What are photons?

Photons get emitted every time when a body has a temperature higher 0 Kelvin (the absolute zero temperature). All bodies, surrounding us (except black holes) at any time radiate. They emit radiation into the surrounding as well as they receive radiation from the surrounding. Max Planck was the physicist who found out that this radiation has to be emitted in small portions, later called quanta and even later called photons. Making some changes in the imagination of how electrons are distributed around the nucleus, it was concluded that electrons get disturbed by incoming photons, by this way gain energy and give back this energy by the emission of photons. And photons not only get emitted from electrons. The nucleus, if well disturbed, emits photons too. Such radiations are called X-rays and gamma rays.

EM radiation is the sum of all emitted photons from the involved electrons, protons and neutrons of a body. All bodies emit infrared radiation; beginning with approx. 500°C they emit visible light, first glowing in red and then shining brighter and brighter. There are some methods to stimulate the emission of EM radiation. It was found out that beside the re-emission of photons there is a second possibility to generate EM radiation. Every time, an electron is accelerated, it emits photons. This explanation helps to understand what happens in the glow filament of an electric bulb. The electrons at the filament are not moving straight forwards, they bump together and running zig-zag. By this accelerations they lose energy and this energy is emitted as photons. Most of this photons are infrared photons, and some of this photons are in the range of the visible light. In a fluorescent tube the electrons get accelerated with higher energy and they emit ultraviolet photons (which get converted into visible light by the fluorescent coating of the glass). Higher energy (with higher velocity) electrons reach the nucleus and the nucleus emits X-rays. As long as the introduced energy is a continuous flow, not one is able to measure an oscillation of EM radiation.

What are EM waves?

Using a wave generator it is possible to create oscillating EM radiation. Such radiations are called radio waves. It was found out that a modified LC circuit in unit with a wave generator is able to radiate and that it’s possible to filter out such a modulated radiation (of a certain frequency) from the surrounding noisy EM radiation.

So the wave generator has a double function. The generator has to accelerate forward and backward the electrons inside the antenna rod and by this the photons of the radio wave get emitted, and the generator makes it possible to modulate this EM radiation with a carrier frequency. It has to be underlined that the frequency of the emitted photons are in the IR range and sometime in the X-ray range. There is an optimal ratio between the length of the antenna rod and the frequency of the wave generator. But of course one can change the length of the rod or one can change the frequency of generator. This changes the efficiency of the radiation to the needed energy input only. To conclude from the length of the antenna rod to the wavelength of the emitted photons is nonsense.

What is the wave characteristic of the photon?

Since the electrons in an antenna rod are accelerated more or less at the same time, they emit photons simultaneous. The EM radiation of an antenna is measurable and it was found out that the nearfield of an antenna has two components, an electric field component and a magnetic field component. This two components get converted in each other, the induce each other. At some moment the transmitting energy is in the electric field component and otherwise the energy is in the magnetic field component. So why not conclude from the overall picture to the nature of the involved photons? They are the constituents which make the radio wave.

• The two components do not induce each other, though it's a common misconception (that's what I've been taught in school as well :-). Because of how wide that misconception is, animations now usually show both the electric and magnetic field in phase, to prevent confusion. Jul 21, 2016 at 9:20
• The final figure here shows the $E$ and $B$ fields oscillating a quarter-turn out of phase. For waves in vacuum that's incorrect; $E$ and $B$ should be in phase.
– rob
Dec 15, 2016 at 19:50
• @HolgerFiedler If the fields are a quarter-turn out of phase, the average value for the Poynting vector is zero and the wave is not transmitting any energy.
– rob
Dec 16, 2016 at 6:51
• @rob Than how the energy transfer in the near field of an antenna work? And how a standing EM wave inside a box work? Dec 16, 2016 at 8:07
• Those would make good follow-up questions; I don't know if I can answer completely in a comment.
– rob
Dec 18, 2016 at 5:42

The reason Maxwell theory cannot describe the photon is that the radiation phenomena is mutual energy phenomena.

It is not a self-energy phenomena. traditional solutions of electromagnetic fields make mistakes here. Mutual energy phenomena includes, mutual energy theorem, mutual energy flow flow theorem, mutual energy principle. All this relates mutual inductance. Self-energy phenomena includes self-energy flow (Poynting vector energy flow), self-energy principle(self-energy flow do not carry energy). All this relates to the self-inductance.

Energy conservation law

How to describe mutual energy radiation phenomena? Maxwell equations 4 formula should be add another formula which is the energy conservation law. Assume there are $$N$$ current sources:$$\boldsymbol{J}_{i}$$, $$i=1,...N$$. The corresponding fields are $$\xi_{i}=[\boldsymbol{E}_{i},\boldsymbol{H}_{i}]$$, One current $$\boldsymbol{J}_{i}$$ will offer another current $$\boldsymbol{J}_{j}$$ some power, $$$$P_{ij}=\iiint_{V}(\boldsymbol{J}_{j}\cdot\boldsymbol{E}_{i})dV\label{eq:1}$$$$

The above is the power current $$\boldsymbol{J}_{i}$$ lost. This power is received by the current $$\boldsymbol{J}_{j}$$.

$$$$P_{ji}=\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j})dV\label{eq:2}$$$$ is the power current $$\boldsymbol{J}_{j}$$ give to the current $$\boldsymbol{J}_{i}$$. When $$\boldsymbol{J}_{i}$$ lost some energy this energy, it will be received by current $$\boldsymbol{J}_{j}$$. Hence, the total energy will not change, that means,

$$$$\intop_{t=-\infty}^{\infty}(P_{ji}+P_{ji})dt=0\label{eq:3}$$$$

Consider all $$N$$ current sources, there is, $$$$\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\intop_{t=-\infty}^{\infty}dt\iiint_{V}(\boldsymbol{J}_{j}\cdot\boldsymbol{E}_{i})dV=0\label{eq:4}$$$$

This formula is self-explanatory (introduced by Shuang-ren Zhao). It should add to the Maxwell equations.

Mutual energy principle Another formula which should also add to Maxwell equation which is the mutual energy principle (introduced by Shuang-ren Zhao)

$$$$-\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{j})\cdot\hat{n}d\Gamma=\sum_{i=1}^{N}\sum_{J=1,j\neq i}^{N}\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j}+\frac{\partial}{\partial t}(\boldsymbol{E}_{i}\cdot\boldsymbol{D}_{j}+\boldsymbol{H}_{i}\cdot\boldsymbol{B}_{j}))dV\label{eq:5}$$$$

The mutual energy principle can be derived from Maxwell equations by adding some conditions. The conditions are the Maxwell equation must established as pairs. In each pair there are solution for transmitting antenna and receiving antenna. Or pair for emitter and absorber. This means assume the receiving antenna and absorber also radiate waves. This also means assume the radiation is a mutual energy phenomena, that receiving antenna and absorber must also join to the radiation theory.

$$R$$ is the set of the solution of retarded wave. $$A$$ is the set of the solution of the advanced wave. $$R\cup A$$ is the set of the solutions of Maxwell's equations. $$R\cap A$$ is the set of the solution the mutual energy principle. $$R\cap A$$ is the solutions of physics. $$R\cap A$$ is possible the solution of physics, but is also possible a invalid solutions.

We also can build the electromagnetic field theory by adding the above mutual energy principle formula, then the above descriptions can be derived from the mutual energy principle.

Assume both energy conservation law and the mutual energy principle are accept by us as two new axioms. From these two laws we can prove that, $$$$\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\intop_{t=-\infty}^{\infty}dt\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{j})\cdot\hat{n}d\Gamma=0\label{eq:6}$$$$ $$\Gamma$$ is the boundary of the volume $$V$$. It can be chosen as big sphere with radius as infinity. This means there should no mutual energy flow go to the outside of our universe. This is clear a correct theorem. In order the above formula as 0, the two electromagnetic fields $$\xi_{i}=[\boldsymbol{E}_{i},\boldsymbol{H}_{i}]$$ and $$\xi_{j}=[\boldsymbol{E}_{j},\boldsymbol{H}_{j}]$$ must be one is retarded wave and another is advanced wave. The retarded wave reaches the surface at a future time. The advanced wave reaches the surface at a past time. The electromagnetic fields will not nonzero in the same time in the surface, hence the surface integral will be 0.

We can assume the current sends the retarded wave as transmitting antenna or emitter. The current sends advanced wave as receiving antenna or absorber. Hence, transmitting antenna and emitter must radiate the retarded wave. The receiving antenna and the absorber must radiate advanced wave.

about the advanced wave, Wheeler and Feynman have the absorber theory. John Cramer has the transactional interpretation of the quantum mechanics.

The mutual energy flow theorem, photon is the mutual energy flow

From the mutual energy principle the mutual energy flow theorem can be derived. Assume $$N=2$$, the mutual energy flow theorem is,

$$$$-\intop_{t=-\infty}^{\infty}dt\iiint_{V_{1}}(\boldsymbol{J}_{1}\cdot\boldsymbol{E}_{2})dV=(\xi_{1},\xi_{2})=\intop_{t=-\infty}^{\infty}dt\iiint_{V_{2}}(\boldsymbol{J}_{2}\cdot\boldsymbol{E}_{1})dV\label{eq:7}$$$$ The flowing is the mutual energy flow: $$$$(\xi_{1},\xi_{2})=\intop_{t=-\infty}^{\infty}dt\iint_{\Gamma}(\boldsymbol{E}_{1}\times\boldsymbol{H}_{2}+\boldsymbol{E}_{2}\times\boldsymbol{H}_{1})\cdot\hat{n}d\Gamma\label{eq:8}$$$$ $$\Gamma$$ is any surface which separates the volume $$V_{1}$$ and $$V_{2}$$. See the following figure for the shape of the mutual energy flow. $$\iint_{\Gamma}(\boldsymbol{E}_{1}\times\boldsymbol{H}_{2}+\boldsymbol{E}_{2}\times\boldsymbol{H}_{1})\cdot\hat{n}d\Gamma$$ is the mutual energy flow. The mutual energy flow is defined in contrast to the self-energy flow: $$\iint_{\Gamma}(\boldsymbol{E}_{1}\times\boldsymbol{H}_{1})\cdot\hat{n}d\Gamma$$ $$\iint_{\Gamma}(\boldsymbol{E}_{2}\times\boldsymbol{H}_{2})\cdot\hat{n}d\Gamma$$. $$(\xi_{1},\xi_{2})$$ is the mutual energy go through the surface $$\Gamma$$ through the mutual energy flow. Mutual energy flow do not decrease like wave. The amplitude of the wave will decrease when it propagates. The mutual energy flow will not decrease when it propagates. The mutual energy flow is very thin when it is radiated or received. The mutual energy will be thick between it its source and sink. Hence, the mutual energy flow looks very like a photon. We can say the photon actually is the mutual energy flow.

According these theory, the self-energy flow or self-energy radiation do not transfer energy in space. Self-energy flow is the normal wave solution of Maxwell equations (the Maxwell equation only for one current source). This wave are canceled by time-reversal waves. there are two kinds of time-reversal waves corresponding to retarded wave and the advanced wave. The following is a figure of photon. The emitter sends the retarded wave, the absorber sends the advanced wave. The retarded wave and the advanced wave either are reactive wave or they collapse back. The mutual energy flow bring the photon energy from the emitter to the absorber.

Wave collapse In quantum mechanics waves collapse, this can be shown as,

In the mutual energy theory, the wave collapse actually is done by a wave backward-collapse process and a mutual energy flow process:

Summary, (1) photon is not a wave, but it is the mutual energy flow. The mutual energy flow is built by the retarded wave sends from an emitter and the advanced wave sends from an absorber. (2) there are 4 waves, retarded wave, advanced wave and two time-reversal waves. All 4 waves cancel each other. However, the mutual energy flow survive. (3) Wave collapse can be described by the two phenomena: The energy is transferred through the mutual energy. The retarded wave and the advanced wave is canceled by the time-reversal waves. If this theory is interesting, the details can be google searched by the keyword mutual energy principle'' or mutual energy flow'', self-energy principle''.