Then the photon will not occupy the ground state, but an excited state in that box, assuming the spectrum allows its energy as one of the allowed levels; otherwise it can only  be described as a superposition of  energy eigenstates in the box, i.e. will not have a definite energy within the box. 

In this discussion what needs to be specified, of course, is the operator, whose eigenstates are involved. For the photon, as an excitation of the electromagnetic field, the operator will be the Helmholtz operator, for example. 

Of course the same statements hold for any particle in a ``trap''. 

Dear Sofia

If the walls are perfectly reflective, why wouldn't the photon bounce around till it re-emerges from the cavity?

Do you need to specify in addition that the entrance gets closed after the photon has entered? (Unless a half-silvered mirror covering the entrance would be satisfactory, the cavity entrance would have to close extremely rapidly, for anything but a very large cavity.)

Regards - Paul

Sofia, first of all you are a bit imprecise in posing this problem. Monochromatic photons with well defined wavelength do not exist.  They can be created in finite wave packets. Nonetheless if you want to somehow create the photon inside such cavity, then, speaking colloquially, such photon  will not "fit" into cavity. The precise explanation for that provided Stam Nicolis.

Moreover the situation is the same as with a classical pulse of light. There are no differences - really. And classically it as a bit simpler to think about this and results are the same. Then of course the energy will not disappear.

So think about a light pulse, or equivalently, the photon wave packet. If it is shorter then a cavity dimension - it will simply bounce from the walls. If it is larger - well then we simply have a interference of it with it self. And as a result, excitations of many excited cavity eigenstates as Stam Nicolis has mentioned.

Dear Paul,

I don't know when the photon enters the box. The time-energy uncertainty says that the more monochromatic is the photon, the longer is its wave-packet.

As to your thought that the photon may eventually get out, yes, I also think of this possibility, but I think that first of all it would be reflected by the mirrors covering the walls. Here is the problem. The linear dimensions of the box are unfit.

Dear Stam and Radosław,

I didn't understand what you say. Let me first insist on a few things. I don't want to create the photon inside the cavity, my problem is that the photon is sent to the cavity and the question is what it would do.

Next, as I answered just now to Paul, the narrower is the frequency spectrum of the photon, the longer is its wave-packet. For sharpening the problem, assume that the wave-packet is at least twice longer than the box, s.t. it can't just bounce from wall to wall. If it enters the box at all (this is a question), it would undergo destructive interference. But the energy conservation doesn't allow that the photon disappear.

Neither do I see a possibility that the photon be split into a couple of photons of lower energies fit to the cavity. The photon has also linear and angular momentum. The latter is equal to ħ. For splitting the photon, all the conservation laws have to be satisfied and I am not sure whether it would be possible.

Then, what would happen with the photon?

To your last point: in the presence of a cavity, there is no real reason why linear momentum should be conserved. Angular momentum would presumably be conserved in a perfectly spherical cavity, but angular momentum also has an orbital part. So these conservation laws do not really cause any serious problems. I would somehow believe that you cannot ``catch'' the photon in such a cavity.

Finally there is, after all, the experimental fact that an atom in a cavity, the natural wavelength of which does not fit the cavity, will indeed not emit as readily.

Dear Sofia, there is no way that photon splits without a real interaction such as in nonlinear medium.

I suggest if you consider a simpler - one dimensional case, such as a Fabry-Perot cavity. All effect will be the same, but simpler to think of. 

First of all if you send a photon (whatever shape) to a perfectly reflective cavity from the outside, then... Photon will be perfectly reflected :) And it will not go inside the cavity. In my previous answer I assumed you somehow are creating a photon inside the cavity (e.g. by placing inside an atom that emits a photon).

If the cavity mirrors are almost perfect such as 99.9% reflection coefficient then the following scenarios can happen:

1) Photon is nearly monochromatic and it is matched to a cavity resonance - photon goes through a cavity.

2) The photon is not matched - then only 0.1% gets inside, whereas 99.9% is reflected. 

Does it help?

To better understand it I recommend you quite simple calculations to do by your own. Consider Fabry-Perot transmitance spectrum and arbitrarily shaped photons spectrum. Then, transmitted spectrum is just combination of these two. Do the inverse Fourier transform to see what passes through the cavity. 

If you want to fit a photon though a small hole as you drawn then a situation is similar as in my previous answer.

https://en.wikipedia.org/wiki/Fabry%E2%80%93P%C3%A9rot_interferometer

Dear All,

the full story is more complex !

Really the Sofia's ideal experiment of a photon entrapped in a box could be very intriguing, but in the present formulation it does not admit a precise answer, as she should be expected, since she has forgotten to specify what are the dynamic constraints for such a problem. Then, let me place Sofia's question inside my quantum gravity theory. Namely I assume that a photon, is considered in a bound state with a surrounding closed material. In such a case the photon assumes a mass-gap and becomes a massive photon. Of course one could ask yet under which conditions such a situation could be conserved. For example in a resonant cavity this could be verified. But whether such circumstances are not verified, then the massive photon can interact with the surrounding material and some quantum reaction could be produced ....

Therefore, the answer to Sofia's question could be a quantum reaction with the surrounding box.

All the best,

Agostino 

Once more ambiguous words are overused. The statement that the photon can occupy a state within the box at all *means* that the photon's state can be expressed in terms of the energy eigenstates of the Hamiltonian that describes the states in the box-and an example is provided by the Helmholtz operator-just like in any waveguide. The statement ``the dimensions of the box are fit/not fit for the photon's wavelength'' (i.e. its energy) *means* that the photon's state is/isn't one of the eigenstates of the Hamiltonian of the box. If it is, the photon will occupy that state. If it isn't, it will be some superposition of the eigenstates of the box. This is a standard exercise in quantum mechanics, for the Ritz variational method, incidentally. 

Formally, outside the box, the photon's state is expanded in a complete set of states, consistent with the boundary conditions;inside the box it's expanded in another set of states and one computes the evolution of the overlap. 

As a related exercise, I suggest solving the following problem: consider a massive particle in 1 dimension. It moves freely and is given at some initial time by a Gaussian wavepacket that is quite broad in position and quite well defined in momentum, so that the wave vector k is very sharply defined. Now at time t=0, suddenly turn on a potential

V(x) = -V0    for   0k^2/(2m), L and 2 pi/k are incommensurable, and when the uncertainty in k is not enough to fix that. I think the particle will generally speaking have a low amplitude of being caught. That is, the particle at the end of the process will be in a coherent superposition of being in- and outside the box, with the amplitude for being outside dominating. In such a case, the energy of the particle is not sharp. If you then only consider the cases in which the particle is found in the box, the energy will vary. That is, of course, not a contradiction, since the energy is only conserved on average in such a setup.

Dear Radosław,

Your comparison of my problem with entering a Fabry-Perot interferometer is a very good idea, it simplifies the judgement.

Now, I see that you infer that the probability that the photon enter the interferometer is very low. Other people conclude the same. To tell you frankly, I also believe that the photon won't enter the cavity (or the Fabry-Perot interferometer).

But I want to draw some conclusion from this, because the experiment is not trivial. I'll tell you tomorrow, now it's very late in my country.

So, thanks a lot, to you and to the other friends, and hope to continue our talk tomorrow.

Best regards to you all,

Sofia

Dear Sofia,

This is a really good question. Simple as it is stated, it turns out hard to answer, but is worth trying to. It took me and my colleagues a 21 pages paper Phys. Rev. D 91, 016005 (2015)  to address a simpler question.

The main issue is the radical change of boundary conditions the photon quantum undergoes when (instantly, or better in an insignificant amount of time) gets into the cavity.

Before entering into the cavity, the photon  is an elementary excitation of the em field in the whole space. It is defined in this way in QFT (I think you are thinking of this case, as a wave packet is only a superposition of elementary excitations), and it is described by a Fock state a(k) |o>. Notice that the a(k) and a(k)^+  operators are defined as those who annihilate or create an elementary excitation of momentum k - IN ALL SPACE R^3 - at an instant of time.

That is, these operators remove or add a normal mode of the wave equation in Minkowski space. Those modes are specified by the equation and the boundary conditions. Only by considering both together,  the allowed wave numbers emerge.

Beyond waves, in quantum physics -i.e. QFT - there are particles; photons in this case. To handle this, we need of a counting tool and the Fock treatment provides us with one, the number operator N(k) = a^+(k) a(k), which counts the number of elementary excitations of momentum k (wave number 1/k) present in the WHOLE SPACE R^3. 

Coming back to your question; before entering into the cavity there is a photon of wave number 1/k as counted by N(k) (or of wave length λ according to your original wording).

The point is that photons inside the cavity are not created or annihilated by  the same operators as before, but by operators b(k_n)^+, b(k_n)  that add or remove one n-th normal mode of the cavity. Accordingly, photons inside the cavity will be counted by the operator b(k_n)^+b(k_n).

Finally, the same quantum state will have different expansions in terms of the R^3 normal modes than in terms of the cavity normal modes. In poor words, what was one free space photon outside may become a bunch of cavity photons inside.

Mathematically this is involved as both photon representations are unitarily inequivalent!

I hope this explanation will serve to add some light to your question. [email protected]

I saw your interests in your Research Gate page

Dear Sofia,

There was a miserable errata in my earlier message.

I wrote :

"

Before entering into the cavity, the photon is an elementary excitation of the em field in the whole space. It is defined in this way in QFT (I think you are thinking of this case, as a wave packet is only a superposition of elementary excitations), and it is described by a Fock state a(k) |o>.....

"

In fact, it is described by a^+(k) |0> !

sorry for this errata!

Juan

Juan León,

I impose on my photon quite draconic restrictions. You see, inside the cavity the spectrum of allowed frequencies is discrete. Let the distance between two such frequencies be Δ. I assume that the source that emits my photon has the line-width smaller than Δ.

Of course, the wave-packet emitted by this source is very long. I don't mind.

But, in this case the reorganization of my photon inside the cavity as a superposition of cavity-allowed frequencies is impossible.

Dear Remi,

May I infer, as bottom line, that you are in favor of the idea that the photon won't enter the cavity? I repeat, the source has a very narrow line-width, which of course makes the wave-packet very long.

Dear friends,

I invite everybody to see my new question, "If the photon doesn't enter the cavity, is it though time-delayed by the cavity?".

That new question is in fact the reason why I asked whether photons enter a cavity for which they are unfit. I thought that they don't enter, but somehow, they test the cavity. My new question refers to this test.

Remi, I didn't understand you. I am familiar with coherent states and quite familiar with squeezed light. But, I don't understand how do you imagine that "cut-in" that you speak of. Please be kind and explain some more phenomenologically. And keep in mind the conservation laws.

Radosław proposed to replace the cavity by a Fabry-Perot interferometer. It's a good idea and let's adopt it, with the modification that the surfaces of the interferometer are coated with silver except for the opening, s.t. the light can leave the apparatus only through the opening.

Given this configuration, can you explain more clearly how do you see that "gradual cut-in" and noise formation of which you talk? (Truly speaking, I believe that the light won't enter the interferometer, but I am not sure.)

You are all artillery-minded. A canon ball exists independantly of its target. It still exists even when it falls at sea. A photon is a null-time particle, and does not exist independantly of its absorber. There is no "and then" for a photon. The frequency, the length, the phase and the polarity are all negociated through the intermediate medium when the transfer begins (even when the transfer fails). The final absorber may be in the cavity : no mirrors are perfect. It may be outside. No one can predict through this math vapourware.

In the real world, mirrors are not mathematical abstractions but are physical devices, with conduction electrons straying at the Fermi speed, and with a not null temperature, thus lots of straying thermal phonons..

In the real world, photons carry a momentum, which reflection excite phonons. At least one phonon per photon, or several interactions with several thermal phonons. So a shift of frequency and a small deviation from the ideal expected angle are unavoidable.

Your incoming photons will excite resonating phonons in the walls of the cavity. So the temperature of the cavity and its walls increases, as much as the number of trapped photons with shifted frequencies grows.

For the outcoming photons, they are are dispersed in direction and in frequency. With such a dispersive device, I wonder which experiment you can really achieve.

Jacques, what if the shift of frequency of which you speak, is not sufficient for shifting the photons to one of the cavity eigen-frequencies?

Beside that, at one or both boundaries between the glass and silver, the electric field E can't be zero, because the beam wavelength does not allow the formation of a standing wave with nodes on these boundaries, see the picture in the attached PDF.

Gauss law says that  ρ/ε = ∇ E, where ρ is the charge density and ε the dielectric constant. Since E ≠ 0, then a charge has to exist on this boundary.

The frequency of a photon, as well as its polarity ans length are not engraved in the marble by the emitter only, but are negociated at the handshake phase by the emitter(s) and the absorber(s), through the impedances of the intermediate media and devices.

The interaction of a photon with one or several phonons in the mirror was not considered in previous versions of TIQM. It should be completed.

For a real experiment, one has to figure the diameter of the hole considering the diameter of the Fermat spindle of each photon, depending on its frequency and the distance between the emitter and the absorber. Example at http://www.agoravox.fr/culture-loisirs/culture/article/contrafactualite-penrose-elitzur-155565

50 cm and \lambda = 0,5 µm yield to 0,3 mm diameter for the Fermat spindle.

Jacques, there are no spectra of elements?

Do you want to say that the lines of the elements that we see in the light received from stars and galaxies, are partly concocted by our detectors?

As to the Hubble law based on the red displacement, can it be a farce of our detectors? 

Are you sure that the physics of our spectrometers is as selective on the spectrum as the Mössbauer effect ? I do not think so.

The curse on astronomers (http://www.deonto-ethics.org/quantic/index.php?title=Emetteurs_chauds_et_%C3%A9vidents,_absorbeurs_discrets_et_incontr%C3%B4lables) has some features in common with the cat who walked alone, in the Just so stories by Rudyard Kipling : relatively few emitters, and all very noticeable (stars, for instance), versus all the universe as absorbers, which only a tiny few are knowable : our sensors. For the cat who walked alone, "all places are equal for me !". For a radiating star "all absorbers are equal for me !". That is a curse...

I encourage you to look at the case of two polarisers between the emitter and the purposed detector. Also to the rotating power of some molecules in solution.

Jacques, I am sorry, but your digressions from courses for astronomers to R. Kipling's cat, don't make your post very readable. People are busy and so am I.

To cut short the story, it's true that a photon emitted by an atom has a line-width, as the excited states of an atom have an energy width. But this is not a solution to my question. I can choose photons from an excited state with narrow line. I can also choose the cavity dimension D so as the distance between the cavity modes be significantly bigger than the line-width.

Best regards

So your photon is a very loooong one. At least decimetric.

Above, we have seen that the minimum diameter of the hole, say b, is millimetric, depending on the wavelength and the distance from the emitter to the final absorber.

So to have your first plating of silver playing a role (the spot not on the hole), the angle of incidence must be at Arctg (b/2D) off the perpendicular to the plates, or farther. So the supposed off-phase distance in the interferometer does not play any role. The reflections continue, until the final absorber is met, in the cavity or outside.

Your problem is a false problem.

It seems difficult to avoid the rise of temperature of the cavity, due to the mean density of photons in it, and the rises of phonons at each reflection.

Jacques, I never indicated figures, "decimetric", "millimetric". Please don't falsify what I said. YOUR photon is decimetric, not mine. I don't permit attributing me words that I didn't say.

I read though the rest of your post. Yes, my problem was wrongly posed and we have in the box a running wave. I am sorry that I neglected this question. I posted the solution to the problem on another question which is a follow-up of this one. I am very thankful to you, since due to you I realized my omission. So, I copy below the solution.

Again thank you.

Dear friends, the problem is solved!

Please open the PDF and see the figure, for understanding what I talk about. I don't promise to answer people that use other notations than in the figure.

As the figure shows, during the repeated reflections of the beam, part of the time the beam returns to the opening. Here, it undergoes partial reflection into the box (i.e. into the glass), and partial transmission to the outer space (i.e. into the air).

However, after a number of reflections, the situation appears that on the lower surface, y = 0, part of the beam falls on silver. This part will continue to run inside the box from silver wall to silver wall, i.e. from y = 0 to y = D, and vice-versa. At a certain time a lateral wall will be encountered, and reflection will happen on it too. Then, after a couple of subsequent reflections, this part of beam will fall again on the opening. Here, again, part of this part will leave the box, and the rest would be reflected into the box, and continue the trip from wall to wall. And so on.

The fact that the thickness D of the box is not equal with an integer number of λ/2, turns out to be irrelevant. Why so? Because we have a running beam, not a stationary one. That means, the wave incident on a silver surface overlaps only partially with the reflected wave. Notice that the beam is not perpendicular to the surfaces. The electric and magnetic fields are contained in the plane perpendicular to the propagation direction, s.t. the electric field has a horizontal component e_x and a vertical component e_y. At the silver surface, the phase of e_x jumps by π, s.t. there is no horizontal electric field in the metal.

You are not so free to choose the diameter of the hole : it must let in the photon, and the width of the Fermat's spindle is not completely unknown.

http://jacques.lavau.perso.sfr.fr/Physique/Microphysique/Fermat.gif

In the vacuum, an acceptable approximation is

z = \sqr{(3 . a . \lambda)} / 4

At a moderate distance of the emitter, respectively the absorber, the angle or the tangent cone is \alpha, where

\alpha^3 < \lambda /(4 R)

to add to the radius of the emitting reaction or the absorbing reaction.

http://jacques.lavau.perso.sfr.fr/Physique/Microphysique_contee.html (work in progress).

It gives reasonable results about the effect of the atmosphere on light coming from stars.

An interesting question related to my point on the escape of the ray after bouncing: does a geometry exists such that there is no escape through the entry hole by bouncing after a finite time and for a finite volume? Not easy 

Yves 

Jacque,

I am so busy right now that I can't afford reading articles. I have to read a complicated material connected with an article that I am planning already for a couple of months. It is quite an unusual article, s.t. I have to prepare it very carefully. Thus, if you explain something, whatever you can do in your own words would be much more useful than indicating articles.

About the width of the hole, I thought it enough big to let pass the transversal section of the incoming beam, with negligible truncation at the hole borders.

However, I understand that we have the same view that the problem is wrongly posed, and in the box remains a running beam. This beam will run forever, loosing of course a fraction by transmission, each time it returns and meets the hole. So, it will undergo a gradual (exponential) weakening. As to your saying that it would heat the walls, thereby loosing even more from intensity, in practice it is true. But if the effect is minor, we neglect it in calculi. It's common to do idealizations in theory, wherever it is reasonable.

With thanks again!

@Yves Pomeau.

First in french between frenchmen : Vous raisonnez en artillerie sur un corpuscule. C'est la coutume, mais en réalisme ça ne vaut rien. Tout photon a un destinataire pour exister.

Then in english for the others : Yves Pomeau thinks in artillery with a corpuscle. Sure, so is the habit. But in the real world, no photon exists without an absorber at the end of its "fly". So you can discuss the geometry and its impedance after having chosen the emitter AND the absorber. Sorry, it is not all comfortable to our tradition of anthropocentrism...

If I understand your point you refer to the measurement process, which needs an absorber. This is OK, but if the absorber is outside the cavity, it does not perturb what you call the photon, and you still have the problem of the photon inside and outside the cavity, solved, as I explained, in the short wave limit and by neglecting the absorption by the walls by geometrical optics, with reflection on the wall + Poincaré recurrence theorem showing that it takes a finite time for the incoming photon to escape through a finite size hole. No anthropocentrism there, just a long tradition of science (going back to Archimedes of Syracuse at least for the law of reflection). 

@ Yves Pomeau.

Insubordinate to propaganda, I have ceased to refer to "measurement process". We came about fourteen, fifteen milliards years too late to dictate that the laws of physics should satisfy the anthropocentrism of Werner Heisenberg, or worse the egocentrism of Eugen Wigner. We know how dearly the biologists had to pay, to conquer the right of building an impersonal science, an impersonal biology. It is a pity that, driven by their anger about the fields of horror of the 1914-1918 war, and driven by their hate of France and the frenchmen in 1925, the Göttingen Knabenphysiker have set up a personal and magic "science". Yes Augustin Fresnel and Joseph Fourier bore the original sin of been born and dead french...

However, in impersonal physics, both emission and absorption reactions have finite though fuzzy diameters, and the Fermat's spindle between them has diameters that are not at your will. Absorption reactions occur sometimes in the reflecting walls, and are not at your will. And even when you think artillery, photons remain null-time particles. Even when you do not think so, the handshake process remains held by the Dirac's equation and its généralizations to other fermions : both orthochrone and retrochrone.

 @ Yves Pomeau.

It is still possible that both the emitter and the absorber are outside, and nevertheless, the travel includes some rebounces in the cavity. You just have to calculate the necessary diameter of the hole for the real length of the real travel. Fortunately as the length increases, the angle of the tangent cone decreases :

\alpha = \sqr{\frac{3.\lambda}{4.a}}

So the minimum travel with just one return journey gives you a majorant of the diameter of the photon, so you can estimate the necessary diameter of the hole.

http://jacques.lavau.perso.sfr.fr/Physique/Microphysique_contee.html

Yves,

I see that you are preoccupied by the problem

"does a geometry exists such that there is no escape through the entry hole by bouncing after a finite time and for a finite volume?"

I gave some thought to your problem - see the attach. I tried a rectangular box - it's a particular case, your problem seems more general. To help you follow the flow of the beam, each round trip in the box is painted with another nuance of blue.

Now, I have no idea whether what I got is a consequence of Poincare's theorem, but, as you can see, the beam indeed returns to the hole. That is due to the fact that the paths become denser and denser, and the diameter of the hole is finite, i.e. it is not infinitely small.

Now, the question remains whether other geometries would permit the beam to travel infinitely, avoiding the hole. I don't think so, but for the moment it's hard for me to say something clear.

dear Sofia, 

Use Poincaré recurrence theorem, take a tiny volume in phase space (ordinary space+ space of orientation of the rays). Poincaré says that, in this space and if the total volume is finite, an initial condition in the tiny box will always return to it. If you draw this box around the hole, replace after  the entrance by a reflecting mirror, it will return to it which means that, just before and if the hole is not covered by a mirror  the ray was incoming on the hole and so has escaped in the absence of mirror. I hope it is clear enough. A fellow mathematician told me ( but I do not believe him) that the time spent inside the box before escaping can be exponential of the inverse of the diameter of the hole. From the way Poincaré Theorem is proved (one of the most clever argument I know), I do not believe this can be correct. 

Best 

Yves

However the minimum diameter of the hole is not at your free fantasy. Remember the Fresnel's physical optics, 1821.

Yves,

you don't need the phase-space as in Poincaré's theorem. Note that the you can use, in our case, the phase velocity, which is constant up to a change in direction by reflection at the walls. Besides that, the theorem speaks of returning to the initial condition. This is not what you want: the initial condition is the beam entering the box through the hole, but you want the beam exit through the hole.

Also, some time ago I noticed a flaw in the proof of Poincaré's theorem, s.t. I am not pleased to resort to it. 

Now, I don't understand why do you need that trick of covering the hole with a mirror. Pay attention that covering the hole is a mechanical motion, it cannot be done instantly. Even if you think that you do it quickly, you draw the shutter upon the hole gradually, covering more and more of the hole, until it is completely covered.

I cover the hole mathematically, but not in reality. If I do that, Poincaré theorem applies because the box has finite volume. Therefore if the ray entered before the hole was covered, it will return to the situation just after the hole was covered, namely just after entering. Now take the cover out and consider the first return after (fake) reflection on the cover: the instant before the return next to the first place (ie just after entering the cavity)  means that, if the hole is uncovered, the ray actually escaped before because the small reflecting mirror was not there and it could only escape. I hope it is not too confuse! If you found a flaw in the great Poincaré recurrence theorem, publish it!

Regards 

Yves

@Yves: your fellow mathematician may have expressed himself unclearly and/or incorrectly: one cannot say anything, to my knowledge, about the distribution of return times without knowing the system in great detail. However, one can say (this is a theorem shown by Kac) that the mean return time can be computed explicitly from the size of the hole and the billiard area (or volume) under the hypothesis that the billiard is ergodic. If it is not, then one can still formally compute the average return time in terms of the phase space volume accessible from the hole, but this may well, in general, be a difficult quantity to estimate.

 As long as you assume that the photon is a dimensionless corpuscle, maybe you raise an interesting math problem, but surely it is not a problem of physics. In physics, Pierre de Fermat, Augustin Fresnel and James Clerk Maxwell woke up and worked before you were born. In real world physics a photon has a length of coherence (evidence by long difference of optical path experiments), and has an evolution of diameter between emitter and absorber. Those length are naturally fuzzy, but are finite.

OK, I know photon are waves. It does not change the fact that, in the limit of geometrical optics you can consider themas geometrical objects. By the way, at the beginning of this discussion I referred to the wave equivalent of this problem, which I think, is related to what are called Gelfand mode, namely eigenvalues with a small imaginary part of the order of the inverse trapping time of a ray in the limit of geometrical optics. All this is well known. 

I have found nothing with "Gelfand mode", only with "Gelfand model", and it is about statistics.

In geometrical Optics, no diffraction gratings exist, no stationary waves, no radiocrystallography, no interferometer, not any of the Young inventions and discoveries, no laser, etc. Pff ! In geometrical Optics, the mirrors are a pure surnatural miracle. GO falls somehow short off the target. Of course, in GO, no photon can exist, can be quantized.

I am severely "minussed" by anonymous people who can not argue.

So I have set up a bilingual list of the Göttingen-København postulates we do not accept any more in transactionnal physics :

http://jacques.lavau.perso.sfr.fr/Physique/postulates.html

Surely, some mistakes of english remain. Please be kind enough to warn me of these.