Of course the photon will enter the cavity-which is expressed by the fact that it will have a non-zero overlap with the eigenstates of the appropriate operator in the cavity. So the ``if'' part of the question is answered in this way. Now if one computes the overlap of the   time dependent state of the photon outside the cavity with, say, its initial state, then the answer is, generically, that it's not possible to assign a sharp time delay, because the photon will be in a superposition of states of definite energy and, thus, of definite time delay with respect to any reference state-an expression of the energy-time uncertainty principle.

Dear Sofia,

It's really an interesting question which I hope I can answer in the attached file. I am considering a somewhat modified situation assuming the approaching particle to be a massive particle (maybe an electron) instead of a photon. In that case one can contend oneself with treating a scalar function as representative of the particle's state. But all steps of the  consideration carry over to the photon case. My short derivation refers only to the time-independent  situation. But you can see how the wave function complies with the fact that the energy of the incoming  particle does not match with one of the cavity's energy eigenvalues. To do the time-dependent case is a lot more demanding. But one can at least give a qualitative answer:  The approaching wave packet first penetrates the hole and is diffracted into the space of the box, which means that it can now be described by a superposition of (packet-shaped) plane waves (partial waves) propagating in different  directions. They are reflected consecutively  at the various walls. Some of them which run in an opposite direction to the incoming primary waves emit portions through the hole thereby suffering a reduction of their amplitude.  The summary effect of this is as if there would be attenuation in the box. By successively being reflected back and forth the partial waves finally build up a quasi-stationary wave field, essentially the one which I derive. After the originally approaching  wave packet has finished its reflection, this wave field in the box starts decaying in a way that is essentially time reversed  to the build-up process. Both processes effect a time delay which you have in mind.

Lothar

Ah, Stam,

I read on your answer that you are busy and don't have time to read details. I said clearly that for answering the present question we need the length of the wave-packet, therefore we use wave-packets of the same form, almost monochromatic and very long, though in each wave-packet we have many photons, not one photon. The state is still |1>, but the emission is strong, s.t. single-photon wave-particles overlap.

The source has an aperture and we can record the time at which we open it and close it, s.t. we have the duration of the wave-packet. If you are displeased with the aperture (mechanic object), then there are other methods of releasing and stopping photon emission so as to allow precise recording of times. No doubt, the wave-packets will have a front end and a back end which are less dense than the body of the packet.

Now, you speak of an overlapping, in the cavity, with the cavity modes. What's our business the cavity modes? We are interested in the overlapping of the wave-packet with itself. That is destructive.

You are right that outside the cavity the direct and reflected beam also overlap and this is a problem, but there are methods to discern between them by modifying the returning beam polarization with a half lambda or quarter lambda plate (you know such things better than I).

So, please read more attentively the text.

Let me once more to stress what I am after: no doubt, the beam "tests" the cavity even if it doesn't enter, i.e. despite the destructive interference. So, do we have the effect that without entering the cavity the arrival time of the front-end at detector is longer than when there is no cavity and no deflection of the beam? Or, alternatively, the beam tests the cavity without loosing time for that? Which of the two is true?

Kind regards.

Dear Lothar,

I am in debt for your PDF, and for your calculus, many thanks. But, I produced a doc-file - hope that you could open, it's Office 2003. For simplicity I represented the box as having a finite thickness and a very big surface, by far much wider than the transversal section of the beam. On one of the surfaces there is an opening through which the beam can enter.

The advantage of the doc-file over PDF is that produces the effect of animation. You can see the multiple reflection of the wave-packet on the silver-coated surfaces. Let's discuss the issue after you see the animation.

Best regards,

Sofia

Dear Sofia,

I am afraid that you are talking about photon or photons but in this new version your question is a simple macroscopic problem ... Really you can use a pulse-laser to send an e.m. wave inside a box trough an hole, of suitable dimension, on some of its wall. Since you assume that the walls of this box are completely reflecting, and the hole remains open, it is sure that after some time, and after a finite numbers of reflections, the laser arrow will go outside the box ...

This finite number of reflections uniquely depends from the initial position of the laser arrow with respect to the box and its hole. For example if the laser arrow is perpendicular to the face where is placed the hole, then its reflection on the opposite face straightly sends the laser arrow outside the box. In any case there is a delay time between the two detectors D and D'. (Incidentally let me remark that source and detectors should be placed on a sphere with centre the hole ...)

However, I insist to think that the more interesting problem is the one arising from your companion question, namely to entrap photons inside a material box. This could be obtained also in your ideal experiment whether the hole should be closed after a suitable pulse-laser arrow should be entered in the box ....

All the best,

Agostino

The lack of precision in stating the premises of the problem is what makes the discussion so ambiguous. However, since the electromagnetic field, in the absence of sources, satisfies a linear equation, therefore any electromagnetic field configuration can be obtained by superposition, it suffices to describe it in terms of the modes that make up the wavepackets. 

Such a wavepacket, outside the box, defines a quantum particle with continuous spectrum. Inside the box, the boundary conditions imply that the spectrum is discrete. 

The statement that the initial condition is the photon is outside the box, simply means that it's described by a wavepacket and the energy eigenvalues belong to the continuous spectrum.  

Inside the box, the wavepacket is expressed as a superposition of the eigenstates of the Hamiltonian defined for the boundary conditions of the box. This Hamiltonian has a discrete spectrum. So the wavepacket, at some initial time, that doesn't matter, due to time translation invariance, that expresses the fact that the Hamiltonians are Hermitian, is expressed in terms of two bases, outside and inside the box. 

If it were monochromatic outside the box, it would have a sharp value of the energy. It would ``fit'' in the box, if that value of the energy corresponded to one of the eigenvalues in the box, whereupon it would evolve, within the box through a phase. Obviously outside the box it would be in a superpostion of incident and reflected states and inside the box in a ``bound state''. The reflection and transmission coefficients are computed the usual way, by imposing continuity conditions. 

If the energy eigenvalue does not correspond to any of the  eigenvalues in the box, in the box it will be in a superposition. 

If it's a wavepacket outside the box, it will be a superposition within the box and a superposition of incident and reflected wavepackets outside the box. 

Since the equations are linear, it suffices to solve the problem for a monochromatic incident photon or electromagnetic wave and obtain the ``physical'' result by superposition. 

This is a standard problem in wave optics and/or quantum mechanics and there's nothing mysterious about it. Everything one wishes can be explicitly calculated in all possible régimes, classical and quantum. 

Dear Stam,

I completely agree with your remarks ... However, what I consider interesting in this type of experiments is the possibility to entrap a photon in such a way that it becomes a massive quantum photon. I do not think that this particular aspect is so standard in QM ...

Regards,

Agostino

A photon in a waveguide, i.e. with one non-compact direction, of course, does acquire a longitudinal polarization and, thus, a ``mass'' of geometric origin. This is well known. the reason is that the boundary conditions break Lorentz invariance. 

Right ! But there is another reason that is not the breaking symmetry to justify the acquisition of mass ... This is the interaction between photon and matter, namely massive particles. This is well justified in my geometric theory of quantum gravity. This mechanism is more general than the breaking symmetry.

This is surely nonstandard in QM !

Regards,

Agostino

In the box in question the excitations don't propagate, so they don't have any well-defined mass. Also, since the box is empty and there are no other fields, there aren't any other particles-just the photon, that entered and, within the box, evolves as a superposition. The reaction of the material of the box is neglected, that's how the box is defined. This is all standard quantum mechanics and is, now, routinely tested in real experiments with trapped photons. Gravity, classical or quantum is not relevant, since the spacetime is flat and the energy-momentum tensor of the radiation can't affect it. One needs a large mass to bend light and the mass of the box isn't that big. 

Indeed, the whole situation can be described by a one-dimensional model, with a reflecting point at x=L and a point that has non-zero transmission and reflection coefficient at x=0. So with an initial condition that the photon is wavepacket localized at some x

Of course Stam !...  You continue to use standard QM. But I suggested to consider more general situations of entrapped photons inside closed materials, in order to go beyond to the Sofia's question ... Please look to my previous post on the companion Sofia's question !

All the best,

Agostino

Dear Agostino and Stam,

I am not sure as you that the problem is classical. You see a beam exiting a laser is coherent light. It is not a Fock state |1>. I supposed that in the wave-packet there are many photons but they are independent, s.t. there are no states |2> or higher. The description of the wave-packet is N|1>, where N does not have to be fixed. The reason behind this choice is that I read somewhere that states |2> may behave as what we call do-photon, which display a frequency twice bigger than the individual photons. I am no specialist about do-photons, I just read of it somewhere, s.t. I want to avoid such a possibility.

Now, I invite both of you to open my attached file. It realizes an animation of the multiple reflection - if one follows the instructions. And here is the issue: on one side of the glass slab the beam meets a silver surface and the amplitude has to be zero, while on the other side the beam meets the opening in the silver coating, s.t. there are a reflection part and a refracted part. Here the beam amplitude doesn't have to be zero. The refracted part exits the cavity. The reflected part remains in the cavity.

However, the beam that meets a silver surface on both sides is destroyed - see the figure. It won't escape as you say, Agostino, it is destroyed.

But such a thing is against the unitarity. The QM doesn't permit that part of a wave-packet be just destroyed, because that means that the total probability to find the particle in space isn't 1 anymore. So, the missing part of the wave-packet has to appear in another place. Where?

Next, for "learning" that the cavity length is unfit, the wave-packet has to travel inside it at least one round-trip. But this round-trip takes time.

Regarding the last point: if one knows the properties of the incident wavepacket, by measuring the interference effects with the reflected wavepacket, one can deduce the properties of the wavepacket in the box, i.e. the properties of the box. Conversely, if one knows the properties of the box, one can compute the interference properties of the superpositions inside and outside for any time. Standard calculations. 

Regarding the previous points, some thought is needed.

Stam,

Boundary conditions don't absorb energy. You say

"If the energy eigenvalue does not correspond to any of the  eigenvalues in the box, in the box it will be in a superposition."

How do you imagine photons of energy E, transforming into photons of different energies E' < E? Where is the rest of the energy? It is absorbed by the boundary conditions?

No, the energy isn't ``absorbed by the boundary conditions''-that statement is meaningless. The correct statement is that, assuming that, outside the box, the photon is in a state of definite energy |E>_out, it's possible to write this state in the basis of eigenstates of its evolution operator, defined in the box: |E>out=Σkck|e>k where the sum runs over the eigenstates in the box-these are countable. The ck=kout as usual. So, if |E>out=|e>m say, then, only the mth eigenstate contributes, the photon fits in the box at that level. If |E>out isn't identical with any of the states, it will be in a superposition, since the states |e>k are a basis in the space of states of the photon, otherwise the whole problem doesn't make sense-the photon can't interact with the box, unless its state can be expressed in the basis of the box. The difference between it fitting and it not fitting is whether one state contributes, or more than one do. In this latter case, once more-this has been said many, many, times now, it's basic quantum mechanics, the photon in the box does not have a definite value of energy. The point is that the Hamiltonian outside the box doesn't commute with the Hamiltonian in the box, so they don't have common eigenstates, that's what happens. Time translation invariance is broken by this  box, therefore energy isn't defined. It's not possible to match the evolution of the photon, before it entered  the box, with the evolution after it entered. The boundary of the box can't be sharply defined, so there is an uncertainty which Hamiltonian to use when and, if they don't commute, there is an uncertainty in the value of the energy. The states that restore unitarity, of course, are the reflected states. One must be quite careful, how one defines the evolution operator. This is, indeed, what happens when one describes tunneling: one has a superposition before the barrier, a superposition in the region of the barrier and the transmitted particle beyond and one imposes continuity and smoothness of the wavefunction. The value of the energy, however, can fluctuate, since it's /.

Stam,

"The boundary of the box can't be sharply defined"

Why?

"The states that restore unitarity, of course, are the reflected states."

Did you see the attached picture? The reflected part is destroyed.

Uncertainty principle means that the photon can't ``tell'' to better than its wavelength, whether it's in or outside of the box. But, once more, if one recalls the treatment of the tunnel effect in elementary quantum mechanics, it's no different. 

The parts aren't destroyed unless they're projected out-if one applies a projector, that, of course, destroys unitarity. But  I was speaking about the particle and the box, not about the other setup.

Stam, did you even see tunneling through a perfectly reflecting mirror?

Also, whatever projector you apply, where goes the difference E-E'?

Sofia, 

the problem with your formulation is that the cavity must have something like a small hole to be receptive to the photon from the outside. Because of this hole, the cavity is different from a closed resonator. The size of the hole, if small,  introduces a small but non zero parameter (I assume it far larger than regular absorption) . This defines the width of the "resonances" you can observe (the precise calculation by solving Maxwell's equations is like the rather standard Hertz resonator) , a width like the radius of the hole. If the frequency of the incident photon is outside of the bandwidth of the resonant cavity, the EM wave does not penetrate inside the cavity except for an exponentially small amount, interesting to get by the way, if it is inside the band, part of its energy goes into the box, with a maximum of the order of the inverse of the radius of the hole for a given incident flux. If you send a single photon (likely not easy, but perhaps possible), you will see a probability of presence of the photon inside the box like the intensity of the classical wave in the box, close to one at resonance, but never equal to one because of energy conservation. The time it takes for the photon to enter the cavity in the resonant case can be estimated simply by an argument of flux across the tiny hole: this time is like the inverse of the hole radius, it takes more and more time to excite the resonance if the flux is small.

Good wishes

Yves   

Dear Yves,

Your answer makes some sense to me. But please assume that the diameter of the hole is wider than the diameter of the transversal cross-section of the beam. Please see the animation in the attached doc-file. If you follow the instruction in the file, you'll see the successive reflexions on the inner walls of the box.

You say,

"If the frequency of the incident photon is outside of the bandwidth of the resonant cavity, the EM wave does not penetrate inside the cavity except for an exponentially small amount" 

Yes, this is the case, but how can the wave "know" that the cavity depth D - see figure - is unfit, if the wave is not repeatedly reflected between the silver coatings (and in consequence also destroyed). These are two opposite effects: being repeatedly reflected and destroyed makes the beam find the cavity unfit, but the destruction of the beam between the silver surfaces doesn't reflect it back into the opening - see the figure.

Sofia, 

I think the point is that, to send a photon to a cavity, the cavity needs to have a hole, small but non zero. This introduces a kind of small dissipation because of the possibility for the radiation to escape through the hole. Therefore the resonance of the cavity has a finite, non zero width of order of the radius of the hole. If the incoming radiation has a frequency in the bandwidth of the resonance, the photon will enter in the cavity. Otherwise it will enter also but with an amplitude mostly concentrated near the home, like for the Hertz resonator in acoustics. This makes an interesting calculation by itself. If there is only one incident photon, its amplitude will split into the one inside the cavity and the reflected photon, depending how far you are from resonance. Classically, the amplitude of the EM field inside the cavity for a continuous illuminating beam wiil be like the inverse of the radius near resonance and it will take a time also of the order of this inverse to build up once the beam begins. I think this problem in its classical version has been addressed, because Helmholtz resonators are often used in various devices to absorb large amplitude unwanted oscillations happening in chemical reactors. All what I said implies that the bandwidth due to the hole is smaller than the interfrequencies space in the cavity without hole. Otherwise a resonance of the cavity is excited but it is not an eigenmode of the cavity, only a mixture of modes depending on the details of the hole-cavity geometry. 

Best 

Yves

Dear Sofia, 

I think you refer to a situation different of the one I have in mind, namely the case of a hole much bigger or at most of the same size as the wavelength. In this case, the bandwidth of the cavity resonances is defined by the inverse time it takes a light ray to escape the resonator through the hole, or being absorbed by the wall. There is no problem then for the time a photon has to probe the cavity to see the resonances, because those resonances are all put together in a band of frequencies since there is no sharply defined eigenmode in this problem because of the hole. Generally speaking, because of the hole, the whole spectrum is made of scattering states from the point of view of quantum mechanics, there is no normalizable eigenstate. If I recall well Russian mathematicians have studied this kind of situation where a scattering state is close to an eigenstate: then the resonance frequency gets a small imaginary part (this could be called, but I'm not certain Gel'fand states).

Best 

Yves

Yves,

It's difficult to discuss things with people which don't read attentively what is written. The hole has a diameter bigger than the width of the transversal section of the beam, not than the beam wavelength. That, in order to limit the diffraction of the incident wave at the borders of the hole. The wave escaping the cavity through the hole (after being reflected by the silvered surface opposite to the hole), yes undergoes some diffraction at the hole border in the direction x - see the figure in my former comment.

Also, neither the glass nor the walls, are absorbing media for the beam's frequency.

When the part of the beam repeatedly reflected into the cavity, advances sufficiently in the direction x s.t. it meets on both surfaces the silver coating, on both these surfaces the beam amplitude has to be zero. This is IMPOSSIBLE, the length D of the cavity in the direction y, does not permit. Thus, between the two silver layers the beam is destroyed - see the figure.

The beam doesn't probe resonances, bandwidth, etc., it probes the cavity length D. The two formulations are quite equivalent, but the latter is the clearest. Speculating on scattering states, Gel'fand states, is futile. One round trip of the wave between the silver layers is enough for finding the cavity unfit.

The problem is the unitarity. It doesn't permit that part of the beam just disappear (as a result of the destructive interference). Then, where goes that part of the beam?

Dear All,

as I said in a previous my post, the interaction photon-cavity is more complex ! Here the cavity is represented, according to Sofia request, by a box, having inside completely reflecting walls, and with an hole at the center of only one face. Let describe this interaction in the framework of the observed quantum super Yang-Mills PDE, say (YM)[i]. Then this reaction is encoded by a nonlinear quantum propagator V that can be pictured as reported in the attached figure. Namely V=V1\bigcup V2\bigcaup V3\bigcup V4. The incident beam is represented by V1 ,  with \partila V1=A\bigcup P1\bigcup B. B represents the place where the incident beam interacts with the box-hole. Therefore a part of the beam is reflected by generating the branch (B,D) of V2, and the other part of the beam goes inside the cavity, namely the branch (B,C) of V2. The branch (B,D) freely continues to propagate outside the box, and the branch (B,C) represents a series of reflections inside the cavity. At some time t2 a part of this beam goes outside the box through the hole, branch (C,E) of V3, and the other one (C,F) remains inside the box. Therefore after a time \delta t = t3-t0 , t3>t2  , the interaction photon-cavity, has produced two types of beam outside the box and another one inside the cavity. During this process the conservation of energy, that always holds for (YM)[i], since this equation is invariant for time translations, allows us to state that the quantum energy content at t0 does not necessarily equal the final quantum energy content at t3, for some singularities of P, where \partial V=A\bigcup P\bigcup N, with N=E\bigcup F\bigcup G. This aspect is very important, since the structure of the nonlinear propagator V can produce a defect of quantum energy between initial and final contitions, but this is just a consequence of the conservation of energy.

This agrees with what Stam posted about 'non commutativity of the Hamiltonians involved in his formulation  ... '.

Let me also add that in this nonlinear quantum propagator V are present quantum tunnel effects, since we have two places of singularity, namely in B and in C.

Of course I skip on details ... Interested readers can look the following my papers:

http://arxiv.org/abs/1205.2894; http://arxiv.org/abs/1205.2894.

All the best,

Agostino

It's useful to reduce the problem, so as to simplify the computations, without eliminating any features. This must be a variation of any number of exercises in quantum mechanics, I seem to recall them in Messiah's book. Since polarization doesn't play any role, work in one dimension. Since relativity doesn't play any role, take a massive particle and, to avoid any fancy integration issues, take a square well potential with, at least, two wells. So V(x) is infinite for x< (-L/2), is equal to 0 for (-L/2)

Of course Stam, you can simulate the Sofia's question with a model in the classical QM (I know very well the Messiah's book ... this was in my heart when I was student ...), but the real situation is more complex and requires a more sophisticated mathematics !

I looked to your quoted arXiv paper. This is surely an interesting paper, but the situation of quantum gravity is more complex ! ... Of course some quantum systems at low energy levels can be approximated with methods of the classical QM ... but nowadays it appears like a game for children.

All the best,

Agostino

Sofia, 

If the hole is bigger than the wavelength, it implies that the cavity is also bigger than this wavelength. Therefore you can apply geometrical optics and the photon/light beam stays  inside the cavity by bouncing elastically on its wall, before it escapes through the entry hole, after a time depending on how many bouncings  are needed for that. In the case of parallelepedic cavities this time is fairly easy to compute. This time yields the resonance width of the cavity, or if you wish to call it so, how much time is necessary for the photon to probe the cavity. This time is also the inverse of the imaginary part of the eigenvalue of the Gelfand mode, of the order of magnitude of the area of the hole.  

Best 

Yves

Stam,

"the condition that the lowest eigenstate can fit - if this doesn't hold, the particle will ``leak'' from the `box' explicitly"

You sweep the problem under the carpet, explicitly. How does the beam learn that the box is unfit? By being reflected between the two silver-coated surfaces of the box - see the figure in the attached PDF.

Here begins the problem. On the silver surface the amplitude of the beam has to be zero. At the upper silver surface that's still possible. However, when hitting the lower silver surface that is no more possible because D is not a multiple of λ/2. So, the beam has to escape somehow from the box. But how? On both sides there is silver. For economizing ideas of tunneling, assume that the silver coating is thick.

Next, if this problematic part of the beam succeeds somehow the escape from the box, it should be delayed because of the round trip inside the glass, shouldn't it?

You talked of the eigenstates of the box. But the beam won't have time to learn what are the eigenstates of the box. Each layer from the beam, that hits the lower silver surface, has a problem to be solved on the spot.

Yves,

the beam doesn't propagate along the diameter of the hole, but almost perpendicularly to the surface of the hole. Open the PDF and you'll see.

Sofia, 

I said nothing about the angle of incidence of the beam wit respect to the hole. If the beam is almost perpendicular to it this is fine with geometrical optics, so the beam bounces on the wall until it reaches again the hole to escape, which defines (by dividing the length of the trip inside the box by C, speed of light) the time spent inside the box and so the inverse width of the resonance inside the cavity. The problem with a hole smaller or of the order of the wavelength and a box much bigger than the wavelength is slightly less obvious to answer. 

Yves

Once more: By measuring everything that can be measured *outside* the box, ``far'' in units of the energy of the probe, that sets the scale, here,  the properties of the box can be determined. In particular: Outside the box what is measured is a superposition of the incident and the reflected beam. If one imposes that the surfaces of the box are perfectly reflecting, then the superposition outside the box exhibits a dependence on the phase shift due to the fact that the beam that's been reflected off the interior wall of the box has a different phase than that thas been reflected off the outer wall of the box. Normalizing this to the intensity obtained from reflection off the oustide wall alone allows the determination of the depth of the box. If the beam doesn't strike the box at normal incidence things get somewhat messier but that's the idea, when the hole is so wide that diffraction can be neglected. If it can't, but the Fraunhofer approximation can be used (i.e. neglecting the spherical waves off the edges of the hole) then nothing much changes qualitatively, the expressions, of course, change but it's still possible to deduce the depth of the box in terms of the wavelength. If one does need to take the spherical waves into account, the computations are more elaborate, the final result has the same meaning.  If one wants to solve the corresponding ``quantum'' problem, nothing changes, in principle, either-it's possible to calculate the superposition and that provides a way to find the depth of the box, normalized to the energy of the incident particle, that sets the scale. Similar considerations apply if the reflection isn't total, etc.

Global energy conservation implies global time translation invariance, so one is computing the stationary state of the system probe+box (in language, appropriate for the quantum problem, the evolution of probe+box is unitary, the system is closed). It's, also, of course, possible to compute the time dependence and the result will be the same, only there's no point in going through all the trouble, since the stationary state is what's of interest: since it's known that nothing will happen until the probe does interact with the box, and one does know what the interaction is.

Yves, why don't you look at the picture in the PDF? People are busy and so am I, answers to something else that I asked are of no interest for me. You say,

"the beam bounces on the wall until it reaches again the hole to escape".

Don't you see that the beam reflected on the upper wall doesn't reach the hole anymore?

Stam,

Here it's not a stationary state in  the sense of a particle in a well. See in the attach a simplified figure. You can see that the beam reflected on the upper (silver-coated) surface, lands directly on the lower silver-coated surface. On these both surfaces the electric field should be zero. But this is impossible because the unfit wavelength of the beam is unfit. The part of the beam caught between silver surfaces can't disappear out of the blue, neither can escape through the hole which was left behind. Nor can it become a superposition of cavity modes because this is a TRAVELLING BEAM, not a stationary beam in a well. So, I repeat, the difficulty pops up INSTANTLY when the beam reflected by the upper surface, reaches the lower surface which is silver-coated.

No doubt that writing Maxwell's equations one would find the solution of the problem. I just venture the idea that at the bottom silver layer, the pattern of the e.m. field may become more complicated. If it is not possible to fulfill the requirement that the electric field be zero in the silver, then could it be that electrons are displaced from the region where the beam touches the lower silver surface? For displaying electrons the photons have to loose some energy, eventually enough for descending to an energetic level fit for the box. Could it be?

That the electric field vanishes on the surfaces doesn't have anything to do with the wavelength of the light wave, but with the fact that the surface is a perfect conductor. The statement is independent of the value of the wavelength.

Stam, what you say? If the wavelength is not fit to the cavity then at the first surface where the problem appears the field in the surface plane can't be zero. That means some local current or charges.

No, of course not. The fact that the wavelength doesn't ``fit the cavity'' means that field in the cavity isn't a standing wave. However the properties of the surface don't depend on that, especially, for the case of conducting surfaces; and for dielectric surfaces, the fact that the field configuration isn't that of a standing wave does not imply dielectric breakdown-which would lead to the appearance of free charges.

Stam, Gauss law says that  ρ/ε = ∇ E, where ρ is the charge density and ε the dielectric constant. If at the boundary between the glass and silver the electric field can't be zero, because, as you say, the wave is not a standing wave with a node on this boundary, then a charge has to exist. 

The electric field in the box will not be *a* standing wave-it will be an infinite superposition of standing waves, of the form sin(nπx/Lx) if Lx is the depth of the box and similar considerations lead to the dependence on the other coordinates (the wave equation is linear and can be solved by separation of variables).

Such a superposition satisfies the boundary conditions on the boundaries of  the box and the coefficient functions  are fixed by imposing the boundary conditions at the hole. This is just the classical counterpart of the quantum particle-the classical wave won't have a well defined wavelength in the box, the quantum particle won't have a well defined energy. 

Stam, I don't want to repeat endlessly that what you say is violation of energy conservation. The states in the superposition of which you speak, are of lower energy that the incident photon. For a photon to pass to a lower energy it has to undergo Compton scattering. On what?

The *total* energy is the only quantity that's conserved, not mode by mode, but over the sum of the appropriate modes.  And the calculation can be simplified, even further: the box is the interval [0,L], the conditions are total reflection at x=L, reflection and transmission at x=0. For x

Now one can ask, what happens, if one imposes k1=k2. Then one finds T1=1  and R1=R and the boundary condition at x=L becomes exp(ik1L)+Rexp(-ik1L)=0, that provides the relation between the reflection coefficient of the box and the size of the box, as probed by the wavenumber k1. 

Stam,

what happens with you? Which total energy? You spoke of a superposition, didn't you? A superposition of modes is of the form A|E1> + B|E2> + C|E3> + . . . i.e. a superposition of modes of single photon. It is not of the form |E1>|E2>|E3>... i.e. splitting of the original photon into several photons. Only in the splitting case you can talk of total energy, E = E1 + E2 + E3, where E is the initial photon energy.

Now, as usually, you don't read all the data of the problem. One day ago I attached a PDF, and invited you to open and see the figure. I attach now an even more detailed version, and I hope that you'll open it. The direction perpendicular to the silver surfaces is y, and the thickness of the box in this direction is D. As to your saying that one has reflection and transmission at x = 0, I really have no idea what you talk about. Look at the figure! The problem is not where the reflected beam falls on the opening, but there where it meets silver on both sides. I say this all the time.

Next, this site offers you the possibility of lower and higher indexing. Why don't you use it? It's difficult to read what you write.

Third thing, if you don't explain your notations, people won't stay to guess, we are all busy. One of your skills is to Teach, to Explain, isn't it?

Best regards!

There's a nice figure, but what matters are the mathematics, in particular what are the expressions for the solutions, taking into account the boundary conditions. The well-known statement, whose proof is the content of  my last message, is that there does exist a solution for any size of the box and for any wavelength of the incident beam, that satisfies the boundary conditions that the amplitude should vanish on the surface of the box and should be  continuous and smooth across the hole. It isn't necessary for the *existence* of such a  solution, that the wavelength of the incident beam be related to the size of the box. If it is, there are consequences about its properties, if there isn't, there are other consequences. So, in the one-dimensional model studied above, for k1=k2 one finds that the phase of the reflection coefficient is π+2k1L; it depends on the wavenumber of the mode and the size of the box, beyond the 180° phase shift due to reflection at the hole. That's how it's possible to deduce the size of the box.

In particular, among the continuum of values that the wavenumber can take in the box, that has a hole (that's what leads to the continuum of values) is the value k2=k1 and it's straightforward to see that, even if k1L isn"t equal to nπ, there do exist non-zero values of reflection and transmission coefficients, that imply, among other properties, energy conservation in that case, too-since the plane waves used, solve the wave equation that, along with the boundary conditions, is consistent with energy conservation.

Dear Sofia,

 hard to believe that a simple problem on a photon and a hole in a box can stir up so much dust! But I think that most of Yves Pomeau’s comments hit the nail on its head. Also from my point of view the problem concerns essentially a boundary problem in the general theory of waves,  and quantum theory plays merely a subordinate role. As Yves pointed out very clearly, the problem you wanted to address, viz. how a hole in a cavity with ideally reflecting walls affects the propagation of a photon approaching that cavity, becomes truly “wavy” if the linear dimensions of the hole are small compared to the photon’s wave length. I have summarized my more detailed comments in two pictures which I append to this message. I noticed from my previous comments that you don’t like the mathematics of boundary problems too much, but without it a discussion runs the risk of derailment. As I already said previously, a complete time-dependent treatment is too involved, but it is clear that the build-up of the wave field inside your “planar resonator” which I derive, happens step by step as a result of successive back and forward reflections of the intruding wave. And it seems to me clear as well that the resonator will add a tail to the wave packet after it has finished its reflection at the outside of the box because of the “reverberation”  due to the successive decay of the wave field inside.

Where do particles come into play, either in quantum mechanics or quantum optics? I shall confine myself to single-particle wave packets.  All that one has to add to the calculation of the classical wave field in quantum optics is merely the insight that the probability density of the photon in question is given by the classical electromagnetic energy density divided by the  energy of the photon: Planck’s constant times its frequency. The associated trajectories are the flow lines of the pointing vector. Of course, the photon of some wave packet follows only one of those Poynting flow lines. In the next realization of the wave packet it will be a different flow line. Where the flow lines are scarce, e.g. within the box, the probability of a photon intruding and leaving the box is small. That means, there will be many wave packets whose associated photon will not enter the box at all.

The situation in quantum mechanics is completely analogous. Here the modulus squared of the wave function yields the probability density of the particle being at the various positions within the propagating wave packet.

Best regards,

Lothar

The statement about the ``associated photon not entering the box at all'' is not correct-since the transmission probability, though small, does not vanish.

Stam, you are stubborn in not looking at the figure. I have no idea what are your quantities L, k1 and k2. I have no wish to make guesses. I asked you to use the symbols used by the figure but you refuse. This is not a dialogue, you talk with yourself. If so, there is no use that I answer you.

As is obvious, L=D. As has been stated repeatedly, k1 = wavenumber outside the box, k2=wavenumber inside the box. Special case: k1=k2. Boundary conditions:  amplitude vanishes on the walls of the box, is continuous and smooth across the hole. For width of hole >> wavelength, normal incidence is relevant. Oblique incidence doesn't change anything significant. Diffraction doesn't change anything regarding the consequences of wavenumber x D not equal to a multiple of π, either. 

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 there is 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, where 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.

Stam, I hope that you'd read my last post, with the solution.

Anyway, I am very thankful for your interest in the same things that interest me. That has big importance. Even when I disagree with you, I appreciate the dialogue with you. And many times I get from you interesting and helpful information.

Now, I asked you in some comment why don't you use the upper index and lower index options. Only now I discovered that they work while editing, but after doing save, they appear as regular text not as they were edited. That's very bad, and I reported to the moderators. They promised to pass the problem to the people in charge with such issues.

My kindest regards!

Dear Sofia,

I am glad that you recognized the solution in your question ! Let me only add that such a solution was just given in a my previous post. I refer to my post where I given a nonlinear quantum propagator dynamically encoding the problem reported in this thread.

All the best,

Agostino

Dear Stam,

Your remark concerning my comment on Sofia’s problem should not be left without a response. You are saying:

The statement about the “associated photon not entering the box at all” is not correct  - since the transmission probability, though small, does not vanish.

I want to give you a simple example which supports my point of view:

Imagine you would direct a beam of sufficiently monochromatic light onto a beam splitter. By definition, the beam splitter divides the incoming electromagnetic current density up into equal portions: one half is directed, let’s say, to the left (path I), the other half to the right (path II).  If the intensity of the incoming light is sufficiently low, and if one has chosen a light source that generates single photons, the beam consists of individual wave packets, each containing a single photon.  At the beam splitter their transmission probability to path I is 50% and 50% to path II. Let 100 of the photons – approaching one by one – run through the beam splitter. You detect 50 of them in path I, the other ones in path II. That means that the latter ones do not enter path I at all although there was a 50% transition probability.  

Best regards,

Lothar

No-*if* you detect 50 of them in path 1,  since the photons are indistinguishable particles it's not possible to state that the ``other'' 50 are in path 2, that's the difference between classical approximations and quantum mechanics.  The detection has *projected* on the 50 photon subspace. 

Dear Agostino,

Why do I need a propagator? The problem was trivial, what was needed, at least for me, was a picture to illustrate that the problem was wrongly posed.

The photon yes enters the cavity if there is an opening. More exactly, on the opening the beam is partly transmitted into the box through the glass, and partly reflected on the glass surface. In continuation, after a couple of round trips, part of beam is reflected on silver on both sides. This part becomes a running beam. For the running beam there is no requirement that D = nλ/2, where n is an integer, s.t. it is not destroyed by destructive interference - the beam reflected on silver does not overlap entirely with the incident beam - see the picture in my answer with the solution.

Bottom line, part of the beam yes enters the cavity, and is reflected from silver surface to silver surface, until it meets again the opening. There, it looses a part by transmission into air - s.t. we get an echo in the detector,after the first return to the reflection - and  the rest continues inside the box until a second echo is produced at a new fall on the opening, and so on.

In all, the effect is a periodic echo, weaker and weaker.  

To Stam and Lothar,

Stam, I also disagree with Lothar, but I fail to understand what you say. Let me put the things in my words, and if and when you have time, please tell me if this is what you meant.

So, Lothar, wave-packets are not classical objects as you say.

The quantum objects - our wave-packets - are NASTY. At the beam-splitter (BS) each incoming wave-packet really splits into two daughter-wave-packets. If the BS is fair (50-50%), each such daughter has 1/2 from the intensity of the parent. Now, the famous wave-particle duality (that until today we can't say that we understand perfectly how it works) arranges the things so that ONLY ONE of the two daughters succeeds to produce a recording in a detector.

That doesn't mean that the other daughter-wave-packet was a fiction, and didn't exist. Yes, I existed. Its existence was PROVED in different ways. But, the simplest proof is that, if you don't disturb the daughters (by placing detectors on them or by interaction with other particles as a consequence of poor vacuum), the two daughters produce an interference pattern if they are brought to cross one another. An interference pattern is not possible unless BOTH daughters are there.

Now, returning to the case that we place detectors on the separate paths of the daughters, why only one produces a detection, it's not yet absolutely clear. There are different opinions, but none of them is sure. More experiments are needed.

Dear Sofia,

any interaction in quantum system must be encoded by a nonlinear quantum propagator when one aims to obtain a correct mathematical formulation. Furthermore, the present system, even if simple, is not so trivial since it presents two quantum tunneling effects. On the other hand the quantum energy balance can be correctly obtained only by considering that the process is encoded by a nonlinear propagator. The proof is in the long story of your question (even if I did not understand the necessity to split it into a companion one).

All the best

Agostino

The statement that wave-particle duality implies that only one of the ``daughter photons'' succeeds in producing a recording in the detector is not correct. The correct statement is that any given photon has a non-zero probability to take one path or the other. If one measures it taking one of the two paths, that destroys the coherence with the other possibility-since it eliminates the other possibility. The theory is understood and experiments have been done, even with ultra-cold atoms, that illustrate the interference effects: e.g. http://www.lkb.ens.fr/Light-scattering-by-cold-atoms,1453?lang=en These test more subtle issues.

To Sofia (Stam might only be interested in a few statements at the end)

Thanks for your comments. As far as the beam splitting is concerned I cannot see any difference in our views. Where do you think we disagree? As an elder theoretician who has taught quantum mechanics for about 30 years I have witnessed the development of some basic ideas in quantum electrodynamics and the discussion of pros and cons of its formalization. My view of quantum mechanical phenomena diverges therefore quite often from the established catechism because I am still familiar with alternatives in contemplating the fundamentals. Often simple, though crucial assumptions which can be stated in simple words, appear nowadays wrapped up in an escapist formal language which obscures the physical mechanism and suggests the impression that one now has a proof and no longer an assumption. Here is now this rather innocent example: Today the younger physicists take it for granted, that wave packets in QED are different from their classical counter-parts. This is also one of your ex cathedra statements. In what respect are they different? And if it were true, how can you show the difference experimentally? When you visit a laboratory where quantum optical experiments are performed you see all these perfectly manufactured equipment pieces: lenses, filters, quarterwave plates, mirrors, polarizing beam splitters, Mach-Zehnder interferometers and so on. All of them are manufactured in optical workshops by exclusively applying the laws of classical electrodynamics. Even if you are firmly convinced that the various objects of QED are manifestly non-classical, I would like to invite you to discuss tentatively some QED-phenomena (atomic spontaneous light emission, photoelectron emission from solids, interference, Compton effect…) by assuming that the involved electromagnetic waves behave classically. All that one has to add to the classical picture is that the waves are always accompanied by a point-like particle which possesses the full electromagnetic energy of the associated wave packet and never breaks up into smaller parts. It has been Einstein’s assumption in 1905 that these “light quanta” as he called them, exist and are point-like. It is therefore suggestive to define the particle’s probability density of being somewhere within the wave packet by the quotient of the electromagnetic energy density divided by its energy: Planck’s constant times the frequency of the wave packet.                                                                                       Einstein’s assumption could never be traced back to a more fundamental level, but  - as stated above – it appears to quite a few people that the formal QED-machinery has exactly achieved that. I love his remark in a letter to his friend Besso in 1951:                                                                                                     “All the years of willful pondering have not brought me any closer to the answer to the question ”what are light quanta”. Today every good-for-nothing believes he should know it, but he is mistaken...”                                                                                            As for my example with the beam splitter: Of course, a classical wave packet splits exactly the way you describe with all the options of interference of the beam portions if one would bring them together some place farther away.. The beam splitter divides the flow of electromagnetic energy up into equal portions. Since, by definition, the associated photon cannot be decomposed, it can only choose path I or II. If there are detectors, just one at the end of each path, they will not show any coincident signals. In the respective experiments the single-photon wave packets are generated such that they are sufficiently far away from each other while they are heading for the beam splitter. They are therefore well  distinguishable by their different positions. Hence, each wave packet undergoes the splitting totally independent from the following one. As I said in my previous message, of those 100 well separated photons there will be on statistical average 50 of them run into path I and 50 into path II. Hence, I can only reiterate: If a photon has chosen to run into path I, for example, it will never occur in path II although there was a 50% probability to do so.              Finally: The experiment shows that there is no effect of measurement on the process of the beam splitting. If one removes one of the detectors, the rate at which the other one fires remains unaffected.

I wish you a nice weekend.

Lothar

Lothar,

your statement

"Since, by definition, the associated photon cannot be decomposed, it can only choose path I or II."

is Bohmian mechanics, it's not what we call Standard quantum mechanics. As you say that you know the different interpretations, you don't need my explanation about the difference between the two.

By the way, there are some expressions that you used and I don't know what's that: "catechism", "ex cathedra". Please be kind and tell me what they mean. Also, you said

"an escapist formal language which obscures the physical mechanism and suggests the impression that one now has a proof and no longer an assumption."

I am not sure what you mean. But I can tell you that in our days students are taught mathematics with so little discussions on phenomenology, that they become totally unable to understand that physics is phenomenology and one has to endeavor to understand how the nature works, not just to write equations. Did you ever hear the words "shut up and calculate"? Many new physicists think that the mathematical model is saint, and how the nature behaves is secondary, i.e. the latter depends on which mathematical model you adopt.

But, to return to Bohm vs. standard QM: both have weak points. The Bohmian mechanics quarrels with the theory of relativity, while the standard QM works with the enigmatic collapse. If one tries to penetrate the heart of the collapse and questions how it works, one is lead to the conclusion that the future is know. Thus, adopting one of these views and ignoring its weaknesses, is wrong.

Only additional experiments can tell us what is the truth.

A nice weekend to you too!

Sofia,

This has been again a very speedy response! Thank you very much. It seems that you hardly sleep.

As for the two semantic stumble blocks “catechism” and “ex cathedra” I must explain that I used the two words in a present-day (figurative) sense: catechism meaning the unquestioned rule book for teaching a specific matter, originally in the context of teaching the Christian faith, “ex cathedra” meaning the way of presenting ideas or convictions before an audience that is not expected to raise questions.

You asked me what I personally think of the nature of light. As I already mentioned in my previous mail: I believe in Einstein, and this is not an irrational affection. I think that this apparently absurd idea of a wave/particle duality– 20 years before Born and Schroedinger  -  has proven in thousands of experiments to be an unquestionable feature of nature. But it has to be discussed in quantum mechanics and in quantum optics in a far more thoughtful and differentiating  way than is usually practiced nowadays by younger colleagues who have been brainwashed as students in “ex cathedra” classes on quantum theory. I have tried to achieve a major clean-up of the various concepts in a synopsis entitled “Quantum Mechanics without Fairy Tales” which is, in part, accessible under arXiv: 0912.3442 entitled “Stochastic Foundation of Quantum Mechanics and the Origin of Particle Spin”. As you can imagine, I have made various attempts to get it published somewhere else in an ordinary journal. This has turned out to be a hopeless enterprise. The “peer reviewers” one has to cope with are all of the kind who know better.

Back to your question concerning the apparently contradictory behavior of light and its concomitant “light quanta”. My opinion rests on this point of view:

Primarily light is generated in an essentially classical way. While an electron, for example an electron in the 2p-state of a hydrogen atom, performs a transition to the 1s-groundstate its quantum mechanical current density oscillates at a frequency  \nu that is given by the energy difference (E2p-E1s)/h where h is Planck’s constant. This current density builds up an oscillating vector potential which is now part of the kinetic energy operator in the time-dependent Schroedinger equation of the hydrogen electron. If one solves this equation one recognizes  that the vector potential forms an outgoing  wave packet which becomes finalized as the hydrogen electron attains the groundstate. The resulting transition time is identical with the one obtained from the very involved standard calculation where the radiation field is quantized.  In my calculation the wave packet possesses the energy h\nu because the electron which gave rise to its existence obeys quantum mechanics. The radiation field (the vector potential wave packet) has a form that is essentially identical with that of a Hertzian dipole. But there is a new feature which cannot be explained within “my” quantum theory of radiation nor by the established QED-version: When the wave packet arrives at an absorber and gets absorbed, the ENTIRE energy of the wave packet disappears and causes an excitation of the absorber by the energy amount E2p-E1s.    And this is true totally independent of the distance between emitter and absorber. Surely, the probability of the absorption by a distant atom decreases as one over distance squared, just as dictated by the Hertzian wave, but otherwise the energy appears to be concentrated almost point-like. If one would keep a single atom in a Paul-trap as Dehmelt and associates did in their paper  PRL 56, p. 2797 (1986), one could excite this atom many times by absorbing linearly polarized photons one by one and successively monitor the spontaneous reemission of photons.  If one would surround this atom by a large set of detectors in a spherically array and display the successive detection events on a corresponding sphere by black markings, one would see their distribution exactly reflecting the electromagnetic energy current density of a Hertzian dipole. All the efforts that have been made since Einstein’s 1905-paper could not resolve the problem why electromagnetic waves behave like this. For miraculous reasons Einstein’s hypothesis proved to be true and underivable.

One is hence led to conclude that the propagation of light follows exactly and  everywhere the laws of classical electrodynamics. This remains to be true when the wavefield contains only very few photons, even when there is only one photon in the wavefield. The probability of the photon being at the various points within that wavefield is given by the electromagnetic energy density divided by the photon energy h\nu. This applies to every conceivable situation: when you have split up the wave packet by a beam splitter into two portions and  let them merge again somewhere farther away, you must first calculate the interference wavefield in a completely classical way and form the energy density of the resulting field. That energy density of the interference field divided by the photon energy gives you the probability density of the photon’s presence at the various points. Where the wavefield vanishes due to interference you cannot find the photon, of course. But in all those wavefields which one usually generates one by one in present-day quantum optical experiments, the associated photon is definitely at one point only. If you want to get a picture of its probability distribution one has to repeat the experiment sufficiently many times.  

So much for Sunday!

Best regards,

Lothar

Dear Lothar,

First of all thank you. So "ex-cathedra" means "dogma", or "believe and don't search". (I like your colorful style, this is why I ask about words not known to me.)

One more thing. Please, I am unspeakably busy, don't send me long posts. I want to give you full attention, but won't be able to read them. For instance, all the paragraph beginning with "Primarily light is generated in ..." and ending with "Einstein’s hypothesis proved to be true and underivable" are things well-known by everybody.

It would be better to keep our talk through messages, it is rather a personal talk.

I noticed that you say "I have made various attempts to get it published somewhere else in an ordinary journal. This has turned out to be a hopeless enterprise." You tell me?! I tried once to publish a work and the editor refused it with arguments totally alien to my work. The "all-knowing" editor didn't even pay me even the minimal honor to read the abstract.

Now, to physics. What you are doing in the paragraph beginning with "One is hence led to conclude", and ending with "sufficiently many times", is de Broglie - Bohm principles. These principles say that there is a wave in which floats a "point", and the point goes where the flux guides it. (By the way, it's flux, what comes to the detector, not energy density.)

You have to know that the de Broglie - Bohm principles CLASH with the relativity. Are you familiar with the so-called "Hardy's paradox"? Hardy's article considered three observers in relative movement with respect to one another. Each one writes the evolution of the wave-function according to the order of the events (passing through beam-splitters, then detection) in his frame. Well, judging exactly as you, they arrive at conclusions that clash. There is NO WAY to reconcile between the three observers, one has to admit that there exist a preferred frame of coordinates - i.e. that only one of the three wave-functions is true in the nature. Since 1992 when Hardy published his article (which also was almost rejected) all the QM community seeks a reconciliation with the relativity. But, total insuccess!!!!!!!!!

You see?! We are not so free to adopt an approach, we pay a price. There is no approach free of contradictions, not even the "ex-cathedra" one. In the standard QM, people sweep under the carpet terrible incompatibilities. This is not honest.

Best regards!

This is about the question I asked in a previous comment: is it possible to design a box with perfect reflection on the inner wall such that an incoming ray cannot escape through the entry hole? The answer is no, if the size of the hole is finite. This is a consequence of Poincaré recurrence theorem. 

Dear  Yves,

you are right ! But there exists also an integral bordism reason. It is just this last motivation that suggested me to state, in a my previous post on this thread, that after a finite time the beam inside the cavity will have a component going outsides the cavity ...

My best regards,

Agostino

Yves,

Did you notice that the question is "If the photon doesn't enter the cavity, . . ."? So if the photon doesn't enter the cavity, do you claim that it is about whether an incoming ray can or not get out of the cavity?"

Does Poincare's theorem says whether a ray that doesn't enter a cavity can or cannot escape from the cavity?

Why don't you pay attention what you write?

Next: I expect people to answer MY question, not ANOTHER question. I would have no objection if you would post an invitation to people to pay attention to an akin question, yours, but at THIS address I expect answers to MY question.

Another thing: Poincare's theorem has to do with the position-momentum phase space. But we have here waves. The applicability of the theorem is not so obvious, you'd better elaborate a bit what you say. And please!, do that at the address of YOUR question.

Best regards!