Photon in a box

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Summary:

what happens to a photon with WL >2D

Main Question or Discussion Point

The eigen wavelengths λn(WL) of EM radiation in box are 2d/n where d is the size of the box.
If I put a photon in a box with WL>2d via an optic cable trough a hole it must reflect on the perfect mirror walls
and be a running wave. Maybe it is possible to decompose it as a set of eigenmodes of the box, though they
are stationary.
This is the first question?
Is this possible?
The second question is: If it is possible, when I measure the energy (E) I would receive some of the eigenmodes E's.
But the initial photon has lower E than the lowest eigenmode. Which is not possible. What happens here?
 

Answers and Replies

  • #2
DrClaude
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Photons are not like massive particles. You cannot "put the photon in the box" if there is no mode in the cavity that will allow a photon of that wavelength.

This is observed experimentally when excited atoms are put in cavities in conditions where there is no cavity mode for the photon that the atom would emit. The atom stays in the excited state.
 
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@DrClaude you answer solves my issue immediately but raises other questions. Sorry I have to ask them.
Then what would the photon do? Reflect back from vacuum? Even stranger it must 'feel' somehow superluminary that there is a 'bad' box ahead and it can have say 10 cm length (the same applies to an atom in a box - How does it 'know' that the box is bad as it did not have radiated?)
And why do you think a massive particle should enter? The reasons in the first post apply to it as well for its wave function.
 
  • #4
f95toli
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A single photon can usually (not always!) be thought of as the "extreme limit" of a classical weak EM pulse,

The situation you are describing is exactly the same as having microwave pulse travelling in a waveguide encountering a microwave cavity, unless the frequency of the pulse matches a mode of cavity the it will indeed be reflected.

Note that the analogue is exact in this case. It is possible to do to a full QM analysis (2nd quantisation of the waveguide etc), but in this case the answer will be identical to what you would get from classical EM.

Also, there is nothing superluminal about the situation, the pulse will travel in the waveguide, it will be reflected as it hits the discontinuity and then travel backwards
 
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  • #5
timmdeeg
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You cannot "put the photon in the box" if there is no mode in the cavity that will allow a photon of that wavelength.
Supposed there is a photon in the box - the mode fits. What happens to the photon if you change the size of the box (e.g the box is a cylinder with piston) such that there is no mode anymore?
 
  • #6
f95toli
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Under normal circumstances it it simply leaks out. That said, it is possible to get non-classical effects if you change the size quickly enough.

This is not exactly the situation you are asking about, but it is somewhat similar
https://www.nature.com/articles/nature10561

There have been lots experiments carried out in this field, typically within the framework of superconducting circuit-QED. You should be able to find quite a few papers using Google scholar

A lot of work in the field of quantum computing is related to using (or at least few) numbers of photons to read out and manipulate qubits (which when using superconducting qubits is nearly always done using cavities). Hence, the theoretical framework needed to analyse these situations is well developed.
 
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  • #7
timmdeeg
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@f95toli I appreciate your quick and clarifying response. Thanks.
 
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A single photon can usually (not always!) be thought of as the "extreme limit" of a classical weak EM pulse,

The situation you are describing is exactly the same as having microwave pulse travelling in a waveguide encountering a microwave cavity, unless the frequency of the pulse matches a mode of cavity the it will indeed be reflected.

Note that the analogue is exact in this case. It is possible to do to a full QM analysis (2nd quantisation of the waveguide etc), but in this case the answer will be identical to what you would get from classical EM.

Also, there is nothing superluminal about the situation, the pulse will travel in the waveguide, it will be reflected as it hits the discontinuity and then travel backwards
How is this not superluminal? Where is it reflected? At the end of the cavity or at the beginning? How knows an atom in a cavity not to radiate?
 
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By the way if the length of the waveguide is not multiple to the wavelength of the photon, will it propagate in it?
 
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f95toli
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How is this not superluminal? Where is it reflected? At the end of the cavity or at the beginning? How knows an atom in a cavity not to radiate?
At the beginning. Again, this is exactly the same as in the classical case (see reflectometry).
I'm not sure why you think the waveguide needs to be a multiple of the wavelength. The photon must of course be propagating in one of the modes of the waveguides (TM, TE or TEM)
As the pulse it hits the start of the cavity there will be a discontinuity. The "probability" of the pulse entering the cavity will depend on how steep the change is (classical EM).

Again, replace "photon" by "weak EM pulse"; the results will be absolutely identical in this case. You can always "convert back" to thinking about photons once you have the classical result.

I suspect you are making the common mistake of thinking of photons as small "wavepackets" with a real size/length; but that is not what a photon is; they are "quantum objects" and it is very hard (perhaps impossible) to visualise them in any meaningful way. Fortunately, the math works out anyway :smile:

In my area we nearly always do the microwave/RF design using classical physics and simulation tools, but when running and analysing the experiments we think in terms of photon numbers (usually average number but sometimes true Fock states) .
 
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  • #11
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There is a discontinuity even if the box is fit for the photon (but it would continue and will not be reflected). So the discontinuity is not the reason for the reflection. The reason are the dimensions of the box. When the photon is reflected at the beginning of the box, how would it know if the box has L=nλ/2 or not.
This information is not available there. It must somehow instantaneously be deciphered by the photon. This is indeed superluminarity.
By the way there is a classical (just historically) problem (unexplaned at least to me) which was noticed by Newton, that the rate of reflection of a glass surface depends on its length. And he measured that for about 300 000 λ or so. This is described by Feynman in QED Strange theory of light.

I think that the waveguide have to be considered also as a box of length L and therefore it must also be multiple of λ for the photon to start propagating. Of course the refraction index must be included in the formula.

I think of photons as point objects inside the probability wave which length is the coherence length of the photon and which depends on the band emitted by the source and can be more than 100 meters. Their width increases with time maybe linearly.
 
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  • #12
Nugatory
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. When the photon is reflected at the beginning of the box, how would it know if the box has L=nλ/2 or not.
This information is not available there. It must somehow instantaneously be deciphered by the photon. This is indeed superluminarity.
You may be misunderstanding what a photon is and how it behaves. The problems you're seeing disappear if you think in terms of a weak EM pulse instead of photons - that is a hint that the problem is that photons don't work the way you're thinking. In particular
I think of photons as point objects inside the probability wave which length is the coherence length of the photon
isn't a very helpful model; any attempt to assign a position to a photon except at the point where an electromagnetic field interacts with something (makes a dot on a piece of film, triggers a photodetector, otherwise exchanges energy andmomentum, ....) is almost guaranteed to mislead.
 
  • #13
f95toli
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There is a discontinuity even if the box is fit for the photon (but it would continue and will not be reflected). So the discontinuity is not the reason for the reflection. The reason are the dimensions of the box. When the photon is reflected at the beginning of the box, how would it know if the box has L=nλ/2 or not.
This information is not available there. It must somehow instantaneously be deciphered by the photon. This is
indeed superluminarity.
If you have a cavity you always -by definition- have a discontinuity; something has to happen to change the waveguide to a cavity, right? The length of a 2-port cavity is defined by a two discontinuities; an input and an output (which do not need to by symmetric)

If you forget about photons for the moment and think about this in terms of a pulse. The pulse will propagate along the waveguide; as it enters the cavity the field distribution starts to change. If pulse contains frequency components that matches one of the modes of the cavity you can start to excite said mode, if not the energy will be reflected back. In real life some energy will of course always leak into the cavity and be dissipated. Again, this is classical EM.

I think that the waveguide have to be considered also as a box of length L and therefore it must also be multiple of λ for the photon to start propagating. Of course the refraction index must be included in the formula.
No, this is not how EM radiation propagates in waveguides. The wavelength of FM radio is about 3m. If you open your radio you will certainly find waveguides much shorter than that, right?

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




I think of photons as point objects inside the probability wave which length is the coherence length of the photon and which depends on the band emitted by the source and can be more than 100 meters. Their width increases with time maybe linearly.
Which as Nugatory has already explained is not at all correct. Thinking about photons as physical "things" in the classical sense is as I mentioned above not a good idea.
 
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  • #14
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That the discontinuity doesn´t play let see what happens if there is no waveguide at all but the photon falls directly on a hole in the box (or a faint laser pulse). Surely there is no discontinuity now. Surely the same will happen. Somehow the front surface ´knows´where the back surface is. This the same as in the above mentioned paradox of Newton. It was not explained by him nor by Feynman. The same is that an atom does not radiate when in an unfit box. This is nonlocality and nonlocality implies superluminal signals.
I don´t think I have a classical view about the photon. As I wrote it is more a wavefunction (in fact probability) with a point substance inside it. I did not say that the substance is localized in a point in the area covered by the wavefunction just that it is inside almost not existing before an interaction.
 
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  • #15
f95toli
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That the discontinuity doesn´t play let see what happens if there is no waveguide at all but the photon falls directly on a hole in the box (or a faint laser pulse). Surely there is no discontinuity now. Surely the same will happen.
No, it is a completely different situation; but not much changes. You are still going from free-space propagation (which is very complicated for a single photon) to something else; and if that "something else" is not matched in some way (using e.g. an antenna) the photon is still most likely to be reflected at the discontinuity.,
Again, this is just classical EM. A large part o RF/MW engineering is trying to get relatively weak EM pulses to go where we want them

Somehow the front surface ´knows´where the back surface is. This the same as in the above mentioned paradox of Newton. It was not explained by him nor by Feynman. The same is that an atom does not radiate when in an unfit box. This is nonlocality and nonlocality implies superluminal signals.
Its been years since I read that book, but I wonder if the problem is that you are taking the path integral (sum over histories) idea too literally? I am certainly no QED expert, but I do know that the different paths do NOT correspond to real photons in any way.

I don´t think I have a classical view about the photon. As I wrote it is more a wavefunction (in fact probability) with a point substance inside it. I did not say that the substance is localized in a point in the area covered by the wavefunction just that it is inside almost not existing before an interaction.
But photons do not have wavefunctions, at least not in the usual QM sense (you can define some functions that behave a bit like a wavefunction but it gets messy).
Again, you shouldn't think of photons as anything with a true "physical size" and the position is undefined. Thinking about photons in this way will inevitable lead you to the wrong conclusion.

I watched a talk by Klaus Molmer a few weeks ago which might be relevant. The talk was recorded
Youtube link

Please note that this is not a "popular" talk and in many places it is assumed that the audience understand that he is being a bit sloppy and know the full story. I think he might be talking about wavefunctions at one point but during the questions he did clarify what he really meant (i.e NOT the conventional wavefunction, it is a state-vector description of light pulses)
 
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No, it is a completely different situation; but not much changes. You are still going from free-space propagation (which is very complicated for a single photon) to something else; and if that "something else" is not matched in some way (using e.g. an antenna) the photon is still most likely to be reflected at the discontinuity
Yes I agree that the medium may be regarded different only in the sense that there are less modes in the cavity than in free space. Then there are less virtual particles in the cavity (as a matter of fact leading to Casimir effect). But I can't see here any mechanism which can stop the propagation of a pulse or photon as it is in classical refection in metals for example (where free electron radiate and extinguish the forward pulse but create reflected pulse). Surely the virtual photons can not behave in anyway anywhere near real electrons. So for me the reflection from the hole in cavity (e.g on the entrance in the cavity) is very strange. Vacuum and cavity vacuum are much different from air and metal.
I can not see any other mechanism but that the wavefunction (or whatever it is) travels to the electrons on the
second wall of the cavity, reflects back (not very classically but only as a probability) and gets out very quickly. The EM field should be so low and quick that will be effectively zero in the cavity. The delay should be almost unmeasurable.
I wonder if the problem is that you are taking the path integral (sum over histories) idea too literally?
Not at all. It has nothing to do with Feynman integrals. He just tells about Newton and refraction. I also reread it 1 or 2 ago.


But photons do not have wavefunctions, at least not in the usual QM sense (you can define some functions that behave a bit like a wavefunction but it gets messy).
That why I have put (in fact probability) in brackets after wavefunction in the previuos post. The photon is not localized before measurement.
 
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