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Author Topic: Direction of Radiation Emitted from Atoms  (Read 22408 times)

Offline lightarrow

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« Reply #25 on: 14/12/2007 20:10:25 »
What I mean by a 'super directive' radiator is ok, I assume, so you have a problem with the other idea(?)
No, I haven't understood why you say that to explain which detector will reveal the photon, you have to introduce high directivity or entanglement. I know what is entanglement, but it's not clear to me how this concept would explain where the photon will be reveald on the screen. I would have called "hidden variables"; did you mean something like that?

Personally, I like Rovelli's (and others) "Relational" interpretation:
http://arxiv.org/abs/quant-ph/0604064
Essentially, it says that you can't separate a quantum system's properties from the quantum system which measures them.
« Last Edit: 14/12/2007 20:25:09 by lightarrow »
 

lyner

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« Reply #26 on: 15/12/2007 00:06:20 »
I suppose what I am saying is that either there is a decision made at the source about the direction it sends the photon or in the choice between detectors as to which one detects it.
An analogy would be a person having two (acoustic) speaking tubes, going to two destinations. He could choose, randomly, to speak down either tube( first option) OR, there could be an electronic circuit connecting (almost instantaneously) the two remote earpieces and controlling which of them was open at any one time (second option).  The result would be, in both cases, that one, but not both of the listeners could hear the message - on a random basis. You would have no way of knowing which system was at work if you couldn't get inside the system and examine it more closely.
I favour the second option because I am not happy with an option which requires extremely non-classical behaviour of a wave. The second has its problems, too, of course but it has precedents. Is that clearer - or do you just not agree with my distinction between the two?

Quote
can't separate a quantum system's properties from the quantum system which measures them.
I guess that could be it, then. The whole system could be, at its simplest, one emitter and two possible detectors. Once one detector has detected the source, it has resolved the uncertainty so the other receiver cannot detect it. In a real situation, this could imply a 'total' web of entanglement for all possible detectors - or everything. . . . but why not?
btw, it seems that your link is restricted to subscribers.
 

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« Reply #27 on: 15/12/2007 02:25:43 »
Forgetting for the moment about the particulate perspective, how does the work from the relativistic perspective?  According to relativity, an EM wave is nothing but a manifestation of a contraction of space happening perpendicular to the direction of motion of a charge - so in which direction is the charge moving?  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?
Can you explain this? (It's new to me!)

Sorry, I probably did not explain myself very well.

Magnetism is a manifestation of special relativity acting on a moving electric field.  Thus electromagnetism is also a manifestation of relativity acting upon an oscillating electric field.

The point is that for special relativity to be applied, you must have something that has a velocity, which means it has a vector (one dimension, not three).  A spherical wave front has no vector.
 

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« Reply #28 on: 15/12/2007 08:17:03 »
Forgetting for the moment about the particulate perspective, how does the work from the relativistic perspective?  According to relativity, an EM wave is nothing but a manifestation of a contraction of space happening perpendicular to the direction of motion of a charge - so in which direction is the charge moving?  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?
Can you explain this? (It's new to me!)

Sorry, I probably did not explain myself very well.

Magnetism is a manifestation of special relativity acting on a moving electric field.  Thus electromagnetism is also a manifestation of relativity acting upon an oscillating electric field.

The point is that for special relativity to be applied, you must have something that has a velocity, which means it has a vector (one dimension, not three).  A spherical wave front has no vector.

Yes, it has.
Spherical wave:
E = (E0/r)ei(kr - ωt)

k = wave vector; it has the direction of the propagating wave in every point of space; its modulus is: k = 2π/λ; λ = wavelenght.

ω = 2πf; f = frequency; in the void: ω = ck

r = position vector; from the source of the wave (source point) to the position where you want to compute the fields (field point).

E0 is a vector with constant modulus and perpendicular to k.
« Last Edit: 15/12/2007 18:32:15 by lightarrow »
 

Offline lightarrow

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« Reply #29 on: 15/12/2007 08:21:58 »
[...]
btw, it seems that your link is restricted to subscribers.
Doesn't it work if you click on "pdf", for example?
Try if you can see the document in .pdf directly from this link:
http://arxiv.org/PS_cache/quant-ph/pdf/0604/0604064v3.pdf
 

Offline lightarrow

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« Reply #30 on: 15/12/2007 10:20:56 »
I suppose what I am saying is that either there is a decision made at the source about the direction it sends the photon or in the choice between detectors as to which one detects it.
An analogy would be a person having two (acoustic) speaking tubes, going to two destinations. He could choose, randomly, to speak down either tube( first option) OR, there could be an electronic circuit connecting (almost instantaneously) the two remote earpieces and controlling which of them was open at any one time (second option).  The result would be, in both cases, that one, but not both of the listeners could hear the message - on a random basis. You would have no way of knowing which system was at work if you couldn't get inside the system and examine it more closely.
I favour the second option because I am not happy with an option which requires extremely non-classical behaviour of a wave. The second has its problems, too, of course but it has precedents. Is that clearer - or do you just not agree with my distinction between the two?
Ok, now it's clear what you intended. However your second explanation seems problematic to me: you can decide to construct the detecting apparatus much later than the signal has been sent (e.g. we constructed light detectors billions of years after than light has been sent from very distant stars). How can you imagine a previous "connection" between source and detector?
 

lyner

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« Reply #31 on: 15/12/2007 11:09:58 »
That's why I made my subsequent remark that everything must be connected to everything else by this quantum-entangling net.
Constructing the detectors would involve connecting to the net at the time of construction. Bringing a detector into existence would involve slotting it into this net; in fact 'existence' as such, would mean connection into the net.
The decision/probability wave would, effectively, be spreading out from the source and the region where the photon would be most likely to be detected would be spreading outwards at what we call the speed of light. The entanglement net would be over / near the surface of this (distorted, because of interaction with matter) sphere.

The more I think about this, the more I see it necessary to have something like this mechanism or else you have to decide that the actual size of a photon is much less than the system it interacts with. If it has larger dimensions, then you still have to explain what happens when it makes a choice between two adjacent atoms. When it has chosen one, what does it do to tell the others that it has chosen? Does it 'swerve' at the last minute towards the one it has chosen?
 

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« Reply #32 on: 15/12/2007 18:12:34 »
That's why I made my subsequent remark that everything must be connected to everything else by this quantum-entangling net.
Constructing the detectors would involve connecting to the net at the time of construction. Bringing a detector into existence would involve slotting it into this net; in fact 'existence' as such, would mean connection into the net.
The decision/probability wave would, effectively, be spreading out from the source and the region where the photon would be most likely to be detected would be spreading outwards at what we call the speed of light. The entanglement net would be over / near the surface of this (distorted, because of interaction with matter) sphere.

The more I think about this, the more I see it necessary to have something like this mechanism or else you have to decide that the actual size of a photon is much less than the system it interacts with. If it has larger dimensions, then you still have to explain what happens when it makes a choice between two adjacent atoms. When it has chosen one, what does it do to tell the others that it has chosen? Does it 'swerve' at the last minute towards the one it has chosen?
Let's see if I have understood: you say that a preexisting net connects the location of the source S to a specific location in space D, so that, if a detector is subsequently put in that location D, that will be the detector which will detect the photon and no others. Did I understand your idea?
 

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« Reply #33 on: 15/12/2007 18:26:26 »
Spherical wave:
E = E0(r/r2)ei(kr - ωt).
Sorry, I should have written:
E = (E0/r)ei(kr - ωt)
where E0 is a vector perpendicular to k.
I have corrected my previous post.
« Last Edit: 15/12/2007 18:30:18 by lightarrow »
 

lyner

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« Reply #34 on: 15/12/2007 21:47:51 »
Quote
if a detector is subsequently put in that location D, that will be the detector which will detect the photon and no others.
That's about what I meant; once a photon has been detected then nothing else can detect it. The uncertainty is only resolved once the photon has actually been detected so the information about its detection must emanate at that time, to everything else, instantly ( or very much quicker than c).
 

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« Reply #35 on: 16/12/2007 05:30:18 »
Spherical wave:
E = E0(r/r2)ei(kr - ωt).
Sorry, I should have written:
E = (E0/r)ei(kr - ωt)
where E0 is a vector perpendicular to k.
I have corrected my previous post.

I think there's an error in that equation.  Spherical waves should be spherically symmetric, and can't depend on a vector r, but only on the radial position, r=|r|.  What's usually called a spherical wave should be written as

E=E0/r  exp[i(kr-ωt)],
(where I think E0 is perpendicular to the outward radial direction at every point, but I'd have to double check).

What you've written is a plane wave traveling in direction k and with an amplitude decaying as 1/r. 
 

Offline lightarrow

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« Reply #36 on: 16/12/2007 18:09:04 »
Spherical wave:
E = E0(r/r2)ei(kr - ωt).
Sorry, I should have written:
E = (E0/r)ei(kr - ωt)
where E0 is a vector perpendicular to k.
I have corrected my previous post.

I think there's an error in that equation.  Spherical waves should be spherically symmetric, and can't depend on a vector r, but only on the radial position, r=|r|.  What's usually called a spherical wave should be written as

E=E0/r  exp[i(kr-ωt)],
(where I think E0 is perpendicular to the outward radial direction at every point, but I'd have to double check).

What you've written is a plane wave traveling in direction k and with an amplitude decaying as 1/r. 

You are right.
I intended to write the equation of a spherical wave wich source is not in the origin of coordinates, but I made a mistake.

If the source of the spherical waves is in r0 and r is the position vector, then the equation should be (I hope it's correct now!):

E = E0/|r - r0| ei(k|r - r0| - ωt)

where |a| indicates the modulus of a vector a.
 

lyner

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« Reply #37 on: 16/12/2007 20:03:25 »
Now you've sorted your vectors out, what do think of the entanglement net thing?
btw, what's wrong with having a vector r?
It's pointing in the right direction and it's the right length,  - what more do you want?
The E vector would be normal to it. Or is my maths too sloppy?
 

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« Reply #38 on: 17/12/2007 04:11:50 »
Now you've sorted your vectors out, what do think of the entanglement net thing?
I just want to make sure I follow you correctly: you're saying that you place two detectors around an atom, and release a photon.  You then entangle the photon and the detectors such that you can get either a photon/detector "hit" on detector 1, or you can get one on detector 2, but not both?

Quote
btw, what's wrong with having a vector r?
It's pointing in the right direction and it's the right length,  - what more do you want?
The E vector would be normal to it. Or is my maths too sloppy?

If you're talking about the spherical/plane wave discussion above, the two cases arrive from different assumptions.  Both solutions start from the Helmholtz equation (which is the wave equation for light of a single frequency).  If you assume nothing in particular about the system and solve it in Cartesian coordinates, the basic components of your field will be plane waves, each of which have a given direction vector k and their polarization is perpendicular to that vector.  You can make fields by adding a bunch of these together.

If you assume your system is spherically symmetric, two of your three dimensions drop out (you can rotate up/down and left/right as much as you want, and nothing will change).  Because of this symmetry, the fields being radiated out from the source are going to look identical in all directions, so they can only depend on the distance, r, you are from the source (a scalar quantity).  The basic solution under this symmetry is the spherical wave that lightarrow explained above.
 

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« Reply #39 on: 17/12/2007 12:33:43 »
Now you've sorted your vectors out, what do think of the entanglement net thing?
You also wrote that detection information will travel faster than light to all the other detectors. How would this happen?

Anyway, I have a different interpretation. Maybe the EM wave is really present in every location of space and so simultaneously on every detector, but only one of them will "click" because in that instant, statistically, it has the (slightly) greatest sensitivity of all of them. Increasing the intensity of light will increase the probability that a detector will "click" and so you will have a greater frequency of detection on the screen = greater number of photons arrived in the unit time.
« Last Edit: 17/12/2007 12:48:07 by lightarrow »
 

lyner

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« Reply #40 on: 18/12/2007 18:59:04 »
But how does one detector know that the photon is not available because another one has just taken all its energy? Th two detectors could be in different galaxies, in different directions but the same distance from the  source. They must be, in some way, aware of each other,  quasi-instantaneously.
I don't have a problem with this idea of entanglement - it's no worse than some of the other ideas in modern Physics. I am warming to it. . . .
« Last Edit: 18/12/2007 19:01:17 by sophiecentaur »
 

lyner

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« Reply #41 on: 18/12/2007 19:30:18 »

If you're talking about the spherical/plane wave discussion above, the two cases arrive from different assumptions.  Both solutions start from the Helmholtz equation (which is the wave equation for light of a single frequency).  If you assume nothing in particular about the system and solve it in Cartesian coordinates, the basic components of your field will be plane waves, each of which have a given direction vector k and their polarization is perpendicular to that vector.  You can make fields by adding a bunch of these together.

If you assume your system is spherically symmetric, two of your three dimensions drop out (you can rotate up/down and left/right as much as you want, and nothing will change).  Because of this symmetry, the fields being radiated out from the source are going to look identical in all directions, so they can only depend on the distance, r, you are from the source (a scalar quantity).  The basic solution under this symmetry is the spherical wave that lightarrow explained above.
Yes - it makes more sense to talk of a spherical wave.

The entanglement idea seems, to me, to be more and more essential for any of this to work at all - if there is to be a transition from quantum - small scale and discrete energy - and classical - large scale and a continuum of  energy.
 

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« Reply #42 on: 18/12/2007 22:23:20 »
But how does one detector know that the photon is not available because another one has just taken all its energy? Th two detectors could be in different galaxies, in different directions but the same distance from the  source. They must be, in some way, aware of each other,  quasi-instantaneously.

I send, from here on Earth, a single photon away in the space. Somwhere, at a great distance, just one of the bilions of billions of potential detectors will receive it. Very strange, it's true. But now I ask you this question: how do you know that you have sent exactly one photon, before detecting it?
 
This is how I see it (and I'm aware I can have an high probability to be wrong):
The photon exists in a detector only. You switch on a light source here and then you decrease its intensity so that just one detector at a time will "click". That's the ONLY thing we know for sure. Where are the photons?
 

lyner

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« Reply #43 on: 18/12/2007 22:36:32 »
Quote
The photon exists in a detector only.
Yes; I see your point.
You can only be sure it WAS there when you saw the detector click.
However, you can very much detect individual gamma ray photons.
You could know that a gamma photon had been emitted by looking for an appropriate  nuclear decay and detecting the beta or alpha emissions, without actually detecting the gamma photon. So I am not sure that you are correct for all cases.

As for your 'single photon' scenario. Why would you need to know it was just one photon? You could be sure there was at least one photon.
In any case, we mustn't fall into the trap of asking "what exactly is happening?" We know there is not a complete answer to this , nor will there ever be.
 

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« Reply #44 on: 18/12/2007 23:40:10 »
Quote
The photon exists in a detector only.
Yes; I see your point.
You can only be sure it WAS there when you saw the detector click.
However, you can very much detect individual gamma ray photons.
You could know that a gamma photon had been emitted by looking for an appropriate  nuclear decay and detecting the beta or alpha emissions, without actually detecting the gamma photon. So I am not sure that you are correct for all cases.

As for your 'single photon' scenario. Why would you need to know it was just one photon? You could be sure there was at least one photon.
In any case, we mustn't fall into the trap of asking "what exactly is happening?" We know there is not a complete answer to this , nor will there ever be.


Your is an interesting critic.
I'm now wondering what can we say about how a detector works in detail. I know that you have to tune its amplification so that it's not too low or too high; in the last case the detector will "click" even with the source switched off!

Someone says this is due to quantum fluctuations of the EM field in the void (don't know if it's the correct explanation, however). In this case maybe the detectors must be tuned to register the same average energy which is sent from the source? So the fact only one detector will click would be a mere consequence of that tuning? I'm in a deep speculation here.
« Last Edit: 18/12/2007 23:42:30 by lightarrow »
 

Offline JP

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« Reply #45 on: 19/12/2007 02:39:08 »
I've thought about this for a while, and I think it's going to be far easier to explain this as collapse of the wave function rather than entanglement.  The problem might lie in trying to talk about photons, which are so strange.  If you're doing the famous 2-pinhole experiment with electrons and you don't put a detector at the pinholes, the electrons "pass through" both pinholes at once (they're in a superposition of a state passing through pinhole 1 and a state passing through pinhole 2).  The mathematical notation for these two states being added is:

a|1>+b|2>

  Now let's say you turn on detectors at the 2 pinholes.  If you detect the electron at a given pinhole, it immediately collapses to a state of only passing through that pinhole.  If you measure it at pinhole 1 (and there's a probability of |a|2=a*a of doing so), it collapses as:

a|1>+b|2> → |1>

The photon's a bit funkier than an electron because you don't start with a particle and quantize it.  You start with a field and define the photon as it's "minimal excitation," but the same formalism should hold.  In other words, you should be able to write the photon as the sum of states hitting all detectors plus some extra states that will not interact with either detector:

a|1>+b|2>+∑|states missing the detectors>

If either detector gets a "hit," then it collapses to the state striking that detector.  In other words, if it hits detector 1, this state collapses as:

a|1>+b|2>+∑|states missing the detectors> → |1>

You can probably manage a description of things using entanglement, but I believe it would be extremely cumbersome to deal with mathematically.  You'd end up having to entangle every possible state of the photon with every possible state of the detector, which is going to be a lot of states.
 

lyner

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« Reply #46 on: 19/12/2007 10:27:41 »
Quote
a|1>+b|2>+∑|states missing the detectors> → |1>
That notation expresses my idea well. The only thing is that it doesn't explain what is happening when it is 'resolved' by one detector..
The fact that  it collapses into   a|1 → |1 needs, somehow, to be communicated to all the detectors which were included in the first expression, instantly (?). The collapse must be instantaneous. I am happy with that if you are.
 

lyner

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« Reply #47 on: 19/12/2007 10:46:59 »
Quote
I'm now wondering what can we say about how a detector works in detail. I know that you have to tune its amplification so that it's not too low or too high; in the last case the detector will "click" even with the source switched off!
You are implying a model which sounds a bit too much like an 'amplifier' for me - I look upon it as a more passive device with, as you say 'tuning' involved.
The idea of resonance was given to me way back in Uni and works for me. I well remember a mechanical experiment with coupled resonators exchanging energy back and forth via their loose coupling; the beat between their natural frequencies corresponding the rate of  the energy passing back and forth.
When I think of an atom emitting a photon, I picture a similar process, with an oscillation ( in the form of an electron as a standing wave) coupling to an em wave (the photon). I picture the reverse process for absorption. Because of the quantum nature of the components of the model, it only happens once, unlike the coupled pendulums. The 'length' of the photon would relate to the time for the energy transfer from atom to em wave, which would, classically, be influenced by the Q factor  (impedance of  reactive components inside the atomic oscillator/ resistance corresponding to the coupling factor).  The Q factor varies for different atoms, I believe; the widths of different spectral lines would show this.
 

Offline lightarrow

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« Reply #48 on: 19/12/2007 13:15:14 »
Quote
I'm now wondering what can we say about how a detector works in detail. I know that you have to tune its amplification so that it's not too low or too high; in the last case the detector will "click" even with the source switched off!
You are implying a model which sounds a bit too much like an 'amplifier' for me - I look upon it as a more passive device with, as you say 'tuning' involved.

My question is: In the case of one "click" at a time of the detector, how do you know that you have detected exactly the energy that has been sent from the source and not a quantum fluctuation of the void?
 

lyner

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« Reply #49 on: 19/12/2007 15:40:22 »
Quote
quantum fluctuation of the void
You have introduced another idea here.
If you are suggesting that an emitted photon's energy is, somehow, absorbed into a general 'bank' of energy, which can turn up somewhere else in a position and time which are related by c, then that's fair enough. It's quite possible that you couldn't tell the difference. This depends on how complete the new idea / model is and how near it fits measurements to-date.

I think you are bringing in an irrelevant argument about effectively  'knowing which photon you detected'. QM doesn't let us assert anything like that but there are plenty of experiments which could link a significant number of received photons with a particular source; take a lamp and a light cell down a coal mine and you can be pretty well sure where the photons all came from! By that I mean that a photon model  which we have been discussing would reasonably indicate that the energy from the source was the energy detected. That's as much as you could say, though. Your alternative model could possibly give an explanation.
What I am aiming for is a model which is as close as possible to a classical one. This  corresponds to a reductionist attitude which characterises most of our Science. On a largely qualitative basis, and without violating much of the quantitative stuff I am familiar with, the model in my mind does this - allowing wave theory, Newton, Fourier and others to apply where possible and for the QM ideas to tend to the classical, in the limit. The only problem is the entanglement idea (just one small step for the brain??!!)
As soon as you enter into the Zero Point Energy world and quantum fluctuations, it gets harder and harder to keep one foot on firm ground (the Science of things we can measure).  Connecting the two worlds is beyond me at the moment but I keep reading.
I may argue very hard at times but I am always open to well founded suggestions.

The main thing that worries me about quantum fluctuations is the open ended possibility of even more loony discussions with fewer and fewer contributors on whom we can rely.

 

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