Why does decoherence not fully solve the measurement problem?

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Why does decoherence not fully solve the measurement problem? I know that must have been discussed here before a lot, maybe someone can me direct to a earlier thread or post that explains it well?

I read some QM texts, but they mostly do not discuss decoherence. I know something with 'definite outcomes' and 'eigenspace selection' troubles the decoherence approach, but never understood what it means...

thank you
 

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  • #2
Fredrik
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I think it does solve it in the context of a many-worlds interpretation that doesn't throw away the Born rule (i.e. not the Everett version), but I don't know if anyone has ever really spelled out all the details.

You will probably find the discussion in Schlosshauer's book enlightening.

Also note that there is no "measurement problem" in the ensemble interpretation. The problem only exists for people who believe that QM describes reality, even at times between state preparation and measurement.

I probably won't try to elaborate much on these things, because I have found that discussions about interpretations are very time consuming, and I'm kind of busy with other things right now.
 
  • #3
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Where can I read how Everett version threw away the Born rule?
Because I am also thinking that Born rule is only an illusion created by consciousness (as very special moment called NOW) and is severily violated in our world (that violation is called antrophic principle).
 
  • #4
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I think it does solve it in the context of a many-worlds interpretation that doesn't throw away the Born rule (i.e. not the Everett version),
I essentially agree, provided that one specifies more clearly what exactly the Born rule states. Namely, if you simply accept the usual meaning of the Born rule in Copenhagen interpretation of QM, then it is questionable whether decoherence is needed to solve the measurement problem in the first place. On the other hand, if you do think that there is a measurement problem, then obviously you are not satisfied with the Copenhagen interpretation, which means that you need to rethink what the Born rule really means.
 
  • #5
SpectraCat
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The "measurement problem" in the context of standard QM generally has two aspects:

1) How does a generic quantum state, which may be superposition of eigenstates of the physical property being measured, resolve itself into a single eigenstate during the measurement process, in a manner that is at least FAPP (for all practical purposes) irreversible?

2) How does the above interaction place a detection device into a "state" (i.e. definite outcome) where its value reflects the eigenvalue of the eigenstate into which the quantum state in part 1 has been resolved?

I believe that the statement "decoherence does not solve the measurement problem" means that decoherence provides a phenomenological explanation for the first point, but provides no insight into the second.
 
  • #6
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Yes.
So point 1 rules out collapse interpetations,
but as point 2 is not resolved we have a choice between different non-collapse interpretations.
 
  • #7
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I thought point 1 was the difficult bit, but point 2 follows reasonably straightforwardly. I can describe my toy model of a quantum measurement if you want (but later because I have to teach perturbation theory now!)
 
  • #8
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So how do you explain point 2?
 
  • #9
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(to peteratcam:) Let me be the first to say if you have a theory of measurement, we're all going to be really interested in it. (Would you start a new thread so we'll know where to look?)
 
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  • #10
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I make no claims at all to have 'solved' the measurement problem. All I have is a toy model which I invented (most probably rediscovered) in order to play around with the question "how do I describe the process of measurement in a fully quantum mechanical way?". In my mind, my toy answers that question to the following extent: it gives a QM mechanism for why you only measure eigenvalues, and why in doing so, you 'collapse' the system of interest into the associated eigenket of the observed operator. (All by unitary evolution of course). My toy assumes QM, so perhaps the whole thing is a bit circular.

I'll put it in a new thread then.
 
  • #11
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The "measurement problem" in the context of standard QM generally has two aspects:

1) How does a generic quantum state, which may be superposition of eigenstates of the physical property being measured, resolve itself into a single eigenstate during the measurement process, in a manner that is at least FAPP (for all practical purposes) irreversible?
This is a common description of the measurement process and I don't believe it's correct. Where do we ever see a pure eigenstate come into existence as a result of a measurement? Let a photon pass through a pinhole so it spreads out. When it hits a photographic plate, it is absorbed. It is absorbed over the whole surface of the plate and ceases to exist. It does not resolve itself into a "position eigenstate of the photon". At the same time, a silver halide crystal undergoes an irreversible phase transition. That doesn't mean the photon got concentrated at the location of the crystal. The crystal didn't need the entire energy of the photon undergo a change of state: it was in a metastable state to begin with and only needed a small distrubance.

What about the photoelectric effect, where the escaping electron did indeed need the total energy of the photon? In that case the exact point where the photoelectron leaves the metal surface is indeterminate, and you can't say the photon was concentrated at a specific location.

What about an electron going through the Stern Gerlach apparatus? It divides into two beams: that is it's "state". When it hits the detector screen the two streams jointly excite a single bound wave function within the screen; at the same time, a crystal on the surface changes state. How do we know that the newly-excited bound wave function doesn't have the same spin that the electron had BEFORE entering the SG apparatus? There is no need to think that the electron, originally in a composition of two spin states, has resolved itself into one state or the other simply because a particular crystal in one branch of the wave stream has changed color.
 
  • #12
SpectraCat
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This is a common description of the measurement process and I don't believe it's correct. Where do we ever see a pure eigenstate come into existence as a result of a measurement? Let a photon pass through a pinhole so it spreads out. When it hits a photographic plate, it is absorbed. It is absorbed over the whole surface of the plate and ceases to exist. It does not resolve itself into a "position eigenstate of the photon". At the same time, a silver halide crystal undergoes an irreversible phase transition. That doesn't mean the photon got concentrated at the location of the crystal. The crystal didn't need the entire energy of the photon undergo a change of state: it was in a metastable state to begin with and only needed a small distrubance.
You are postulating your own interpretation there, and it requires some support. In particular, the statements:

"(the photon) is absorbed over the whole surface of the plate and ceases to exist"

and

"the crystal didn't need the entire energy of the photon undergo a change of state: it was in a metastable state to begin with and only needed a small distrubance"

cannot simply be asserted without any evidence or justification ... well I suppose they *can*, but it will be hard to convince people you are correct.

What about the photoelectric effect, where the escaping electron did indeed need the total energy of the photon? In that case the exact point where the photoelectron leaves the metal surface is indeterminate, and you can't say the photon was concentrated at a specific location.
The location of the exact point is unpredictable, but it still leaves from a single point, so I guess I don't understand your logic there ...

What about an electron going through the Stern Gerlach apparatus? It divides into two beams: that is it's "state". When it hits the detector screen the two streams jointly excite a single bound wave function within the screen; at the same time, a crystal on the surface changes state. How do we know that the newly-excited bound wave function doesn't have the same spin that the electron had BEFORE entering the SG apparatus? There is no need to think that the electron, originally in a composition of two spin states, has resolved itself into one state or the other simply because a particular crystal in one branch of the wave stream has changed color.
Here you are just wrong ... the Stern-Gerlach experiment was initially done on silver atoms ... these atoms were deposited on a glass plate, so their positions are not in doubt. The atoms clearly "resolved themselves into one state or another" when they reached the plate. Furthermore, SG experiments cannot be performed in the typical way with electrons, since they are charged particles, and the Lorentz force will overcome the interaction with the magnetic moment.
 
  • #13
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You are postulating your own interpretation there, and it requires some support. In particular, the statements:

"(the photon) is absorbed over the whole surface of the plate and ceases to exist"

and

"the crystal didn't need the entire energy of the photon undergo a change of state: it was in a metastable state to begin with and only needed a small distrubance"

cannot simply be asserted without any evidence or justification ... well I suppose they *can*, but it will be hard to convince people you are correct.
My point about ceasing to exist is simply that at no verifiable moment in time did the photon exist in anything resembling a "position eigenstate". I think I am correct in believing there is really no such thing as a position eigenstate for photons.

About thermodynamics of silver halide: again, this should be quite uncontroversial. True, I don't have a reference, but I believe I've got the arrow pointing the right way. I've never heard of anyone being able to regenerate used photographic film by gently heating it or some such method.


The location of the exact point is unpredictable, but it still leaves from a single point, so I guess I don't understand your logic there ...
I put this example in as a contrast with the phtographic plate to show that in situations where you have a verifiable amount of energy absorbed in the detection process, your position measurement can become a little vague. It's not the the location of the exact point is "unpredictable": in fact it's unmeasurable, or at least it's certainly not measured in any version of the experiment that I know of. Even your assertion that it leaves from a single point is not easily verifiable.



Here you are just wrong ... the Stern-Gerlach experiment was initially done on silver atoms ... these atoms were deposited on a glass plate, so their positions are not in doubt. The atoms clearly "resolved themselves into one state or another" when they reached the plate. Furthermore, SG experiments cannot be performed in the typical way with electrons, since they are charged particles, and the Lorentz force will overcome the interaction with the magnetic moment.
Good one. I forgot about the silver atoms. But this wasn't my example: my example was the Stern Gerlach experiment performed on electrons, and you haven't exactly dealt with it. Well, mabye you have after all, to the extent that you've challenged me to apply the same logic to silver atoms. Fair enough:

What do we actually observe in the Stern Gerlach experiment? A jet of silver effuses (learned that word from peteratcam's explanation in another thread!) from an oven, splits in two passing through a magnet, and two cloudy patches develop on a plate. All this is in accordance with the unitary time-evolution of the wave function. Where does the "collapse" occur? I know the answer: just send a single silver atom through the apparatus and watch for it to appear at one spot or the other. The problem is this is a very difficult experiment to do, and I'm not sure it's ever been done. People assume that the behavior of the bulk material can be broken down into individual atomic events. That's an assumption and it is not easy to verify experimentally.
 
  • #14
SpectraCat
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My point about ceasing to exist is simply that at no verifiable moment in time did the photon exist in anything resembling a "position eigenstate". I think I am correct in believing there is really no such thing as a position eigenstate for photons.

About thermodynamics of silver halide: again, this should be quite uncontroversial. True, I don't have a reference, but I believe I've got the arrow pointing the right way. I've never heard of anyone being able to regenerate used photographic film by gently heating it or some such method.
What does your earlier comment have to do with silver halide thermodynamics? You hypothesized that it was in a metastable state, where it didn't need the full energy of the photon to react. This is contrary to the normal physical description for how photographic emulsions work ... there is an (effectively) irreversible chemical reaction (the reduction of silver ions to metallic silver), which can only happen when an electron is promoted from the valence band of the silver halide salt, into the conduction band. The reason these emulsions can be used to record visible light images is because the band gap is accessible to energies of photons in the visible spectrum. If it were accessible to lower energy photons, then you would be able to record infrared images also, which is not possible.

So, your hypothesis starts out on (very) thin ice, and that is why it needs to be supported by appropriate references.

I put this example in as a contrast with the phtographic plate to show that in situations where you have a verifiable amount of energy absorbed in the detection process, your position measurement can become a little vague. It's not the the location of the exact point is "unpredictable": in fact it's unmeasurable, or at least it's certainly not measured in any version of the experiment that I know of. Even your assertion that it leaves from a single point is not easily verifiable.
Of course it's not unmeasurable ... there are velocity map imaging techniques that can record how photoelectrons are emitted from molecules after UV excitation ... these could certainly be applied to metal surfaces to look at photoelectrons .. that is a *way* easier experiment. The question as to whether they will be emitted from "single points" is a question of resolution. The emitted electrons will have wave character, so their emission points will be somewhat uncertain, but that uncertainty should be small .. at least on the order of microns, and perhaps much smaller.

Good one. I forgot about the silver atoms. But this wasn't my example: my example was the Stern Gerlach experiment performed on electrons, and you haven't exactly dealt with it. Well, mabye you have after all, to the extent that you've challenged me to apply the same logic to silver atoms. Fair enough:
No I "dealt with it" by pointing out that SG experiments can't be carried out on electrons, because of the Lorentz force.

What do we actually observe in the Stern Gerlach experiment? A jet of silver effuses (learned that word from peteratcam's explanation in another thread!) from an oven, splits in two passing through a magnet, and two cloudy patches develop on a plate. All this is in accordance with the unitary time-evolution of the wave function. Where does the "collapse" occur? I know the answer: just send a single silver atom through the apparatus and watch for it to appear at one spot or the other. The problem is this is a very difficult experiment to do, and I'm not sure it's ever been done. People assume that the behavior of the bulk material can be broken down into individual atomic events. That's an assumption and it is not easy to verify experimentally.
Again, you are hypothesizing in a seemingly nonsensical manner. There is never any "bulk" silver, or even silver clusters, flying through the apparatus ... atom sources are well understood, and can be tuned so the cluster flux is effectively zero, and only atoms are emitted. So, whatever builds up on the plate builds up an atom at a time.

But you don't have to believe me about such things, consider the atom interferometer that has been built using SG magnets: http://www3.interscience.wiley.com/journal/107632668/abstract?CRETRY=1&SRETRY=0

Also, with modern detectors, it is certainly possible to detect single atoms in a position sensitive way ... for example they could be resonantly ionized using a laser with a small cross section, and the resultant charged particles detected. Here is a link to a delayed choice SG experiment done using single metastable hydrogen atoms. http://quantmag.ppole.ru/Articles/Lawson_p5042_1.pdf
 
  • #15
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What does your earlier comment have to do with silver halide thermodynamics? You hypothesized that it was in a metastable state, where it didn't need the full energy of the photon to react. This is contrary to the normal physical description for how photographic emulsions work ... there is an (effectively) irreversible chemical reaction (the reduction of silver ions to metallic silver), which can only happen when an electron is promoted from the valence band of the silver halide salt, into the conduction band. The reason these emulsions can be used to record visible light images is because the band gap is accessible to energies of photons in the visible spectrum. If it were accessible to lower energy photons, then you would be able to record infrared images also, which is not possible.
Thanks, Spectracat; this is actually a better explanation of the process than I've seen. I used the words "spontaneity" and "thermodynics" to indicate what you have called an "irreversible chemical reaction". I think it comes to the same thing.

It's significant in my explanation because I distinguish two kinds of detection events:

1) the energy required for the detection process is supplied entirely by the particle being detected.

2) there is available energy in the detection mechanism to drive the detection process.

From your explanation, the photographic plate falls in category 2, which was my original point: since the all the energy of the photon wasn't required to drive the phase transition, you can't conclude that it was ever localized at the position of the crystal.


Of course it's not unmeasurable ... there are velocity map imaging techniques that can record how photoelectrons are emitted from molecules after UV excitation ... these could certainly be applied to metal surfaces to look at photoelectrons .. that is a *way* easier experiment. The question as to whether they will be emitted from "single points" is a question of resolution. The emitted electrons will have wave character, so their emission points will be somewhat uncertain, but that uncertainty should be small .. at least on the order of microns, and perhaps much smaller.
My original point was that in the photoelectric experiment, you don't normally identify where the photon hit the plate. You are speculating that the experiment could be modified to pinpoint the source of the emitted electrons. I am speculating that you can't.


No I "dealt with it" by pointing out that SG experiments can't be carried out on electrons, because of the Lorentz force.
I'm no experimentalist, but this doesn't seem like an insurmountable obstacle. Theoretically you could place the ordinary SG apparatus in a room surrounded by a giant magnet pointing the opposite way, so there was nothing left of the field except the gradient.



Again, you are hypothesizing in a seemingly nonsensical manner. There is never any "bulk" silver, or even silver clusters, flying through the apparatus ... atom sources are well understood, and can be tuned so the cluster flux is effectively zero, and only atoms are emitted. So, whatever builds up on the plate builds up an atom at a time.
You are awfully picky about my choice of descriptive words. I used the word "bulk" in context to contrast it with "one-atom-at-a-time", and re-reading my initial post it seems to me this is pretty clear from context. You make it sound like I imagine great globs of molten silver flying through the air.


But you don't have to believe me about such things, consider the atom interferometer that has been built using SG magnets: http://www3.interscience.wiley.com/journal/107632668/abstract?CRETRY=1&SRETRY=0

Also, with modern detectors, it is certainly possible to detect single atoms in a position sensitive way ... for example they could be resonantly ionized using a laser with a small cross section, and the resultant charged particles detected. Here is a link to a delayed choice SG experiment done using single metastable hydrogen atoms. http://quantmag.ppole.ru/Articles/Lawson_p5042_1.pdf
The question is: does the spin state "collapse" when you measure it? The SG experiment does not answer this question because it doesn't look at single atoms. You prepare a beam in an arbitrary orientation, put it through the magnets and you get two spots on a plate. You have therefore measured the spin of the beam to the extent that it is, say, 40% up and 60% down based on the intensity of the two spots. It is, so to speak, a "bulk proerty" of the beam. You never actually observe the spin state collapsing.

Again, you are speculating that the experiment could be redesigned so that a single silver atom with, say, a y-oriented spin, is sent through a z-oriented SG magnet; and that at the "moment of measurement, its spin suddenly jumps into either the +z or -z state. I am speculating that you can't.
 
  • #16
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Thanks, Spectracat; this is actually a better explanation of the process than I've seen. I used the words "spontaneity" and "thermodynics" to indicate what you have called an "irreversible chemical reaction". I think it comes to the same thing.

It's significant in my explanation because I distinguish two kinds of detection events:

1) the energy required for the detection process is supplied entirely by the particle being detected.

2) there is available energy in the detection mechanism to drive the detection process.

From your explanation, the photographic plate falls in category 2, which was my original point: since the all the energy of the photon wasn't required to drive the phase transition, you can't conclude that it was ever localized at the position of the crystal.




My original point was that in the photoelectric experiment, you don't normally identify where the photon hit the plate. You are speculating that the experiment could be modified to pinpoint the source of the emitted electrons. I am speculating that you can't.




I'm no experimentalist, but this doesn't seem like an insurmountable obstacle. Theoretically you could place the ordinary SG apparatus in a room surrounded by a giant magnet pointing the opposite way, so there was nothing left of the field except the gradient.





You are awfully picky about my choice of descriptive words. I used the word "bulk" in context to contrast it with "one-atom-at-a-time", and re-reading my initial post it seems to me this is pretty clear from context. You make it sound like I imagine great globs of molten silver flying through the air.




The question is: does the spin state "collapse" when you measure it? The SG experiment does not answer this question because it doesn't look at single atoms. You prepare a beam in an arbitrary orientation, put it through the magnets and you get two spots on a plate. You have therefore measured the spin of the beam to the extent that it is, say, 40% up and 60% down based on the intensity of the two spots. It is, so to speak, a "bulk proerty" of the beam. You never actually observe the spin state collapsing.

Again, you are speculating that the experiment could be redesigned so that a single silver atom with, say, a y-oriented spin, is sent through a z-oriented SG magnet; and that at the "moment of measurement, its spin suddenly jumps into either the +z or -z state. I am speculating that you can't.
I can only comment on one point... the "magnet pointing the other way" idea is FAR better in theory than practice. In practice, I don't believe the Lorentz force could be eliminated to anyone's satisfacting without introducing unwanted elements.

The rest, well... does a spin state EVER collapse? These experiments support, and represent the progress of SQM where other interpretations fail to do anything but keep pace. Yes, you can critique any one experiment, but it's the growing preponderance of the evidence that is the issue.
 
  • #17
SpectraCat
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Thanks, Spectracat; this is actually a better explanation of the process than I've seen. I used the words "spontaneity" and "thermodynics" to indicate what you have called an "irreversible chemical reaction". I think it comes to the same thing.

It's significant in my explanation because I distinguish two kinds of detection events:

1) the energy required for the detection process is supplied entirely by the particle being detected.

2) there is available energy in the detection mechanism to drive the detection process.

From your explanation, the photographic plate falls in category 2, which was my original point: since the all the energy of the photon wasn't required to drive the phase transition, you can't conclude that it was ever localized at the position of the crystal.
No, how on earth did you come to that conclusion? The way that the irreversible chemical reaction happens is that the *entire* energy of the visible photon is absorbed, resulting in the promotion of an electron from the valence band to the conduction band. The resulting electron then reduces a silver ion to a silver atom. This is an explicitly local process, otherwise the entire principle of photography would be invalid ... oh by the way, your eyes wouldn't work either if the process of photon detection wasn't local. The counter-examples are myriad and obvious ... I have no idea where you are coming up with your "interpretations".


My original point was that in the photoelectric experiment, you don't normally identify where the photon hit the plate. You are speculating that the experiment could be modified to pinpoint the source of the emitted electrons. I am speculating that you can't.
Dammit, I am *not* speculating, so watch it .. this is my research field ... velocity map imaging is a well-documented technique that deals precisely with the issue that you raised. You should darn well read up on it before you place yourself on equal footing with me in this area and accuse me of being speculative.

I'm no experimentalist, but this doesn't seem like an insurmountable obstacle. Theoretically you could place the ordinary SG apparatus in a room surrounded by a giant magnet pointing the opposite way, so there was nothing left of the field except the gradient.
Whatever, AFAIK, this experiment hasn't been done .. there is no reason why it wouldn't work in the theoretical case that you mention. So, what was the interpretation you put forward for the electron experiment? And don't make it dependent on a single detection method (i.e. phosphor screen) .. it needs to be equally relevant irrespective of the physical measurement technique (e.g. photo electron multiplier tube, channeltron electron multiplier, charge coupled device, etc.). You also need to allow for separate, independent detectors to be placed at the two output of the split beam .. AFAICS, that will completely invalidate your initial analysis where the electron "jointly excites" two spots on the screen, but only one shows up, or whatever you were trying to say.

You are awfully picky about my choice of descriptive words. I used the word "bulk" in context to contrast it with "one-atom-at-a-time", and re-reading my initial post it seems to me this is pretty clear from context. You make it sound like I imagine great globs of molten silver flying through the air.
I have no idea what you were postulating .. it was not clear from context.

The question is: does the spin state "collapse" when you measure it? The SG experiment does not answer this question because it doesn't look at single atoms. You prepare a beam in an arbitrary orientation, put it through the magnets and you get two spots on a plate. You have therefore measured the spin of the beam to the extent that it is, say, 40% up and 60% down based on the intensity of the two spots. It is, so to speak, a "bulk proerty" of the beam. You never actually observe the spin state collapsing.
Uh-huh ... and what is your postulated counterexample, that the silver atoms somehow hit both spots on the screen? Please ... this experiment is well understood and described in the context of the wavefunctions of single particles becoming entangled with the distinct spatial paths of the spin-up and spin-down components of the beam. This entanglement is then resolved into a single state via decoherence when the wavefunction interacts with the detector.

Again, you are speculating that the experiment could be redesigned so that a single silver atom with, say, a y-oriented spin, is sent through a z-oriented SG magnet; and that at the "moment of measurement, its spin suddenly jumps into either the +z or -z state. I am speculating that you can't.
Ok, you are really starting to irritate me with the completely unfounded accusations of speculation ... did you even look at the articles I posted? One of them expressly deals with "single atom mode" for the SG apparatus discussed. I am talking about reference-supported science here, and you are hypothesizing half-baked alternate theories that are completely unsupported by anything, and generally mis-represent at least one known aspect of physics.
 
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I guess that means we're not friends anymore. :>(
 
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No, how on earth did you come to that conclusion? The way that the irreversible chemical reaction happens is that the *entire* energy of the visible photon is absorbed...
But an "irreversible chemical reaction" doesn't need energy to drive it; it releases energy. So the fact that the reaction proceded to completion is no evidence at all that the "entire" energy of the photon was absorbed at the location of the spot.

...resulting in the promotion of an electron from the valence band to the conduction band. The resulting electron then reduces a silver ion to a silver atom. This is an explicitly local process, otherwise the entire principle of photography would be invalid ... oh by the way, your eyes wouldn't work either...
Sure they would. The rate of events is still proportional to the local field intensity in my model, the same as in yours.
 
  • #20
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But an "irreversible chemical reaction" doesn't need energy to drive it; it releases energy. So the fact that the reaction proceded to completion is no evidence at all that the "entire" energy of the photon was absorbed at the location of the spot.



Sure they would. The rate of events is still proportional to the local field intensity in my model, the same as in yours.
What does the reversability of a chemical reaction have to do with it being exo- or endo- thermic? Energy release is a function of the nature of the reaction, not its reversability.
 
  • #21
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What does the reversability of a chemical reaction have to do with it being exo- or endo- thermic? Energy release is a function of the nature of the reaction, not its reversability.
I don't think I said that the process was exothermic, although I'm guessing it would be. Spectracat might be able to tell us one way or the other. The point is with an irreversible reaction you don't need to supply external energy in order to make it happen. To speed it up maybe, but that's another question.

The free energy of the reaction products is lower than the free energy of unilluminated crystal, so there is no evidence that the entire energy of a photon had to be concentrated at the point of reaction.
 
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I don't think I said that the process was exothermic, although I'm guessing it would be. Spectracat might be able to tell us one way or the other. The point is with an irreversible reaction you don't need to supply external energy in order to make it happen. To speed it up maybe, but that's another question.

The free energy of the reaction products is lower than the free energy of unilluminated crystal, so there is no evidence that the entire energy of a photon had to be concentrated at the point of reaction.
I should have been clearer, and you're right, it could be a reaction that is purely luminescent, etc. That said, there are certainly irreversable reactions which DO require energy, such as the transition of a metal to a plasma. I understand the point you're trying to make however, and this isn't Chemistry Forums so I'll leave it be.

I'm still unconvinced by this "point of reaction" or detector-centric rather than quanta-centric view. I've heard it a lot, but it just seems to shift the issue rather than solve it, and as SpectraCat keeps saying everything about this is vague and unverified. I need more than analogies or toy models to be sold on this, when there is (as Cat has said) a lot of well established evidence in this field.
 
  • #23
SpectraCat
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I don't think I said that the process was exothermic, although I'm guessing it would be. Spectracat might be able to tell us one way or the other. The point is with an irreversible reaction you don't need to supply external energy in order to make it happen. To speed it up maybe, but that's another question.
No, this is an endo-ergic reaction, that is, it requires energy to go from the reactant (ground state silver halide) to the product (silver metal). Those thermodynamics are easily verified. The energy to make the reaction go is supplied by the photon. Once the electron is promoted to the conduction band, which takes the entire energy of the photon, the reaction can then occur, not before.

The free energy of the reaction products is lower than the free energy of unilluminated crystal, so there is no evidence that the entire energy of a photon had to be concentrated at the point of reaction.
No, that is not right .. the free energy is lower than the silver halide crystal plus the conduction band electron. The electron promotion has to happen first, or else there is not enough energy for the reaction to go. As I have said, that is why photographic film is sensitive to light in the visible spectrum ... if it were just made of thermodynamically metastable crystals, as you have hypothesized, then the wavelength of the photon wouldn't matter, would it?

Look, you have way more problems with your model than just the chemistry being backwards. You have to explain how it is possible to form an image in the first place, if the process of excitation that exposes the film doesn't depend on the position of the incident photon. Your speculation is that the photon is somehow absorbed globally, and some random silver halide crystal spontaneously turns into silver, but that the locations of the two events are not connected. This is demonstrably at odds with the electronic and optical physics, as well as the chemistry involved in the problem, as it is conventionally understood. So, if you want to have a snowball's chance to convince anyone, you need to address each of the points one by one, which you have not done. You have just made some guesses about what you think might be happening, which I have shown to be incorrect. I love to argue, so when I have time I post responses, but eventually it gets tiresome when you don't seem to be paying serious attention to what I am saying.
 
  • #24
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No, this is an endo-ergic reaction, that is, it requires energy to go from the reactant (ground state silver halide) to the product (silver metal). Those thermodynamics are easily verified. The energy to make the reaction go is supplied by the photon. Once the electron is promoted to the conduction band, which takes the entire energy of the photon, the reaction can then occur, not before.



No, that is not right .. the free energy is lower than the silver halide crystal plus the conduction band electron. The electron promotion has to happen first, or else there is not enough energy for the reaction to go. As I have said, that is why photographic film is sensitive to light in the visible spectrum ... if it were just made of thermodynamically metastable crystals, as you have hypothesized, then the wavelength of the photon wouldn't matter, would it?

Look, you have way more problems with your model than just the chemistry being backwards. You have to explain how it is possible to form an image in the first place, if the process of excitation that exposes the film doesn't depend on the position of the incident photon. Your speculation is that the photon is somehow absorbed globally, and some random silver halide crystal spontaneously turns into silver, but that the locations of the two events are not connected. This is demonstrably at odds with the electronic and optical physics, as well as the chemistry involved in the problem, as it is conventionally understood. So, if you want to have a snowball's chance to convince anyone, you need to address each of the points one by one, which you have not done. You have just made some guesses about what you think might be happening, which I have shown to be incorrect. I love to argue, so when I have time I post responses, but eventually it gets tiresome when you don't seem to be paying serious attention to what I am saying.
The only way his position makes sense is with a Pilot Wave, or some kind of complex or massive consciousness bias. I don't see that being feasible.

@conway: Can I ask why you want to believe this, or do believe this compared to the formalism? Your notion would invalidate all current optical physics and more, so... where's the upside of this interpretation?
 
  • #25
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No, this is an endo-ergic reaction, that is, it requires energy to go from the reactant (ground state silver halide) to the product (silver metal). Those thermodynamics are easily verified.
Yes, I don't see why not. I might be able to verify them myself if you would be kind enough to write out the chemical reaction in full.


No, that is not right .. the free energy is lower than the silver halide crystal plus the conduction band electron.
Of course it is. If the free energy of the products is lower than the free energy of the reactants, (which is all I claimed) then of course the free energy of the products is lower than the free energy of the reactants plus a random photon.


Look, you have way more problems with your model than just the chemistry being backwards. You have to explain how it is possible to form an image in the first place...
Not really. When I started this discussion I said "let the photon pass through a pinhole so it spreads out across the area of the photographic plate". In that situation there is of course no image.
 

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