Why does decoherence not fully solve the measurement problem?

[conway]

I stand by exactly what I said. You're telling me that an irreversible chemical reaction can't be endothermic?

No. I said that reversible implies spontaneous, not exothermic. And therefore, based on Spectracat's calling the photographic process "irreversible", it was reasonable for me to conclude that it proceded with a release of free energy. Assuming that Spectracat meant what he said. What do you think he meant?

As for the rest, I wasn't trying to analyze you, but rather see what aspect of an alternate theory appeals to you so greatly that formalism is unappealing? I don't agree with him, but Demystifier regularly makes a case for dBB, and I understand why; he sees the theory as being more appealing with fewer contradictions with our observed reality. As the theory is empirically identical to SQM, it's clearly a matter of choice right now. In your case, I wish to understand, what is your dBB, or do you generally hold with formalism/TCI 'with major reservations' (as I believe many do, myself included)?

I don't like the wave function collapse. I thought that would have been clear from my very first post. And I find that many of the most common examples used to demonstrate the supposed collapse of the wavefunction "(from) a superposition of eigenstates of the physical property being measured, ... into a single eigenstate during the measurement process", as Spectracat put it, are false examples. Including the specks of silver on the photographic plate.

 

[Frame Dragger]

No. I said that reversible implies spontaneous, not exothermic. And therefore, based on Spectracat's calling the photographic process "irreversible", it was reasonable for me to conclude that it proceded with a release of free energy. Assuming that Spectracat meant what he said. What do you think he meant?

I assumed he meant "irreversible" or "reversible" in the sense that it is used in chemistry, and logically.

I don't like the wave function collapse. I thought that would have been clear from my very first post. And I find that many of the most common examples used to demonstrate the supposed collapse of the wavefunction "(from) a superposition of eigenstates of the physical property being measured, ... into a single eigenstate during the measurement process", as Spectracat put it, are false examples. Including the specks of silver on the photographic plate.

I get that, but SpectraCat just put a broadside into your view...

 

[conway]

I know it must appear that way to you, but surely you know that people with my type of obsession don't let such small obstacles get in their way. The fact of the matter is that I have devised a cunning and amusing argument to demonstrate how his figures show just the opposite. I only hesitate to post it because I have been cautioned several times already in this thread for irritating behavior.

 

[SpectraCat]

There's one more point I'd like to clear up, because I took a lot of criticism for it in this thread. On two occasions Spectracat referred to the photographic process as "irreversible". I responded that if it was irreversible, it didn't need any input of energy from the photon in order to procede to completion. I was roundly ridiculed for this:



And by Frame Dragger as well:



I was even doubting myself for a while there and so I looked up the correct usage of the term. It seems I was right after all. In chemistry, an irreversible reaction is one that procedes essentially to completion once it has begun, being driven by its own available free energy. Lighting a match would be an example. A reversible reaction is qualitativley different in that while it may procede in the positive direction, it eventually comes to rest at an equilibrium concentration. The textbook example is the synthesis of ammonia from hydrogen and nitrogen.

So when Spectracat said that the photographic process was an irreversible reaction, unless he meant to support my claim he was simply making an incorrect use of the terminology.

*Sigh* ... I was actually giving you credit for a more nuanced understanding of the concept of reversibility in the chemical sense. Go back and look at my post .. I said it was *effectively* irreversible ... look at my later post where I explained what I mean about chemical irreversibility in more detail. {EDIT: I just checked, and apparently I never posted what I wrote on this .. at least I can't find it now. Anyway, it was simply that all chemical reactions are reversible in principle, and it is only secondary effects that make them irreversible. For example, it is often just simple diffusion of the products away from the reaction site that practically rules out the reverse reaction. In other cases, there might be a secondary reaction that binds up one of the products, making it inaccessible for the reversal of the reaction. For activated processes where there is a reverse barrier, there may be rapid energy dissipation out of the products, so that there is a very small probability of them "finding their way" to the reverse reaction before the internal energy drops below the level where it is possible. Highly exothermic reactions, like your match example, are very good approximations of irreversible processes, because the energy barrier for the reverse reaction is so high that there is little (although not zero) probability of observing the reverse reaction.} I was actually assuming that you understood some or all of that, at least implicitly. Furthermore, I was consistent about my description of the thermodynamics ... I said from the start that the products were higher in energy than the reactants .. that was the whole point of my criticism of your hypothesis. Just as a reminder, when the energy (strictly speaking the Gibbs free energy) of the products is higher than that of the reactants, we are describing a thermodynamically non-spontaneous reaction.

So it should have been impossible for you to assume that I meant in any way the basic, general interpretation of irreversibility that you have quoted above. For crying out loud, think about what you are saying! If that definition held in this example, then the entire silver halide crystal would be reduced to silver metal, and possibly the surrounding ones as well. But that doesn't happen ... a *single* silver atom is produced for each photon absorbed. In fact, as I showed you in the link I posted, this atom is unstable, which means that the reaction isn't really irreversible at all (as I also explained), but that there is a non-negligible change that the reaction will spontaneously go in the reverse direction .. which *is* the thermodynamically favored direction.

And by the way ... it didn't just "turn out" that I was right in this case .. I wasn't guessing. I knew this stuff in advance ... I was not speculating, ever. However you refused to take my word for any of it, or look stuff up on your own, so it was just due to your incredible stubbornness that I ended up posting all the gory details. The primary reason I did this is that PF is a database of knowledge, and it behooves all of us to do our best to keep incorrect statements to a minimum, acknowledge and retract them when they are identified by others, and make it clear when we are making "educated guesses", or just wild speculations. The primary issue I have with you is that you put your own half-baked speculations on an equal footing with the more carefully developed, researched and referenced positions of others, and that could be potentially quite confusing to someone who is looking through these threads in hopes of learning something.

So, I don't mind having these discussions (much), or helping you develop your ideas where I can (I will eventually get around to posting on the two-electron well thing again), but I just wish you would be a little more accepting of criticism when it is backed up by solid logic and facts, and not just assume that you are correct because you can't immediately see the point behind my arguments or those of others. I guess it would be helpful if you asked more questions and made fewer declarative statements when you are trying to develop these ideas based on your own admittedly speculative thinking. That's just a suggestion, and you are of course free to carry on as you see fit.

 

[conway]

So do you want to see my thermodynamic analysis or not?

 

[SpectraCat]

So do you want to see my thermodynamic analysis or not?

Anything you can back up with some facts is fine by me. If think you can "spin" my analysis another way, then am at least a little interested to see your idea.

 

[SpectraCat]

I don't like the wave function collapse. I thought that would have been clear from my very first post.

And I find that many of the most common examples used to demonstrate the supposed collapse of the wavefunction "(from) a superposition of eigenstates of the physical property being measured, ... into a single eigenstate during the measurement process", as Spectracat put it, are false examples. Including the specks of silver on the photographic plate.

Please say that you find them unconvincing instead of false, ok? Isn't that more accurate? The fact is that I showed in excruciating detail why your argument is wrong for the photograph .. you choose not to accept it, for whatever reason. You want to squeeze out the last shred of possibility that your idea has *any* leg to stand on, let alone convincing credibility.

Look, the silver halide crystals (grains) in question are microscopic ... 2-30 microns in average size, depending on the film quality. So, the "uncertainty" of the location where the photon was detected, due to the need for "ion trap sites" in the grains that you mentioned, is limited to that size range. Call a grain a pixel (that is typical), and consider how many pixels are illuminated in the Airy-disk image attached below (copied from hyperphysics site http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/cirapp2.html" ). So, the diffraction pattern extends over dozens or hundreds of pixels, and you are talking about position uncertainty on the scale of a single pixel ... that seems pretty insignificant in terms of "delocalization", compared to how much the diffraction has "spread out" (your words) the photon's wavefunction based on the image.

Furthermore, the fact that the location of the silver atom on the grain *might be* different then the precise site where the photon was absorbed is not really evidence of the "delocalization" you have hypothesized. It is much more likely that it just indicates that the photoelectron diffused or was ejected away from the site where it was created (i.e. the photon absorption site). That explanation is far more consistent with the current understanding of the physics and chemistry involved in this process, which has been developed, attacked, critically analyzed, and at least provisionally accepted by the greater scientific community over several decades.

But that's not good enough for you for some reason ...

 

[SpectraCat]

You asked for this... remember that.

Look, it's not that he has nothing interesting to say .. quite the opposite .. some of his other threads have been quite interesting, and he has a different way of looking at things, which should not be under-valued. All I ask is a little refinement of his presentation, indicating clearly which parts are supported by facts/references, which are unsupported but carefully analyzed, and which are just whimsical notions that he hasn't thought through. Those are the standards I try to adhere to .. and most regular posters do the same.

A little more acceptance of constructive criticism would be nice too, but I can understand why that might come hard

 

[conway]

Anything you can back up with some facts is fine by me. If think you can "spin" my analysis another way, then am at least a little interested to see your idea.

OK, but you aren't going to like this...

You have presented the following equation in support of your position:

AgBr ----> Ag + ½ Br2 ΔH = +99 kJ/mol

You point out that the reaction is endothermic and therefore cannot procede without the input of energy from a photon.

It’s a good argument, and it should be a good argument because it’s my argument. I was the one who suggested, in the face of some initial ridicule, that we could settle the question by looking at the thermodynamics of the process. It turns out that I was right to look at the thermodynamics but wrong to think that we would settle the question in this way.

The Gibbs Free Energy is, of course, the parameter which normally tells us if a reaction procedes to the right or to the left. You have used the enthalpy instead in presenting your numbers; but no matter. The correction for the entropy is in any event rather small; only a little more than 3 kJ/mol in this case. Nothing decisive.

The important factor you have neglected is concentration. The Gibbs Free Energy equation gives us the change in free energy only when the reactants and their products are present in stoichiometric ratios. In the photographic process, a typical crystal may have, after being exposed to light, only a literal handful of silver atoms out of trillions. It is apparent that the silver and silver halide species are very far from their stoichiometric proportions, and therefore a more careful analysis is required.

The themodynamically correct method must be to treat the crystal as a solid solution of silver and silver bromide. In its initial state, the crystal is 100% AgBr. Is the formation of a single silver atom thermodynamically spontaneous, or does it require a net input of free energy?

For convenience, we will double the reaction to clear fractions:

2AgBr -------------> 2Ag + Br2 ΔG = +192 kJ

We can calculate the familiar equilibrium constant from first-year chemistry with the formula

K = exp(-ΔG/RT)

With RT = 2.2 kJ (approx) we get

$$k=exp(-87) = 10^{-38}$$

With this information we can write the chemical equilibrium equation:



$$K = 10^{-38} = \frac{[Ag]^2[Br_2]} {[AgBr]^2}$$


This equation easily solves for the equilibrium concentration of Ag being approximately one part in ten trillion (10^-13). This is within an order of magnitude or so of typical concentrations in an actual exposed crystal. It is apparent that for an initially pure crystal of AgBr, the spontaneous creation of trace amounts of silver is thermodynamically favored. The stimulation of the incident light wave (I’m going to stop calling it a photon) merely speeds up the process, but is not strictly necessary energetically.

We can do the same calculation a different way. Use the Gibbs Free Energy to calculate the conversion of one part per trillion (10^-12) of silver. It comes to 192 nanojoules. This is the input of free energy required to drive the process. But this assumes that the two species are unmixed. In fact, we ought to treat the silver as being in solution. Then we can show that, within the accuracy of this calculation, the needed energy is available from the entropy of mixing.

The entropy of mixing is given by the formula

ΔS = nRm*ln(m)

where n is the number of moles and m is concentration of the mixed species. Multiplication by T gives you the free energy of mixing:

ΔG = T ΔS = nRTm*ln(m)

For two moles at a concentration of one part per trillion, the free energy comes to 121 nanojoules; and make no mistake, it is in the right direction to drive the reaction forward. It is true that with the numbers I have chosen we are just a little short of the 192 nanojoules we said we needed, but remember we haven’t yet accounted for the contribution from the mixing of the bromine. In any case, you only need to drop the concentration another factor of ten to tilt the reaction decisively to the right.

The case becomes even more convincing (convincing to me, you understand: I know you're still not convinced) when we recall that the crystal in its pure form is considered to be a poor photodetector. In practise the material must be doctored by the addition of impurities, dislocations, and what are called “electron traps” to become really effective. It’s not hard to imagine (OK, for me to imagine) that the energy needed at the trace concentrations we are dealing with comes at least partly if not in large measure from the “doping” of the crystal. In other words, there is plenty of chemical energy available to drive the transition from silver bromide to silver without needing the energy of a photon. This is in line with my original description of the process as proceding from a metastable state to one of lower energy.

 

[SpectraCat]

OK, but you aren't going to like this...

You have presented the following equation in support of your position:

AgBr ----> Ag + ½ Br2 ΔH = +99 kJ/mol

You point out that the reaction is endothermic and therefore cannot procede without the input of energy from a photon.

It’s a good argument, and it should be a good argument because it’s my argument. I was the one who suggested, in the face of some initial ridicule, that we could settle the question by looking at the thermodynamics of the process. It turns out that I was right to look at the thermodynamics but wrong to think that we would settle the question in this way.

The Gibbs Free Energy is, of course, the parameter which normally tells us if a reaction procedes to the right or to the left. You have used the enthalpy instead in presenting your numbers; but no matter. The correction for the entropy is in any event rather small; only a little more than 3 kJ/mol in this case. Nothing decisive.

The important factor you have neglected is concentration. The Gibbs Free Energy equation gives us the change in free energy only when the reactants and their products are present in stoichiometric ratios. In the photographic process, a typical crystal may have, after being exposed to light, only a literal handful of silver atoms out of trillions. It is apparent that the silver and silver halide species are very far from their stoichiometric proportions, and therefore a more careful analysis is required.

The themodynamically correct method must be to treat the crystal as a solid solution of silver and silver bromide. In its initial state, the crystal is 100% AgBr. Is the formation of a single silver atom thermodynamically spontaneous, or does it require a net input of free energy?

For convenience, we will double the reaction to clear fractions:

2AgBr -------------> 2Ag + Br2 ΔG = +192 kJ

We can calculate the familiar equilibrium constant from first-year chemistry with the formula

K = exp(-ΔG/RT)

With RT = 2.2 kJ (approx) we get

$$k=exp(-87) = 10^{-38}$$

With this information we can write the chemical equilibrium equation:



$$K = 10^{-38} = \frac{[Ag]^2[Br_2]} {[AgBr]^2}$$


This equation easily solves for the equilibrium concentration of Ag being approximately one part in ten trillion (10^-13). This is within an order of magnitude or so of typical concentrations in an actual exposed crystal. It is apparent that for an initially pure crystal of AgBr, the spontaneous creation of trace amounts of silver is thermodynamically favored. The stimulation of the incident light wave (I’m going to stop calling it a photon) merely speeds up the process, but is not strictly necessary energetically.

We can do the same calculation a different way. Use the Gibbs Free Energy to calculate the conversion of one part per trillion (10^-12) of silver. It comes to 192 nanojoules. This is the input of free energy required to drive the process. But this assumes that the two species are unmixed. In fact, we ought to treat the silver as being in solution. Then we can show that, within the accuracy of this calculation, the needed energy is available from the entropy of mixing.

The entropy of mixing is given by the formula

ΔS = nRm*ln(m)

where n is the number of moles and m is concentration of the mixed species. Multiplication by T gives you the free energy of mixing:

ΔG = T ΔS = nRTm*ln(m)

For two moles at a concentration of one part per trillion, the free energy comes to 121 nanojoules; and make no mistake, it is in the right direction to drive the reaction forward. It is true that with the numbers I have chosen we are just a little short of the 192 nanojoules we said we needed, but remember we haven’t yet accounted for the contribution from the mixing of the bromine. In any case, you only need to drop the concentration another factor of ten to tilt the reaction decisively to the right.

The case becomes even more convincing (convincing to me, you understand: I know you're still not convinced) when we recall that the crystal in its pure form is considered to be a poor photodetector. In practise the material must be doctored by the addition of impurities, dislocations, and what are called “electron traps” to become really effective. It’s not hard to imagine (OK, for me to imagine) that the energy needed at the trace concentrations we are dealing with comes at least partly if not in large measure from the “doping” of the crystal. In other words, there is plenty of chemical energy available to drive the transition from silver bromide to silver without needing the energy of a photon. This is in line with my original description of the process as proceding from a metastable state to one of lower energy.

Interesting, it is wrong, for a few reasons, but it is fairly well thought out and presented. I don't have time to present a full rebuttal right now, but here is a brief summary of the main problem:

You assume the silver halide crystals are 100% pure to start with .. why? Based on your analysis, isn't it more logical to assume they are in the equilibrium that you have predicted? After all, that chemical equilibrium is in play during the crystal formation process. So all the silver atoms that should be formed to keep the system in equilibrium are already present in the crystals, and so there is no thermodynamic driving force from the system being out of equilibrium, as you have predicted.

 

[conway]

Interesting, it is wrong, for a few reasons, but it is fairly well thought out and presented. I don't have time to present a full rebuttal right now, but here is a brief summary of the main problem:

You assume the silver halide crystals are 100% pure to start with .. why? Based on your analysis, isn't it more logical to assume they are in the equilibrium that you have predicted? After all, that chemical equilibrium is in play during the crystal formation process. So all the silver atoms that should be formed to keep the system in equilibrium are already present in the crystals, and so there is no thermodynamic driving force from the system being out of equilibrium, as you have predicted.

Then if the system is at equilibrium it is at least neutral with respect to displacement by a few silver atoms in either direction. In addition, the post-cooling stress dislocations might provide some sites of lower energy to act as traps.

 

[SpectraCat]

Then if the system is at equilibrium it is at least neutral with respect to displacement by a few silver atoms in either direction. In addition, the post-cooling stress dislocations might provide some sites of lower energy to act as traps.

I don't understand what you are getting at with the above comments ... if the system is at/near equilibrium, then the driving force you have hypothesized could supply "metastability" (your term), is basically zero. So all the energy is going to have to come from that pesky photon.

With regard to the traps ... *lower* energy sites aren't going to help you with your case. You need things to be in a *higher* energy state so that there is some latent energy to drive the reaction (assuming I understand your position).

 

[Frame Dragger]

Conway, you do yourself NO services by leading with a couple of pages of vague arguments. If you had posted #44 earlier, I wouldn't have thought you were a crank, SpectraCat would still have hair, and you wouldn't have had to work so hard to be understood.

I finally understand why you believe what you do about the detector being the "source" of the 'dot' and not the photon. Your chemistry looks good, but as SpectraCat has said, you need to NOT be in equilibrium, and these traps need energy to spare. Given that, I'm not sure those "traps" would do much trapping in the context of the crystal at (otherwise) equilibrium.

Assuming however that they did, and you meant to say "higher" energy states in the traps, I'm still unclear how this is a BETTER or more convincing explanation than "the photon strikes, a "dot" may appear". In your view, the "incident light wave" helps to disturb matters, but is not necessary? Wouldn't the silver spontaneously develop in the absence of added energy then? This reminds me of early star formation from 'dust' clouds; seemingly clear, but the issue of just WHAT supplies the "bump" to start the collapse is still shockingly debatable.

 

[conway]

Conway, you do yourself NO services by leading with a couple of pages of vague arguments. If you had posted #44 earlier, I wouldn't have thought you were a crank, SpectraCat would still have hair, and you wouldn't have had to work so hard to be understood.

I have to interpret this as one more instance of fault-finding on your part. In general I try to stick to the physics and not to respond to these little personal swipes (although I suppose even drawing attention to them constitutes a form of response). I just wonder that it doesn't occur to you people that I might find your ways as peculiar as you seem to find mine.

However, in this case I feel I have to respond to your specific point because it echoes something Spectracat said in a discussion last month, to the effect of "why did you ask the question if you already know the answer?" In this case, it should be pretty clear from the discussion that until two days ago I didn't have the thermodynamics at all; I didn't know what the reaction was and it was a bit of a shock to find out how endothermic it was. So I had to come up with post 44 on the fly. I don't see how you could expect me to have posted it earlier when I didn't think it up until yesterday.

I finally understand why you believe what you do about the detector being the "source" of the 'dot' and not the photon. Your chemistry looks good, but as SpectraCat has said, you need to NOT be in equilibrium, and these traps need energy to spare. Given that, I'm not sure those "traps" would do much trapping in the context of the crystal at (otherwise) equilibrium.

Assuming however that they did, and you meant to say "higher" energy states in the traps, I'm still unclear how this is a BETTER or more convincing explanation than "the photon strikes, a "dot" may appear". In your view, the "incident light wave" helps to disturb matters, but is not necessary? Wouldn't the silver spontaneously develop in the absence of added energy then?

My argument does not and cannot possibly explain the fine details of the photographic process. What I claim to have done is made a prima facie case for the possible existence of local sites within the crystal where the reduction of silver is thermodynamically favored. I hope that I have made the scenario sufficiently plausible that is becomes worthwhile to consider the question "what if the reduction of silver does not require the complete energy of a photon?"

You ask why the silver wouldn't develop spontaneously in these circumstances. Again, I have to remind you that the crystals used in photography are complex engineered structures for which we are not going to be able to fully analyze all the details. The best I can do right now is argue by a very rough analogy, which you may or may not find relevant. But here it is:

You have an electron in a potential well (surprise, surprise!). A finite potential well. Nearby there is another well, a deeper but still finite well, unoccupied. We want to get the electron from well A to well B. The potential barrier is 1 eV; so the electron won't go there of its own accord. But in the presence of a photon, the electron may be driven into the free state, from whence it may be captured by the second hole. We can even put both potential wells inside a big box, so the "free" state doesn't actually dissipate off to infinity but remains confined for long enough to effectively mediate between the two wells.

In this analogy, the electron doesn't go from hole A to hole B even though hole B is thermodynamically favored, because it can't overcome the energy barrier. That's my model of the photographic process. In the paragraph above I've applied my model to the traditional explanation of the photography reaction. You can see two things: in the traditional picture, the process requires the entire energy of the photon to make it work; and it works even if well B is at a much higher potential then well A.

My picture of the process procedes a little differently, but first I'd like some feedback as to whether this simplified model of two potential wells inside a big box will be considered an acceptable simplification to allow us to discuss how photography works.

 

[conway]

As for the rest, I wasn't trying to analyze you, but rather see what aspect of an alternate theory appeals to you so greatly that formalism is unappealing? I don't agree with him, but Demystifier regularly makes a case for dBB, and I understand why; he sees the theory as being more appealing with fewer contradictions with our observed reality. As the theory is empirically identical to SQM, it's clearly a matter of choice right now. In your case, I wish to understand, what is your dBB, or do you generally hold with formalism/TCI 'with major reservations' (as I believe many do, myself included)?

Not to go off on a tangent, but in the meantime I have to say I liked that "what is your dBB". Nicely put. And I didn't really answer it except partially in the negative. What I like is an actual process that takes place in real time (and I won't necessarily object to Cramer's processes in reverse time either except I don't quite know how to use them yet).

 

[Frame Dragger]

I'm not trying to take a swipe, I'm just kind of blunt. As for the physics, I think the main issue with your argument is: electron in potential well -> lower well = needing a HIGHER energy state in the silver, not a lower one as you described when you laid it all out on the last page.

As for why it wouldn't spontaneously act, there is always the possiblity that the electron will tunnel. The other issue I have is what SpectraCat has mentione: this seems to be a scenario in which no image could develop, just "static". I know that you're saying this is an advanced application for a partial theory (and I respect that), but it seems to be a large hole. The energy issue first and foremost however...

Btw, thanks for giving me the best answer you have (unfinished personal processes are just that after all) as to your interpretation. It helps me understand where you're coming from, and I'm willing to look at QM from any angle as long as I can return to instrumentalism.

 

[conway]

I'd like to respond but you haven't quite made it clear to me whether you are willing to discuss this question in terms of the simplified potential well model that I proposed or whether you are going to insist that I deal with the full-blown chemistry of silver bormide.

 

[Frame Dragger]

I'd like to respond but you haven't quite made it clear to me whether you are willing to discuss this question in terms of the simplified potential well model that I proposed or whether you are going to insist that I deal with the full-blown chemistry of silver bormide.

Oh, now that I see what you're getting at, yes please explain what you feel comfortable with. I don't say that I'll agree, but I won't be looking at you as though you've gone cross-eyed either.

So, my issue with your potential well model is that your "step" involves a higher energy state devolving, which makes sense. What doesn't make sense, is why that doesn't occur as a result of a photon's energy being "dumped" into the system. I understand that we're not talking about photographic plate here, but a model; go for it, I'd like to hear.

 

[conway]

So, my issue with your potential well model is that your "step" involves a higher energy state devolving, which makes sense.

I'm honestly having trouble parsing this sentence: if it makes sense, why do you have an issue with it?

What doesn't make sense, is why that doesn't occur as a result of a photon's energy being "dumped" into the system. I understand that we're not talking about photographic plate here, but a model; go for it, I'd like to hear.

OK, here's how it goes.

If we can think about the potential wells for the time being, this is the case I want to analyze. There is one electron and there are three states: electron in well A (source), electron in well B (target), and electron in the “condcution band” (C). The system starts off with the electron in state A.

According to the traditional picture, when a photon is incident on the system, there is a probability the electron will go from A to C. Once it is in C, there is a probability that it drop back down either to A (where it started) or to B. Once it drops into B it is trapped, whether B is at a higher or lower potential than A. (I know so far this is just a model, but for the sake of reference, that’s your speck of silver.)

I’m working a different picture. I allow the system to be driven by an oscillating electric field that couples A and C. After a time we may find the electron in a superposition of 90% A and 10% C. Since C and B are also coupled, the wave function may evolve further so that the electron finds itself in a superposition of 90% A, 10% C and 1% B. (Due to rounding, percentages may not add to exactly 100.)

The next step is where I haven’t yet been able to fill in the details. If you can help me out here I am willing to share the Nobel Prize with you. I turn off the electric field and watch what happens. What I am looking for is a mechanism whereby the electron can drain out of A into B via C. You can call it “quantum siphoning”.

There is plenty of energy available to drive the process, just as there is in an ordinary siphon. The problem is that when C couples to B, it is true that the probability drains from C to B which is what we want. And energy is released which we would like to use in order to replenish C from A, to keep the process moving. Just like an ordinary siphon. The probelm is that the energy at B is released in the form of radiating electromagnetic energy, and it is not clear to me how I recapture it with good efficiency at A, especially if A is relatively far from B. And I absolutely need to recapture it, because remember I've turned off the electric field.

That’s my problem. Perhaps you will agree that the solution I’m looking for is vaguely transactional in Cramer’s sense. There might be other ways of making it work but I haven’t figure it out yet.

 

[Frame Dragger]

Hmmm... It sounds like a mixture of the TI and quantum tunneling in reverse. I'll be honest, I need time to think about this. I understand the traditional picture well enough, but... without an incident photon to drive the reaction, what's to keep the superpositions from evolving in a radically different way so that there is no connection between the "oscillating field" and the behaviour of the electron?

Still, just keeping to the problem you've raised, I have no CLUE how you could recpature whatever is released at B, unless you posit that somehow the coupling of A and C initially sets the stage for a complete return of the energy emitted at B. I'm not an expert, but... that seems impossible. Really... really impossible. In fact, the idea of doing what you propose is a bit like finding a 100% efficient blanket for a fusion plant (to produce Tritium) to capture neutrons... you can't afford a miss.

Of course, if the well's themselves could have some kind of entangled state that is as yet unsuspected...

Hm... I don't think I should spend my share of the nobel cash. ;)

 

[zonde]

The next step is where I haven’t yet been able to fill in the details. If you can help me out here I am willing to share the Nobel Prize with you.

I am quite willing to participate in discussion that is interesting for me without any promise of Nobel Prize. But entirely different matter would be if you will be about to invent hyperdrive and would be promising to share a trip to Alpha Centauri.

I turn off the electric field and watch what happens. What I am looking for is a mechanism whereby the electron can drain out of A into B via C. You can call it “quantum siphoning”.

There is plenty of energy available to drive the process, just as there is in an ordinary siphon. The problem is that when C couples to B, it is true that the probability drains from C to B which is what we want. And energy is released which we would like to use in order to replenish C from A, to keep the process moving. Just like an ordinary siphon. The probelm is that the energy at B is released in the form of radiating electromagnetic energy, and it is not clear to me how I recapture it with good efficiency at A, especially if A is relatively far from B. And I absolutely need to recapture it, because remember I've turned off the electric field.

Let's assume that medium for waves has nonlinear properties so that ordinary wave becomes very sensitive to small perturbations and is unstable. However soliton waves are quite stable.
In that case there can be situation that is quite reversed from not being able to recapture energy to situation where it is quite hard to get rid of energy when the radiation can not form soliton wave.

Maybe simpler way to look at that is to assume two electron and one positron superposition. Positron should not escape somewhere very fast so it is quite reasonable that it can be recaptured annihilating one of the electrons.

 

[conway]

I'm glad you're enjoying the discussion but although I have to confess I don't know what solitons are, I think you're probably off target by invoking them in this instance.

The interaction between the free and bound waves of a potential well is fairly unambiguous and straightforward in the Schroedinger representation, and it is nothing more exotic than ordinary electromagnetic radiation. You can see this most easily by inspecting a superposition of the first and second bound states of the well, and watching how the probability shifts back and forth from left to right with the passage of time. It's exactly equivalent to a small classical antenna and in fact the rate of energy loss is given by the classical antenna formulas.

The waves in question when dealing with the free states are not all that different, and the radiation is still governed by antenna equations. This is the energy that is dissipated when the free state evolves into the bound state, and this is the energy that we I have spoken about wanting to recapture.

 

[SpectraCat]

Hi conway,

I still don't have a lot of time to spend on PF right now, but I just wanted to give a brief comment on your latest model. One issue that I see with it is that you are modeling the initial transition that drives the electron into the conduction band as if it were between discrete states. In fact it is not, the electron starts out in the valence band and is promoted to the conduction band upon photon absorption. These bands are comprised of very closely spaced quantum levels that very nearly approximate a local continuum of states. Therefore, I don't think you can use the simple model you have proposed for understanding the coupling of the photon. I am not really that strong with band-theory, but if you just look at the spectra width of these bands, which are quite broad with respect to discrete molecular bands, you will find that the lifetimes are very short ... I expect that they are too short to allow the kind of coherent state preparation you are trying to set up, because of decoherence.

So, the electron will stay in the conduction band ... but it very quickly evolve into a superposition of states that doesn't have much spectral overlap with the "hole" it left behind (which is also a broad superposition), so the probability of stimulating the reverse transition is very low. You might be able to look up some information on this if you search "stimulated electron-hole recombination". I will check it myself later, but I have to get back to work.

 

[conway]

But my model is simply a three state system; there's no decoherence or anything like that going on. We can argue about how well my model reflects the photographic process, but I thought we were going to suspend that argument for the time being.

If I can explain the collapse of the wavefunction even within my simplified model, I think it would be worth something. We can then turn to the photographic process and see how it impacts on that question. We're just not there yet.

 

[Frame Dragger]

Hmmm... I'm going to research "Stimulated electron-hole recombination" and see if that helps to illuminate the subject, thanks SpectraCat.

@Conway: True, but maybe the reason you're stuck is that there is no model which allows the recapture you're looking for... at least, not that can be used to model the photographic process. If your model can never be expanded to describe physical procesess, then I fail to see the value in explaining collapse within such restrictions if they are known to be non-physical.

EDIT: I know that we're going back to the photographic process here, but I think this is interesting. http://arxiv.org/ftp/physics/papers/0405/0405035.pdf I don't believe it can help you, but it is... well... it's worth the read.

 

[conway]

@Conway: ...maybe the reason you're stuck is that there is no model which allows the recapture you're looking for... at least, not that can be used to model the photographic process. If your model can never be expanded to describe physical procesess, then I fail to see the value in explaining collapse within such restrictions if they are known to be non-physical.

If that's the way you feel I guess I'm going to have to withdraw my offer to share the Nobel Prize money with you. Anyhow, I don't exactly need your help anymore because I think I figured it out. But it was nice at least for a while to not be treated as a total crank, if only for a couple of days.

 

[conway]

Maybe I was a little hasty when I said I didn't need to share my Nobel Prize money with you guys. What happened is I figured out how to make my Quantum Siphoning work. You can't do it with just two potential wells; you need millions of them, which is what you actually have in a crystal. I wrote it all up on my blogsite (you can find it if you just google "Quantum Siphoning"). But I'm having a little snag getting anyone to seriously look at my paper. I submitted it to the American Journal of Physics and it was rejected within 48 hours (!). And I don't have the endorsement needed to put it up on arXiv.org. So it looks like I could use a co-author to help me out here.

 

[yoda jedi]

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

it can be solved by a nonlinear quantum mechanics (SQM not, and without MWI).

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.

RIGHT !
clever insight

called ψ-complete view, unlike ψ-epistemic.
as for ψ-epistemic which quantum states are solely representative of our knowledge.

 

[cesiumfrog]

only a literal handful of silver atoms

(I don't think that word means what you think it means.)

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

Conway, your idea was basically that the energy of the photon is spread over the entire surface that it is shone at, and that localised detection doesn't actually prove the entire photon to have become localised again?

Lets say we have a single detector with a large input-area, that registers once for each time a photon is shone at it. Now let's move the detector twice as far away from the source, and replace this single detector with a panel of four independent detectors (each identical to the first).

According to standard physics, each time we trigger a photon emission, exactly one of the detectors will respond.

According to your "interpretation", each time: there is a 42% chance of exactly one detector responding, a 32% chance of none responding, 21% chance of two both responding to the same photon, nearly a 5% chance of the photon being detected in three different places, and 0.4% chance that all the detectors respond to the photon.

Do you stand by this?

 

[cesiumfrog]

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.

What about if one of the two beams is directed through a second Stern Gerlach apparatus, why shouldn't you then expect a total of three beams?

 

[conway]

(I don't think that word means what you think it means.)


Conway, your idea was basically that the energy of the photon is spread over the entire surface that it is shone at, and that localised detection doesn't actually prove the entire photon to have become localised again?

Lets say we have a single detector with a large input-area, that registers once for each time a photon is shone at it. Now let's move the detector twice as far away from the source, and replace this single detector with a panel of four independent detectors (each identical to the first).

According to standard physics, each time we trigger a photon emission, exactly one of the detectors will respond.

According to your "interpretation", each time: there is a 42% chance of exactly one detector responding, a 32% chance of none responding, 21% chance of two both responding to the same photon, nearly a 5% chance of the photon being detected in three different places, and 0.4% chance that all the detectors respond to the photon.

Do you stand by this?

I made a mistake when I said "let a photon pass through a pinhole". It wasn't exactly a mistake because I never like to use the word photon, but I thought for the sake of brevity I could get away with it in context. I should have said "let an amount of electromagnetic energy on the order of a single photon pass through a pinhole". I didn't mean to imply that it's possible to shoot a single photon just like you'd shoot a pea through a straw.

I know there are people who claim to do magical things with single photon sources, but I don't know much about these things. I suspect it's harder to do than you think; in particular, I doubt that the experiment you've described has been carried out in exactly the way you describe it, although I've heard there are related experiments which claim to show the thing that you're driving at.

I do know something about more traditional light sources, like thermal sources and lasers. Laser soursces are easiest to analyze because the photon statistics are Poisson. If you can possibly frame your objections to my "interpretation" in terms of laser or thermal light I might be able to deal with them.

 

[collinsmark]

Hello conway,

I've been following this thread with great interest. But I have some questions.

I made a mistake when I said "let a photon pass through a pinhole". It wasn't exactly a mistake because I never like to use the word photon, but I thought for the sake of brevity I could get away with it in context. I should have said "let an amount of electromagnetic energy on the order of a single photon pass through a pinhole". I didn't mean to imply that it's possible to shoot a single photon just like you'd shoot a pea through a straw.

It almost sounds like you are saying that when measured, photons do not behave as containing discrete packets of energy. Can your model handle the experimental evidence involving the photoelectric effect?

In other words, could your model distinguish between a laser shining through a small hole with a given frequency and intensity (intensity at the photographic paper) and a different laser at half the frequency with twice the intensity (maybe intensity isn't the right word here. Perhaps "twice the number of photons" might be better)? (Assume the experiment is set up such that the hole size and laser intensities are such that the same diffraction pattern is produced, and the overall electromagnetic power hitting the paper is the same in both cases -- only the frequency is different).

According to your model, since the energy is evenly distributed across the photographic paper, and since the energy is equal in both cases, how would your model predict/explain the different behavior between the two cases confirmed by experimental evidence (photoelectric effect)?

I know there are people who claim to do magical things with single photon sources, but I don't know much about these things. I suspect it's harder to do than you think; in particular, I doubt that the experiment you've described has been carried out in exactly the way you describe it, although I've heard there are related experiments which claim to show the thing that you're driving at.I do know something about more traditional light sources, like thermal sources and lasers. Laser soursces are easiest to analyze because the photon statistics are Poisson. If you can possibly frame your objections to my "interpretation" in terms of laser or thermal light I might be able to deal with them.

What if the photon source is swapped with an electron beam shooting through a crystal lattice. Electrons are subject to diffraction and interference too (such as in the electron version of the double-slit experiment). And as we know, electrons can expose photographic paper too.

According to your model, would the energy of an electron also be spread across paper (via its spread out wave function)? What about its charge? Is the charge localized when measured, or is it spread out too? Perhaps a more pertinent question is looking at it in terms of cesiumfrog's request, with the 4 identical detectors at twice the distance. In this configuration, would your model predict that if a single electron is ejected from the source, that there is a chance that multiple electrons could be detected?

 

[conway]

I don't really have a theory other than the idea of combining Schroedinger's equation and Maxwell's equations. It's surprising how many things you can explain with just this combination without worrying about things like "photons". In theory this should be pretty mainstream but it doesn't seem to be all that widespread. Even at the highest levels it seems you find people who aren't familiar with some of the basic pictures. For example, if you write the Schroedinger equation for the hydrogen atom and take a superposition of the s and p states, you get a tiny classical antenna. Everything the hydrogen atom does electromagnetically can pretty much be explained in terms of the properties of this antenna. You ask me if my theory can explain this or that...so for starters, I have to ask you if you recognize this picture of the hydrogen atom?

 

[collinsmark]

Hello conway,

I'm not sure I follow you here.

I am familiar with the the energy eigenstates of an electron in a hydrogen atom, and the concept of superposition of states. If I had to, I could dust off my old Griffiths book or something and calculate up the wave-function equations for an electron being in a superposition of an s and p state (such as $$\Psi = \frac{1}{\sqrt{2}}|(n=1, l=0, m_l=0, s= 1/2)> \ + \ \frac{1}{\sqrt{2}}|(2, 1, 0, 1/2)>$$, or whatever energy eigenstates are chosen). But conceptually, I can imagine the results. If the wave-function is in a superposition of two energy eigenstates, the expectation value and phase will oscillate back and forth in some way, which might resemble a rotation, or it might resemble simple harmonic motion (in some ways), or perhaps a time-varying bimodal distribution like two pistons in an engine, depending on which energy eigenstates are part of the superposition.

I've also studied radio frequency communications theory, and I'll give you that there are surprising similarities in the mathematics between it and QM, with all the Fourier transforms, Bessel functions, and the like.

But I don't get the connection to the antenna. Even though the wave-function's expectation value may be varying with time due to the superposition of states, the atom is not radiating electromagnetic energy due to this. If it was, QM would have serious conservation of energy problems. If left completely isolated, hydrogen atoms would widdle away to nothing (assuming that it stayed in the superposition of states indefinitely [i.e same thing as saying no photons were released or absorbed]). So I think I can recognize the picture of an electron's wavefunction in a superposition of energy eigenstates in a hydrogen atom, but no, I don't see how that relates to a classical antenna.

[Edit: my knowledge of quantum electrodynamics (above and beyond non-relativistic quantum mechanics) is presently rather sparse, but from what I can gather, I am not presently aware of electrons radiating energy when being in a superposition of states, even though the expectation value of the wave-function may oscillate.]

 

[Frame Dragger]

@collinsmark: My limited understand of QED (which was recently bolstered by researcing it after being roundly spanked here) would indicate that they would oscillate, but not radiate, as you say.

@conway: I don't see how that forms a classical antenna... I think I'd need to see the math or a model. The thing is, if it DID act as an antenna, collinsmark would be right... we'd have a universe of radiation and nothing else because the electron would crash into its hydrogen neuclus.