A simple QM experiment analysis question

In summary: The observer is classical - measurement collapses the wave function, etc.I'm not sure what you mean by that?Yes. The result of the casual/classical (wave function collapse upon the interaction with the classical detector) analysis is trivial. What I'm more interested in is the analysis of the whole system (including the detectors, counters, heater) from the QM perspective (where the classical behavior emerges from the quantum via the decoherence).
  • #1
dmtr
182
0
Can somebody help analyzing the outcome of the following thought
experiment:
1) A simple setup with a photon source, half-silvered mirror and two
photon detectors with counters (on each path - A and B);
2) We adjust the setup to get an equal distribution between the
counters on paths A and B;
3) Now we modify the setup in such a way that upon a detection of a
photon on path A some free energy is being 'spend' (so path A now is
more entropic than B).

The question - would (3) modify the distribution?

-- Dmtr
 
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  • #2
What are you getting at; why would you expect the result to change?

(Conventionally, what goes on at the macroscopic/classical level isn't going to interfere back here, and you'd probably get a causality paradox otherwise, but this reminds me of interesting questions about why sugar molecules are found in position/chirality eigenstates rather than with definite energies..)
 
  • #3
Well, it is clear to me, that the experiments (2) and (3) are different. And I don't know if the results of these experiments would be different or the same. That is a good enough reason for a question.

(Yes. From the casual/classical point of view the outcome is going to be the same for 2 and 3. Unfortunately this point of view is an approximation and I'm not sure if it would be good enough to analyze this thought experiment. And by the way - consider the regular classical analysis of such an experiment versus the classical analysis based on the action principle. Wouldn't they contradict each other?)

So what I'm really interested in is the answer from the QM perspective.

-- Dmtr
 
  • #4
Just to be on the same page, here is the schematic illustration of the experiment:

Dc987-exp00.png


-- Dmtr
 
  • #5
How about you try to set up a variation of experiment 2, where you publish the results of counters A&B in a scientific journal. A week afterward, I'll read the journal and smash one vase (or run my space heater for an extra minute, flash a light, or some other entropy-increasing action) for each count you record having got from detector A. In other words, you've unknowingly carried out experiment 3. If there was a difference between experiment 2 & 3, then you would have been able to already tell from the results you yourself later published (and if I happen to be on Mars, then this will have communicated my decision to you far faster than the speed of light; or maybe I'll change my mind, causing history to rewrite). Make sense? (The problem is that the signal leaving a detector is, by definition, macroscopic and classical.)

Entropy might play a part in choosing the preferred basis line along which a quantum state collapses, but it cannot affect which side of that line the state collapses to.

dmtr said:
consider the regular classical analysis of such an experiment versus the classical analysis based on the action principle. Wouldn't they contradict each other?
I'm not sure what you mean by that?
 
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  • #6
Yes. The result of the casual/classical (wave function collapse upon the interaction with the classical detector) analysis is trivial. What I'm more interested in is the analysis of the whole system (including the detectors, counters, heater) from the QM perspective (where the classical behavior emerges from the quantum via the decoherence).

-- Dmtr
 
  • #7
Do you mean you want to treat more of the apparatus as "microscopic" (i.e., governed by quantum mechanics)?

Then what will your entropic process be? Since QM is time reversible, where will you source the information (randomness) to inject into that stream of energy. (What form of energy could you even define, microscopically, as having high or low entropy?) And still, I don't see anything interfering with the results unless the beam-splitter outputs are provided an opportunity to remix? It seems to me like the apparatus that you're describing isn't sufficient for exploring the effect in question.
 
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  • #8
Yes. I want to treat the apparatus as "microscopic". Entropic processes would be: "observer doing the measurement", "apparatus generating the entropy in the environment (apparatus entangling with the environment?)". Does that sounds right?

The observer is classical - measurement collapses the wave function, etc.

-- Dmtr
 
  • #9
dmtr,

how about this variation. The setup is done with two detectors as before, then one is removed. The process of detection and displaying something on the instruments will have different entropic impact from allowing the light to escape.
 

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  • Dc987-exp04.png
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  • #10
Mentz114,
Yes, your version is similar, but I'm not sure if it is simpler to analyze... I'll think about it over the weekend.

-- Dmtr
 
  • #11
dmtr said:
Just to be on the same page, here is the schematic illustration of the experiment:



-- Dmtr

Maybe I am nitpicking now, but there is a path missing from your schematic. If want to do a full QM analysis you need to include the vacuum "entering" from the top; a beam splitter always has two inputs and two outputs.

It is still not clear what you mean by "entropic process"? Do you simply mean that the system includes some form of ohmic bath?
 
  • #12
f95toli said:
It is still not clear what you mean by "entropic process"? Do you simply mean that the system includes some form of ohmic bath?

In the real-life setup that would be an electric heater. But I guess in this thought experiment it can be zeroing the memory (which requires a corresponding increase in the entropy of the environment / Landauer's principle).

f95toli said:
Maybe I am nitpicking now, but there is a path missing from your schematic. If want to do a full QM analysis you need to include the vacuum "entering" from the top; a beam splitter always has two inputs and two outputs.

So there would be some extra decoherence due to the vacuum noise, right? And it could slightly diminish differences between the counters values in the experiment (3)? Well, I don't know if there would be any differences in the first place...

-- Dmtr
 
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  • #13
Ok. Let's make things more interesting. Here is my conceptual analysis of the experiment (3). First we should forget that the system is classical and remember that it is really in the superposition of |path A> and |path B>. And what is important here is the resulting number of states.

Now the resulting number of states:
* for path A it would be: {a lot of decoherence states of the photomultipliers, counters, heater}
* for path B it would be: {a lot of decoherence states of the photomultipliers, counters}

So it looks like the path A would have slightly larger number of states associated with it. As a result we would have slightly larger probability of finding the apparatus in the state A, and if repeated many times slightly larger number in the counter A.

How larger? Well.. that depends of how much decoherence associated with photomultipliers and counters is going on. So "publishing the results of each outcome in a scientific journal." can really screw things :)

Ok. Now can somebody please either refute my argument or give some rough quantitative estimates?

-- Dmtr
 
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  • #14
Would it be something like:

[tex]N_{A}/N_{B} = e^{S_{A} - S_{B}} [/tex]

where [tex]N_{A}[/tex], [tex]N_{B}[/tex] are the counter values and [tex]S_{A}[/tex], [tex]S_{B}[/tex] - total entropy in the space-time light cone (entropy 'visible' to the observer)?


-- Dmtr
 
  • #15
Short answer is no, 50-50. Can you cite any reference that even suggests otherwise?
 
  • #16
cesiumfrog said:
Short answer is no, 50-50. Can you cite any reference that even suggests otherwise?

How can I derive this answer from quantum mechanics?

No. I can't cite any references. The experiment is too stupid to be in the textbooks. It is obvious that the answer is 50-50. Otherwise there are all kinds of paradoxes: providing that you have a largish heater, subjective FTL communication is possible, you can break causality, speed up quantum computations, there are all kind of observer effects, etc, etc.

-- Dmtr
 
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  • #17
dmtr:

It seems to me something in the future won't affect the process that takes place at the beam splitter. If all wave equation collapses ( for want of a better term ) were governed by future entropy considerations, life would be non-existent.
 
  • #18
dmtr said:
How can I derive this [obvious] answer from quantum mechanics?
Use the Born rule rather than attributing equal probability to every term/"microstate". The state prepared by the beam-splitter is independent of what follows it (as such elements only affect the later evolution of parts of the total quantum state, and in a manner that conserves probability).
 
  • #19
cesiumfrog said:
Use the Born rule rather than attributing equal probability to every term/"microstate". The state prepared by the beam-splitter is independent of what follows it (as such elements only affect the later evolution of parts of the total quantum state, and in a manner that conserves probability).

Ok. Thank you. I'll need some time to digest that.

Mentz114 said:
It seems to me something in the future won't affect the process that takes place at the beam splitter.

I don't think this is true. Remember, the quantum mechanics is time reversible.

Mentz114 said:
If all wave equation collapses ( for want of a better term ) were governed by future entropy considerations, life would be non-existent.

That's a very vague statement. And in fact the exact opposite can be true, life is a highly entropic process and thus would be more favorable process.
 
  • #20
Here is another diagram, illustrating the entropy exchange with the environment:

Dc987-exp01.png


-- Dmtr
 
  • #21
cesiumfrog said:
Use the Born rule rather than attributing equal probability to every term/"microstate".

Well. I have the right to apply the Born rule to the whole apparatus, right?

cesiumfrog said:
The state prepared by the beam-splitter is independent of what follows it (as such elements only affect the later evolution of parts of the total quantum state, and in a manner that conserves probability).

Yes, but only if I'm observing the photons coming out of the beam-splitter. If I am doing that, according to the Copenhagen interpretation the measurement would collapse the wave function and the application of the Born rule will give me probabilities.

But I'm not doing that. I'm observing the counter values. And a lot of scattering photons, generated by photomultipliers, counters, heater.

-- Dmtr
 
  • #22
Ok. Here is the key question. Would you agree that any system have tendency to drift to the state of maximum entropy?

-- Dmtr
 
  • #23
dmtr said:
Ok. Here is the key question. Would you agree that any system have tendency to drift to the state of maximum entropy?
That sounds like a loaded question - I'm not guaranteed to win at the casino purely because I can't otherwise pay next quarter's heating bill. In other words, ordinary thermodynamics does not specify the manner nor rate of entropy increase, only that the universe as a whole will tend in the direction of heat death.
 
  • #24
dmtr:

when the ice-box in my fridge cools to below room temperature, heat has been forced to move from a colder to a hotter region, against the thermodynamic 'trend'. According to your logic this is impossible. One could argue that if we consider the planet, not just my fridge, then the thermodynamic trend has not been violated. In either case, quantum choices were made that locally went against the global trend.
 
  • #25
cesiumfrog said:
That sounds like a loaded question - I'm not guaranteed to win at the casino purely because I can't otherwise pay next quarter's heating bill.

No, I wouldn't think so. But in fact I think that observers are governed by the same basic laws. And for a 'quantum' random event, the subjective observer history with the maximum probability would be where the observer creates the maximum entropy in the environment.

So sorry. No probable casino winnings. Especially if there is an alternative history in which you perform some work, earn some money and pay next quarter's heating bill anyway. That history would create a lot more entropy in the environment.

-- Dmtr
 
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  • #26
Mentz114 said:
In either case, quantum choices were made that locally went against the global trend.

I was not saying that all the photons would chose path A. But OK. Let's analyze the extreme scenario.

Let's assume that we've made some nice time-reversible photomultipliers and counters. So:
SP = 0;
SC = 0;

Now the only possible way of interaction with the environment is the heater. Everything else is time reversible. With the time flow entropy can only increase (it can not even stay the same!). So the only logical conclusion here is that with the time flow all the photons would chose the path A.

A completely reversible photomultiplier+counter can be replaced by a quantum harmonic oscillator. And the light source with an another oscillator with some initial energy. Here is an illustration:
Dc987-exp02.png


-- Dmtr
 
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  • #27
dmtr said:
With the time flow entropy can only increase (it can not even stay the same!).
Huh? Why not? (This was the point of my previous post.)
 
  • #28
cesiumfrog said:
Huh? Why not? (This was the point of my previous post.)

You are right. I'm not precise enough with the words here. Still, my point was that with the time flow any passive system can only increase the entropy of the environment. And you can consider any system to be a passive system (assuming that you take the initial internal energy of the system as zero).

An example you've given with the fridge. What you have there is a rather complex passive system scattering light :) It takes light in (electromagnetic waves) it spits light out (heat). And with the time flow it increases the entropy of the environment.

Offtopic: As far as I can see this entropy exchange (entanglement with the environment in the quantum world) in fact is what we subjectively call the time flow. But that's only my guess. And I wouldn't bet on it. What I would bet on is that observers are not really different from fridges :) in terms that you can consider an observer to be a passive system.

-- Dmtr
 
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  • #29
dmtr:

To calculate the wave function of a system I need to find the Hamiltonian of the system, then solve Schroedingers equation. Now I am in a position to work out the eigenvalues and the probabilities associated with them. This procedure gives results which agree with experiment. There is explicit entropy term in this calculation.

Are you suggesting one should include an environmental part in the Hamiltonian which will affect the final probabilities ? In the density matrix formalism it is possible to explain the 'selection' of the observed eigenstate by doing this - but it doesn't affect the probabilities calculated from the wave function.
 
  • #30
Mentz114 said:
dmtr:

To calculate the wave function of a system I need to find the Hamiltonian of the system, then solve Schroedingers equation. Now I am in a position to work out the eigenvalues and the probabilities associated with them. This procedure gives results which agree with experiment. There is explicit entropy term in this calculation.

Are you suggesting one should include an environmental part in the Hamiltonian which will affect the final probabilities ? In the density matrix formalism it is possible to explain the 'selection' of the observed eigenstate by doing this - but it doesn't affect the probabilities calculated from the wave function.

AFAIK it doesn't affect the probabilities because of the infinities, right? If you use that approach, the single photon scattering in the environment over the infinite time would give the same infinite entropy release to the environment as two photons scattering, right? So you would have [tex]S_{A} = S_{B} \rightarrow \infty[/tex]. I think there is at least one boundary condition - we should only consider the entropy, produced by the apparatus in the space-time light cone of the observer. I don't know if it is infinite, or not.

I'm not sure what the answer is. I'm a CS graduate and I only had a few overview courses on quantum mechanics and quantum information theory. My guess is that just calculating the number of states from the entropy would give us approximate counter values:

[tex]N_{A}/N_{B} \approx e^{\frac{S_{A} - S_{B}}{k}}[/tex]

I also think that he extreme case of [tex]S_{C} \rightarrow 0[/tex], [tex]S_{P} \rightarrow 0[/tex], [tex]S_{H} \rightarrow \infty[/tex] shows us very clearly that the heater 'sucks' photons from the path B. So I wouldn't be surprised if the switching on a largish heater in the real life would diminish the [tex]N_{B}[/tex] increase rate.

It also makes sense, that by introducing the potential to release a lot of entropy you create a potential well. And the equilibrium moves towards it.

Here is a question for you: what if we will couple 'the computation is complete' state of a quantum computer with a large release of entropy to the environment? Would it speed up the computation?

-- Dmtr
 
  • #31
In a thermodynamic ensemble, counting microstates is a valid procedure, but it's not meaningful in single quantum events like when a photon encounters a beam splitter.

The overlap between QM and thermodynamics seems to be important only when there's entanglement. See

http://arxiv.org/abs/0905.2562

Code:
H.Casini, M.Huerta

(Submitted on 15 May 2009 (v1), last revised 7 Oct 2009 (this version, v2))
Abstract: In this review we first introduce the general methods to calculate the
 entanglement entropy for free fields, within the Euclidean and the real time formalisms.
 Then we describe the particular examples which have been worked out explicitly in two,
 three and more dimensions.
 
  • #32
Mentz114 said:
In a thermodynamic ensemble, counting microstates is a valid procedure, but it's not meaningful in single quantum events like when a photon encounters a beam splitter.

Unfortunately we don't have a single quantum event of a photon encountering a beam splitter. We have a single photon encountering a beam splitter and causing a lot of entropy in the environment (via the interactions with the photomultipliers, heater, etc).

The whole point of this thought experiment is to illustrate how the probabilities on the outcome of a single quantum event can be influenced by the future entropy considerations.

The fact that they are influenced is trivial. Simply consider the setup with the quantum harmonic oscillators from my message "Oct21-09 08:10 PM".

Mentz114 said:
The overlap between QM and thermodynamics seems to be important only when there's entanglement. See http://arxiv.org/abs/0905.2562

Correct me if I wrong, but it sounds like that you are thinking that the entanglement is a small and unimportant factor. Some "spooky action at the distance". That disappear almost instantly in any noticeable environment temperature. Yes. Clean "EPR-paradox" entanglement in fact disappears very fast. But only because of the other entanglements with the environment. Decoherence is entanglement with the environment. See a Wikipedia article on the decoherence, or any other generic introduction on the subject i.e.: http://arxiv.org/abs/quant-ph/9803052. Theoretically the entangled states can stay at any temperatures at any distance for an unlimited period of time. This is trivial from the unitarity principle in the QM. There have been experiments (reported in the Nature/Nature Physics, I can locate you the articles), directly confirming that the entangled states can stay at large distances (miles) over the large periods of time.

So yes. I would very much agree that "the overlap between QM and thermodynamics seems to be important only when there's entanglement". Only I would say that the entanglement is very very important. And if the Second Law is simply the effect caused by the Decoherence with the time flow, the entanglement would be the only important factor in the thermodynamics.

-- Dmtr
 
  • #33
The whole point of this thought experiment is to illustrate how the probabilities on the outcome of a single quantum event can be influenced by the future entropy considerations.
I'm unconvinced. A thought experiment proves nothing.
 
  • #34
Mentz114 said:
I'm unconvinced. A thought experiment proves nothing.

Yep. Exactly my thinking. Proves nothing. A thought experiment is only useful to illuminate some aspects of a theory or hypothesis.

Unfortunately I can only do a very limited testing (with the equipment I have access to). Most likely, what I'll be able to do, is to place some boundaries on the S values, not to see the effect itself. I'm trying to realize the experiment on a very simple setup made out of a 106 bits/sec optical QRNG device. In the current setup I have ~3 µW 'extra' free energy spending per single photon. (That would be roughly the equivalent if you shine a laser pointer on the beam splitter and drain 50 GigaWats extra on one path.)

-- Dmtr
 
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