# A simple QM experiment analysis question

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:

-- Dmtr

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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.)

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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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.

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 $$S_{A} = S_{B} \rightarrow \infty$$. 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:

$$N_{A}/N_{B} \approx e^{\frac{S_{A} - S_{B}}{k}}$$

I also think that he extreme case of $$S_{C} \rightarrow 0$$, $$S_{P} \rightarrow 0$$, $$S_{H} \rightarrow \infty$$ 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 $$N_{B}$$ 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

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.

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".

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

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.

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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