
#19
Feb113, 07:41 AM

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#20
Feb113, 08:19 AM

P: 684

I'm still wondering how exactly this is related to the observable entropy increases. If we start with two pure states, I guess that any fundamental interaction that leads to maximal entanglement should begin to disentangle the systems afterwards. So I would expect an oscillating entropy for the systems (in classical mechanics, no entropy change arises from such a situation). Which of course would call for an explanation why our observations always take place in the rising entropy domain. 



#21
Feb113, 08:25 AM

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#22
Feb113, 08:45 AM

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P: 4,491

A Bornruleindependent motivation for doing the partial trace is the fact that the evolution of the resulting object (reduced density matrix) does not depend on the whole Hamiltonian, but only on the Hamiltonian for the subsystem. 



#23
Feb113, 09:00 AM

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P: 4,491

The good question is why the direction in which entropy increases is everywhere the same, i.e., why it is not the case that entropy increases in one subsystem and decreases in another? The answer is that it is interaction between the subsystems which causes them to have the same direction of the entropy increase: http://arxiv.org/abs/1011.4173v5 



#24
Feb113, 09:16 AM

P: 684





#25
Feb113, 09:27 AM

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P: 4,491

A much more refined analysis of that stuff, but still very nontechnical, is the book: H. Price, Time's Arrow and Archimedes Point Another good related nontechnical book is: D. Z. Albert, Time and Chance There is also a good nontechnical chapter on that in: R. Penrose, The Emperor's New Mind 



#26
Feb213, 12:30 AM

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#27
Feb213, 07:34 AM

P: 1,657





#28
Feb213, 07:53 AM

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#29
Feb213, 08:08 AM

P: 1,657

The collapse interpretation says that initially the system is in some state [itex]\vert \Psi\rangle[/itex]. You perform an experiment to measure some observable with eigenvalues [itex]\lambda[/itex] and corresponding eigenstates [itex]\vert \Psi_\lambda\rangle[/itex] (for simplicity, assume nondegeneracy). Then the results are that afterward: For every value of [itex]\lambda[/itex], there is a probability of [itex]\vert \langle \Psi \vert \Psi_\lambda\rangle \vert^2[/itex] that the system is in state [itex]\vert \Psi_\lambda\rangle[/itex] This is captured by the density matrix formalism as the transition [itex]\vert \Psi \rangle \langle \Psi \vert \Rightarrow \sum_\lambda \vert \langle \Psi \vert \Psi_\lambda\rangle \vert^2 \vert \Psi_\lambda \rangle \langle \Psi_\lambda \vert[/itex] 



#30
Feb213, 01:56 PM

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P: 8,004

I guess thinking about it classically, Demystifier's argument must be right. Measurement gives us more information, which is a reduction in entropy. Entropy increases when we forget, according to Landauer's exorcism of Maxwell's demon.
I guess what's not obvious to me is  how much coarse graining do we need, since the partial trace in getting the reduced density matrix is a form of coarse graining? 



#31
May2713, 01:11 AM

P: 83

Then Von Neumann Entropy and the Shannon Entropy are the same if we average over a time not so little so the relaxing time of the subsystems but little in comparation with the relaxation time of the complete system.
When finished the decoherence proccess, diagonal terms of density operators are null, then entropy grew, but when collapse matters, the entropy can decrease, in the extremal case to a pure state again, or a mix with no more entropy like before start decoherence proccess. If the measurement is not ideal, entropy can increase, obviously not to the entropy of the final of decoherence proccess, because then there is no information gained, and not measurement at all. 



#32
May2713, 01:37 AM

P: 83

Of course, the wave function is different if we considerate the collapse or not, and the post evolution is too. But the situation is different, if you consider the measurement realized, the posterior wavefunction give the probability conditionated to the measurement result, if you donīt considerate the collapse and considerate the total evolution, you only obtain the probability of measurements, not conditionate to no result of a measurament. All the problems are solved considering that wavefunction is only an instrument to calculate the probabilities of a measurement, in relation with the previous information, and not a real state of the system. The measurement problem doesnīt exist with this consideration, and collapse is perfectly ok




#33
May2713, 09:03 AM

P: 1,657

Something that I've thought about that in some ways makes the informationtheoretic view more palatable to me, personally, is the "consistent histories" approach. You give up on describing what the state of the world is and the dynamics of that world state, and instead view the object of interest to be histories of observations. Quantum mechanics tells us the relative probabilities of those histories. 



#34
May2713, 01:40 PM

P: 158

What I can't understand is why there seems to be so much resistance to this obvious solution. Yes, it involves including advanced field solutions (with negative energies), but you can't get away from that in the relativistic domain anyway. So the basic message is that in order to solve the measurement problem, you need to take into account relativistic processes (emission and absorption) in a direct action picture (basic field propagation is time/energy  symmetric). This is very natural, since it is the most general theoretical formulation. Note that you can regain the empirical asymmetry of radiative processes with appropriate boundary conditions. Then the existence of those BC become theoretically falsifiable, which makes it a stronger theory methodologically  in contrast to the standard approach in which an asymmetric field theory is simply assumed ad hoc. See my new book on TI, in particular Chapters 3 and 4, for how TI solves the measurement problem. Sorry the book is rather pricey, but you can get it at many libraries and on interlibrary loan if interested. Also I will provide specially discounted, autographed copies for students with documented finanicial hardship. Contact me through my website to apply for this discount. http://transactionalinterpretation.o...tocontactme/ 



#35
May2713, 11:57 PM

P: 83

I dontīknow details about the Bohmian interpretation. But there is no QFT theory for this interpretation. I donīt know if it is beacuse it is not possible or it is not clear that it is possible, but it is a pending question.




#36
May2813, 12:00 AM

P: 83

I donīt understand the problem with Bell theorem for the informational interpretation of wavefunction. The only thing that it no solves is that it imposses a minimun limit for the measurement, but no the real duration of measurement ( maybe it implies to study measurement apparatus), and the possible degeneration in measurement, that can provokes an increased entropy from pure state just before the decoherence start 


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