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I might be raising measurement problem in quantum physics in different words, but I will ask the question.

Quantum decoherence has been proposed by proponents as a theory that eliminates all weird issues involving quantum mechanics, such as Schrodinger's cats and measurement problem.

So in quantum decoherence theory, system (example: a particle) becomes "measurable" when system interacts with another system or environment with certain conditions and consequences.

But this seems to raise issues of how systems can ever be interacting "objectively". After all, every system described by wavefunctions that only give probability for possible measurements in general (of course there are cases that when met with proper basis and knowledge of wavefunction, definitive measurement can be guaranteed). This for me seems to suggest even before quantum decoherence, two systems that will interact have been connected by some wavefunction that describes the chance of quantum decoherence. Or we need some kinds of observers to either bring effects of quantum decoherence.

If the question seems a bit confusing, the context in which the question is being asked is how quantum decoherence theory affects quantum mechanics's measurement problem.

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2 Answers 2

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First of all, it is indeed correct to model decoherence as a system interacting with what is called the "environment". Basically you have a joint CLOSED (unitary) evolution of system+environment, after which you discard the environment (technically called a partial trace), and you are left with the state of the system. Your "observer" can be taken as part of the environment with which the system interacts.

Now, to experimentally detect decoherence, all you have to do is to perform the experiment repeatedly, measure the in various bases, and collect the statistics. At the end of the day you can perform what is called "state tomography", that is, reconstruct the state of the system from your measurement statistics, and there you will see that the off-diagonal elements of the density matrix tend to be small (which is an indication of decoherence).

In principle, you DON'T want decoherence for micro tasks such as quantum computing. Decoherence is what makes our macroscopic world be as it is, that is, behave classically.

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  • $\begingroup$ But then this suggests that mixed-state wavefunction already exists before any process of measurement. Let us give Schrodinger's cat example. This would suggest that cat and measurement apparatus and cat are both described by a mixed-state wavefunction even before measurement. Is this correct view of quantum decoherence idea of measurement problem? (So no interaction) $\endgroup$
    – user50374
    Commented Jul 22, 2014 at 14:51
  • $\begingroup$ You do not need mixed states, can still describe the evolution starting with a pure state. And the environment is always assumed to start in a pure state (if mixed, then you can "purify" it by enlarging the environment, without any observable consequences. $\endgroup$
    – vsoftco
    Commented Jul 22, 2014 at 14:53
  • $\begingroup$ Yeah. I wasn't placing too much importance in mixed state. So this seems to lead to the idea of both cat and measuring apparatus connected by some wavefunction already before measurement, right? Just want to check. $\endgroup$
    – user50374
    Commented Jul 22, 2014 at 14:55
  • $\begingroup$ Yes, you can think about it this way: initially, the cat and the apparatus are decoupled, then you turn on the interaction for some time (which in effect measures the system), then you discard the environment and look at the cat's system. So yes, in order to be able to talk about apparatus, measurements etc, you need to have your system interacting with the respective object, i.e. apparatus. The whole measurement process can be thought of as an interaction between 2 systems. $\endgroup$
    – vsoftco
    Commented Jul 22, 2014 at 14:58
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A simple answer is that if you have a pure state and it interacts with a system then it entangles with that system. If you measure your quantum system it will not be in a pure state anymore it will be in a mixed state. That is how you will know.

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  • $\begingroup$ But I thought you could only measure pure states. It's just that when they're in a mixed state, you have a chance of measuring some other pure state too. Is this not correct? $\endgroup$
    – Brandon Enright
    Commented Oct 22, 2016 at 1:06
  • $\begingroup$ I think you are confusing a mixed state with a superposition state anyways not every quantum state has a wave function or pure state representation. If we could only measure pure states then certain quantum states could not be measurable.In general what we have is a density matrix and there is a measurement theory for that. But my answer still stands if you had a pure state it would entangle with the environment, just the system's evolution would not be unitary and so could not be in a pure state anymore and thus you would know there was an environment. $\endgroup$
    – Amara
    Commented Oct 22, 2016 at 13:47

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