# Decoherence relationship to the measurement problem

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1. Feb 24, 2014

### Feeble Wonk

Decoherence relationship to the "measurement problem"

I have heard the argument that "decoherence" in quantum states causes the quantum collapse to occur, and that this solves the "measurement problem". But I'm still left with a nagging question... Does decoherence only produce quantum collapse when there is an observation made of the quantum system with sufficient specificity to identify the decoherence. If so, how can that possibly be considered a solution to the "measurement problem"?

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2. Feb 24, 2014

### atyy

Decoherence does not solve the measurement problem. At the end of decoherence one gets a reduced density matrix that is an improper mixture, whereas to solve the measurement problem it has to be a proper mixture. The treatment of an improper mixture as a proper mixture is an additional postulate equivalent to introducing wave function collapse.

3. Feb 24, 2014

### Jilang

I am a bit unclear as to how one distinguishes between a proper and an improper mixed state. Is there an experiment that differentiates them?

4. Feb 24, 2014

### StevieTNZ

See GianCarlo Ghirardi's thought experiment in his book "Sneaking a Look at God's Cards".

5. Feb 24, 2014

### Jilang

Thanks Stevie, have you got a page number for that?

6. Feb 24, 2014

### Feeble Wonk

If I can, I'd like to return to my previous question that I'm still confused by. Is observation/measurement required with sufficient specificity to identify quantum decoherence to cause the entangled quantum system to collapse? Does the entire entangled quantum system remain in superposition to the extent that observable detail allows, regardless of potential decoherent states "below" observable detail?

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7. Feb 24, 2014

### atyy

If one measures observables confined to the subsystem, then no experiment can distinguish proper and improper mixed states. However, if one can measure the system and the environment then proper and improper mixed states can be distinguished. http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf (see section 1.2.3 on p10)

Regardless of what one measures, decoherence does not solve the measurement problem because the system as a whole is still in a superposition of measurement outcomes. It does not make sense to say there is an outcome on the subsytem, but no outcome on the environment.

Decoherence does not solve the measurement problem. For decoherence to solve it for "practical purposes", we have to make the assumption that an improper mixed state is a proper mixed state.

http://arxiv.org/abs/quant-ph/0312059
As Pessoa (1998, p. 432) puts it, "taking a partial trace amounts to the statistical version of the projection postulate."

https://www.amazon.com/Exploring-Quantum-Cavities-Photons-Graduate/dp/0198509146 (see "The quantum-classical boundary" on p80)

What does decoherence solve? A nice example is why some molecules are chiral. Also, it shows the emergence of a "pointer basis". For solutions to the measurement problem, one has to look to de Broglie-Bohm theory, or perhaps many-worlds (among other possibilities).

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8. Feb 24, 2014

### Feeble Wonk

So... If I understand what you are saying... the quantum state of any system exists in a superposition state to the degree allowed by observable measurement. And in that system there will exist decoherent potential states that will remain in superposition until such time as there is a observable event and/or measurement that dictates a collapse to one of those states. Is that correct?

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9. Feb 24, 2014

### bhobba

That's the point - no experiment can.

That's why decoherence explains APPARENT collapse.

Thanks
Bill

Last edited: Feb 24, 2014
10. Feb 24, 2014

### bhobba

It doesn't do what Stevie claims.

Its an experiment to determine the difference between a superposition and a mixed state.

http://books.google.com.au/books?id...ng a look at god's cards appendix 15a&f=false

It's an experiment to determine the difference between |final> = 1/root 2 (|V, A+> + |H, A->) and an ensemble ie a mixed state. |final> is a pure state which is different from an ensemble which is a mixed state and is expressed as operators - not vectors.

I posted this in another thread but will repeat it here:

The issue is states are not elements of a vector space as some books, especially those at the intermediate level like Griffiths, will tell you. They are in fact positive operators of unit trace defined by the general form of the Born Rule. To really grasp it you need to see the two axioms of QM as detailed by Ballentine in his text.

1. To each observation there corresponds a Hermitian operator whose eigenvalues give the possible outcomes of the observation.
2. There exists a positive operator of unit trace P such that the expected outcome of the observation associated with the observable O is E(O) = Trace (PO) - this is the Born Rule in its most general form. By definition P is called the state of the system.

In fact the Born Rule is not entirely independent of the first axiom, as to a large extent it is implied from that via Gleason's Theroem - but that would take us too far afield - I simply mention it in passing.

Also note that the state, just like probabilities, is simply an aid in calculating expected outcomes. Its not real like say an electric field etc. In some interpretations its real - but the formalism of QM is quite clear - its simply, like probabilities, an aid in calculation.

By definition states of the form |x><x| are called pure. States that are a convex sum of pure states are called mixed ie are of the form ∑ pi |xi><xi| where the pi a positive and sum to one. It can be shown all states are either pure or mixed. Applying the Born rule to mixed states shows that if you have an observation whose eigenvectors are the |xi><xi| then outcome |xi><xi| will occur with probability pi. Physically one can interpret this as a system in state |xi><xi| randomly presented for observation with probability pi. In such a case no collapse occurs and an observation reveals whats there prior to observation - many issues with QM are removed. Such states are called proper mixed states.

Pure states, being defined by a single element of a vector space, can be associated with those elements and that's what's usually done. Of course when you do that they obey the vector space properties so the principle of superposition holds ie if |x1> and |x2> are any two pure states a linear combination is also a pure state. This is what is meant by a superposition. Note it deals with elements of a vector space not convex sums of pure states when considered operators - they are mixed states. This means the state 1/root 2 |x1> + 1/root 2 |x2> is a pure state and is totally different from the mixed state 1/2 |x1><x1| + 1/2 |x2><x2|.

Now what decoherence does is transform a superposition like |x> = 1/root 2 |x1> + 1/ root 2 |x2> into a mixed state like X = 1/2 |x1><x1| + 1/2 |x2><x2|. When that is done it can be interpreted as a proper mixed state which solves many of the issues with collapse etc.

The thought experiment Stevie refers to determines the difference between a pure state |x> represented by a vector and the mixed state X represented by an operator. Doing an observation on either for |x1> and |x2> gives exactly the same result. However if you observe it for different things you can tell the difference, as the thought experiment demonstrates. An improper mixed state and a proper mixed state are exactly the same mathematically and represented by the operator X - there is no way to tell the difference.

Thanks
Bill

Last edited: Feb 24, 2014
11. Feb 24, 2014

### atyy

I don't understand exactly what you are saying, but yes, that seems essentially correct.

12. Feb 24, 2014

### bhobba

Cant say I understand what you are saying.

The easiest way to understand it is as tracing over the environment:
http://physics.stackexchange.com/qu...ake-the-partial-trace-to-describe-a-subsystem

Whats going on is the system being observed and the environment become entangled, which via the math detailed in the link leads to the system being observed being in an effective (another word is improper) mixed state. It looks exactly the same as a proper mixed state. A proper mixed state is a state that is randomly presented for observation and if it was a proper mixed state collapse would have occurred. But since it only LOOKS like a proper mixed state it only LOOKS like collapse has occurred - it really hasn't - which is why the word APPARENT is used.

However there is nothing stopping assuming it actually occurred there, Von-Neumann showed it could be placed anywhere, and in doing so many issues are solved.

That said, and without going into the details - issues do remain - but of course research is ongoing.

Thanks
Bill

13. Feb 24, 2014

### Maui

This setup was invented so that people would believe that it's the environement that does the collapse. It doesn't(experiments prove this).

The whole setup is flawed from the beginning as it assumes that improper mixed states are the same as proper mixed states. They are not!

Then you add contrived structures that you label 'environment' which are foreign to qm. Of course decoherence is guaranteed to take place once you introduce that which all interpretations have as their endgoal - classicality.

Ok it's not religion, but if someone said it was, i wouldn't argue.

Last edited: Feb 24, 2014
14. Feb 24, 2014

### atyy

With respect to the measurement problem, I think decoherence is interesting for the many-worlds interpretation, which is a potential solution to the measurement problem. One problem with many-worlds is that the branches are ambiguous because a state can be decomposed in many ways. Decoherence by choosing a "pointer" basis could help that. It still needs some finessing, since decoherence is never perfect. There are very interesting ways to do the finessing in David Wallace's "The Emergent Multiverse" https://www.amazon.com/The-Emergent-Multiverse-according-Interpretation/dp/0199546967

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15. Feb 25, 2014

### Maui

You still have to assume that improper mixed states are the same as proper one so that there be interactions and world splitting. This injects philosophy so it cannot be science.

Last edited by a moderator: May 6, 2017
16. Feb 25, 2014

### bhobba

Its not philosophy that there is no experiment that can tell the difference.

Until there is a way to tell the difference between interpretations experimentally its purely a personal choice based on all sorts of thing of which philosophy is just one.

Thanks
Bill

17. Feb 25, 2014

### Maui

Experiments prove that the world is quantum and not classical. You should not pretend you are explaining the quantum to classical transition by stating the obvious - that you always observe a classical world.

Yes, that I agree with. As long as you label your decoherence pet theory as another philosophical interpretation, we are on the same page. But you are presenting it as a significant progress on the measurement problem and that is.... er.... yes, ridiculous.

Last edited: Feb 25, 2014
18. Feb 25, 2014

### atyy

Why do you say that? It's not obvious to me since the pointer basis is selected without having to assume that improper and proper mixed states are the same.

19. Feb 25, 2014

### Maui

How is the pointer basis selected? If you do not assume they are one and the same(for all practical purposes) you will get no decoherence but pure states.

Last edited: Feb 25, 2014
20. Feb 25, 2014

### eloheim

This seems as relevant a place as any to ask a question I've had. What would it mean to "observe a quantum world?" I mean we see non-classicality in experiments all the time. I assume when people say the above they mean macroscopic non-classicality? Or how would the world appear exactly, so that we knew it was the case?