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1. Aug 15, 2015

### greypilgrim

Hi,

I've been reading some (very) basic texts on decoherence and have some questions about it:

1. What is the difference between decoherence and dephasing?

2. Assuming decoherence, can we completely do away with collapse theories (i.e. theories in which wavefunction collapse must be postulated)?

3. Could decoherence deal with a completely isolated system (no idea if this is experimentally possible, but assuming it is)? Or will decoherence then be induced by vacuum fluctuations?
Or what happens if we measure very quickly after the preparation of a pure state, such that no considerable decoherence could have taken place?

4. If I understand correctly, decoherence explains how the density operator of a system transforms from a pure to a mixed state by interaction with its environment. So if we look at a system without environment, i.e. the quantum state of the whole universe, what can we say about that? Is it pure and remains pure, or could it be mixed? What difference would it make?

5. The many-worlds interpretation in combination with decoherence is pretty popular. But where exactly does many-worlds come in? Decoherence transforms a quantum superposition into a state that can completely be represented with classical probability theory, i.e. statistical physics, where, as far as I know, no interpretational questions arise. So where do we still need interpretations?

Thanks!

2. Aug 15, 2015

### Zcs

Hi there.

For your 5th question; decoherence occurs for near macro systems (i.e. systems with many particles or many degrees of freedom), hence when you're investigating a micro-system (which has no significance with respect to realism in Copenhagen interpretation) you still encounter the problem of wavefunction collapse. Operational Quantum Mechancs deal with this by just disregarding the reality of wavefunction, hence the collapse is just a formal problem and has no physical implication. Many-worlds interpretation allows one to deal with collapse, simultaneously claiming that the wavefunction does represent a kind of reality for micro-systems as well (though not a very commonsensical one).

3. Aug 15, 2015

### bhobba

Dephasing is associated with many kinds of decoherence but not all. The difference is technical and I cant explain it at the lay level.

What I am going to tell you is from a standard textbook by an acknowledged expert in the field. Everything I say can be found in that book:
https://www.amazon.com/Decoherence-Classical-Transition-Frontiers-Collection/dp/3540357734

The answer is no. One needs an additional postulate or even postulates depending on what one considers reasonable to assume. Again it is technical. The measurement problem has 3 parts. It is reasonable to assume decoherence solves the first two although you will get long threads discussing it on this forum. I will not get into that because it will simply lead to a rehash of what has been hashed out before. It is the view of the textbook - and will leave it at that. The bit it doesn't explain, colloquially, is why do we get any outcomes at all, or at a technical level how an improper mixed state becomes a proper one.

Obviously not because to know anything about the system you must interact with it in which case its not isolated. That does not mean decoherence cant occur inside the system.

Decoherence transforms a superposition into an improper mixed state. Each part is interpreted as a separate world. If you dont know what an improper mixed state is I really cant explain it at the lay level. Suffice to say each part is basically a possible outcome.

Past threads on this topic often degenerate into long drawn out discussions of certain technical aspects associated with this reasonableness thing I mentioned. I will not be drawn into it. I simply reiterate what I said is standard textbook stuff.

Thanks
Bill

4. Aug 15, 2015

### atyy

According to this terminology, would GRW solve what you are calling the "measurement problem" (for at a restricted domain of non-relativistic quantum mechanics)?

5. Aug 15, 2015

### bhobba

Of course it would. It provides a mechanism for the improper mixed state to become a proper one.

Thanks
Bill

6. Aug 15, 2015

### greypilgrim

But classical statistical physics can also be described with improper mixed density operators, can't it? Assume we have a classical gas with Maxwell-Boltzmann distribution and measure the velocity of a single particle. Why don't we need an interpretation for this measurement? Couldn't it equally be possible that each velocity makes up a separate world?

This is very confusing to me – before I read about decoherence, I interpreted the separate worlds as the summands that make up a pure superposition. But with decoherence, the separate worlds are somehow the constituents of a mixed state, which in this case is just a classical convex combinations of states. However in classical statistical physics nobody asks for interpretations, why not?

7. Aug 15, 2015

### bhobba

Of course it can't. You need to study what an improper mixed density operator is. Its 100% quantum. Its the convex sum of pure states as projection operators. Its a quantum state - not classical.

I have a bit of difficulty here. I have zero idea why you would make such a claim because when you know what a quantum state is, and a mixed state is a quantum state, you know it has no classical analogue. I urge you to study the detail. Using as little math as possible the following gives the detail:
http://quantum.phys.cmu.edu/CQT/index.html

Please take the time to study it before going any further. It will take time and effort - but really there is no short cut.

Classical physics uses classical probability theory - QM is an extension of that:
http://www.scottaaronson.com/democritus/lec9.html

One can interpret classical probability theory in a many world type way - but it would be a very weird view. Its just as weird in QM but the mathematics is very elegant and beautiful. You need to study the math to understand why.

Thanks
Bill

Last edited: Aug 15, 2015
8. Aug 15, 2015

### atyy

But isn't collapse spontaneous in GRW? What is the "mechanism" for collapse?

9. Aug 15, 2015

### greypilgrim

Let's assume a qubit in the pure state
$$\left|\psi\right\rangle=\alpha\left|0\right\rangle+\beta\left|1\right\rangle$$
or as density operator
$$\left|\psi\right\rangle\left\langle\psi\right|=\left|\alpha\right|^2 \left|0\rangle\langle0\right|+\left|\beta\right|^2 \left|1\rangle\langle1\right|+\alpha\beta^* \left|0\rangle\langle1\right|+\alpha^*\beta \left|1\rangle\langle0\right|\enspace.$$
Decoherence transforms this to
$$\rho=\left|\alpha\right|^2 \left|0\rangle\langle0\right|+\left|\beta\right|^2 \left|1\rangle\langle1\right|\enspace.$$

But this is exactly the same as if we take a classical bit (not qubit), represent 0 and 1 as elements of an orthonormal basis on a 2D Hilbert space (i.e. identify 0 with $\left|0\right\rangle$ and 1 with $\left|1\right\rangle$) and use a classical probabilistic source which gives 0 with probability $p_0=\left|\alpha\right|^2$ and $p_1=\left|\beta\right|^2$. Using density operator notation doesn't change the fact that this is a classical mixture.

Identifying classical states with convex sums of orthogonal projectors on Hilbert spaces is often done in quantum information theory and that way we can write classical statistical mechanics completely analogously to QM, i.e. observables as Hermitian operators and expectation values as traces. And that's what confuses me. Before decoherence, we have a pure mixture with no classical analogue. After decoherence, we end up with a mixed state that can equally be obtained mixing orthogonal states with a classical probability generator.

10. Aug 15, 2015

### bhobba

They postulate a non linear stochastic term in the wave-function. This isn't a thread about GRW - if you want to discuss it I think that requires a new thread. But I wont be contributing to it because my knowledge of it is not detailed.

Thanks
Bill

11. Aug 15, 2015

### bhobba

It isn't the same as classical probability.

A state in classical probability is always a mixed state represented by a vector of positive numbers that sum to one.

A mixed state in QM is not a vector - it is a positive operator of unit trace that operates on a complex vector space - they are entirely different things.

To make sense of a mixed state in QM you must invoke the Born Rule. To make sense of the vector of classical probability its in the very definition of probability itself - each element of the vector gives the probability of that event as defined in the Kolmogerov axioms.

What many worlds does is interprets that mixed state to make sense. The interpretation of classical probability is in its very definition.

If I write the probability vector [1/2,1/2} then it says event 1 occurs with probability 1/2 and event 2 occurs with probability 1/2.

If I write 1/2|a><a| + 1/2 |b><b| it says a big fat nothing until you apply the Born Rule. And what happens depends entirely on the observable used in the Born Rule.

To break this impasse MW makes an assumption. It assumes |a><a| and |b><b| are separate worlds. That means in one world you get |a><a| and the other |b><b|. If I didn't make that assumption then I cant say anything without knowing the observable.

In QM particles do not have definite position and momentum - what it has depends on what observable you observe with. In statistical physics it has a definite position and momentum - regardless of what you observe with.

Thanks
Bill

Last edited: Aug 15, 2015
12. Aug 15, 2015

### bhobba

You cant. In classical statistical mechanics particles have definite momentum and position - in QM they don't - unless you want to go to something like BM. You can never get, for example, interference effects in statistical mechanics.

Thanks
Bill

13. Aug 16, 2015

### Jimster41

I don't understand this sentence? Probably a terms issue, but I immediately picture rogue waves. I can imagine they can be represented by a classical wave equation. Is a collection of potentially interfering probabalistic ensembles not also a valid description?

14. Aug 16, 2015

### greypilgrim

Hi.

I'm not saying that the states that make up the mixture are classical, but the mixture is, i.e. there are no off-diagonal interference terms.

Consider following two scenarios:

Scenario 1: A source emits the pure quantum state $\left|\psi\right\rangle=\alpha\left|0\right\rangle+\beta\left|1\right\rangle$. If I understand decoherence correctly, it will eventually transform this state into the mixed state
$$\rho=\left|\alpha\right|^2 \left|0\rangle\langle0\right|+\left|\beta\right|^2 \left|1\rangle\langle1\right|\enspace,$$
as I wrote in #9. Many-worlds now claims that there will be two separate worlds.

Scenario 2: We now have two sources, one always emits the state $\left|0\right\rangle$, the other one always $\left|1\right\rangle$. A classical (!) randomness device now chooses the first source with probability $p_0=\left|\alpha\right|^2$ and the second one with $p_1=\left|\beta\right|^2$. We end up with the exact same state
$$\rho=\left|\alpha\right|^2 \left|0\rangle\langle0\right|+\left|\beta\right|^2 \left|1\rangle\langle1\right|\enspace.$$

Since this is classical probability, many-worlds does not apply (and no other interpretation, the question of interpretation doesn't even come up). However, the states $\rho$ are exactly the same in both scenarios. So what makes the difference?

15. Aug 16, 2015

### bhobba

It isn't. That's the key point. Such is an in interpretive assumption.

This has been gone over and over in many threads - its the difference between an improper and proper mixed state:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

See section 1.2.3.

If after reading that its still not clear then I must leave it to someone else.

Thanks
Bill

Last edited: Aug 16, 2015
16. Aug 16, 2015

### bhobba

I am speaking of statistical mechanics - not wave mechanics.

Thanks
Bill

17. Aug 17, 2015

### bhobba

That's correct - so? Different physical situations. Its well known you cant tell the difference from the state - but they were prepared differently.

Thanks
Bill

18. Aug 17, 2015

### greypilgrim

But it's possible to interpret any proper mixed state as an improper one, by means of purification.

Conversely, the essay you mentioned explicitly says that in order to fully solve the measurement problem, we need the assumption that we can interpret the mixed states we get from tracing out the environment (i.e. improper mixtures) as proper ones ("ignorance interpretation", p. 37).

19. Aug 17, 2015

### bhobba

It doesn't matter how you twist and turn an improper mixed state is not a proper one by its very definition. By definition an improper mixed state is any mixed state that's not a proper one.

So?

The situation is this. In the ignorance ensemble interpretation you assume, somehow, the improper mixed state is a proper one. One way that could happen is if BM is correct, another is GRW - it simply doesn't specify - that's where ignorance comes from.

Thanks
Bill

20. Aug 17, 2015

### greypilgrim

I'm not twisting anything. All I'm saying (and that's not only me but basically the whole field of quantum information theory) that proper and improper mixed states are mathematically and operationally fully equivalent and indistinguishable. Hence, any attempt to differentiate the two of them must inelegantly come in as a postulate.

How can we ever know if a given mixed state is truly proper, i.e. it's not part of an entangled state?

Assuming our world is indeed governed by quantum laws and the wavefunction of the universe is pure (and the universe doesn't interact with anything, so that its time evolution is unitary), can proper mixed states even exist?