What Are the Key Questions Surrounding Quantum Decoherence?

[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!

 

[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).

 

[bhobba]

What is the difference between decoherence and dephasing?

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

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

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/dp/3540357734/?tag=pfamazon01-20

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.

Could decoherence deal with a completely isolated system (no idea if this is experimentally possible, but assuming it is)? 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?

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 can't occur inside the system.

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?

Decoherence transforms a superposition into an improper mixed state. Each part is interpreted as a separate world. If you don't know what an improper mixed state is I really can't 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

 

[atyy]

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.

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

 

[bhobba]

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

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

Thanks
Bill

 

[greypilgrim]

Decoherence transforms a superposition into an improper mixed state. Each part is interpreted as a separate world.

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?

 

[bhobba]

But classical statistical physics can also be described with improper mixed density operators, can't it?

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

 

[atyy]

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

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

 

[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.

 

[bhobba]

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

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 won't be contributing to it because my knowledge of it is not detailed.

Thanks
Bill

 

[bhobba]

$$\rho=\left|\alpha\right|^2 \left|0\rangle\langle0\right|+\left|\beta\right|^2 \left|1\rangle\langle1\right|\enspace.$$

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 can't 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

 

[bhobba]

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,

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

 

[Jimster41]

You cant. You can never get, for example, interference effects in statistical mechanics.

Thanks
Bill

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?

 

[greypilgrim]

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.

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?

Thanks in advance!

 

[bhobba]

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.

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

 

[bhobba]

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?

I am speaking of statistical mechanics - not wave mechanics.

Thanks
Bill

 

[bhobba]

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?!

That's correct - so? Different physical situations. Its well known you can't tell the difference from the state - but they were prepared differently.

Thanks
Bill

 

[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).

 

[bhobba]

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

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.

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).

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

 

[greypilgrim]

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.

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?

 

[bhobba]

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.

They are mathematically and operationally equivalent.

But are prepared differently.

That's all there is to it. Nothing more to be said.

can proper mixed states even exist?

Obviously they can exist.

Thanks
Bill

 

[greypilgrim]

But as I stated in the second part of #20, assuming the universe is in a pure state, is it possible to prepare a proper mixed state? The only source of true randomness would be an improper mixed state, wouldn't it?

 

[bhobba]

But as I stated in the second part of #20, assuming the universe is in a pure state, is it possible to prepare a proper mixed state? The only source of true randomness would be an improper mixed state, wouldn't it?

I have zero idea why you are arguing something so obvious. For example if GRW was true collapse prepares proper mixed states all the time - and that's just one possibility.

Also look around you - there are many many ways proper mixed states can be prepared eg using a dice. That the universe is in a pure state - assuming such is possible which is controversial - changes nothing of that.

Thanks
Bill

 

[greypilgrim]

I'm talking about many-worlds. I'm trying to avoid any collapse, that's why I assumed the time evolution of the universe to be unitary (as far as I understand GRW, its collapses are non-unitary as they are in other collapse interpretations).

I can't see how you possibly could obtain "proper probability" in such a universe. Let's take the example of the throw of dice:
Either the dice is not entangled and fully non-interacting with its environment, then it evolutes unitarily and the outcome is not random.
Or it is entangled with its environment or becomes so by interaction, then its reduced state is an improper mixed state.

Is there a third option that eventually makes the dice end up in a proper mixed state?

 

[bhobba]

I can't see how you possibly could obtain "proper probability" in such a universe. Let's take the example of the throw of dice:

The dice is a classical object. Its a fact classical objects exist.

What I suspect concerns you is the factorisation issue. I will not go into that again - its been done to death. You can find the threads yourself.

Thanks
Bill

 

[bhobba]

I'm talking about many-worlds. I'm trying to avoid any collapse,

The modern version of MW is very tight and rigorous. Its very theorem proof, theorem proof. It can't be faulted once you accept its premises - but it can't be explained simply - there are many proofs that are complex and pages long. You must study the detail - and it will take time:
http://users.ox.ac.uk/~mert0130/books-emergent.shtml

Just as an example it doesn't even make use of observations explicitly - its like decoherent histories and relies a lot on the concept of history to boot up many other concepts. In fact Gell-Mann describes Decoherent Histories as MW without the MW's.

Thanks
Bill

 

[greypilgrim]

Well now we're moving in circles. The interpretation issues come up when we try to find the link between quantum laws and a macroscopically classical world. Trying to explain why we see classical objects, i.e. why interference terms vanish, i.e. why a throw of dice has a definite outcome.

And if decoherence now states that interference terms don't vanish but just become very very small (and MWI says that nothing collapses, i.e. no non-unitary evolution), well, then classical objects don't exist. They can be approximated classically to a very (very very very) good precision, but this doesn't change the fact that randomness extracted from such objects is eventually still quantum, hence mixed states relying on that randomness must be improper.

What do you mean by the factorisation issue?

Thanks for the book reference, I'm trying to get a copy.

 

[bhobba]

And if decoherence now states that interference terms don't vanish but just become very very small (and MWI says that nothing collapses, i.e. no non-unitary evolution), well, then classical objects don't exist. They can be approximated classically to a very (very very very) good precision, but this doesn't change the fact that randomness extracted from such objects is eventually still quantum, hence mixed states relying on that randomness must be improper.

I don't understand why when something is way below detectability the fact its not zero is an issue. Obviously it isn't.

What do you mean by the factorisation issue?

https://www.physicsforums.com/threads/why-does-nothing-happen-in-mwi.822848/

Thanks for the book reference, I'm trying to get a copy.

I have a copy and enjoyed reading it. I am not a believer in MW but for me it was good seeing its rigorous development. Its modern basis is not what you might think. That after decoherence the parts of a mixed state is a separate world is only a rough idea of its complete elucidation. It also helped me understand decoherent histories better.

Thanks
Bill

 

[greypilgrim]

I don't understand why when something is way below detectability the fact its not zero is an issue. Obviously it isn't.

Well in this case it makes a big (at least interpretational) difference: It determines the existence of non-unitary processes, of a classical world, of proper mixed states and of probabilistic processes where the world doesn't separate.

If the off-diagonal terms become zero (and I mean 'zero' zero) by some (necessarily non-unitary) process, there exists classical randomness which can be used to create proper mixed states. Hence, there are probabilistic measurements that do NOT lead to separate worlds.

If the off-diagonal terms only become very small (but nonzero), there is no need for non-unitary processes. A pure superposition remains pure and any randomness derived from it to create mixed states actually leads to entanglement, so those mixed states are improper. Hence ANY probabilistic process leads to separate worlds.

 

[bhobba]

Well in this case it makes a big (at least interpretational) difference

I disagree. If its way below detectability then FAPP its zero.

Thanks
Bill

 

[greypilgrim]

Well, FAPP any discussion about interpretations is moot.

As I said, in the first case a throw of dice would not lead to separate worlds, in the second it would, even if the off-diagonal terms are way below detectability. Why is this no big interpretational difference?

 

[cube137]

I have zero idea why you are arguing something so obvious. For example if GRW was true collapse prepares proper mixed states all the time - and that's just one possibility.

Also look around you - there are many many ways proper mixed states can be prepared eg using a dice. That the universe is in a pure state - assuming such is possible which is controversial - changes nothing of that.

Thanks
Bill

Bill. You said a cat can never be in pure state.. and pure state only applied to electrons, buckeyball and few others. Why did you say above that "That the universe is in a pure state - assuming such is possible which is controversial..." Why didn't you just say categorically the universe is never a pure state because even a cat can't be pure state? What do you mention controversial.. how can the universe even be conjectured as being a pure state. Maybe you meant because the universe was a close system.. meaning likewise if the cat had oxygen tank and could breath inside a totally shielded box.. it is possible to be in pure state? But what part would have interferences in the particles of the cat (or the universe if you think the possibility the universe can be in pure state because you use the words (assuming such is possible which is controversial )? Sorry for quoting from this old thread. but just want to directly question something you mentioned here and the thread title is appropriate for it too anyway.

 

[bhobba]

"That the universe is in a pure state - assuming such is possible which is controversial..."

That contradicts nothing I have said.

I mentioned previously 'Modelling a system as pure is done not because its actually like that - its done to have a tractable model'

Then I asked you to think about it. I suggest you do the same here.

Nugatory wrote:

If you're going to dig as deeply into the formalism as you want to, you're going to have to learn the math - there is no other way to get to where you want to be. Atyy's link is very good, but it is written for people who have already been through a no-kidding college-level introduction to quantum mechanics, where the basic notion of states as vectors in a Hilbert space is taught. Only after you've been through that will you be ready to take on the density matrix formalism.ten in the left/right basis but not when written in the up/down basis; #2 is the other way around. All four of these states will have off-diagonal elements in one basis or the other.

Once you have done that then we can chat about if the universe is in a pure state or not. There is an issue if the concept of state is itself applicable to the entire universe - it turns out to be very interpretation dependant. But delving into that requires knowing what a state is in a technical sense.

Thanks
Bill

 

[cube137]

That contradicts nothing I have said.

I mentioned previously 'Modelling a system as pure is done not because its actually like that - its done to have a tractable model'

What is meant by tractable model?

Then I asked you to think about it. I suggest you do the same here.

Nugatory wrote:Once you have done that then we can chat about if the universe is in a pure state or not. There is an issue if the concept of state is itself applicable to the entire universe - it turns out to be very interpretation dependant. But delving into that requires knowing what a state is in a technical sense.

Thanks
Bill

Just a clue.. I'm reading the paper and I understood Nugatory stuff. What interpretation (is it Many worlds) in which the universe is in pure state? But how can all the particles in the universe form interference? maybe pure state is not really about interferences but more of trace 1 where the vector in Hilbert space is not in any axis (any component)?

 

[bhobba]

What is meant by tractable model?

Its obvious - they have the meaning as per a dictionary. In mathematical modelling, and physics is a mathematical model, you often make simplifying assumptions to actually solve problems. Its done all the time - so frequently, and so obviously, its seldom explicitly mentioned.

Just a clue.. I'm reading the paper and I understood Nugatory stuff. What interpretation (is it Many worlds) in which the universe is in pure state? But how can all the particles in the universe form interference? maybe pure state is not really about interferences but more of trace 1 where the vector in Hilbert space is not in any axis (any component)?

Google is your friend:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

To discuss it you need to know what a state preparation procedure is, and what a filtering type observation is. When you can explain that, in your own words, not with a link, then we can discuss it. And please start a new thread - this is departing from the threads topic.

Thanks
Bill