Decoherence and apparent collapse

1. Feb 11, 2016

durant35

Can anybody explain what is the difference between apparent collapse and collapse? Ive read that decoherence leads to apparent collapse but what does that mean, is there an eigenstate after decoherence but its not known why did it occur. I dont understand it all.

2. Feb 11, 2016

Staff: Mentor

Its to do with the difference between proper and improper mixed states.

You will find a number of threads discussing it here if you do a search. I participated in a number of them. Unfortunately I failed miserably. I really tried - but explaining it at the lay level simply can't be one - at least I can't do it.

I have already given this link to you. It explains it clearly:
http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

It is however technical. To understand it you must learn the technicalities. Here the jig is up - lay explanations are not possible.

At the lay level here is the book to get:
https://www.amazon.com/Where-Does-The-Weirdness-Mechanics/dp/0465067867

Thanks
Bill

Last edited by a moderator: May 7, 2017
3. Feb 12, 2016

zonde

If you have ensemble of systems in proper mixed state each member is in definite state (either one or the other). End result of collapse is such a proper mixed state.
But if ensemble of systems is in improper mixed state each member is in superposition of states. But all these superpositions of separate members look the same in position representation while they are different in other representations. This improper mixture corresponds to apparent collapse.

4. Feb 12, 2016

Staff: Mentor

That's incorrect.

There is no difference between proper and improper mixed states as far as QM is concerned. Its purely how it was prepared.

Thanks
Bill

5. Feb 12, 2016

naima

States are described by density matrices. When you receave a beam described by a diagonal density matrix, if you think that the state is attached to particles, you can say that:
1) you receave a probabilistic mixture of different pure states (proper mixture)
2) All the particles are in the same impure state (improper mixture)
As no possible experiment can answer to this question, One can say that it is the same thing or that that there is a false prerequisit.

6. Feb 12, 2016

Staff: Mentor

I don't think it's incorrect so much as imprecise. I think we're tripping over the problems of using natural language here, as there are important mathematical differences between the two - although QM ends up extracting the same statistical prediction from two interestingly different beasts so they're the same as far QM is concerned.

But what I think even more is that it would be sooooo nice if the QM forum would go 24 hours without setting off the big red alarm bell in the Mentor's Central Command and Control HQ (which is located in the blast-proof and radiation-hardened basement of Avengers' Mansion, if you're wondering).

Last edited: Feb 12, 2016
7. Feb 12, 2016

Staff: Mentor

Bhobba's recommendation of the Lindley book is excellent.

You might also want to take a look at atyy's comments in this thread. Some comfort with the density matrix formalism is required, but that's going to be true of any real answer to your question. Unfortunately I don't know of any gentle introduction to this formalism (although I'm sure there's one out there somewhere - it's not that hairy and scary a topic).

8. Feb 12, 2016

naima

in that thread Atyy wrote:
"In an improper mixed state, each system is in the same pure state."

What is for him the definition of a pure state?
For me a pure state can have a density matrix with only one non null element and diagonal density matrices cannot.

9. Feb 12, 2016

Staff: Mentor

If I'm understanding you properly, your definition is equivalent to saying that a pure state can be represented by a ray in Hilbert space instead of with a density matrix. If so, you're good. But the word "can" is important here; whether the density matrix of a pure state has one or several non-null elements depends on the basis that you're using.

10. Feb 12, 2016

naima

It remains that i do not understand atyy's sentence in the link you gave.

11. Feb 12, 2016

durant35

I know this escapes strict physics formalism, but I can't wrap my head around the mechanism of decoherence. If everything is observed by everything else that should mean the absence of superpositions on a macro level, but while reading Zurek's paper he mentions that decoherence time is extremely short but he mentions that our the superpositions in our neurons are destroyed quickly. Shouldn't it be the case that there are no superpositions at all on a macro level, not in an instant since observing causes apparent collapse. This has got me so mind boggled that I look at people and don't understand why are they alive and are they in a superposition for a split second that I can't perceive since decoherence isn't instanteneous. I hope for a good answer because this is driving me crazy.

12. Feb 12, 2016

A. Neumaier

You can safely remain sane - there is nothing to be understood except that all the talk is tremendously exaggerated half-understood science.

Quantum-mechanical states are not objective properties of objects but descriptors for the (quantum-)statistical behavior of many similar objects.

Thus it shouldn't drive you more crazy than knowing that the average human being is half male and half female, or that the average number of people visiting your favorite restaurant each month is not an integer but has a fractional part.

13. Feb 12, 2016

Staff: Mentor

Will try and help with that.

Thanks
Bill

14. Feb 12, 2016

stevendaryl

Staff Emeritus
What "wave function collapse" amounts to is a hypothetical process by which a single possibility is randomly chosen out of a superposition of possibilities. So a system goes from being in a superposition of "state A" and "state B" to a state that is purely "state A" (or purely state B) when you perform a measurement on it. When a coin has a 50% chance of being heads, and 50% chance of being tails, and you take a peek at it and find that it's heads, you would normally assume that it must have been heads all along, and your looking at it just confirmed this fact. But in quantum mechanics, there is a distinction between a "mixed state" and a "superposition". A mixed state is when the system is actually in one state or another, but you just don't know which, and you use probabilities to describe your uncertainty. When a system is in a superposition of two states, it is not correct to think of the system as really in one state or the other, we just don't know which. It's more like vector addition: If you combine motion to the north with motion to the west, then you get motion to the northwest. If someone is traveling northwest, it isn't that there is a 50/50 chance that he is traveling north and a 50/50 chance that he is traveling west.

Practically speaking, what is the difference between the two situations: a superposition versus a mixed state? Well, the difference has to do with "interference". When you compute probabilities for a future state based on the current state, in the mixed case, you just compute a weighted average of the probabilities for each state separately. If the state is a superposition, you have additional terms, "interference" between the two possibilities.

So what is decoherence? Roughly speaking, it's when the effect of the environment (the rest of the universe) messes with the interference patterns to make it impossible to notice interference effects. So after decoherence, there is no longer a measurable difference between superpositions and mixtures. At this point, we may as well assume that the wave function has "collapsed" to a specific state, but we just don't know which one. As far as the mathematics is concerned, there is no testable difference. But with decoherence, there is no actual collapse--the complete system (including the environment) is still in a superposition, but it's impossible to detect the interference effects that are the hallmark of superpositions.

15. Feb 12, 2016

stevendaryl

Staff Emeritus
I don't think that comparison is quite accurate. If you take the simplest case of quantum superposition, which is superposition between an electron that is spin-up in the z-direction and an electron that is spin-down in the z-direction, then that superposition is (with the right coefficients) a state that is pure spin-up in the x-direction. There is a big difference between "The electron is either spin-up in the z-direction, or spin-down in the z-direction, I just don't know which" and "The electron is spin-up in the x-direction".

16. Feb 12, 2016

A. Neumaier

In terms of density matrices, a state is pure if its density matrix has rank 1 and hence can be written as $\rho=\psi\psi^*$ with a normalized wave function $\psi$. Expressed equivalently, it cannot be expressed as a convex linear combination of other density matrices. This has nothing to do with being diagonal or not, or with zero entries. Both $\rho=\pmatrix{1 & 1\cr 1 &1}$ and $\rho=\pmatrix{1 & 0\cr 0 &0}$ are pure, while $\rho=\pmatrix{1 & 1\cr 1 &2}$ and $\rho=\pmatrix{1 & 0\cr 0 &1}$ are mixed.

Now atyy apparently uses improper and proper synonymous with pure and mixed, while many others - e.g., http://arxiv.org/pdf/quant-ph/0109146.pdf - use improper for a state that is prepared by mixing pure states, and proper for a state that appears as a restriction of a pure state of a bigger system to a subsystem (which usually results in a mixed state).

17. Feb 12, 2016

A. Neumaier

But this has nothing to do with what I asserted, completely accurately:
spin-up or spin-down in a given direction are not properties of the electron before a corresponding measurement, except if they are prepared that way by a suitable preparation.

In the Copenhagen interpretation they are properties after the measurement, which is there regarded as a re-preparation in a definite collapsed state. The interaction with the detector changes the old state in an uncontrolled way into the new, definite state. Only in this definite state one can talk about a property of the electron. Before measurement, the state is just a statistical summary of possible outcomes. It does not matter whether the state is pure or mixed.

In the ensemble (= minimal = statistical) interpretation states are anyway properties of the prepared ensemble only, not of a single electron.

And the last sentence of my post,
was obviously only a classical illustration that, compared to the real things, statistics have slightly different, superficially weird, properties. It was not meant to give a quantitatively correct account of quantum phenomena. For the latter we have the operator formalism.

18. Feb 12, 2016

stevendaryl

Staff Emeritus
I think I can explain Atyy's quote using a pair of entangled electrons:

The spin-part of the wave function for a pair of electrons with total spin zero looks like this:

$\frac{1}{\sqrt{2}}(|U\rangle |D\rangle - |D\rangle |U\rangle)$

It's a superposition of two possibilities: (1) The first electron is spin-up, and the second electron is spin-down, or (2) the other way around.

Now, if one of the electrons is unobservable (it's lost down a black hole, or something), we can describe the other, observable electron as being in a mixed state of being 50% spin-up and 50% spin-down by tracing out the degrees of freedom due to the lost electron. But unlike a proper mixed state, the probabilities are not due to ignorance about the true state of the electron, but are due to the mathematical device of tracing out unobservable degrees of freedom.

Mathematically, a two-component system will have a density matrix of the form $\rho_{i \alpha j \beta}$ where $i, j$ are indices referring to the first subsystem, and $\alpha, \beta$ are indices referring to the second subsystem. Tracing out the second subsystem means coming up with an effective one-system density matrix $\rho'_{ij}$ defined by:

$\rho'_{ij} = \sum_\alpha \rho_{i \alpha j \alpha}$

For any experiment that only involves the first subsystem, $\rho'$ makes exactly the same predictions as $\rho$.

Every density matrix can be written (possibly in more than one way) in the form:

$\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j |$

where $|\psi_j\rangle$ is a column matrix and $\langle \psi_j|$ is its hermitian conjugate, a row matrix.

A pure state is just one that can be written using just one such term (that is, all the $p_j$ are equal to zero except for one). Equivalently, a pure state is one where $\rho^2 = \rho$.

19. Feb 12, 2016

stevendaryl

Staff Emeritus
I was only referring to your analogy of a typical human as being half male and half female. That is not an accurate analogy for quantum superpositions.

20. Feb 12, 2016

durant35

But are we as macroscopical objects in a superposition of states at any given instant? And the objects around us? It is so confusing.

21. Feb 12, 2016

A. Neumaier

???
What then is the true state of the electron? We are completely ignorant about it, except for what is encoded in its mixed state., which gives the correct probabilities, just as in any other case where a mixed state appears. Thus the density matrix encodes everything that is not due to ignorance. (The only ''true state'' assumed is that of the 2-electron system. It is a state of a completely different system!)

However, when we measure the other particle and assume its outcome to be known, the state of the 2-particle system changes into a tensor product state, and the mixed state collapses to a pure state. Again the density matrix encodes everything that is not due to ignorance.

Thus distinguishing between proper and improper mixtures is completely spurious.

22. Feb 12, 2016

stevendaryl

Staff Emeritus
As I said in an earlier message, a composite system can be in a superposition, but when you focus on a subsystem, that subsystem may (and often will) look like a mixed state. A macroscopic system is always part of a larger system (the rest of the universe), so when you look at the macroscopic system alone, it will look like a mixed state, not a superposition. The significance of this is that a mixed state can be interpreted as "the system is actually in this state or that state, we just don't know which", which is the situation we have with classical (non-quantum) systems. So for all practical purposes, we can reason as if macroscopic systems are classical; any uncertainty is just due to lack of knowledge about the detailed state.

So I would say that macroscopic systems are almost never in superpositions (unless you get so macro as to include the entire universe, maybe).

What's special about macroscopic systems in this regard? It's just that macroscopic systems interact with the rest of the universe much stronger than microscopic systems. It's these interactions that destroy superpositions and produce (effectively) mixed states.

23. Feb 12, 2016

stevendaryl

Staff Emeritus
If you prepare an electron that is spin-up in the x direction, it is just not correct to say that "It's either spin-up in the z-direction, or spin-down in the z-direction, we just don't know which".

Well, I don't agree with that.

24. Feb 12, 2016

A. Neumaier

It depends on what one assumes as given when creating the statistical model of the situation, and whether the macroscopic object is regarded as part of the situation or as part of its context. Nothing to worry about.

No serious application of quantum mechanics works with macroscopic superpositions. It is only Gedankenexperiments that talk about them, in the hope of shedding light on foundational issues. But all light that could be shed on it in this way was shed long ago, and since then it just confuses people rather than enlightens them.

if you like horror you can go to horror movies. If you like quantum weirdness you can read the popular literature about it. Thus if you like to remain sane you should regard macroscopic superpositions as akin of monsters created to entertain rather than as reality.

25. Feb 12, 2016

A. Neumaier

This is complete nonsense. (Though one can often read it, this statement is never made use of in actual predictions. Hence it survives unchecked though it makes no sense at all.)

In the situation you describe, the system is in the mixed state if a state can be assigned to it at all. It is definitely not in one of the pure states contained in the mixed state - the notion of containment is ill-defined. Assuming it to be in a pure state when the state is mixed gives wrong statistical conclusions!