Interpretation of quantum mechanics in cosmology

[atyy]

In Mukhanov's Physical Foundations of Cosmology p348, he writes

"Decoherence is a necessary condition for the emergence of classical inhomogeneities and can easily be justified for amplified cosmological perturbations. However, decoherence is not sufficient to explain the breaking of translational invariance. It can be shown that as a result of unitary evolution we obtain a state which is a superposition of many macroscopically different states, each corresponding to a particular realization of galaxy distribution. Most of these realizations have the same statistical properties. Such a state is a close cosmic analog of the “Schroedinger cat.” Therefore, to pick an observed macroscopic state from the superposition we have to appeal either to Bohr’s reduction postulate or to Everett’s many-worlds interpretation of quantum mechanics. The first possibility does not look convincing in the cosmological context."

Is many-worlds the favoured interpretation in cosmology, or are there other options?

 

[jedishrfu]

Sixty symbols YouTube channel has a recent video on QM, the various interpretations and views of physicists today:

 

[Demystifier]

Is many-worlds the favoured interpretation in cosmology, or are there other options?

There are certainly many other options, but let me mention only two.

Copenhagen-like: There is no universe as a whole. There are only our local observations of light in a telescope, which we INTERPRET as a "big universe".

Bohmian: The wave function of the universe is real, there is no collapse, but what we observe is not the wave function itself. We observe objects (classical-like particles or fields) which are guided by the wave function.

 

[atyy]

There are certainly many other options, but let me mention only two.

Copenhagen-like: There is no universe as a whole. There are only our local observations of light in a telescope, which we INTERPRET as a "big universe".

Bohmian: The wave function of the universe is real, there is no collapse, but what we observe is not the wave function itself. We observe objects (classical-like particles or fields) which are guided by the wave function.

I like these ideas. I wonder why Mukhanov doesn't mention them.

 

[Sonderval]

@Demystifier
Isn't there a (non-local) collapse of the wave function in Bohmian mechanics? After all, it is usually accounted as non-local realistic theory?

 

[Demystifier]

@Demystifier
Isn't there a (non-local) collapse of the wave function in Bohmian mechanics? After all, it is usually accounted as non-local realistic theory?

Bohmian mechanics is nonlocal, but not due to a collapse. What is nonlocal is the equation of motion that governs the motion of particles.

 

[Demystifier]

Is many-world interpretation really better than Copenhagen interpretation in the cosmological context?

The standard argument that it is goes as follows:
- Copenhagen interpretation requires an EXTERNAL OBSERVER.
- But for the whole Universe there is nothing external to it.
- Therefore there is no external observer for the Universe.
- The many-word interpretation (MWI) does not require an external observer.
- Therefore MWI is better.

Here I want to criticize that argument. More precisely, I want to point out that even though MWI does not require an external observer, it does require something very similar and equally problematic. Therefore, in the cosmological context, MWI is not better than Copenhagen interpretation.

Here is my argument, analogous to the standard argument above:
- To explain why the macro world looks "classical", MWI needs to explain the illusion of collapse.
- To explain the illusion of collapse without observers (and without true collapse), MWI proposes the wave-function splitting.
- To explain the splitting (instead of simply postulating it), MWI needs to refer to decoherence.
- To explain decoherence, it needs an EXTERNAL ENVIRONMENT.
- But for the Universe as a whole, there is no external environment.
- Therefore for the Universe as a whole, there is no decoherence.
- Therefore for the Universe as a whole, there is no wave function splitting.
- Therefore for the Universe as a whole, there is no illusion of collapse.
- Therefore MWI cannot explain why the Universe as a whole looks "classical".

Of course, someone who still likes MWI will point out that nobody ever observes the Universe as a whole. We always observe only parts of the Universe (even if very big ones), and MWI with environment-induced decoherence can explain that.

But even if that is true*, this makes MWI not better than Copenhagen. Namely, completely analogously, an adherent of the Copenhagen interpretation may point out that since we never observe the Universe as a whole, there is still room for an observer external to that part of the Universe we do observe.

So to argue that MWI is better than Copenhagen, it is not valid to use quantum cosmology as an argument. Perhaps there are other valid arguments for preferring MWI over Copenhagen, but there are no valid arguments based on quantum cosmology.

*For a more general critique of MWI se also
https://www.physicsforums.com/blog.php?b=4289

 

[Sonderval]

@Demystifier
You are of course right - there is a non SGL-change in the wave function of a subsystem, but no collapse of the whole wave function of the universe. Thanks.

 

[martinbn]

I don't understand what the problem with CI is. The external observer/apparatus/classical object is needed only when measurements/observations are considered. I don't see the problem with talking about the state of the universe. You cannot measure the whole universe, there will be always degrees of freedom that will be fixed, so there will always be something external when it comes to measurements.

 

[Demystifier]

I like these ideas. I wonder why Mukhanov doesn't mention them.

Mukhanov is an expert for cosmology, including some quantum aspects of it, but he is not an expert for foundations and interpretations of QM.

 

[Demystifier]

I don't understand what the problem with CI is. The external observer/apparatus/classical object is needed only when measurements/observations are considered. I don't see the problem with talking about the state of the universe. You cannot measure the whole universe, there will be always degrees of freedom that will be fixed, so there will always be something external when it comes to measurements.

Exactly! See also my post #7 above.

 

[kith]

- To explain why the macro world looks "classical", MWI needs to explain the illusion of collapse.
[...]
- Therefore MWI cannot explain why the Universe as a whole looks "classical".

I guess that many worlders would ask why we should assume that the universe as a whole looks classical. So I don't think your argument applies.

I agree that the external observer requirement is not really a problem for Copenhagen. But doesn't the real problem occur way before we try to describe the universe? I'm thinking about the classical setting of an observer being observed. If an observer A measures a system which is in the state |ψ> = Ʃ_n|ψ_n>, collapse occurs and we end up in one definite eigenstate |a_n>|ψ_n>. However, an external observer B, who measures the combined system A+system finds that it is in a superposition Ʃ_n|a_n>|ψ_n>. So the formalism with collapse gives two different states for the same situation. I thought the conclusion of Copenhagen was that it simply isn't valid to treat the observer quantum mechanically.

 

[stevendaryl]

Bohmian mechanics is nonlocal, but not due to a collapse. What is nonlocal is the equation of motion that governs the motion of particles.

Something that never made any sense to me about Bohmian mechanics, which maybe is just because I missed a key point, is the relationship between the wave function and particle positions.

The assumption made by Bohmian mechanics is that particles have definite positions at all moments. They obey a deterministic equation of motion which depends not only on classical forces, but also on a "quantum potential" related to the wave function. The distribution of particle positions is assumed to be related to the square of the wave function. Then there is a consistency condition which is easily provable, which is that if the particles start off distributed according to the square of the wave function, and their positions evolve deterministically according to the quantum equation of motion, then the distribution will be related to the square of the wave function at all future times, as well.

That's all perfectly fine. But now, suppose we are talking about the quantum mechanics of a SINGLE particle. What can it possibly mean to say that its location is distributed according to the square of the wave function?

 

[stevendaryl]

Here is my argument, analogous to the standard argument above:
- To explain why the macro world looks "classical", MWI needs to explain the illusion of collapse.

- To explain the illusion of collapse without observers (and without true collapse), MWI proposes the wave-function splitting.
- To explain the splitting (instead of simply postulating it), MWI needs to refer to decoherence.
- To explain decoherence, it needs an EXTERNAL ENVIRONMENT.
- But for the Universe as a whole, there is no external environment.
- Therefore for the Universe as a whole, there is no decoherence.
- Therefore for the Universe as a whole, there is no wave function splitting.
- Therefore for the Universe as a whole, there is no illusion of collapse.
- Therefore MWI cannot explain why the Universe as a whole looks "classical".

The problem with this argument is that you haven't said what it means to say that the Universe as a whole looks "classical". What would it "look" like if it didn't look classical?

This absolutely isn't a problem for MWI, at least I don't see how it is. If the universe as a whole is in a superposition of macroscopically distinguishable states, then when an astronomer takes a look at the sky, his brain will similarly be in a superposition of macroscopically distinguishable states. You can say that we don't observe this, but that remark only makes sense if you can say how someone could possibly observe it.

 

[Demystifier]

But now, suppose we are talking about the quantum mechanics of a SINGLE particle. What can it possibly mean to say that its location is distributed according to the square of the wave function?

This is a probability distribution. Probability makes sense even for a single particle.

If that's not clear enough, then consider analogy with classical probability. Suppose we are talking about classical mechanics of a SINGLE coin. What can it possibly mean to say that the probability of tail and probability of head are both equal to 1/2?

If you are not sure whether probability has anything to do with a single object, then see also
https://www.physicsforums.com/blog.php?b=4330

 

[stevendaryl]

This is a probability distribution. Probability makes sense even for a single particle.

If the particle has a definite position, then what sense does it make to say that it has (for example) a 50/50 chance of being here or there? It can certainly make sense in terms of subjective probability; I might not know whether the particle is here, or there, and I might give both possibilities equal weight. But that's a fact about the state of my knowledge, not about the system.

In Bohmian mechanics, the wave function is assumed to be objective. It's not relative to an observer's knowledge. So it doesn't make any sense to say its square is my subjective notion of probability.

If that's not clear enough, then consider analogy with classical probability. Suppose we are talking about classical mechanics of a SINGLE coin. What can it possibly mean to say that the probability of tail and probability of head are both equal to 1/2?

It makes perfect sense as a subjective measure of probability. As far as I know, heads and tails are equally likely, so I assign them each probability 1/2. But if the dynamics of the coin are deterministic, it doesn't make any sense to say that, objectively, the probability is 1/2.

 

[atyy]

If the particle has a definite position, then what sense does it make to say that it has (for example) a 50/50 chance of being here or there? It can certainly make sense in terms of subjective probability; I might not know whether the particle is here, or there, and I might give both possibilities equal weight. But that's a fact about the state of my knowledge, not about the system.

In Bohmian mechanics, the wave function is assumed to be objective. It's not relative to an observer's knowledge. So it doesn't make any sense to say its square is my subjective notion of probability.
It makes perfect sense as a subjective measure of probability. As far as I know, heads and tails are equally likely, so I assign them each probability 1/2. But if the dynamics of the coin are deterministic, it doesn't make any sense to say that, objectively, the probability is 1/2.

As I understand, Bohmian mechanics puts all the uncertainty into the initial distribution. So for a single particle, it would be the distribution of initial conditions over many trials. So it becomes a problem analogous to that of equilibrium statistical mechanics. Since statistical mechanics is not believed to be fundamentally true, Bohmiann mechanics suggests that quantum mechanics is not fundamentally true either (in that sense BM is not an interpretation).

 

[Demystifier]

It can certainly make sense in terms of subjective probability; I might not know whether the particle is here, or there, and I might give both possibilities equal weight. But that's a fact about the state of my knowledge, not about the system.

Exactly!

In Bohmian mechanics, the wave function is assumed to be objective. It's not relative to an observer's knowledge.

Correct.

So it doesn't make any sense to say its square is my subjective notion of probability.

It doesn't make sense to say that its square IS PROBABILITY BY DEFINITION. But it still makes sense that probability happens to be equal to its square "by accident", due to specific dynamics.

 

[bohm2]

More precisely, I want to point out that even though MWI does not require an external observer, it does require something very similar and equally problematic...
Here is my argument, analogous to the standard argument above:
- To explain why the macro world looks "classical", MWI needs to explain the illusion of collapse.
- To explain the illusion of collapse without observers (and without true collapse), MWI proposes the wave-function splitting.
- To explain the splitting (instead of simply postulating it), MWI needs to refer to decoherence.
- To explain decoherence, it needs an EXTERNAL ENVIRONMENT.
- But for the Universe as a whole, there is no external environment.
- Therefore for the Universe as a whole, there is no decoherence.
- Therefore for the Universe as a whole, there is no wave function splitting.
- Therefore for the Universe as a whole, there is no illusion of collapse.
- Therefore MWI cannot explain why the Universe as a whole looks "classical".

I followed this and it makes sense to me but here is a paper that tries to argue that this may not be a problem for MWI. I didn't fully understand his arguments:

A generic quantum state is a superposition of macroscopically different universes, a variant of Schrödinger’s cat at the cosmic level. Again, the question arises whether the non-observation of such configurations can be understood. In the case of quantum mechanics, decoherence by the irreversible interaction with the environment leads to the emergence of classical behaviour. But what is the situation in cosmology? After all, the Universe as a whole has no environment. The key to an answer is provided by the observation that decoherence is a process in configuration space. This is the high-dimensional (probably infinite-dimensional) space on which the total wave function is defined. It is in this space where the separation between system (relevant degrees of freedom) and environment (irrelevant degrees of freedom) must be made.

Emergence of a classical Universe from quantum gravity and cosmology
http://rsta.royalsocietypublishing.org/content/370/1975/4566.full.pdf

 

[stevendaryl]

It doesn't make sense to say that its square IS PROBABILITY BY DEFINITION. But it still makes sense that probability happens to be equal to its square "by accident", due to specific dynamics.

I just don't understand what it can mean to say that the square of the wave function just happens to be equal to the probability. Subjective probability is not unique. Different people with different information assign different subjective probabilities to events. So the square of the wave function happens to be equal to whose subjective probability?

The other confusion I have about Bohmian mechanics is how the wave function responds to measurements. In von Neumann style quantum mechanics, detecting a particle causes the particle's wave function to collapse to a delta function (or at least, to a highly localized function of position). How does that happen, or does it, in the Bohmian theory?

 

[Demystifier]

I just don't understand what it can mean to say that the square of the wave function just happens to be equal to the probability. Subjective probability is not unique. Different people with different information assign different subjective probabilities to events. So the square of the wave function happens to be equal to whose subjective probability?

Let me use an example. You have two different buckets, the smaller one with surface equal to A1=1/3 m^2, and the bigger one with surface equal to A2=2/3 m^2. These surfaces are objective properties of buckets; they are not probabilities. But now suppose you are trying to catch a single rain drop with these buckets and suppose that you know that one of them caught the drop. But you don't know which one did it. What is the probability P1 (P2) that it was caught by the first (second) bucket? It is easy to see that the probabilities are determined by the buckets surfaces, i.e., that P1=1/3, P2=2/3.

Rougly speaking, you may think of wave function as a "bucket", the "surface" of which is equal to |psi|^2, and the role of which is to hold the "drop".

I am sure that one can make an even better analogy, but hopefully this one is sufficient to answer your question.

The other confusion I have about Bohmian mechanics is how the wave function responds to measurements. In von Neumann style quantum mechanics, detecting a particle causes the particle's wave function to collapse to a delta function (or at least, to a highly localized function of position). How does that happen, or does it, in the Bohmian theory?

Let me use the bucket analogy again. In von Neumann theory, one of the buckets simply disappears, i.e., the configuration consisting of two buckets collapses to a single bucket. In Bohmian theory no bucket disappears, but the drop ends up in one bucket only, so the other bucket becomes irrelevant for the drop.

 

[Demystifier]

Bohm2, with such a view of MWI of Kiefer, I agree that cosmology is not a particular problem for MWI. But my point is that, with such a view, cosmology is also not a particular advantage of MWI.

 

[stevendaryl]

Let me use an example. You have two different buckets, the smaller one with surface equal to A1=1/3 m^2, and the bigger one with surface equal to A2=2/3 m^2. These surfaces are objective properties of buckets; they are not probabilities. But now suppose you are trying to catch a single rain drop with these buckets and suppose that you know that one of them caught the drop. But you don't know which one did it. What is the probability P1 (P2) that it was caught by the first (second) bucket? It is easy to see that the probabilities are determined by the buckets surfaces, i.e., that P1=1/3, P2=2/3.

That makes perfect sense. The objective property, surface area, can, in certain circumstances, be equal to some specific person's subjective probability. That was my point about Bohmian mechanics. If the wave function is objective, and probability is subjective, then they can't be equally except for some specific person's state of knowledge. Whose?

Let me use the bucket analogy again. In von Neumann theory, one of the buckets simply disappears, i.e., the configuration consisting of two buckets collapses to a single bucket. In Bohmian theory no bucket disappears, but the drop ends up in one bucket only, so the other bucket becomes irrelevant for the drop.

But the other bucket ISN'T irrelevant in Bohmian theory. In Bohmian theory, the wavefunction relates to particle positions in two ways:

1. It (purely by chance) happens to be equal to the probability distribution on particles.
2. The wave function provides a "quantum potential" which deterministically affects the motion of particles.

When you detect a particle to be in some specific bucket, rather than another, that changes the subjective probability associated with the particle. But does it also affect the wave function that gives rise to the quantum potential?

 

[Demystifier]

That makes perfect sense. The objective property, surface area, can, in certain circumstances, be equal to some specific person's subjective probability. That was my point about Bohmian mechanics. If the wave function is objective, and probability is subjective, then they can't be equally except for some specific person's state of knowledge. Whose?

Whose knowledge? Of any person who knows the wave function but does not know the position of the particle.

When you detect a particle to be in some specific bucket, rather than another, that changes the subjective probability associated with the particle.

True.

But does it also affect the wave function that gives rise to the quantum potential?

No it doesn't.

 

[stevendaryl]

Whose knowledge? Of any person who knows the wave function but does not know the position of the particle.

But there are gradiations of information between "knowing the position" and "knowing nothing about the particle other than the wave function". For instance, someone may know that the particle is within some region without knowing its exact position.

In answer to my question: "But does it [a measurement] also affect the wave function that gives rise to the quantum potential?"

No it doesn't.

If that's the case, then Bohmian mechanics makes different predictions, in principle, than von Neumann interpretation. The von Neumann interpretation assumes that after a measurement, the wave function is a delta-function (or otherwise very localized), and evolves accordingly. If the Bohmian interpretation holds that a measurement does not affect the wave function, then the predictions for subsequent measurements will be different, because they are using different wave functions.

 

[Demystifier]

But there are gradiations of information between "knowing the position" and "knowing nothing about the particle other than the wave function". For instance, someone may know that the particle is within some region without knowing its exact position.

Yes, in principle it is possible, and in that case the probability for that observer will not be simply |psi|^2.

If that's the case, then Bohmian mechanics makes different predictions, in principle, than von Neumann interpretation. The von Neumann interpretation assumes that after a measurement, the wave function is a delta-function (or otherwise very localized), and evolves accordingly. If the Bohmian interpretation holds that a measurement does not affect the wave function, then the predictions for subsequent measurements will be different, because they are using different wave functions.

Yes, you are right that it is possible in principle. But there are good arguments that it is still not possible in practice:
http://arxiv.org/abs/quant-ph/0308039

Nevertheless, there are people who hope that it might somehow be possible even in practice:
http://arxiv.org/abs/1001.2758