Question regarding the Many-Worlds interpretation

[stevendaryl]

No, that was a criticism of Hobson's approach. It's also relevant in the context of this thread because people keep attempting to appeal to Gleason's theorem in order to recover the Born probabilities in the MWI, but this fails for a similar reason, namely that proper and improper mixtures are not the same thing.

In a way, getting a proper from an improper mixture is what the measurement problem is all about. Collapse interpretations solve it by fiat: someone snips their fingers, and out comes the desired proper mixture. Many people find this dissatisfying, and with good reason. But then proposing a solution that ends up depending on the very same sleight of hand is no progress at all.

That's absolutely not true. The progress is that you do away with a nonphysical collapse hypothesis. I agree that there are conceptual difficulties with MWI, but what the use of "improper" mixtures shows is that there is really no evidence that any particular measurement collapses the wave function. So there is no evidence that macroscopic objects (even measurement devices and observers) can't be treated quantum mechanically.

I certainly agree that there is still mystery involved in the interpretation of quantum mechanics, measurements and probabilities and all that. But I don't consider the use of mixed states to be still a mystery.

 

[bhobba]

Well, you didn't really give me something to understand, you merely gave a reference.

I beg to differ. As I said in that thread there is a key assumption in your argument:

Basically you are assuming it possesses those properties SIMULTANEOUSLY globally - which the situation doesn't require. The complete analysis in given in the reference and hinges on the concept of framework used in that interpretation - basically one is free to choose frameworks as long as they are consistent - and a framework exists where it has both those properties locally - but not globally.

You even said it in that thread - assuming an epistemic interpretation that you left out in what you posted here.

Thanks
Bill

 

[S.Daedalus]

I think you miss the point of the use of decoherence in saying that proper and improper mixed states are observationally indistinguishable. There is a mathematical difference between the two, because the improper mixed state contains "interference terms" that are absent in the proper mixed state. But in order to observe these interference effects, you have to perform a measurement that has different outcomes (or different probabilities of outcomes) if the interference terms are present.

No. That's the point of Alice and Bob's story: they only make the local measurements, which don't detect interference terms and look just like the states actually were in a mixed state. But upon comparing their measurement records, they will notice that they always got the same results---i.e. that they are perfectly correlated---while the prediction on the belief of proper mixedness is that they would not observe any correlation at all.

That's not to say that for all practical purposes, you can consider decohered states to be effectively mixed, since as you say the relevant measurements are pretty much impossible to perform; but this is not just a FAPP question: if you want to appeal to Gleason in order to get probabilities in the MWI, then you require that the identification can be made exactly; in any other case, Gleason just doesn't talk about probability in the MWI.

I don't think your mathematical demonstration is correct. I don't think you combine mixed states that way.

Of course you do.

 

[S.Daedalus]

I beg to differ. As I said in that thread there is a key assumption in your argument:

Even that quote just directs me to the reference. I repeat: what is it that prohibits Alice and Bob from just writing down their results and comparing them?

 

[stevendaryl]

No. That's the point of Alice and Bob's story: they only make the local measurements, which don't detect interference terms and look just like the states actually were in a mixed state. But upon comparing their measurement records, they will notice that they always got the same results---i.e. that they are perfectly correlated---while the prediction on the belief of proper mixedness is that they would not observe any correlation at all.

That's not to say that for all practical purposes, you can consider decohered states to be effectively mixed, since as you say the relevant measurements are pretty much impossible to perform; but this is not just a FAPP question: if you want to appeal to Gleason in order to get probabilities in the MWI, then you require that the identification can be made exactly; in any other case, Gleason just doesn't talk about probability in the MWI.

Of course you do.

The article specifically says "If A and B are two distinct and independent systems then $\rho_{AB}=\rho_{A}\otimes\rho_{B}$ which is a product state."

They don't elaborate on what "independent systems" means, but in fact, that rule is only valid if you ignore entanglement. It's absolutely incorrect in the case we're talking about.

 

[S.Daedalus]

They don't elaborate on what "independent systems" means, but in fact, that rule is only valid if you ignore entanglement. It's absolutely incorrect in the case we're talking about.

Yes, that's the point I'm trying to get across: attaching an ignorance interpretation to their local states is exactly ignoring the entanglement between them, and erroneously stipulating that they can treat the system as being in a definite, but unknown, state---which is simply not the case if you have entanglement. It's exactly the error made in the argument that proper and improper mixtures can be treated the same!

 

[kith]

Seems that I am the only one where "the penny did not drop".

I'm in the same boat. I find it difficult to get to the core of the issue and even to ask the right questions.

What I get is the assertion that probabilities either don't make sense (but I don't get how replacing them by amplitudes is sufficient) or if we insist on having them, they follow from Gleason's theorem (but I don't get if and what additional assumptions are made in this case).

 

[kith]

Locally, that's true, but once Alice and Bob get together and compare their measurement results, or a measurement is carried out on both parts of the system, you get results that falsify the idea that the parts of the system are in some definite state, and we just don't know which---correlations that we can't account for with such a model in the first case, and interference results in the second. These are perfectly valid observations, so I don't see how it's true that the two states are 'observationally equivalent'.

I think if you want to argue that improper and proper mixtures are the same in the framework of the MWI, you have to look at it from the other viewpoint: every mixture is improper. For seemingly proper mixtures, it is just to complicated to track the correlations between the system and the large and uncontrollable environment .

 

[S.Daedalus]

I think if you want to argue that improper and proper mixtures are the same in the framework of the MWI, you have to look at it from the other viewpoint: every mixture is improper. For seemingly proper mixtures, it is just to complicated to track the correlations between the system and the large and uncontrollable environment .

Yes, I believe that's a valid point of view. But of course, the issue of proper mixtures only arose because of the attempt to appeal to Gleason's theorem in order to get the Born rule in the MWI (I know, we're getting deeper and deeper down the rabbit hole here), for which (I have argued) you need proper mixtures (i.e. states that really are in a given subspace---otherwise, the measure on subspaces Gleason gives just has nothing to say).

 

[kith]

Yes, I believe that's a valid point of view. But of course, the issue of proper mixtures only arose because of the attempt to appeal to Gleason's theorem in order to get the Born rule in the MWI (I know, we're getting deeper and deeper down the rabbit hole here), for which (I have argued) you need proper mixtures (i.e. states that really are in a given subspace---otherwise, the measure on subspaces Gleason gives just has nothing to say).

I think the main point of mfb was that we shouldn't talk about probabilities at all in the MWI. In this case, it is meaningless to talk about the Born rule or it's derivation. If we insist to talk about probabilities we have to make additional assumptions (?) which allow us to derive the Born rule via Gleason's theorem.

I am much more interested in discussig if and how we get and verify predictions from QM using the MWI without talking about probabilities. I think it is crucial to understand this viewpoint first and then look at how it is connected to the probabilistic picture.

 

[bhobba]

Even that quote just directs me to the reference. I repeat: what is it that prohibits Alice and Bob from just writing down their results and comparing them?

Nothing - but how does that imply the proper mixed state you wrote down?

Basically your supposed proper mixed state is wrong. The proper mixed state depends on who does the observing and it only contains two terms.

Thanks
Bill

 

[stevendaryl]

Yes, that's the point I'm trying to get across: attaching an ignorance interpretation to their local states is exactly ignoring the entanglement between them, and erroneously stipulating that they can treat the system as being in a definite, but unknown, state---which is simply not the case if you have entanglement. It's exactly the error made in the argument that proper and improper mixtures can be treated the same!

I think you're wrong. Using $\rho_{AB} = \rho_A \otimes \rho_B$ is the same sort of assumption as using $P(X \wedge Y) = P(X) \times P(Y)$ to compute joint probabilities. It's correct under the assumption that $X$ and $Y$ are independent random variables, but not in general.

It's not the fact that they are "improper" mixtures that prevents you from combining density matrices that way; it's the fact that you are treating systems as independent that have a common history. Probabilities due to ignorance are NOT independent for systems that share a common history.

 

[S.Daedalus]

I think the main point of mfb was that we shouldn't talk about probabilities at all in the MWI. In this case, it is meaningless to talk about the Born rule or it's derivation. If we insist to talk about probabilities we have to make additional assumptions (?) which allow us to derive the Born rule via Gleason's theorem.

I am much more interested in discussig if and how we get and verify predictions from QM using the MWI without talking about probabilities. I think it is crucial to understand this viewpoint first and then look at how it is connected to the probabilistic picture.

I think this is what mfb meant with his 'hypothesis testing', but I'm afraid the point is lost on me---as I said, I can certainly form the hypothesis that relative frequencies are asymptotically distributed according to Born, test it, and become convinced it's right; I can also form the hypothesis that there are green apples, test it, and become convinced of it. Both merely means that the MWI is consistent with that hypothesis, but in neither case does it then follow from it, so it only posits something which is not explained within the MWI.

Nothing - but how does that imply the proper mixed state you wrote down?

The proper mixed state is implied by Alice and Bob's believe that they can attach an ignorance interpretation to their local states. The fact that they will observe perfect correlations once they compare their measurements falsifies this belief.

I think you're wrong. Using $\rho_{AB} = \rho_A \otimes \rho_B$ is the same sort of assumption as using $P(X \wedge Y) = P(X) \times P(Y)$ to compute joint probabilities. It's correct under the assumption that $X$ and $Y$ are independent random variables, but not in general.

It's not the fact that they are "improper" mixtures that prevents you from combining density matrices that way; it's the fact that you are treating systems as independent that have a common history. Probabilities due to ignorance are NOT independent for systems that share a common history.

No, it's the fact that Alice and Bob apply an ignorance interpretation to their states that makes me write down the state in that way---that's just what an ignorance interpretation means: they belief that their state is actually in either of the states $|0\rangle$ or $|1\rangle$; from this alone, it follows that they must hold the global state to be the mixture I wrote down.

I'm really not sure where the disconnect lies. To me, 'improper mixtures aren't proper mixtures' is nothing but a 40-some year old cut-and-dried textbook result, preached to every student of QM who finds out that if he traces out the measured system, what's left looks like a statistical mixture of distinct apparatus states, and then thinks to have solved the measurement problem. The argument is given in many different ways by different authors---first, perhaps, by d'Espagnat (though it was probably known earlier), or in the textbook by Hughes, and so on. Quick googling brought up this by Mittelstaedt, and many other statements like it. I don't really think there's any controversy around this issue.

 

[bhobba]

It's not the fact that they are "improper" mixtures that prevents you from combining density matrices that way; it's the fact that you are treating systems as independent that have a common history. Probabilities due to ignorance are NOT independent for systems that share a common history.

I also think he is misunderstanding what decoherence says. Whoever does the observation does the decoherence - it is at that point it becomes an improper mixed state and it is exactly the same as the proper one - mathematically that is.

Thanks
Bill

 

[stevendaryl]

No, it's the fact that Alice and Bob apply an ignorance interpretation to their states that makes me write down the state in that way.

But what you wrote is NOT correct, under the ignorance interpretation of mixed states.

As it says in the article:

$\rho_{A} = tr_A \rho_{AB}$
$\rho_{B} = tr_B \rho_{AB}$

From these two definitions, it does not follow that
$\rho_{AB} = \rho_A \otimes \rho_B$
except in special cases.

You can't, in general, combine subystem density matrices that way, regardless of whether the mixtures arose from "ignorance".

 

[Jazzdude]

Decoherence adherents, unless they are being disingenuous like the paper cited before, do not claim it solves the measurement problem. What they claim is its non issue because its observationally the same as a proper mixture and gives the appearance of wavefunction collapse.

I understand very well what you are claiming. But even that "proper and improper mixtures are observationally indistinguishable" requires the measurement postulate, or something equivalent. Because otherwise ensembles cannot be expressed in terms of density matrices.

Prove me wrong by deriving the density matrix formalism for ensembles without referring in any way to any form of the measurement postulate.

Cheers,

Jazz

 

[Jazzdude]

I never claimed that - in fact I don't even know what you mean by that.

Then please let's discuss what you don't understand. Just saying that it's not true won't get us any farther.

To be clear my claim is that decoherence solves the preferred basis problem as stated on page 113 of the reference I gave before by Schlosshauer. He gives 3 issues the measurement problem must solve:

1. The preferred basis problem
2. The problem of non observability of interference
3. The problem of why we have outcomes at all.

The statement he makes is my position:
'it is reasonable to conclude decoherence is capable of solving the first two, whereas the third problem is linked to matters of interpretation'

Yes, I know Schlosshauer's publications. And what I'm saying is that 1) is very questionable for reasons I gave earlier, 2) is a valid conclusion unless we gain a possibly more complete understanding of observation that invalidates is and 3) is absolutely out of reach of the decoherence framework because it doesn't even look at single outcomes.

But you add 4) that decoherence gives a behavioristic explanation for the world we see by stating that ensembles and reduced states are indistinguishable by experiments. That takes what we know experimentally about observation (namely that it only results in probabilities and we can therefore encode ensembles more densely by putting them in a form compatible with the measurement rule) to explain observation. That is circular.

Cheers,

Jazz

 

[bhobba]

Prove me wrong by deriving the density matrix formalism for ensembles without referring in any way to any form of the measurement postulate.

How can I prove you wrong for something I am not claiming. The claim is - it don't matter - not that the measurement postulate isn't still there.

Even with proper mixed states you have the measurement postulate - its still there - its just that it conforms to our intuition of having the value prior to observation and you don't have this collapse because its revealing what's already there. It simply means you can interpret in a more reasonable manner that APPEARS to solve the measurement problem.

Thanks
Bill

 

[Jazzdude]

How can I prove you wrong for something I am not claiming. The claim is - it don't matter - not that the measurement postulate isn't still there.
Even with proper mixed states you have the measurement postulate - its still there - its just that it conforms to our intuition of having the value prior to observation and you don't have this collapse because its revealing what's already there. It simply means you can interpret in a more reasonable manner that APPEARS to solve the measurement problem.

So you're saying, in a setting where you have decoherence AND the measurement postulate is valid, you get something that explains how a decohered system looks like a mixture, and that explains all practical aspects of measurement. Why use decoherence at all? This follows from only the measurement postulate already.

Cheers,

Jazz

 

[bhobba]

But you add 4) that decoherence gives a behavioristic explanation for the world we see by stating that ensembles and reduced states are indistinguishable by experiments.

Sigh.

I say it gives the APPEARANCE that the world behaves in way that conforms more readily to our intuition and hence gives the APPEARANCE of solving the problem.

Thanks
Bill

 

[stevendaryl]

I understand very well what you are claiming. But even that "proper and improper mixtures are observationally indistinguishable" requires the measurement postulate, or something equivalent. Because otherwise ensembles cannot be expressed in terms of density matrices.

Prove me wrong by deriving the density matrix formalism for ensembles without referring in any way to any form of the measurement postulate.

Which postulate are you calling the "measurement postulate"? There are basically four assumptions in the standard, collapse interpretation of quantum mechanics:

1. Observables correspond to Hermitian operators.
2. When you measure an observable, you get an eigenvalue of the corresponding operator.
3. The probability of getting eigenvalue $o$ is the absolute square of the wave function projected onto the eigenstate corresponding to $o$.
4. Afterwards, the system is in the eigenstate corresponding to the eigenvalue obtained.

The use of density matrices certainly doesn't require assumption 4. It's not clear to me that it actually requires assumption 2, either.

 

[bhobba]

So you're saying, in a setting where you have decoherence AND the measurement postulate is valid, you get something that explains how a decohered system looks like a mixture, and that explains all practical aspects of measurement. Why use decoherence at all? This follows from only the measurement postulate already.

I am in Australia where its 1.00 am and I stayed up to watch a big election we just had - its been decided so I will heading off to bed soon.

The reason you use decoherence is you can think the world is revealed by measurement. If it was an actual mixed state that would be the case.

I am 100% sure you know there are a number of interpretations that use decoherence in their foundations and they all do it for various reasons. I adhere to the decoherence ensemble interpretation as mentioned in the link previously. The advantage is it allows me to think it has the property prior to observation. Its a variation on the usual ensemble interpretation. The high priest of that interpretation Ballentine believes its a crock of rubbish because its not required. It purely a matter of what appeals.

Anyway off to bed.

Thanks
Bill

 

[Jazzdude]

I adhere to the decoherence ensemble interpretation as mentioned in the link previously. The advantage is it allows me to think it has the property prior to observation. Its a variation on the usual ensemble interpretation. The high priest of that interpretation Ballentine believes its a crock of rubbish because its not required. It purely a matter of what appeals.

You used the same argument in the context of an MWI discussion however. Any reference to the measurement postulate is not allowed here to draw any conclusions. That's the context I use for pointing out that your argument is not valid.

Good night,

Jazz

 

[stevendaryl]

I am in Australia where its 1.00 am and I stayed up to watch a big election we just had - its been decided so I will heading off to bed soon.

This is off-topic, but how did it go? The predictions were that Labor was going to lose big.

 

[S.Daedalus]

But what you wrote is NOT correct, under the ignorance interpretation of mixed states.

As it says in the article:

$\rho_{A} = tr_A \rho_{AB}$
$\rho_{B} = tr_B \rho_{AB}$

From these two definitions, it does not follow that
$\rho_{AB} = \rho_A \otimes \rho_B$
except in special cases.

You can't, in general, combine subystem density matrices that way, regardless of whether the mixtures arose from "ignorance".

Yes, you can't in general do that because doing so means ignoring the entanglement between the two states. But that is exactly what it means to attach an ignorance interpretation to their local states for Alice and Bob, as it means taking their respective systems to be in either the state $|0\rangle$ or $|1\rangle$, while just not knowing which one applies. But if this is the correct description, there can be no entanglement.

In other words: if you're saying that Alice and Bob can validly apply an ignorance interpretation, you are saying that their combined state is in the mixture I wrote down. Consider coming at this from the other way, equiprobably creating the states $|0\rangle$ and $|1\rangle$ locally. Presumably, you'd then agree with me that the global state is as I wrote it down, no? But then an 'ignorance interpretation' just means that you can consider a state as being just so constituted; the two say exactly the same thing. That Alice and Bob can consider their local states to be either $|0\rangle$ or $|1\rangle$ is quite simply only true if the state $\rho_{AB}$ as I have given it obtains.

 

[stevendaryl]

Yes, you can't in general do that because doing so means ignoring the entanglement between the two states.

It depends on what you mean by "entanglement". Do you consider classical cases where $P(A \wedge B) \neq P(A) \times P(B)$ to be "entanglement"?

But that is exactly what it means to attach an ignorance interpretation to their local states for Alice and Bob, as it means taking their respective systems to be in either the state $|0\rangle$ or $|1\rangle$, while just not knowing which one applies. But if this is the correct description, there can be no entanglement.

I don't agree with that. Bob believes that his subsystem is in state 0 or state 1, and he doesn't know which. Alice believes that her subsystem is in state 0 or state 1, but she doesn't know which. It doesn't follow that for the composite system, that all four states:

$|00>, |01>, |10>, |11>$

are possible.

 

[tom.stoer]

I think it is all written in the thread now. It would be pointless to repeat it.

perhaps it is ... hidden in 300 posts ...

 

[S.Daedalus]

I don't agree with that. Bob believes that his subsystem is in state 0 or state 1, and he doesn't know which. Alice believes that her subsystem is in state 0 or state 1, but she doesn't know which. It doesn't follow that for the composite system, that all four states:

$|00>, |01>, |10>, |11>$

are possible.

This is exactly what follows. Honestly, I don't think this is going to get any more productive; I'll just leave you with what Timpson says on the issue (I'm not entirely certain this link will stay valid; it's on page 253 of his book Quantum Information Theory and the Foundations of Quantum Mechanics, starting with 'When $|\Psi\rangle_{12}$ is an entangled state...' and ending with 'Thus reduced states of an entangled system cannot be given an ignorance interpretation'). The book is from 2013, so I will simply accept it as accurately reflecting current knowledge.

 

[stevendaryl]

In other words: if you're saying that Alice and Bob can validly apply an ignorance interpretation, you are saying that their combined state is in the mixture I wrote down. Consider coming at this from the other way, equiprobably creating the states $|0\rangle$ and $|1\rangle$ locally. Presumably, you'd then agree with me that the global state is as I wrote it down, no? But then an 'ignorance interpretation' just means that you can consider a state as being just so constituted; the two say exactly the same thing. That Alice and Bob can consider their local states to be either $|0\rangle$ or $|1\rangle$ is quite simply only true if the state $\rho_{AB}$ as I have given it obtains.

I don't agree. Let's take a classical example: I take two identical envelopes, and put $10 in one, and $20 in the other. I shuffle the envelopes and hand one to Alice and one to Bob. Then Alice would describe the state of her envelope as "It holds $10 with probability 1/2, and it holds $20, with probability 1/2". Bob would describe the state of his envelope the same way. But if they are trying to describe the composite state of the two envelopes, they can't just take the "product". In both cases, a mixed state arises from ignorance, but the two uncertainties are not independent. There is zero probability that both Alice and Bob have $20.

 

[stevendaryl]

This is exactly what follows.

That's not true about CLASSICAL probabilities, which are all due to ignorance (or can be interpreted that way).

 

[S.Daedalus]

That's not true about CLASSICAL probabilities, which are all due to ignorance (or can be interpreted that way).

Read the Timpson quote I provided: "The true state of the $N$-party system would then simply be the tensor product of each of these true states for subsystems, or a convex combination of these if there were further correlations between them (emphasis mine)." The part in italics takes care of the possibility of there being classical correlations present; while you're right that you could in principle thus account for the simple correlation measurements I've described, you could not violate a Bell inequality, for example, as there would still be no entanglement present.

 

[stevendaryl]

Read the Timpson quote I provided: "The true state of the $N$-party system would then simply be the tensor product of each of these true states for subsystems, or a convex combination of these if there were further correlations between them (emphasis mine)." The part in italics takes care of the possibility of there being classical correlations present; while you're right that you could in principle thus account for the simple correlation measurements I've described, you could not violate a Bell inequality, for example, as there would still be no entanglement present.

Definitely, the Bell inequality gives a limit to how far you can push an "ignorance" explanation.

 

[S.Daedalus]

Definitely, the Bell inequality gives a limit to how far you can push an "ignorance" explanation.

But then, that just means that there is no ignorance interpretation---or are you saying that there's one as long as you don't make Bell tests?

 

[mfb]

perhaps it is ... hidden in 300 posts ...

And you think the next 300 will be better?

But then, that just means that there is no ignorance interpretation---or are you saying that there's one as long as you don't make Bell tests?

Ignorance interpretations need something more than simple ignorance. The de-Brogle-Bohm is an example where probabilities just arise from our ignorance (of the particle positions), but it is nonlocal.

 

[Jazzdude]

Definitely, the Bell inequality gives a limit to how far you can push an "ignorance" explanation.

Bell hardly restricts ignorance, it merely shows that you cannot recover local realism using ignorance. In fact, I believe that ignorance of some sorts is what causes random measurement outcomes. And I don't mean something like Bohmian Mechanics.

Cheers,

Jazz