Question regarding the Many-Worlds interpretation

[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

 

[mfb]

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.

Is this backed by some established interpretation, or just a feeling?
I think Bohmian mechanics is exactly the type of ignorance that could help (unlike the early ideas to explain it as ignorance in a classic theory) if you really want ignorance to explain quantum effects.

 

[lugita15]

mfb, if it's not too much trouble, could you respond to Tom's questions in post #254? I think that might clarify much of the confusion.

 

[Jazzdude]

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.

Let's see what we need to construct the density representation.

By definition, an ensemble is a list of states with the probability of finding them. Only one of the states is realized, but we do not know which one. So our fundamental ensemble is the set \{(|\psi_n\rangle,p_n):n\in\mathbb{N}\} where the |\psi_n\rangle are a family of (in general nonorthogonal) states and p_n are the associated probabilities. Now suppose that our Hilbert space is only 2 dimensional, that is we have a basis \{|0\rangle,|1\rangle\}.

A density operator in this Hilbert space is a hermitian 2x2 matrix. Clearly, this matrix encodes much less information than the (possibly very long) list of states in the ensemble. Since there are infinitely many possible realized (mutually non-orthogonal) states in this Hilbert space, there is no way of reducing the information down to a density matrix without throwing some of the information away.

The idea for reduction is that not all of that information is observable when we perform a measurement according to the measurement postulate. Let's pick an orthonormal measurement basis \{|a\rangle,|b\rangle\} and perform a measurement. Depending on which of the ensemble states is realized, we get an outcome |a\rangle with the measurement probability \langle\psi_n|a\rangle\langle a|\psi_n\rangle and the corresponding expression for |b\rangle. Since all single ensemble states are mapped to the same two measurement outcomes, we can collect the results and multiply both probabilities, the measurement outcome probability and the ensemble state probability. That means we have |a\rangle with probability \sum_n p_n \langle\psi_n|a\rangle\langle a|\psi_n\rangle and |b\rangle with probability \sum_n p_n \langle\psi_n|b\rangle\langle b|\psi_n\rangle. Both probabilities sum to 1 if the states are normalized and the ensemble proabilities sum to 1. That means \sum_n p_n \langle\psi_n|a\rangle\langle a|\psi_n\rangle+\sum_n p_n \langle\psi_n|b\rangle\langle b|\psi_n\rangle=1. We can rewrite this as \mathrm{tr}_{a,b}\sum_n|\psi_n\rangle p_n \langle\psi_n| = 1, which works in any basis, not just the indicated measurement basis. Let's call the matrix we trace over \rho. We can also read off the probability expressions, that for a projector P onto any subspace the associated probability of finding a measurement outcome in this subspace is \mathrm{tr}(\rho P). It's straight forward to find the expectation value of any operator A from here to be \langle A \rangle = \mathrm{tr}(\rho A). Other similar results follow.

So we have seen, that \rho encodes all possible measurement outcomes in any possible measurement basis, with potentially a lot less required information than encoded in the original ensemble description. But we had to use assumptions from the measurement postulate for this. Explicitly in your labeling I used 1) when I assumed that the measurement basis was orthogonal and again for the expectation value, where I also used 2). The remaining 3) and 4) were used in the measurement process of the single ensemble states.

Now you may know of a way to derive the same result without making these references. I would be curious to learn how. But unless such a way exists, the use of a density matrix requires the measurement postulate.

Cheers,

Jazz

 

[Jazzdude]

Is this backed by some established interpretation, or just a feeling?
I think Bohmian mechanics is exactly the type of ignorance that could help (unlike the early ideas to explain it as ignorance in a classic theory) if you really want ignorance to explain quantum effects.

Just to make sure I am understood correctly, I don't think ignorance is the way to any "classical" explanation of quantum theory. But I think we can at least get rid of the randomness. My statement is also founded on much more than a feeling. I am specifically referring to http://arxiv.org/abs/1205.0293, where the ignorance lies in the parts of the universe any kind of observer cannot interact with because of the locality of interactions.

Cheers,

Jazz

 

[mfb]

@lugita15: sure.

1) MWI is talking about branches and relies on decoherence to identify them, but is not able to count them or to derive a corresponding measure

It is a pointless attempt to count them. It is as meaningful as (correctly!) counting "I will win in the lottery XOR I will not win in the lottery" as 2 different results. What does that number of 2 tell us?
It has been shown that there is just one consistent, context-independent measure. What else do you want for a derivation?

2) My simple question regarding the "probability being in a certain branch" which I can identify via a result string seems to become meaningless

Without probabilities, there are no probabilities, indeed.

3) I still have the feeling that my concerns regarding the "missing link" between the experimentally inaccesable top-down perspective of the full Hilbert space with all its branches and the accessable bottom-up approach restricted to the branch I am observing right now have not been understood

I think that is right.

4) We have the above mentioned statistical frequencies, but I learn that MWI does not provide the corresponding probabilities - that there are no probabilities at all

MWI does not need probabilities.

5) It is often claimed that the Born rule can derived, but what does it mean if there are no probabilities?

There is no need for the Born rule.
If you want to add something like a "probability based on ignorance" (the interpretation itself does not need this at all), Gleason's theorem tells you you have no other choice than the Born rule.

 

[stevendaryl]

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?

I think we're arguing at cross-purposes here. According to MWI, all mixed states are "improper" in the sense that they AREN'T simply due to ignorance. According to a "collapse" interpretation, if a wave function has already collapsed, but you haven't checked to see what state it collapsed to, then the system is in a "proper" mixed state. There is no practical way to decide between these two interpretations. The ignorance interpretation is ALWAYS wrong, according to MWI.

 

[stevendaryl]

MWI does not need probabilities.

There is no need for the Born rule.

I'm not sure what you mean by "need" here. The fact is that we find empirically that the Born rule accurately predicts relative frequencies for repeated, independent measurements. This is an empirical fact that I would think requires explanation. If you don't have the Born rule, then what kind of empirical support can there be for quantum mechanics? Well, I guess there is some: the eigenvalues for observables, maybe.

 

[mfb]

I'm not sure what you mean by "need" here. The fact is that we find empirically that the Born rule accurately predicts relative frequencies for repeated, independent measurements. This is an empirical fact that I would think requires explanation. If you don't have the Born rule, then what kind of empirical support can there be for quantum mechanics? Well, I guess there is some: the eigenvalues for observables, maybe.

You can do hypothesis testing, and the "right" hypotheses survive everywhere where the measurement results look like they follow the Born rule. Apparently we are in a branch where this was true in the past (within some variation).
Asking "why" is as meaningful as asking why we are on the 3rd planet around a specific main-sequence star and not on some other habitable planet, or asking why you are stevendaryl and not someone else.

 

[stevendaryl]

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

Yes, you're right.

 

[stevendaryl]

You can do hypothesis testing, and the "right" hypotheses survive everywhere where the measurement results look like they follow the Born rule. Apparently we are in a branch where this was true in the past (within some variation).
Asking "why" is as meaningful as asking why we are on the 3rd planet around a specific main-sequence star and not on some other habitable planet, or asking why you are stevendaryl and not someone else.

That's pretty unsatisfying. We could just as well throw out all of quantum mechanics, and say that every measurement returns some real number--which one is completely undetermined. We just happen to live in a possible world in which those real numbers seem to be predicted by the Born rule.

 

[mfb]

That's pretty unsatisfying. We could just as well throw out all of quantum mechanics, and say that every measurement returns some real number--which one is completely undetermined.

No, that would be completely different.

We just happen to live in a possible world in which those real numbers seem to be predicted by the Born rule.

We just happen to be in a likely world in probabilistic interpretations. That is exactly the same statement, as I don't care about branches with a small measure. They are just not interesting.

 

[stevendaryl]

No, that would be completely different.
We just happen to be in a likely world in probabilistic interpretations. That is exactly the same statement, as I don't care about branches with a small measure. They are just not interesting.

They are interesting if we happen to find ourselves in one of them.

 

[lugita15]

You can do hypothesis testing, and the "right" hypotheses survive everywhere where the measurement results look like they follow the Born rule. Apparently we are in a branch where this was true in the past (within some variation).

But just because it was true (with some variation) in the past, what causes it to continue to be likely to be true (with some variation) in the future? Or do you think we aren't justified in our confidence that the Born rule will continue to hold (with some variation) in the future?

 

[mfb]

They are interesting if we happen to find ourselves in one of them.

What does "if" mean? We do not.

But just because it was true (with some variation) in the past, what causes it to continue to be likely to be true (with some variation) in the future?

See my post about hypothesis testing.

 

[Quantumental]

mfb: are you claiming that you don't actually think that the multiverse in mwi obeys Born Rule, but that we happen to be on a branch where our past seems to confirm the Born Rule, but in reality this is just a illusion?