A In what sense does MWI fail to predict the Born Rule?

  • Thread starter Thread starter Derek P
  • Start date Start date
  • Tags Tags
    Born rule Mwi
  • #61
Okay fair enough, thanks for the discussion everybody, I think I need to read a bit more and return with more specific questions.
 
Physics news on Phys.org
  • #62
DarMM said:
Okay fair enough, thanks for the discussion everybody, I think I need to read a bit more and return with more specific questions.

Look into Gleason.

Thanks
Bill
 
  • Like
Likes Derek P
  • #63
bhobba said:
Look into Gleason.

Thanks
Bill
I know the proof of Gleason's theorem, but it has never genuinely helped me comprehend the MWI arguments as it comes from a very different direction, Wallace argues that his proof is a separate line of argumentation to Gleason. I think I need to read Wallace's book in full perhaps.
 
  • #64
stevendaryl said:
In the "typical" history, the relative frequency for heads is 0.5. In an atypical history, maybe the relative frequency for heads is 0.7.
But these are histories that can be many times generated and compared, so that one can tell what is typical. For ''worlds'', this is impossible - so calling a world typical or atypical if only one is known is kind of weird.
stevendaryl said:
In the many-worlds interpretation, there will be many versions of my notebook.
There are all imaginable versions of your notebook, and calling some of them typical is prejudice based on the few notebooks in the single world you have access to.
 
Last edited:
  • Like
Likes jbergman
  • #65
stevendaryl said:
There will be "worlds" where those expected frequencies occur, and worlds where they don't. The first type of world will be happy with quantum mechanics, and the other type will not be happy with it.
So you call the ones conforming to the laws observed in our world typical, because they are similar to our worlds, and worlds that do not, atypical. I believe the correct word for the notion you have in mind is not ''typical'' but ''like the one we actually observe''. Thus it is tautological that in these worlds we observe Born's rule.
DarMM said:
I would imagine the answer is because it (approximately) holds in "most" worlds.
The problem is that ''most'' has no meaning unless you have a means to actually estimate the numbers. With access to only a single world, it is as if we had been shown only the result a single throw of a die (say, 2), without recourse to the mechanism of generating it (involving symmetry and labels 1,...,6), and have to deduce from that the probability law for casting dice.
 
  • #66
Derek P said:
in what sense does MWI fail to predict the Born Rule?
The real question is: In what sense does MWI predict Born's rule? I cannot see any coherent argument.
 
  • Like
Likes jbergman and Stephen Tashi
  • #67
A. Neumaier said:
So you call the ones conforming to the laws observed in our world typical, because they are similar to our worlds, and worlds that do not, atypical.

No, that's not what I was saying. A world is typical or not relative to a proposed law of physics depending on whether the relative frequencies in that world correspond to the probabilities derived from that theory.
 
  • #68
stevendaryl said:
No, that's not what I was saying. A world is typical or not relative to a proposed law of physics depending on whether the relative frequencies in that world correspond to the probabilities derived from that theory.
Yes, this is exactly what I was paraphrasing.

You know exactly one world and what is typical there. Based on this you propose a law of which you know that it holds in this particular world to some approximation. Then you postulate that among all other worlds, those are typical that behave like the single example you know of. It is no surprise that this way of proceeding predicts, no matter with which world you start, that this world satisfies the laws you started with.

If the world under consideration permitted only experiments where the result is always 0.707, this would be the observed law of physics. Now you might propose for arbitrary worlds the law of physics that the result is a constant ##p\in [0,1]##. Then a world would be called typical if ##p\approx 0.707##. To make you theory predictive in the MWI sense, you just need to postulate in addition that the worlds are distributed such that ##p^2## is uniformly distributed in [0,1] and that our world is randomly drawn from it. Now - abrakadabra - you can find an easy derivation of the law of physics in our particular world by doing elementary statistics:
The given world represents the mean value of all possible worlds. By improving the statistics of the proposed collection of worlds, i.e., by requiring that these worlds are even more like the given world, one can make the probability of finding the given world overwhelmingly large.

Of course ''possible'' and ''probable'' is only according to your prejudiced assumptions about what constitutes a possible world. Nothing at all depends on any of these worlds actually existing, or on the true existence or nonexistence of many other worlds that are mostly completely different from the given world.

Thus you assumed what you wanted to derive/explain, just disguised in a lot of mystery about alternative worlds.
 
  • Like
Likes PeterDonis
  • #69
I suppose I have one specific question.

We can roughly divide C*-algebras into a few classes based on a two axes of distinction. The first being commutative and non-commutative. The second being the factor type:
$$I_{n},\quad I_{\infty},\quad II_{n},\quad II_{\infty},\quad III_{\lambda}$$

Discrete Classical Probability theory is commutative ##I_{n}##, qubit systems are non-commutative ##I_{n}##, normal non-relativistic QM is Type ##I_{\infty}##. Type ##II## algebras, depending on if commutative or non-commutative, are classical or quantum statistical mechanics (in the thermodynamic limit).
Non-Commutative Type ##III_{1}## has been shown by Fredenhagen and Longo to be the algebra involved in QFT, making QFT in a sense the most general probability theory possible.

Given that Many-Worlds keeps the same formalism as Quantum Mechanics, I was wondering if in the literature there is a discussion of the reasons the theory is isomorphic to a probability theory (and in QFT's case, the unique most general one) given that it is deterministic? It seems odd to me that in a theory that is about the deterministic evolution of the universe, with probability only arising from local subjective viewpoints, the deterministic evolution would have the mathematical form of a probability theory. Is this discussed anywhere? (I've looked, but people here might have better knowledge)
 
  • #70
Derek P said:
I think it is a very awkward attempt to reinstate the idea of probability as a real property of the system. I don't know why anyone would want to do this as MWI seems to predict observation frequencies perfectly well without defining instrinsic probability or whatever you want to call it. But does it? This of course is why I started this thread

Without assuming the concept of probability, how would you define being successful at predicting observational frequencies?

If we use the concept of probability, there are familiar ways to define what it means to be successful at predicting observational frequencies - namely that the prediction method predicts a frequency that has a high probability of being the actual probability. However, what definition can we make without the concept of probability?

There have been attempts to found probability theory on actual frequencies, such as the "collectives" of Richard von Mises. However, I'm not aware of any that meet the modern standards of mathematical rigor.

There is the fundamental problem that "frequency" is a concept (initially) defined for finite numbers of events. For infinite collections of events, frequency must be defined as a limit. To define that limit, some way of considering only a finite number of events from the infinite collection must be specified.

Taking pains to speak only of frequencies, how do we decide if MWI predicts the Born Rule? It has to be something like "On the the most frequent branches ( i.e. the most frequent "wolds") , the frequency of events observed in a repeated experiment is approximately the frequency given by the Born Rule." The delicate part of that argument is how to define what finite sets of branches are used in computing the frequency of branches.

There is a non-circular and non-trivial aspect to the above argument. It is not self-evident that there is a single set of weights than can be used in defining how we pick finite sets of branches to use in defining their frequency that also works to produce the frequencies of events observed in experiments within the frequent branches. If such a set of weights exists, how do we know it is unique? If the weights exist and are unique, we still have to show they correspond to the numbers (for probabilities) given by the Born Rule.

The above type of argument using only frequencies can be disparaged as "branch counting". However, without taking the notion of probability as fundamental, I don't see any alternative approach.
 
  • #71
DarMM said:
It seems odd to me that in a theory that is about the deterministic evolution of the universe, with probability only arising from local subjective viewpoints, the deterministic evolution would have the mathematical form of a probability theory.
Isn't this just the result of classifying ##C^*## algebra factors? Or do you ask specifically why MWI assumes ##I_\infty##?
 
  • #72
A. Neumaier said:
Yes, this is exactly what I was paraphrasing..

Well, your "paraphrase" substituted a completely different meaning. It isn't what I meant. It isn't what I said. I didn't say that because that's not what I meant.
 
  • #73
stevendaryl said:
Well, your "paraphrase" substituted a completely different meaning. It isn't what I meant. It isn't what I said. I didn't say that because that's not what I meant.
But that's how your arguments sound to me. The words you are using convey nothing significant, since you simply postulate a lot of unseen worlds and a probabilistic law for them to justify with a weak argument only what you know already about the world you see.
 
  • #74
A. Neumaier said:
But that's how your arguments sound to me. The words you are using convey nothing significant, since you simply postulate a lot of unseen worlds and a probabilistic law for them to justify with a weak argument only what you know already about the world you see.

No, that wasn't what I was doing, at all. When you're confused about what someone means, ask questions. There is no point in responding to you if you just ignore what I say and pretend I said something else.
 
Last edited:
  • #75
Stephen Tashi said:
Taking pains to speak only of frequencies, how do we decide if MWI predicts the Born Rule? It has to be something like "On the the most frequent branches ( i.e. the most frequent "wolds") , the frequency of events observed in a repeated experiment is approximately the frequency given by the Born Rule."

Probability in quantum mechanics is definitely not a matter of "counting branches". You can have just two branches, and that doesn't mean that they have equal weight.

If you prepare an electron in a state that is spin-up in the z-direction, and then measure its spin along an axis that is 5o away from the z-axis, you'll get spin up with probability .998 and spin-down with probability .002. In a Many-Worlds type interpretation, there are now two branches, but they're not equally weighted.
 
Last edited:
  • #76
A. Neumaier said:
Isn't this just the result of classifying ##C^*## algebra factors? Or do you ask specifically why MWI assumes ##I_\infty##?
I think the question I'm asking might be more on the philosophical side after some thought. It was basically that the mathematical structure of any quantum theory is that of a C*-algebra, i.e. a generalised probability theory. In MWI, probability is not a fundamental component of the theory and yet the theory has the mathematical structure of a probability theory. I was wondering if there was any discussion of this odd mathematical structure, given the non-probabilistic nature of the theory.
 
  • Like
Likes akvadrako
  • #77
A. Neumaier said:
The real question is: In what sense does MWI predict Born's rule? I cannot see any coherent argument.
Really? I would understand if you had said "valid argument" but to deny the existence of a coherent argument is patently ridiculous.
 
  • #78
stevendaryl said:
Probability in quantum mechanics is definitely not a matter of "counting branches". You can have just two branches, and that doesn't mean that they have equal weight.
Yes, if you take the usual notion of "probability" as fundamental, but my remarks are directed at attempts that begin with the MWI sans probabilities and attempt to substitute actual frequencies for probabilities. If you take the MWI, you can have unequal weights, but these have no direct interpretation as probabilities - at least not without a lengthy proof. If you want to deal only with actual frequencies, you have to figure out a way to define your frequencies in such a manner that branches that have higher weights are "frequent". For example, you could assume there are more copies of them.
 
  • #79
DarMM said:
Just another question, doesn't decoherence already require the Born rule, to permit tracing over the environment? Hence without the Born Rule, how do you show the state vector is of essentially Schmidt form to permit the clear branching structure without the Born Rule?
No. Decoherence does not require the Born Rule! It's a physical process that does not resolve a state as a proper mixture anywhere.
I may have failed to understand your question about the Schmidt form but I would assume that the answer lies in the fact that we define the subsystems - the original system, the detector, the environment, the observer - and each of these has its own state space. So the state space must be factorizable and the state is then a sum of product states.
 
  • #80
Derek P said:
No. Decoherence does not require the Born Rule!
How is Tracing justified without the Born Rule?
 
  • #81
DarMM said:
How is Tracing justified without the Born Rule?
No idea. Why would you want to?
 
  • #82
Stephen Tashi said:
Yes, if you take the usual notion of "probability" as fundamental, but my remarks are directed at attempts that begin with the MWI sans probabilities and attempt to substitute actual frequencies for probabilities. If you take the MWI, you can have unequal weights, but these have no direct interpretation as probabilities - at least not without a lengthy proof. If you want to deal only with actual frequencies, you have to figure out a way to define your frequencies in such a manner that branches that have higher weights are "frequent". For example, you could assume there are more copies of them.
Well there are. What you are calling a branch is a branch of the original superposition - the state of the system under observation. It's not a world.
 
  • #83
Derek P said:
No idea. Why would you want to?

Well, it's the most general form of a quantum prediction. If you have a state that is described by a certain density operator ##\rho##, and you have an operator ##A##, then the expected value of ##A### is ##tr(\rho A)##.
 
  • #84
Derek P said:
No idea. Why would you want to?
Tracing is required to obtain decoherence and tracing requires the Born Rule. Hence decoherence does require the Born rule.

I'm reading Wallace's book right now, so I'll see what he says.
 
  • #85
DarMM said:
Tracing is required to obtain decoherence and tracing requires the Born Rule. Hence decoherence does require the Born rule.
I'm reading Wallace's book right now, so I'll see what he says.
stevendaryl said:
Well, it's the most general form of a quantum prediction. If you have a state that is described by a certain density operator ##\rho##, and you have an operator ##A##, then the expected value of ##A### is ##tr(\rho A)##.
You're both missing something. The Born Rule and tracing are not assumed in MWI, they are derived from the unitary model. Indeed it is hard to imagine how a treatment that concludes with quantum predictions could fail to reproduce the Projection Postulate - except that it would no longer be a postulate but a theorem. But let's not be too ambitious - this thread is about the Born Rule, not the entirety of MWI's mathematical underpinnings. It can be tackled by explicitly expanding the global state as a sum of many ket products. It is a result, not an assumption. And of course once the Born Rule is established, you can hit the density matrix with it if that's what floats your boat.
 
Last edited:
  • #86
Derek P said:
You're both missing something. The Born Rule and tracing are not assumed in MWI, they are derived from the unitary model. Indeed it is hard to imagine how a treatment that concludes with quantum predictions could fail to reproduce the Projection Postulate - except that it would no longer be a postulate but a theorem.

He wasn't saying that tracing was assumed. He was asking how it was justified. A theorem is a good justification, if there is a theorem.
 
Last edited:
  • #87
DarMM said:
tracing requires the Born Rule
Actually, tracing and decoherence do not involve measurement, hence tracing is strictly speaking an additional assumption independent of the Born rule in its conventional formulation. The latter is a statement about measurement results, but things not yet measured have no results, hence the conventional form of Born's rule is inapplicable.
 
  • Like
Likes bhobba and Derek P
  • #88
stevendaryl said:
He wasn't saying that tracing was assumed. He was asking how it was justified. A theorem is a good justification, if there is a theorem.
Okay. So let's pretend there isn't a theorem and tracing is not justified. You can (I believe) derive the Born rule by expanding the state as a sum of components. Of course you will actually be taking a trace but you won't be relying on a theorem to give you permission.
 
Last edited:
  • #89
bhobba said:
Look into Gleason.

Thanks
Bill
You need to change your signature to "Every serious student of QM needs a copy of Ballentine and to Look into Gleason" !
 
  • #90
DarMM said:
However, if it's not branch-counting, what is it?
No, it's state counting.
 

Similar threads

Replies
69
Views
13K
  • · Replies 25 ·
Replies
25
Views
8K
Replies
6
Views
14K
  • · Replies 150 ·
6
Replies
150
Views
23K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 45 ·
2
Replies
45
Views
12K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
62
Views
10K