- 10,876
- 423
Yes. Without that, we have a model that might possibly be correct, but doesn't qualify as a theory and shouldn't be called "many-worlds interpretation" since it doesn't tell us what the worlds are.
Dmitry67 said:Please explain. I have random generator with 2 outcomes: Frequent (probability 99.9999%) and Rare (0.0001%). After a trial I have 2 branches: F and R (with the different 'measure of existence').
The question is, why I almost always end in F branch? That claim is naive from the MWI point of view. But denying it is denying any predictive power of QM.
dmtr said:Because of the symmetry and unitarity. Because of the claim that if the outcomes are symmetric there is an equal probability to see each outcome. And the claim that the sum of probabilities should add to 1.
It's not sufficient to just pick a basis. The first thing we need to do is to decompose the omnium/universe into subsystems. Mathematically this corresponds to expressing the Hilbert space as a tensor product of two (or more) Hilbert spaces: \matcal H=\mathcal H_1\otimes\mathcal H_2. Then we can consider bases for those two spaces. Let's say that \{|\psi_\mu\rangle\} is a basis for \mathcal H_1 and that \{|\phi_\alpha\rangle\} is a basis for \mathcal H_2. Then we can define |\psi_\mu,\phi_\alpha\rangle=|\psi_\mu\rangle\otimes|\phi_\alpha\rangle. This definition ensures that \{|\psi_\mu,\phi_\alpha\rangle\} is a basis for \mathcal H.Dmitry67 said:Ok.
There is only one Bird, but infinitely many Frogs.
So it is required to translate Bird->Frog, then what is a choice of a Frog (preferred basis)?
What kind of basis should be used? All possible? Some subset? Based on what criteria?
Dmitry67 said:Or probably you don't realize that the verb "to see" needs additional clarification in the muti-history theory where basic can be define anyway you want.
Dmitry67 said:F and R are symmetric from the 'number of observers', but not symmetric from 'measure of existence' approach.
Plus Copenhagen Interpretation (CI) or MWI are indistinguishable from each other except one is deterministic the other probabilistic. So that basically means that MWI loses out as it has nothing to distinguish it from CI?
dmtr said:I do realize that. It is likely that you'll need an additional postulate. Boundary of a boundary principle.
RUTA said:Why BBP?
dmtr said:It's a guess, based on the criteria that the postulate should:
* have something to do with mathematical identity;
* be 'fundamental enough'.
Also BBP is a very safe guess. :)
Dmitry67 said:So, what you are saying is "Mathematics is not enough. There must be also a way to translate Birds view into Frogs view". Is this correct?
RUTA said:Are you working on anything relating quantum and BBP? Do you have any papers on this topic? Read 0908.4348 to see why I ask.
dmtr said:Sorry. I'm not a physicist. I've studied the 'basic modern minimum' - CM, SR/CFT, SM, GR, QM, but that's pretty much it. This QFT interpretation paper is interesting, but it is also a way over my head.
Still, I really like the idea of how they interpret the "probability amplitudes" as "symmetry amplitudes". It's just change of wording, but it makes a lot of sense. I couldn't get if they use the MWI approach (many subjective histories) or some kind of superdeterminism (one history). And how they use BBP I couldn't understand at all.
dmtr said:Still, I really like the idea of how they interpret the "probability amplitudes" as "symmetry amplitudes". It's just change of wording, but it makes a lot of sense.
dmtr said:I couldn't get if they use the MWI approach (many subjective histories) or some kind of superdeterminism (one history). And how they use BBP I couldn't understand at all.
Yes it is.pellman said:Demystifier, is the argument about the Born rule similar to what I am saying here https://www.physicsforums.com/showthread.php?t=230032&highlight=many+worlds ?
Demystifier said:Yes it is.
You are right. However, most literature on that issue is not very easy to read. My intention is to present and discuss this issue in a simpler way, so that all physicists can easily understand it.pellman said:Isn't the failure of the MWI to predict the statistics a rather glaring problem? I would think some MWI proponent has covered this. Surely we're not covering new ground here?
pellman said:Very cool.
Isn't the failure of the MWI to predict the statistics a rather glaring problem? I would think some MWI proponent has covered this. Surely we're not covering new ground here?
The question is: WHY would likelihood be monotonically related to the absolute value of the amplitude?hamster143 said:If you allow that every branch has an amplitude and the "likelihood" is in some way monotonically related to the absolute value of the amplitude, you can naturally deduce that the relationship has to take form of the Born rule, P_i = |c_i|^2 / \sum_k |c_k|^2.
Demystifier said:You are right. However, most literature on that issue is not very easy to read. My intention is to present and discuss this issue in a simpler way, so that all physicists can easily understand it.
Well, I didn't say that I will write a paper. But I didn't say that I will not write it either.pellman said:Looking forward to reading it. If you do write that paper, shoot me a PM or email when its done.
Demystifier said:By the way, what do you think about my analogy in #71?
In other words, by introducing an additional "hidden" variable (called Jack) that lives in one of the worlds, the Born rule attains a natural explanation. Needless to say, Jack is an analog of the pointlike particle in the Bohmian interpretation.
I do have an answer, but it's not that one. I think both the question and that answer are unscientific at best, and nonsense at worst. The question is asking for the correct definition of something undefined. That can't ever make sense. And that answer asserts the existence of a "limit" which is "defined" by a description of an act that can't be performed. That's just more nonsense.Hurkyl said:Let me use another classical analogy.
Suppose we have an ordinary six-sided die. I ask the question "are the sides equally likely?" What does that question really mean?
Of course, I'm sure you have a ready answer for that, based on the classical frequentist interpretation of probabilities -- mutter something about propensities, repeated experiments, argue that, for each positive epsilon, the propensity of "The proportion of times '1' comes up is in (1/6 - \epsilon, 1/6 + \epsilon)" converges to 1 as the number of experiments goes to infinity, and then we go and do controlled empiricial trials until we're satisfied we're close close enough to infinity to believe the answer.
Hurkyl said:I offer the following theorem:
Let \psi be a state a Hilbert space H. Let \psi^{(n)} := \psi \otimes \psi \otimes \cdots \otimes \psi (n times).
Let M_{n, p, \epsilon} be a "toy frequency counting operator" on H^{(n)} \otimes \mathbf{C}^2. (C2 is the space of pure qubit states) -- for some basis {em} of H, M is given by
- If the proportion of the a_k equal to 1 is in (p - \epsilon, p + \epsilon), then
M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m \oplus 1\rangle- Otherwise,
M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle
We can construct a function f_{p, \epsilon, \psi} that maps each positive integer n to the state space of a qubit given by taking the partial trace of M \left(\psi^{(n)} \otimes |0\rangle) -- heuristically, this function tells us the (probably mixed!) quantum state of the result of toy measurement if we do n trials.
Finally, define P(e_n ; \psi) = p if and only if, for every positive epsilon, we have \lim_{n \to +\infty} f_{p, \epsilon, \psi}(n) is the pure state |1\rangle.
Theorem: P as defined above is given by the Born rule.
I don't think that's correct. This is what some Wikipedia article said, right? I suspect that whoever came to that conclusion believed that every system is always in an eigenstate. (An idea which as you know has been thoroughly disproved by Bell inequality violations). It seems to me that all Hartle did was to prove that for each eigenstate |k>, there's an operator that has every |s\rangle\otimes|s\rangle\otimes\cdots (where |s> is an arbitrary state) as an eigenvector with eigenvalue |\langle k|s\rangle|^2.Count Iblis said:The argument by Hartle allows you to replace the Born rule by the weaker rule that says that measuring an observable of a system if the system is in an eigenstate of that observable, will yield the corresponding eigenvalue with certainty.
No, but 100/2=50.pellman said:Not entirely sure I follow it. Is the analogy such that it corresponds to an observation/experiment in which result BIG HOUSE is 50/2 times more likely than result LITTLE HOUSE?
Not necessarily, but why not to mention it if is related to MWI and to the Born rule.pellman said:Are we going for the Many-worlds-Bohmian interpretation now?
Right. That's what I meant.Demystifier said:No, but 100/2=50.
Not necessarily, but why not to mention it if is related to MWI and to the Born rule.
Hurkyl said:Let \psi be a state a Hilbert space H. Let \psi^{(n)} := \psi \otimes \psi \otimes \cdots \otimes \psi (n times).
Let M_{n, p, \epsilon} be a "toy frequency counting operator" on H^{(n)} \otimes \mathbf{C}^2. (C2 is the space of pure qubit states) -- for some basis {em} of H, M is given by
- If the proportion of the a_k equal to 1 is in (p - \epsilon, p + \epsilon), then
M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m \oplus 1\rangle- Otherwise,
M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle = M|e_{a_1}, e_{a_2}, \ldots, e_{a_n}, m\rangle
We can construct a function f_{p, \epsilon, \psi} that maps each positive integer n to the state space of a qubit given by taking the partial trace of M \left(\psi^{(n)} \otimes |0\rangle) -- heuristically, this function tells us the (probably mixed!) quantum state of the result of toy measurement if we do n trials.
Finally, define P(e_n ; \psi) = p if and only if, for every positive epsilon, we have \lim_{n \to +\infty} f_{p, \epsilon, \psi}(n) is the pure state |1\rangle.
Theorem: P as defined above is given by the Born rule.
Fredrik said:I don't think that's correct. This is what some Wikipedia article said, right? I suspect that whoever came to that conclusion believed that every system is always in an eigenstate. (An idea which as you know has been thoroughly disproved by Bell inequality violations). It seems to me that all Hartle did was to prove that for each eigenstate |k>, there's an operator that has every |s\rangle\otimes|s\rangle\otimes\cdots (where |s> is an arbitrary state) as an eigenvector with eigenvalue |\langle k|s\rangle|^2.
It's natural to interpret this operator as a frequency operator when it acts on a tensor product of eigenstates, but it doesn't seem to have a natural interpretation when it's acting on |s\rangle\otimes|s\rangle\otimes\cdots with |s> arbitrary. The only way I can justify the interpretation as a frequency operator is to use the Born rule, i.e. the rule he's trying to prove.
I would also say that Hartle used the Born rule before that, when he assumed that the Hilbert space of a system is the tensor product of the Hilbert spaces of its subsystems. I don't think that assumption can be justified without appealing to the Born rule.
So my opinion is that Hartle's article is completely useless. If I'm wrong, I'd like to know it, so feel free to try to convince me. Post #18 in the other thread (linked to above) has a direct link to the article. My comments are in #13.
They don't address the issues I mentioned, so they're probably not aware of them. If they're right about the need for a discrete state space, they've just found another problem with an article that was already useless. It also seems to me that this article is much worse than Hartle's. Their summary of Hartle's argument looks like complete crackpot nonsense to me.Count Iblis said:There is another discussion of Hartle's arguments http://arxiv.org/abs/hep-th/0606062" where it is suggested that you need to introduce a discrete Hilbert space to fix the derivation.
How does that justify the use of the tensor product, or the interpretation of Hartle's operator as a frequency operator?Count Iblis said:About the assumptions, I don't think one assumes the Born rule, only the L^2 norm.
You get a deterministic model if you just remove the Born rule, but it's not a theory since we don't have a way to interpret the mathematics as predictions about results of experiments. It also doesn't include anything about "worlds", since the Born rule is what tells us that there's something in the mathematics that we can think of as worlds. It doesn't even include a description in terms of subsystems, since the Born rule is what justifies the use of the tensor product.Count Iblis said:The MWI is supposed to be a deterministic theory, so talking about probabilities in the postulates is unnatural.
I haven't seen any reason to believe that. I haven't even seen an attempt to prove it. I just have a vague memory of seeing that claim in a Wikipedia article that I wasn't able to find when I tried yesterday, and that used Hartle's article as a reference. But I don't think Hartle's results could be described that way even if his argument had been valid.Count Iblis said:Instread, you can attempt to replace them by statements about certainties, like that measuring a system in an eigentate will yield a certain outcome with certainty.
There are already no such sectors in the individual system Hilbert space, so they're clearly talking about the tensor product of infinitely many copies of that space. The removal of the zero norm "vectors" that get included by accident when we take the tensor product of infinitely many copies of a Hilbert space is necessary to ensure that the result is a Hilbert space. It doesn't have anything to do with the Born rule.Count Iblis said:This combined with cutting away zero norm sectors of Hilbert space gives you the Born rule.
Demystifier said:The minimal set of assumptions defining MWI is:
1. Psi is a solution of a linear deterministic equation.
2. Psi represents an objectively real entity.
PTM19 said:To avoid being ruled out MWI requires a lot of additional postulates
I addressed those claims in my reply to you in the thread in the philosophy forum (and also earlier in this thread).PTM19 said:The point being that MWI is in no sense a minimal interpretation - on the contrary it is the most extremely baroque of interpretations, as it necessitates belief in an immense number of unobservable universes,
True, but the same thing holds for every claim that some theory describes what actually happens. This is because experiments can't tell us anything except how accurate a theory's predictions are.PTM19 said:what's more it cannot even be falsified making it unscientific.
If there's anything I'm sure about here, it's that this claim is completely wrong.PTM19 said:MWI is the most extreme violation of Ockham's razor principle one can imagine.
See my comment in the philosophy forum.PTM19 said:The minimal interpretation is an ensemble interpretation...
Hurkyl said:![]()
Those two axioms don't even say anything about one world, because they don't make any predictions about (probabilities of) results of experiments.PTM19 said:For example it has to postulate the existence of parallel universes and that those universes are unobservable. The concept and existence of parallel universes certainly does not follow from those two axioms, it has to be postulated in their interpretation.
RUTA said:The problem I have with MWI is that you can’t know whether your physics is legit. Suppose the “real” experimental distribution is 50-50, someone occupies a universe that is 100-0 and someone else 0-100 (with others at all possible distributions in between). So, theoretical physicists in those universes (and most others in between) who predict the “real” 50-50 distribution will not get tenure in physics, but will have to try for tenure in philosophy or religious studies. Physicists in those universes who predict the empirical, but “wrong,” results are heralded and rewarded. I don’t understand why anyone in physics would subscribe to the MWI ontology.
There are already no such sectors in the individual system Hilbert space, so they're clearly talking about the tensor product of infinitely many copies of that space. The removal of the zero norm "vectors" that get included by accident when we take the tensor product of infinitely many copies of a Hilbert space is necessary to ensure that the result is a Hilbert space. It doesn't have anything to do with the Born rule.
dmtr said:I don't see any problem, as long as the symmetric apparatus gives you symmetric probabilities (probability amplitudes to be precise).
RUTA said:You don't know if it's symmetric or not. You can't know. All you know is the history of your particular branch. You don't know if the statistics you observe are the overall "correct" statistics, but you can guess they're probably not.
dmtr said:But I do observe the "correct" statistics. And it corresponds nicely to the symmetric apparatus. In the example above there is roughly the same number of '0's and '1's in the original sequences. I can also see that the SUM: 1 1 1 1 0 0 0 1 2 1 contains more '1's than '0's (or '2's). This corresponds nicely to the: '50/50' world is more probable than 100/0 (or 0/100).
And it doesn't look like this "correct" statistics excludes MWI in any way. Why I-who-saw-"1" should consider myself any better from I-who-saw-"0"? From the symmetry considerations I shouldn't.
RUTA said:You're missing the point. Your results are only one path in the bifurcated tree. You have no idea where your particular history resides in the tree because you only have access to your single history.
This is simply wrong. The existence of parallel universes DOES follow from those two axioms. But you cannot understand it without understanding the theory of decoherence (which, by the way, is not only a theory, but also an experimental fact.)PTM19 said:For example it has to postulate the existence of parallel universes and that those universes are unobservable. The concept and existence of parallel universes certainly does not follow from those two axioms, it has to be postulated in their interpretation.
Demystifier said:The existence of parallel universes DOES follow from those two axioms.
Hurkyl said:Let's say I'm using coordinates and arithmetic as a means of studying Euclidean geometry.
Would you say that this does, or does not, require more assumptions than just the postulates of Euclidean geometry?
Would you say that the arithmetic properties of coordinates do, or do not, follow from the axioms of Euclidean geometry
The theory of real numbers turns out to be equivalent to the theory of Euclidean geometry, with one direction working pretty much just as you described.yossell said:But off the top of my head, I'm not sure how much coordinate geometry can be recovered in this way.