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

  1. May 2, 2018 #1
    Please restrict your answers to criticisms of derivations of the Born Rule that are generally accepted by proponents of MWI. Please provide a verbal description of the issue where possible so that people like myself who are certainly graduate-plus* but rusty as hell have a chance of seeing what the point is.

    *plus nearly half a century in my case :biggrin:

    So then, in what sense does MWI fail to predict the Born Rule?
     
  2. jcsd
  3. May 3, 2018 #2

    PeterDonis

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  4. May 3, 2018 #3
    Like PeterDonis said, you have to assume something. Take the simplest case possible, a fully symmetric wavefunction with 2 equal-weighted branches which have no interaction terms. How can you say the probability of finding yourself (or a particle) on either branch is 50% ?

    Using only the axioms of unitary QM it doesn't seem possible.
     
  5. May 3, 2018 #4
    Well I read the first page which included a rapturous endorsement of the maths by no less than @bhobba. Of course he may have changed his mind since then. In addition @stevendaryl, @atyy, @vanhees71 to name but a few had their say, but I didn't see anything at all suggesting that the usual derivation is flawed. Should I continue?
     
  6. May 3, 2018 #5

    PeterDonis

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    What do you think is "the usual derivation"? Can you give a reference?
     
  7. May 3, 2018 #6

    DarMM

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    As far as I understand there is no totally clear derivation of the meaning of non-uniform probabilities using purely unitary quantum theory with no projection postulates. Zurek's proof relies on hard to justify assumptions about subspaces of states and assumes some notion of probability to begin with, as it starts with a Schmidt state as the post-measurement state.

    The Oxford style decision theory proofs only prove that in a many worlds situation one can construct a utility function whereby an agent is permitted to act as if the Born rule were true. However this doesn't show the utility function to be unique, nor is it obvious that it is the most sensible one. Even more strongly, just because I might act in a way that satisfies the axioms of decision theory, doesn't mean anything for the real observed frequencies of measurement results recorded by lab equipment. Also the Oxford proofs do too much handwaving at certain points about how the branching structure works in order to force the proof.
     
  8. May 3, 2018 #7
    <shrug> My bad. "that there was no unflawed derivation" then.
     
  9. May 3, 2018 #8

    PeterDonis

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    I'm not sure if there's much point if we don't have any specific derivation to discuss.
     
  10. May 3, 2018 #9

    Stephen Tashi

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    In what sense does any approach to QM derive the Born Rule? - as opposed to taking it as an assumption.

    Perhaps that's a good topic for another thread.
     
  11. May 3, 2018 #10

    stevendaryl

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    Well, @bhobba likes to mention a theorem (Gleason's Theorem) that shows that if you've already decided that the quantum state gives probabilities for measurement outcomes, then under some pretty minimal assumptions, it turns out that the square of the amplitude is pretty much the only natural choice.
     
  12. May 3, 2018 #11

    stevendaryl

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    I'm not sure what's the best place to look for the proof, but here's one:

    https://arxiv.org/pdf/quant-ph/9909073.pdf

    I think that, in a certain sense, the interpretation of the square of the wave function as a probability seems baked into the very definition of a Hilbert space. The fact that ##\int |\psi(x,t)|^2 dx = 1## is certainly very suggestive that ##|\psi|^2## can be interpreted as a probability density.
     
  13. May 3, 2018 #12

    Mentz114

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    In my understanding the Born rule has two parts
    1. The square of the amplitude is probability
    2. Doing a suitable measurement on a system (applying ##\hat{A}_m##) results in the ##m##th eigenstate of ##\hat{A}_m## with probability ##|\alpha_m|^2##

    I think MWI runs into trouble with the second using the original 'naive' splitting protocol.

    The problem is that if we follow a particular line of splits for a lot of splits and count, frequencies don't all match the expected ones. For instance for 100 coin tosses.

    This can be fixed easily by tinkering with the splitting protocol and introducing a (non-physical) memory ##M## for each instance of our observer.
    P is the probability of getting a 1 on each trial of a binary process with outcomes '0' and '1'.
    Let the splitting be controlled by the quantity ##M/S## where ##M## is the count of '1' splits and ##S## is the number of splits in the history.
    The rule is
    if M/S < P split into 1,1
    if M/S = P split into 1,0
    if M/S > P split into 0,0.

    Now the limiting probability estimate M/S → P as required.

    But the sequences are not random !
    If P=1/2 the sequence becomes alternating 0's and '1's. This non-randomness happens for any value of P.
    So, this is not a good solution ( no coconut). I think the reason is that MWI robs QM of doing anything that can give random results because that requires quantum indeterminacy, which is gone.
    My feeling is that (even after one attempt to fix this) that one must have indeterminacy to make correct predictions.

    [ I may be conflating indeterminacy and non-unitarity. It is the aim of MWI to remain unitary, but it seems that this inevitably loses randomness in some fundamental way]
     
    Last edited: May 3, 2018
  14. May 3, 2018 #13

    Nugatory

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    They don't, but when they introduce the reduction of the wave function as an assumption they can incorporate the Born rule into that assumption. The difficulty for MWI is that MWI rejects any reduction postulate, so has to find the Born rule in unitary evolution.
     
  15. May 4, 2018 #14
    MWI proponents reject reduction but most don't reject using an additional assumption from which they can derive the Born rule; that's more often something critics claim. It's clear the unitary axioms are not enough to even talk about probability and as a result MWI has no advantage over other interpretations in this regard. This says it better than I can:

    In defence of the self-location uncertainty account of probability in the many-worlds interpretation
    Kelvin J. McQueen and Lev Vaidman. February 13, 2018

     
    Last edited: May 4, 2018
  16. May 4, 2018 #15

    Stephen Tashi

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    In that paper we read:

    ....
    If one successfully appends the Born Rule to the Many Worlds Interpretation by taking the viewpoint of a single observer in the MWI, doesn't that say that a physicist may as well take the collapse viewpoint ? - seeing as the physicist is a single observer.

    The verbal interpretations of MWI seem to take for granted that observers that are different (due to being on different branches) cannot be aware of each other. Intuitively that seems self evident, but how is that shown mathematically? How is "A is not aware of B " translated into a precisely defined physical process?
     
  17. May 4, 2018 #16
    That does seem valid and as another way of saying it, Copenhagen can be derived from the MWI by considering just a single viewpoint, at least practically because of decoherence. But I guess not exactly due to Wigner's friend-type experiments. If I experience collapse, but I'm part of a larger isolated experiment, then it's still possible an outside experimenter will undo that collapse by causing me and other versions of me to interfere, causing the probability of some futures to become zero.

    Even not regarding that very difficult experiment (at least with people), with many worlds you can analyse the viewpoint of multiple physicists together. If you try to do that with Copenhagen in a single world, you'll get inconsistencies.

    Do you mean after decoherence, how it "not aware of" defined? I think about it as no longer interfering and can be written as separate terms that evolve independently.
     
    Last edited: May 4, 2018
  18. May 4, 2018 #17
    Do you mean MWI or is this some other species of interpretation that I don't recognise?
     
  19. May 4, 2018 #18
    Thanks, fixed.
     
  20. May 4, 2018 #19
    Right at the beginning. Note. I very strongly disagree that worlds separate at measurement, let alone before. They separate at decoherence. Before that, the two pre-world states are orthogonal. They absolutely cannot affect each other or be aware of each other. Separation of worlds is something completely different.
     
    Last edited: May 4, 2018
  21. May 4, 2018 #20

    Stephen Tashi

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    I can imagine such an un-collapse happening from the viewpoint of an "outside" observer C who can observe how observer B undoes the collapse of observer A. But isn't observer A really two (or more) different observers from the viewpoint of MWI and observer C ?

    I have difficulty digesting arguments that treat complex systems (such as human observers) as simple mathematical structures - such as superimposed waves or Hilbert Space rays in QM ( or points on a world line in the theories of relativity). Perhaps this is conceptual limitation on my part, but it would please me to hear details of how decoherence can implement the common language notion of "A is not aware of B"

    A realistic model of a human observer portrays him as a aggregation of matter, some of which "comes and goes" as far as what we consider being "part of" the observer. The matter has more properties than position and momentum. If we take the view that matter and everything else is really waves in various types of fields then how do you define which parts of the fields are observer A and which parts are observer B -especially in MWI where the notion of "observer A" seems only to make sense at a single instant in time. (An instant later, observer A splits up.)

    A simple mathematical question about relations is: If E is aware of S and A is aware of S then is E necessarily aware of A? In the common language notion of "is aware of", I'd say No. So if earthling E is aware of star S and alien A is aware of star S then earthling E may not be aware of alien A.

    If "is aware of" is replaced by "is entangled with" , then shouldn't the answer be Yes?

    In another thread, I asked whether using "rational agents" to introduce probability into MWI assumes these are actual agents or are they merely conceptual agents - analogous to a conceptual "unit test mass" that we imagine in order to define a force field. I think this is a meaningful question due to the complexity of implementing actual agents.
     
  22. May 4, 2018 #21
    In this case, A is like the particle in a delayed-choice experiment, right? So you normally don't say there are 2 particles, just that it's in a superposition of two states.
    One thing that's probably in between is considering the whole experiment running in a quantum computer, and the observers as simple preprogrammed agents with a few bits of storage.
    Maybe sticking with entanglement instead of awareness is just easier. Does anything else matter?
     
  23. May 4, 2018 #22
    So would I as they would be unaware of each other even if there were no decoherence (assuming that A and B are pre-worlds (my term) - states that will evolve into separate worlds through decoherence in due course)..
    You don't define what an observer is. You define what an observer observes. Hence the common statement "an observer can be anything that interacts with the system". The system has properties.
    No. The idea of awareness presumably includes events happening to A which then alters B. This is absolutely impossible with entanglement alone.
    You might well ask. 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 - someone here had cast doubts on whether MWI actually does derive the Born Rule. I don't see the possible inability of MWI to define something it has no need of as a defect or even surprising. I am a very simple person. Vague too or so I'm told.

    On the subject of which may I respectfully ask that people here stick to the topic.It's already getting diluted beyond any usefulness. Thanks!
     
  24. May 4, 2018 #23

    DarMM

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    There's no problem with intrinsic probabilities nor the feeling that Many Worlds needs to include them. Rather the problem is precisely the first part of your sentence, MWI doesn't seem to predict observed frequencies very well.

    Attempts to derive the Born Rule, like Zurek's, are attempts to show that expected values for frequencies within a single world match those given by textbook Quantum Mechanics.

    That is, for a state:
    $$\sqrt{\frac{1}{3}}| a \rangle + \sqrt{\frac{2}{3}}| b \rangle$$

    That your odds of being in a world where "b" is observed are ##2/3##. However there is no clear proof of this from MWI as of 2018.
     
  25. May 4, 2018 #24
    First of all thank you very much for such a definitive statement. If the alleged problems with deriving the Born Rule apply strictly to frequencies then the question becomes very straightforward. There are a number of versions of a very simple proof such as this one from Michael Price. I would not like to get too tied into one person's semi-popular presentation but those who say MWI can't produce the Born Rule should be able to point to where Price, and many others with similar arguments, go wrong.
     
  26. May 4, 2018 #25

    PeterDonis

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    I don't see how this is a proof of anything.

    First, his argument relies on the B state being degenerate; but it is easy to write down states with unequal amplitudes for A and B terms, neither of which is degenerate. (The easiest is to do it with qubit states.)

    Second, his argument relies on there being a simple integer ratio between the two amplitudes. But of course the vast majority of states do not obey that restriction. (In fact, it's simple to show that the pairs of amplitudes that do are a set of measure zero in the space of all possible pairs of amplitudes, and therefore, by this person's own argument, do not exist.)

    Third, to address the obvious objection to the above, I don't buy his hand-waving argument for how his "proof" extends to the case of non-degenerate states or non-rational amplitude ratios.

    But finally, even leaving all that aside: counting "numbers of worlds" and squaring their relative amplitudes and saying that those are the probabilities of orthonormal "worlds" is not proving the Born rule: it's assuming it. His "proof" is just arguing in a circle.

    And finally finally, he misstates what DeWitt proved in his 1970 paper. DeWitt did not prove that "the norm of the worlds in which the Born rule is violated vanishes". He only showed that, in the limit of an infinite number of observations on an ensemble of identically prepared systems, the difference between the relative frequency of a particular outcome and the square of the amplitude of that outcome vanishes. That is not at all the same thing.
     
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