> On 5 Sep 2020, at 20:11, 'Brent Meeker' via Everything List > <everything-list@googlegroups.com> wrote: > > > > On 9/4/2020 11:27 PM, Bruce Kellett wrote: >> On Sat, Sep 5, 2020 at 3:52 PM 'Brent Meeker' via Everything List < >> <mailto:everything-list@googlegroups.com>everything-list@googlegroups.com >> <mailto:everything-list@googlegroups.com>> wrote: >> On 9/4/2020 10:18 PM, Bruce Kellett wrote: >>> On Sat, Sep 5, 2020 at 2:42 PM 'Brent Meeker' via Everything List < >>> <mailto:everything-list@googlegroups.com>everything-list@googlegroups.com >>> <mailto:everything-list@googlegroups.com>> wrote: >>> On 9/4/2020 7:02 PM, Bruce Kellett wrote: >>>> On Sat, Sep 5, 2020 at 11:29 AM 'Brent Meeker' via Everything List < >>>> <mailto:everything-list@googlegroups.com>everything-list@googlegroups.com >>>> <mailto:everything-list@googlegroups.com>> wrote: >>>> >>>> But the theory isn't about the probability of a specific sequence, it's >>>> about the probability of |up> vs |down> in the sequence without regard for >>>> order. So there will, if the theory is correct, be many more sequences >>>> with a frequency of |up> near some theoretically computed proportion |a|^2 >>>> than sequences not near this proportion. >>>> >>>> >>>> The theory is about the probabilitiies of observations. The observation in >>>> question here is a sequence of |up> / |down> results, given that the >>>> probability for each individual outcome is 0.5. If the theory cannot give >>>> a probability for the sequence, >>> >>> It can. But QM only predicts the p=0.5. To have a prediction for a >>> specific sequence HHTTHHHTTHTHTH... you need extra assumptions about >>> indenpendence. >>> >>> Sure. And independence of the sequential observations is clearly implied by >>> the set up. >>> And given those assumptions your theory will be contradicted with near >>> certainty. >>> >>> Why? >> >> The probability of getting any given entry in the sequence is 1/2, so the >> probability of getting the whole sequence right is 1/2^N . >> >> I thought I had said that quite clearly. And that that is true for any one >> of the possible 2^N different sequences. >> >> >>> Which is why I say the test of QM is whether p=0.5 is consistent with the >>> observed sequence in the sense of predicting the relative frequency of H >>> and T, not in the sense of predicting HHTTHHHTTHTHTH... >>> >>> >>> I am not attempting to predict a particular sequence. >> >> That's what you seemed to reply when I said QM was only predicting the >> relative frequency of H within the sequence. If you now agree with that, >> then you will also agree that there will many sequences with a relative >> frequency of 0.5 for H and given any epsilon the fraction of such sequences >> repetitions with 0.5-epsilon<frequency(H)<0.5+epsilon goes 1 as N->oo. >> Which is what we mean by confirming the QM prediction of 0.5. >> >> You are off on the wrong track. I am not disagreeing with this. It is just >> that this is not what I am talking about. In the single world, stochastic >> case, it is, as Albert said, true that as N goes to infinity, all sequences >> converge in probability to the relative frequency of 0.5. But that is not my >> point. >> >> >>> All that I have said is that the probability of any such sequence in N >>> independent trials is 1/2^N. And that is simple probability theory, which >>> cannot be denied. >> >> >> >> Which is what you have said above, and I agree. >> >>>> then multiply the probabilities for each particular result in your >>>> sequence of measurements. The number of sequences with particular >>>> proportions of up or down results is irrelevant for this calculation. >>>> >>>> Again, you are just attempting to divert attention from the obvious result >>>> that the Born rule calculation gives a different probability than expected >>>> when every outcome occurs for each measurement. In the Everett case, every >>>> possible sequence necessarily occurs. This does not happen in the genuine >>>> stochastic case, where only one (random) sequence is produced. >>> >>> In the Everett theory a measurement of spin up for a particle prepared in >>> spin x results in two outcomes...only one is observed. If that is enough to >>> dismiss Everett then all the this discussion of probability and the Born >>> rule is irrelevant. >>> >>> >>> I have no idea what you are talking about! Nothing like that was ever >>> suggested. Everett predicts that in such a measurement, both outcomes >>> obtain -- in separate branches. >> >> As I understand your argument you're saying Everett is falsified because, no >> matter what N is, it predicts a branch HHHHHHHHHH...H which...What? Is >> wrong? Doesn't occur? Is inconsistent with the Born rule (it isn't)? Is >> not observed? >> >> No, listen carefully. Everett predicts that such a sequence will certainly >> occur for any N. In other words, the probability of the occurrence of such a >> sequence is one. Whereas the Born rule, as we both now seem to agree, >> predicts that the probability for the occurrence of such a sequence is >> 1/2^N. It is the fact that Everett and the Born rule predict different >> probabilities for the same sequence that is the point -- not that either >> predicts the impossibility of such a sequence. It is the predicted >> probabilities that differ, not the sequences. >> >> And if you have a theory that predicts two different values for some result, >> then your theory is inconsistent. Everett and the Born rule are inconsistent >> because they predict different probabilities for this sequence of N |up>s in >> N trials (or any other particular sequence, for that matter. Even though >> that latter point seems to have confused you!) > > But you are not using Everett's theory. You're strawmanning Evertt. You're > saying that since Everett says some sequence occurs he is predicting it with > probability 1. But that's only predicting that it occurs in evolution of the > wave function. It's not a prediction of the QM probability that is being > tested. And it's not following thru on Everett's interpretation that > connects the theory to observation. It's imposing your idea of how it > connects to observation; essentially cutting off Everett's interpretation > part way thru. > > Everett's theory is deterministic so it's not relevant to criticize it for > "predicting probability 1" when it predicts all the results. I agree with > you that you can't get a probability out of a deterministic theory unless you > put in some additional postulate...like ignorance or coarse graining...and > that's exactly what Everttian's do. They say that the branches are an > ensemble and you have some probability of being the observer in one of the > ensemble...an ignorance based probability measured by either branch counting > or weighting of branches. I think this is a kind of cheat, since it is not > simply a consequence of Schroedinger's equation.
Right. It is more a consequence of the theory of mind used in the definition of what is an observer (a machine with memory). Everett used “naive mechanism”, which works very well in this setting, but eventually, a definition of machine requires the Church-Turing-Post-Kleene thesis, and then the wave itself has to be retrieved from all computations. Mechanism provides a testable explanation for the appearance of the physical laws. It extends Darwin in a many histories interpretation of arithmetic on which all sound universal (oracular) machine converges. That explains quanta and qualia from elementary arithmetic (or any universal machinery phi_i). We can’t do better. Elementary arithmetic + induction can already prove that elementary arithmetic is not obtainable by simpler notions (up to Turing-equivalence). > On the other hand, Gleason's theorem is a consequence. So once you cheat > enough to introduce the probability concept, It is not cheating. It is the acknowledgement of the universal machine that she does not know which computations support them. > getting to Born's rule is just a matter of making up a story you like. ? The corresponding theorem of Gleason is still lacking for arithmetic, but the fact that the universal machine get the quantum logic where needed illustrates that the "Gleason theorem" of Gleason will be plausibly directly applicable. > > So my view is that once you've developed decoherence theory and you've shown > that the reduced density matrix is diagonalized, you might as well then > bite-the-bullet and postulate that the theory is probabilistic. Then the > math (Gleason's theorem) forces the interpretation that those diagonals are > the probabilities of results. Then "everything happens" is just a story > attempting to back-fill a picture of how you got there based on ignorance > (self-locating uncertainty). OK > There are some people who can't abide probabilistic theories and will invent > fantastic worlds in order to have a deterministic ensemble which then must be > reduced by ignorance to agree with observation. They then feel they've made > great progress because they think their theory is deterministic. It is ontological attachment. It is the same error than saying that “God made it” is an explanation, instead of a problem to solve. Bruno > > Brent > >> >> >> If you just say it predicts something which is not observed; then my point >> is that it always predicts outcomes that are not observed unless P=1. >> >> >> Whether the sequence is observed or not was never the point. Although, in >> Everett, there is always one observer of the sequence of all |up>s. This may >> occur with the Born rule, but not inevitably. The probabilities differ, >> which was the actual point. >> >> >> Brent >> >>> But the probability of this is one. Repeat N times. N time one is still >>> just one. >> >> >> I did not say that very well. I mean one multiplied by itself N times, or >> 1^N = 1. >>> There is nothing more to it than that. I think you are being desperate in >>> your attempts to play 'advocatus diaboli'. The point is that the Born rule >>> is inconsistent with Everett. >>> >>> Bruce >> -- >> You received this message because you are subscribed to the Google Groups >> "Everything List" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to everything-list+unsubscr...@googlegroups.com >> <mailto:everything-list+unsubscr...@googlegroups.com>. >> To view this discussion on the web visit >> <https://groups.google.com/d/msgid/everything-list/CAFxXSLRErSGWg9pM08ph5ahTg%3Du9HYXu9WMKo4zHDtzvwm0W9A%40mail.gmail.com?utm_medium=email&utm_source=footer>https://groups.google.com/d/msgid/everything-list/CAFxXSLRErSGWg9pM08ph5ahTg%3Du9HYXu9WMKo4zHDtzvwm0W9A%40mail.gmail.com >> >> <https://groups.google.com/d/msgid/everything-list/CAFxXSLRErSGWg9pM08ph5ahTg%3Du9HYXu9WMKo4zHDtzvwm0W9A%40mail.gmail.com>. > > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-list+unsubscr...@googlegroups.com > <mailto:everything-list+unsubscr...@googlegroups.com>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/5F53D4E6.8010702%40verizon.net > > <https://groups.google.com/d/msgid/everything-list/5F53D4E6.8010702%40verizon.net?utm_medium=email&utm_source=footer>. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/5CD2A091-39C3-46D8-96E0-AF0B9E7FA5A0%40ulb.ac.be.
