On 4/26/2022 7:01 PM, Bruce Kellett wrote:

On Wed, Apr 27, 2022 at 11:35 AM smitra <smi...@zonnet.nl> wrote: On 27-04-2022 03:11, Bruce Kellett wrote: > On Wed, Apr 27, 2022 at 10:32 AM smitra <smi...@zonnet.nl> wrote: > >> On 27-04-2022 01:37, Bruce Kellett wrote: >> >>> I think you >>> should pay more attention to the mathematics of the binomial >>> distribution. Let me explain it once more: If every outcome is >>> realized on every trial of a binary process, then after the first >>> trial, we have a branch with result 0 and a branch with result 1. >>> After two trials we have four branches, with results 00, 01,10,and >>> 11; after 3 trials, we have branches registering 000, 001, 011, 010, >>> 100, 101, 110, and 111. Notice that these branches represent all >>> possible binary strings of length 3. >>> >>> After N trials, there are 2^N distinct branches, representing all >>> possible binary sequences of length N. (This is just like Pascal's >>> triangle) As N becomes very large, we can approximate the binomial >>> distribution with the normal distribution, with mean 0.5 and standard >>> deviation that decreases as 1/sqrt(N). In other words, the majority of >>> trials will have equal, or approximately equal, numbers of 0s and 1s. >>> Observers in these branches will naturally take the probability to be >>> approximated by the relative frequencies of 0s and 1s. In other words, >>> they will take the probability of each outcome to be 0.5. >>> >> >> The problem with this is that you just assume that all branches are >> equally probable. You don't make that explicit, it's implicitly assumed, >> but it's just an assumption. You are simply doing branch counting. > > The distinctive feature of Everettian Many worlds theory is that every > possible outcome is realized on every trial. I don't think that you > have absorbed the full significance of this revolutionary idea. There > is no classical analogue of this behaviour, which is why your lottery > example is irrelevant. I spelled out the sequences that Everett > implies in my earlier response. These clearly must have equal > probability -- that is what the theory requires. QM without collapse does not require equal probabilities. Branches are not a fundamental concept of the theory. You just put this in by hand. > It is not an assumption on my part -- it is a consequence of Everett's basic idea. Everett's (or for that matter any other person's) ideas cannot be the basis for doing physics in a rigorous way. Your argument is not based on QM without collapse, you are making ad hoc assumptions about branching when branching isn't a fundamental process in QM. > So there is no branch counting involved. That is just another red > herring that you have thrown up to distract yourself from the cold > hard logic of the situation. > You just presented an elaborate presentation involving N branching steps and counted all 2^N branches as equal. That's branch counting and it's known to not be compatible with QM. The MWI can be taken to be QM without collapse and this is known to be a consistent theoryIt would seem that you are claiming that QM without collapse is notbased on Everett's ideas. If you claim that such a theory exists andis consistent, then you really should present that theory, and pointout that it has nothing to do with Everett, or with obtaining everyoutcome of a trial on different branches.My impression is that you do not have any worked-out theory -- youjust throw arbitrary objections to my working through the consequencesof Everett's approach to quantum mechanics. I have shown that manyproblems exist with Everettian QM. If you agree, and are prepared,with me, to throw out Everett, then we agree, and there is nothingmore to be argued about (at least, until you present some differentcomplete theory).

`I think Everett's idea was just to get rid of wave-function collapse and`

`instead assert that the apparently incompatible results of an experiment`

`were just different entanglements of one's brain/instrument with`

`different superposed components of the state of the system measured.`

`This is all consistent with the Copenhagen interpretation, except in CI`

`all but one of the orthogonal components is discarded. Decoherence has`

`cast some light on why the components quickly become orthogonal and why`

`they become orthogonal only in certain bases.`

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