On Sun, May 8, 2022 at 11:32 AM smitra <smi...@zonnet.nl> wrote:
> On 05-05-2022 01:15, Bruce Kellett wrote:
> > On Thu, May 5, 2022 at 5:27 AM smitra <smi...@zonnet.nl> wrote:
> >
> >> On 04-05-2022 01:49, Bruce Kellett wrote:
> >>>
> >>> I have not introduced any concept of probability. The 2^N branches
> >>> that are constructed when both outcomes are realized on each of N
> >>> Bernoulli trials are all on the same basis.
> >>
> >> If you ignore the amplitudes in the states, and that means modifying
> >> QM into something else.
> >
> > QM does not assume that all branches exist equally. In Everett you
> > have already modified QM into something else.
> >
> > The Schrodinger equation is insensitive to the amplitudes. You get the
> > same set of 2^N branches from the Schrodinger equation, whatever
> > amplitudes you have. The weights of these branches certainly depend on
> > the amplitudes: if there are n zeros in the set of N trials, there are
> > N-n ones. The weight of the corresponding binary string is a^n
> > b^(N-n), but without further assumption, this plays no role in the
> > future development of the state or in the interpretation of the binary
> > string. If you interpret it as the probability of the string, you
> > again have a conflict, since all binary strings are constructed on an
> > equal basis, the natural probability for each is 2^{-N}.
>
> There is no conflict whatsoever with assuming the Born rule and the
> Schrodinger equation. The "construction on an equal basis" is not at all
> implied by the Schrödinger equation.
>
It is when you take the SE to imply that all possible outcomes exist on
each trial. That gives all outcomes equal status.
Bruce
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