On Tue, May 3, 2022 at 10:11 PM smitra <smi...@zonnet.nl> wrote:
> On 28-04-2022 07:51, Bruce Kellett wrote:
> > On Thu, Apr 28, 2022 at 3:24 PM Brent Meeker <meekerbr...@gmail.com>
> > wrote:
> >
> >> On 4/26/2022 5:32 PM, smitra wrote:
> >>
> >>> On 27-04-2022 01:37, Bruce Kellett wrote:
> >> Changing the weights of the components in the superposition does not
> >> change the conclusion of most observers that the actual probabilities
> >> are 0.5 for each result. This is simple mathematics, and I am amazed
> >> that even after all these years, and all the times I have spelled this
> >> out, you still seek to deny the obvious result. Your logical and
> >> mathematical skill are on a par with those of John Clark.
> >>
> >> It's indeed simple mathematics. You apply that to branch counting to
> >> arrive at the result of equal probabilities.
> >
> > I have not used branch counting. Please stop accusing me of that.
> >
>
> You are considering each branch to have an equal probability when there
> is no logical reason to do so, and when that's also being contradicted
> by QM.
>
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. There is no probability involved. The branches
are all equivalent by construction.
I think you are being confused by the presence of coefficients in the
expansion of the original state: the a and b in
|psi> = a|0> + b|1>
The linearity of the Schrodinger equation means that the coefficients, a
and b, play no part in the construction of the 2^N possible branches; you
get the same set of 2^N branches whatever the values of a and b. Think of
it this way. If a = sqrt(0.9) and b = sqrt(0.1), the Born rule probability
for |0> is 90%, and the Born rule probability for |1> is 10%. But, by
hypothesis, both outcomes occur with certainty on each trial. There is a
conflict here. You cannot rationally have a 10% probability for something
that is certain to happen. This is why some people have resorted to the
idea that there are in fact an infinite number of branches, both before and
after the measurement. What the measurement does is partition these
branches in the ratio of the Born probabilities. But this is just a
suggestion. There is nothing in the Schrodinger equation, or in quantum
mechanics itself, that would suggest that there are an infinite number of
branches. In fact, that idea introduces a raft of problems of its own --
what is the measure over this infinity of branches? What does it mean to
partition infinity in the ratio of 0.9:0.1? What is the mechanism
(necessarily outside the Schrodinger equation) that achieves this?
You are concerned that a collapse introduces unknown physics outside the
Schrodinger equation. You will have to be careful that your own solution
does not introduce even more outrageous physics outside the Schrodinger
equation. Collapse, after all, has a perfectly reasonable mechanism in
terms of the flashes of relativistic GRW theory.
My conclusion from this is that Everett (and MWI) is inconsistent with the
Born rule. So your idea of QM without collapse but with the Born rule, is
simply incoherent. There can be no such theory that is internally
consistent.
> >>> So, the conclusion has to be that one should not do branch
> >>> counting. The question is then if this disproves the MWI. If by
> >>> MWI we mean QM minus collapse then clearly not. Because in that
> >>> case we use the Born rule to compute the probability of outcomes
> >>> and assume that after a measurement we have different sectors for
> >>> observers who have observed the different outcomes with the
> >>> probabilities as given by the Born rule.
> >
> > In which case the Born rule is just an additional arbitrary
> > assumption: it is not part of the Schrodinger equation. Your theory of
> > QM minus collapse is not well-defined. You simply take whatever you
> > want from text-book quantum mechanics, with no regard to the
> > consistency of your model.
> >
>
> QM includes the Born rule. QM minus collapse is just that: QM minus
> collapse. It's not QM minus collapse minus the Born rule.
>
>
> >>> You then want to argue against that by claiming that your argument
> >>> applies generally and would not allow one to give different
> >>> sectors unequal probabilities. But that's nonsense, because you
> >>> make the hidden assumption of equal probabilities right from the
> >>> start.
> >
> > I simply assume the Schrodinger equation. Then, following Everett, we
> > take it to be deterministic, so that all branches occur on every
> > trial. Since it is deterministic, there is no concept of probability
> > inherent in the Schrodinger equation, and I do not assume any
> > definition of probability. So the branches occur as they occur, there
> > is no assumption of equal probability. It is just that the
> > construction means that all 2^N branches occur on the same basis and
> > necessarily count equally in the overall branching picture.
> >
>
> Why do they necessarily count equally? What is the meaning of the
> wavefunction? Why don't the amplitudes matter?
>
The amplitudes don't matter because the Schrodinger equation is insensitive
to these amplitudes. The Born rule is simply an imposition from outside --
it is not derivable from the SE itself. The amplitudes matter only once you
assume that the theory is probabilistic, then the amplitudes, through the
guessed Born rule, give you a measure of these probabilities. But the SE
itself, as interpreted by Everett, is deterministic, not probabilistic.
There are no probabilities in the SE. So before you introduce probabilities
and the Born rule, the amplitudes do not make any difference. Probabilities
were introduced in order to connect quantum mechanics with the experimental
evidence in one world. And probabilities make sense only in this context.
Strictly, the wave function makes sense only as a way to predict and
calculate probabilities. The idea of wave function realism is just
metaphysics -- there is no experimental evidence for such an idea.
>>> There is nothing in QM that says that branches must count equally,
> >>> and the lottery example I gave makes it clear that you can have
> >>> branching with unequal probabilities in classical physics.
> >
> > As I have said, there is no classical analogue of an interaction in
> > which all outcomes necessarily occur. So your lottery example is
> > useless. There is no concept of probability involved in any of this.
> >
>
> The lottery example I gave clearly is a classical example in which all
> outcomes necessarily occur.
That is a matter of interpreting what an outcome is. In normal parlance,
the outcome of a lottery is the drawing of a winning ticket. There is only
one draw, one winning ticket, one outcome. The other possible outcomes
(which do not occur) are represented by other draws with different winning
tickets. So, as in all classical cases, there is never a situation in which
all possible outcomes occur.
You could take the view that not having the winning ticket is as much an
outcome as winning the lottery. This is a bit contrived, but it does allow
you to say that all outcomes are realized in that all numbered tickets
exist, even though only one of them wins. The probability of winning is
then reduced to branch counting, and the concept of an outcome is reduced
to a triviality. Whereas, in the case where the winning ticket is the only
outcome of relevance, the probability of winning is given solely by the
number of tickets on issue. It has nothing to do with branch counting.
Your reasoning does not involve any QM at
> all, you just apply it to the MWI. Your argument goes through also in
> case of the lottery example, in which case it leads to an obviopusly
> wrong conclusion. So, it's your reasoning that's at fault not the MWI
> taken to be QM minus collapse.
>
My reasoning does involve QM in an essential way. A quantum state is a
vector in Hilbert space that can be expanded in terms of some set of basis
states. If these basis states are pointer states, stable under decoherence,
and each is the eigenstate of some operator with some eigenvalue, then the
set of all possible outcomes of a trial is the set of all these eigenvalues
(assumed, for convenience, to be distinct). These are distinctive quantum
concepts that have no classical analogues.
Bruce
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