On 04-05-2022 01:49, Bruce Kellett wrote:
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.
If you ignore the amplitudes in the states, and that means modifying QM
into something else.
There is no probability
involved. The branches are all equivalent by construction.
What you are constructing is not the result of QM.
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.
Of course you can. The lottery example shows that even in classical
physics you can imagine this happening. If a million copies of you are
made and one will win a lottery whole the rest won't then you have one
in a million chance of experiencing winning the lottery, even though
both outcomes of winning and losing will occur with certainty. One has
to distinguish between the bird's eye and frog's eye view in a setting
where there are copies of observers.
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.
There is the Born rule. If one throws away the Born rule then one has to
specify another model that explains where it comes from. But of you just
throw away the Born rule and don't replace that woth anythong else, then
you are obviously not going to reproduce the same results.
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?
That simply means that there is as of yet no good model for QM without
the Born rule.
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.
If objective collapse exists then one should be able to demonstrate that
in an experiment. There are as of yet no experimental results that
suggests that a collapse mechanism exists.
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.
That's based on assuming a model for the MWI that by construction is
faulty.
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 Schrodinger equation keeps track of the amplitudes. That's actually
the whole point of the Schrodinger equation. It tells you how the
amplitudes assigned to states change over time.
The Born rule is simply an imposition
from outside -- it is not derivable from the SE itself.
The Lorentz force equation is not derivable from the Maxwell equations
either.
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.
The multiverse is deterministic, what observers observe isn't/
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 no evidence for a collapse interpretation either, an collapse
requires a new physical mechanism that has yet to be observed.
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.
Not of we make copies f the same person and consider a lotter draw for
those copies. Then all outcomes occur and yet there is only a small
probability of winning the lottery.
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.
This changes when you consider copies of the same observer.
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.
Yes, but you can't omit the amplitudes. The reasoning you use to do that
is not based on QM and that reasoning can be applied to classical cases
too where it then also gives the wrong results.
Saibal
Bruce
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