On 28-04-2022 07:51, Bruce Kellett wrote:
On Thu, Apr 28, 2022 at 3:24 PM Brent Meeker <meekerbr...@gmail.com>
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
change the conclusion of most observers that the actual
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
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
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
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?
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. 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.
Yes, there's nothing in QM that says the branches must count
equally. But there's also nothing in the evolution of Schroedingers
equation that they must count as _a^2_ and _b^2_. Of course IF they
are probabilities then it follows from Gleason's theorem that they
follow the Born rule. But in that case you have reintroduced almost
all the philosophical problems of the Copenhagen interpretation.
When exactly does this splitting occur? Can the split be into
irrational numbers of branches? A splitting is in some particular
basis and not in other bases. What determines the pointer basis?
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
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 view this discussion on the web visit