On 4/26/2022 5:32 PM, smitra wrote:
On 27-04-2022 01:37, Bruce Kellett wrote:
On Tue, Apr 26, 2022 at 10:03 AM smitra <[email protected]> wrote:
On 24-04-2022 03:16, Bruce Kellett wrote:
A moment's thought should make it clear to you that this is not
possible. If both possibilities are realized, it cannot be the
case
that one has twice the probability of the other. In the long run,
if
both are realized they have equal probabilities of 1/2.
The probabilities do not have to be 1/2. Suppose one million people
participate in a lottery such that there will be exactly one winner.
The
probability that one given person will win, is then one in a
million.
Suppose now that we create one million people using a machine and
then
organize such a lottery. The probability that one given newly
created
person will win is then also one in a million. The machine can be
adjusted to create any set of persons we like, it can create one
million
identical persons, or almost identical persons, or totally different
persons. If we then create one million almost identical persons, the
probability is still one one in a million. This means that the limit
of
identical persons, the probability will be one in a million.
Why would the probability suddenly become 1/2 if the machine is set
to
create exactly identical persons while the probability would be one
in a
million if we create persons that are almost, but not quite
identical?
Your lottery example is completely beside the point.
It provides for an example of a case where your logic does not apply.
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.
But it shows why you can't use branch counting. There's no physical
mechanism for translating the /a/ and /b/ of /|psi> = a|0> + b|1>/ into
numbers of branches. To implement that you have put it in "by hand"
that the branches have weights or numerousity of /a /and /b/. This is
possible, but it gives the lie to the MWI mantra of "It's just the
Schroedinger equation."
Brent
The important point to notice is that this result of all possible
binary sequences for N trials is independent of the coefficients in
the binary expansion of the state:
.
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. 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.
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. 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.
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?
Brent
Saibal
Bruce
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