On 10/16/2025 4:59 PM, Alan Grayson wrote:
On Thursday, October 16, 2025 at 2:26:20 PM UTC-6 Brent Meeker wrote:
On 10/16/2025 3:41 AM, Alan Grayson wrote:
On Wednesday, October 15, 2025 at 7:52:07 PM UTC-6 Alan Grayson
wrote:
On Tuesday, October 14, 2025 at 10:24:42 PM UTC-6 Brent
Meeker wrote:
On 10/14/2025 12:20 PM, Alan Grayson wrote:
On Monday, October 13, 2025 at 10:28:30 PM UTC-6 Brent
Meeker wrote:
On 10/13/2025 5:04 AM, Alan Grayson wrote:
On Sunday, October 12, 2025 at 11:50:58 PM UTC-6
Brent Meeker wrote:
On 10/12/2025 10:18 PM, Alan Grayson wrote:
On Sunday, October 12, 2025 at 10:37:32 PM
UTC-6 Brent Meeker wrote:
If there's no collapse then every possible
sequence of results is observed in some
world and the relative counts of UP v.
DOWN in the ensemble of worlds will have a
binomial distribution. So for a large
numbers of trials those worlds in which
UPs and DOWNs are roughly equal will
predominate, regardless of what the Born
rule says. So in order that the Born rule
be satisfied for values other than 50/50
there must be some kind of selective
weight that enhances the number of
sequences close to the Born rule instead
of every possible sequence being of equal
weight. But then that is inconsistent with
both values occuring on every trial.
Brent
Why does Born's rule depend on collapse of wf? AG
Where did I say it did?
Brent
The greatest mathematicians tried to prove Euclid's
5th postulate from the other four, and failed; and
the greatest physicists have tried to dervive
Born's rule from the postulates of QM, and failed;,
except for Brent Meeker in the latter case. You
claimed it in the negative, by claiming that
without collapse, Born's rule would fail in some
world of the MWI. An assertion is just that, an
assertion. Can you prove it using mathematics? AG
Sure. Consider a sequence of n=4 Bernoulli trials.
Let h be the number of heads. Then we can make a
table of the number of all possible sequences bc
with exactly h heads and with the corresponding
observed proportion h/n
h bc h/n
0 1 0.0
1 4 0.25
2 6 0.5
3 4 0.75
4 1 1.0
So each possible sequence will correspond to one of
Everett's worlds. For example hhht and hthh belong
to the fourth line h=3. There are sixteen possible
sequences, so there will be sixteen worlds and a
fraction 6/16=0.3125 will exhibit a prob(h)~0.5.
But suppose it was an unfair coin, loaded so that
the probability of tails was 0.9. The possible
sequences are the same, but now we can apply the
Born rule and calculate probabilities for the
various sequences, as follows:
h bc h/n prob
0 1 0.0 0.656
1 4 0.25 0.292
2 6 0.5 0.049
3 4 0.75 0.003
4 1 1.0 0.000
So most of the observers will get empirical answers
that differ drastically from the Born rule values.
The six worlds that observe 0.5 will be off by a
factor of 1.8. And notice the error only becomes
greater as longer test sequences are used. The
number of sequences peak more sharply around 0.5
while the the Born values peak more sharply around 0.9.
Brent
Sorry, I don't quite understand your example? What has
this to-do with collapse of the wf and the MWI? Where is
collapse implied or not? How is Born's rule applied when
the wf is discrete? AG
You wrote, "...claiming that without collapse,/Born's
rule would fail in some world of the MWI/....Can you
prove it using mathematics?" So I showed that in MWI,
which is without collapse, 6 out of 16 experimenters will
observe p=0.5 even in a case in which the Born rule says
the likelihood of p=0.5 is 0.049. Of course your
challenge was confused since it is not Born's rule that
fails. Born's rule is well supported by thousands if not
millions of experiments. Rather it is that MWI
fails...unless it includes a weighting to enforce the
Born rule. But as Bruce points out there is no mechanism
for this. If the experiment is done to measure the
probability (with no assumption of the Born rule) then
there are 16 possible sequences of four measurements and
6 of them give p=0.5 and 6/16=0.375, making p=0.5 the
most likely of the four outcomes. What this has to do
with collapse of the wave function is just that the Born
rule predicts the probabilities of what it will collapse
to. So (assuming MWI) there are still 6 of the 16 who
see 2h and 2t but somehow those 6 experimenters have only
a small weight of some kind. Their existence is kind of
wispy and not-robust.
Brent
I didn't mean to imply that Born's rule is violated. But what
you need to do IMO, is show how Born's rule is applied to
your assumed events as seen without collapse in some world of
the MWI. Otherwise, you just have a set of claims without any
proof of their validity. AG
You say Born's rule will do this or that, but you don't say
exactly HOW it will do this or that. AG
I only wrote "... the Born rule says..." and "... the Born rule
predicts..." If you don't understand how a mathematical formula
can "say" or "predict" I can't help you.
Brent
To use Born's rule, you need a wf.
Not if you already know the probability of |1> and |0> which values I
just assumed. Do you need me to take the square roots and write down
the corresponding wave function, 0.949|0> + 0.316|1>
What is the wf one gets from your h-t scenarios? That is, how do you
calulate Born's rule in your scenario. Why is this so hard to
understand?
For who?
if we have two ways to do the calculation, with collapse and
no-collapse in this-world, and we get different answers, then the MWI
is falsified (assuming that Born's rule give the correct answer). We
can share the prize. AG
No because those aren't the only two possibilities. In fact advocates
of MWI also use the Born rule as a "weight" for the various worlds, but
brushing under the rug the fact that this weight is just the probability
of that world happening. They don't like that because they want all the
worlds to happen, so they think of it as the probability that you
experience that world...even though you experience all of them.
Brent
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