On Thursday, October 16, 2025 at 6:55:27 PM UTC-6 Brent Meeker wrote:
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>
*So, IMO, we need a computer simulation which systematically tests a huge
number of probabilities, and their wf's, to determine any difference
between collapse and no-collapse interpretations. I suspect the latter will
fail Born's rule in every case, falsifying the no-collapse interpretation.
Also, one need to do this experiment in this-world only, since the worlds
of the MWI are indistinguishable. AG *
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.
*How can we experience all the worlds? We only experience one world, this
world. AG *
Brent
--
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 [email protected].
To view this discussion visit
https://groups.google.com/d/msgid/everything-list/6a179446-4a6b-4460-869b-2fc8b094ffc8n%40googlegroups.com.
