On 10/18/2025 10:43 PM, Alan Grayson wrote:
Why does Born's rule depend on collapse of wf? AGWhere 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? AGSure. 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*Any particular reason you labeled second column as bc? AG *
Yes, it's an abbreviation.
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? AGYou 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. AGI 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>*Is this wf for the biased coin? For the unbiased, I would expect the multiplying parameters would be the same and equal to .5. AG *
No, that would be 0.707 for each. Brent
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. AGNo 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 --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/385098a8-4b20-43c1-92aa-cfa41a6cf015n%40googlegroups.com <https://groups.google.com/d/msgid/everything-list/385098a8-4b20-43c1-92aa-cfa41a6cf015n%40googlegroups.com?utm_medium=email&utm_source=footer>.
-- 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/45f509c7-3d8f-4711-bcb0-a45ad4bf9585%40gmail.com.
