Is Born's rule satisfied in MWI?

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Alan Grayson

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Oct 12, 2025, 9:56:05 PMOct 12
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Correct me if I'm mistaken, but as far as I know the wf has never been observed; only the observations of the system it represents. This being the case, in a large number of trials. Born's rulle will be satisfied regardless of which interpretation an observer affirms; either the MWI with no collapse of the wf, or Copenhagen with collapse of the wf. That is, since we can only observe the statistical results of an experiment from a this-world perspective, and we see that Born's rule is satisfied, so I don't see how it can be argued that the rule fails to be satisfied if the MWI is assumed. I think the same can be said about the other worlds assumed by the MWI, namely, that IF we could measure their results, the rule would likewise be satisfied.AG

Brent Meeker

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Oct 13, 2025, 12:37:32 AMOct 13
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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

On 10/12/2025 6:56 PM, Alan Grayson wrote:
Correct me if I'm mistaken, but as far as I know the wf has never been observed; only the observations of the system it represents. This being the case, in a large number of trials. Born's rulle will be satisfied regardless of which interpretation an observer affirms; either the MWI with no collapse of the wf, or Copenhagen with collapse of the wf. That is, since we can only observe the statistical results of an experiment from a this-world perspective, and we see that Born's rule is satisfied, so I don't see how it can be argued that the rule fails to be satisfied if the MWI is assumed. I think the same can be said about the other worlds assumed by the MWI, namely, that IF we could measure their results, the rule would likewise be satisfied.AG --
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Alan Grayson

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Oct 13, 2025, 1:18:13 AMOct 13
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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 

Brent Meeker

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Oct 13, 2025, 1:50:58 AMOct 13
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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

On 10/12/2025 6:56 PM, Alan Grayson wrote:
Correct me if I'm mistaken, but as far as I know the wf has never been observed; only the observations of the system it represents. This being the case, in a large number of trials. Born's rulle will be satisfied regardless of which interpretation an observer affirms; either the MWI with no collapse of the wf, or Copenhagen with collapse of the wf. That is, since we can only observe the statistical results of an experiment from a this-world perspective, and we see that Born's rule is satisfied, so I don't see how it can be argued that the rule fails to be satisfied if the MWI is assumed. I think the same can be said about the other worlds assumed by the MWI, namely, that IF we could measure their results, the rule would likewise be satisfied.AG --
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Alan Grayson

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Oct 13, 2025, 8:04:01 AMOct 13
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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

Brent Meeker

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Oct 14, 2025, 12:28:30 AMOct 14
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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


On 10/12/2025 6:56 PM, Alan Grayson wrote:
Correct me if I'm mistaken, but as far as I know the wf has never been observed; only the observations of the system it represents. This being the case, in a large number of trials. Born's rulle will be satisfied regardless of which interpretation an observer affirms; either the MWI with no collapse of the wf, or Copenhagen with collapse of the wf. That is, since we can only observe the statistical results of an experiment from a this-world perspective, and we see that Born's rule is satisfied, so I don't see how it can be argued that the rule fails to be satisfied if the MWI is assumed. I think the same can be said about the other worlds assumed by the MWI, namely, that IF we could measure their results, the rule would likewise be satisfied.AG -- 
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Alan Grayson

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Oct 14, 2025, 3:20:02 PMOct 14
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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

Brent Meeker

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Oct 15, 2025, 12:24:42 AMOct 15
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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



On 10/12/2025 6:56 PM, Alan Grayson wrote:
Correct me if I'm mistaken, but as far as I know the wf has never been observed; only the observations of the system it represents. This being the case, in a large number of trials. Born's rulle will be satisfied regardless of which interpretation an observer affirms; either the MWI with no collapse of the wf, or Copenhagen with collapse of the wf. That is, since we can only observe the statistical results of an experiment from a this-world perspective, and we see that Born's rule is satisfied, so I don't see how it can be argued that the rule fails to be satisfied if the MWI is assumed. I think the same can be said about the other worlds assumed by the MWI, namely, that IF we could measure their results, the rule would likewise be satisfied.AG -- 
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Alan Grayson

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Oct 15, 2025, 9:52:07 PMOct 15
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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 colapse in some world of the MWI. Otherwise, you just have a set of claims without any proof of their validity. AG 

Brent Meeker

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Oct 16, 2025, 1:46:02 AMOct 16
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I did exactly that already.  Born's rule is used to assign a weight to each world, which I listed above.  I can't force you to pay attention.

Brent

Alan Grayson

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Oct 16, 2025, 6:04:03 AMOct 16
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Someday you might become a great teacher, but that day has yet to arrive. Better yet, since the MWI is a no-collapse interpretation, can you use say the double slit experiment in this-world, to show that without collapse the MWI is falsified, that it will give a different result than collapse in double slit scenario?  AG 

Alan Grayson

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Oct 16, 2025, 6:41:28 AMOct 16
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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 

Alan Grayson

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Oct 16, 2025, 12:00:47 PMOct 16
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To be more explicit, Born's rule using the normed square of a wf to get a probability. How do you get from your discrete coin tossing scenario, to a wf, in order to show a discrepancy between collapsed and non-collapse probability results? AG 

Brent Meeker

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Oct 16, 2025, 4:26:20 PMOct 16
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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

Alan Grayson

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Oct 16, 2025, 7:59:04 PMOct 16
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To use Born's rule, you need a wf. 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? 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 

Brent Meeker

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Oct 16, 2025, 8:55:27 PMOct 16
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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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Alan Grayson

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Oct 17, 2025, 12:30:12 AMOct 17
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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

Brent Meeker

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Oct 17, 2025, 1:13:05 AMOct 17
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Why don't you ask somebody who believes in MWI, instead of me?

Brent

Alan Grayson

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Oct 17, 2025, 8:42:31 AMOct 17
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So you're not interested in possibly falsifying the MWI? Your attitude is puzzling. 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 
Why don't you ask somebody who believes in MWI, instead of me?

Because you structured your scenario as if multiple worlds can make your measurements. But AFAICT, that's not what the true believers claim. Anyway, doing all measurements in one world, this world, seems sufficient to possibly falsify the interpretation. IMO, it needs to be falsified, so this false path to reality can finally be put in the dust bin of history. AG

Brenta

Brent Meeker

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Oct 17, 2025, 7:01:59 PMOct 17
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But it can't be falsified if you add the Born rule to it, which advocates of MWI do.  They just apply it to what they call "self-locating uncertainty", which I think is double-talk for "the only world that happened".

If you think it can be falsified, write out the experiment that will do so.

Brent

Alan Grayson

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Oct 17, 2025, 10:54:30 PMOct 17
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On 10/17/2025 5:42 AM, Alan Grayson wrote
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 

So you're not interested in possibly falsifying the MWI? Your attitude is puzzling. 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 
Why don't you ask somebody who believes in MWI, instead of me?

Because you structured your scenario as if multiple worlds can make your measurements. But AFAICT, that's not what the true believers claim. Anyway, doing all measurements in one world, this world, seems sufficient to possibly falsify the interpretation. IMO, it needs to be falsified, so this false path to reality can finally be put in the dust bin of history. AG
But it can't be falsified if you add the Born rule to it, which advocates of MWI do.  They just apply it to what they call "self-locating uncertainty", which I think is double-talk for "the only world that happened".

If you think it can be falsified, write out the experiment that will do so.

Brent

Yes, now that I understand your coin tossing model, I believe I know how it can done, and when I write it up, I'll ask AI to write the program and do the calculation. I think if we work only in this-world, we might be able to show that the collapse model, which we know gives the right result, will differ from the no-collapse model. Do you agree that if no-collapse is falsified in this world, this is sufficient for the proof that it's nonsense? AG

Brent Meeker

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Oct 17, 2025, 11:38:50 PMOct 17
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Depends on what you mean by "falsified".  QM is a probabilistic theory, so simply showing it is wrong about one particular experiment can be marked up to statistics.  But if you mean statistically falsified at a high confidence level, sure.

Brent

Alan Grayson

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Oct 19, 2025, 1:43:20 AMOct 19
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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

Any particular reason you labeled second column as bc? AG 

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>

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 

Brent Meeker

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Oct 19, 2025, 2:29:18 AMOct 19
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On 10/18/2025 10:43 PM, Alan Grayson wrote:

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

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? 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>

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. 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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Alan Grayson

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Oct 19, 2025, 8:15:35 AMOct 19
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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


By the above paragraph, it seems you've already falsified the MWI, except that you could claim that's what no-collapse yields in this-world. I don't see any reason for claiming each sequence is observed in different worlds.
Any particular reason you labeled second column as bc? AG 
Yes, it's an abbreviation.

What does bc stand for? AG 

Brent Meeker

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Oct 19, 2025, 9:43:45 PMOct 19
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Branch count

Liz R

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Oct 20, 2025, 3:17:31 AMOct 20
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If we assume that there are distinct universes which branch, then the Born rule isn't going to be satisfied (at least, not without some sort of contrived epicycles) in any situation where the probabilities aren't in some integer ratio, e.g. if they're irrational. If on the other hand we assume that there is a continuum of identical universes that is partitioned by a measurement (as David Deutsch suggests in "The Fabric of Reality") then the partitioning can be as finely divided as you like. However, continua are possibly problematic in actual physical systems, like infinities, that is to say, not realistic (because they effectively involve dividing by (uncountable?) infinity). The idea that spacetime can't be infinitely warped - that singularities are unphysical - is related to the idea that continua can't exist. I assume a theory of quantum gravity would not allow either.

In the absence of continua the Born rule can only be satisfied in a multiverse if all measurements split the universes into some integer ratio. This seems rather arbitrary - a measurement with a 1% chance of result X and 99% of result Y produces 100 branches (99 indistinguishable from each other), while a measurement with a 50-50 chance produces 2.

A multiverse has philosophical appeal - the string landscape answers the question "why these laws of physics?" while the quantum multiverse answers the question "why this history?" However as far as I know there is no strong scientific (testable, refutable, etc) evidence for either.

Alan Grayson

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Oct 20, 2025, 5:18:21 AMOct 20
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On Sunday, October 19, 2025 at 6:15:35 AM UTC-6 Alan Grayson wrote:
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


By the above paragraph, it seems you've already falsified the MWI, except that you could claim that's what no-collapse yields in this-world. I don't see any reason for claiming each sequence is observed in different worlds. AG

You seem very close to proving that the no-collapse interpretation, aka MWI, gives very wrong results, but I see no interest in publishing it. Why not expand your argument and publish it? AG
Any particular reason you labeled second column as bc? AG 
Yes, it's an abbreviation.?

What does bc stand for? AG 


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>

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.

How is that calculation done? TY, AG 

Brent Meeker

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Oct 20, 2025, 3:55:50 PMOct 20
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Measurements will always produce integer ratios as estimates of probabilities.  If the estimate is sufficiently close to the predicted value it is considered consistent with the Born rule.  We can never measure exactly real number values.

Brent
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Liz R

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Oct 20, 2025, 7:48:42 PMOct 20
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True, because we can only measure to finite precision. However that doesn't deal with what might be called the "1% problem".

Alan Grayson

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Oh, I see. The square must be .05. ... Indulge me on this. When I studied QM, we akways used S's equation to solve for the wf, and it was always a real valued complex function. So it was simple to find its norm using complex conjugates. But I don't know how to find the norm for a linear sum of bras. I'd like to see how its done. AG

Brent Meeker

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Oct 21, 2025, 1:26:47 AMOct 21
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On 10/20/2025 7:23 PM, Alan Grayson wrote:


On Monday, October 20, 2025 at 3:18:21 AM UTC-6 Alan Grayson wrote:
On Sunday, October 19, 2025 at 6:15:35 AM UTC-6 Alan Grayson wrote:
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


By the above paragraph, it seems you've already falsified the MWI, except that you could claim that's what no-collapse yields in this-world. I don't see any reason for claiming each sequence is observed in different worlds. AG
There's no unique sequence "in this world" because there's no unique "this world" in MWI.

Brent

You seem very close to proving that the no-collapse interpretation, aka MWI, gives very wrong results, but I see no interest in publishing it. Why not expand your argument and publish it? AG
Any particular reason you labeled second column as bc? AG 
Yes, it's an abbreviation.?

What does bc stand for? AG 


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>

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.

How is that calculation done? TY, AG 

Oh, I see. The square must be .05. ... Indulge me on this. When I studied QM, we akways used S's equation to solve for the wf, and it was always a real valued complex function. So it was simple to find its norm using complex conjugates. But I don't know how to find the norm for a linear sum of bras. I'd like to see how its done. AG

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. 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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Alan Grayson

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Oct 21, 2025, 1:52:15 AMOct 21
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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
By the above paragraph, it seems you've already falsified the MWI, except that you could claim that's what no-collapse yields in this-world. I don't see any reason for claiming each sequence is observed in different worlds. AG
There's no unique sequence "in this world" because there's no unique "this world" in MWI.

Brent

IMO this is ridiculous. How can you disprove the MWI when you accept its foolish claim of many worlds? All that's required is to show that the no-collapse hypothesis gives wrong results compared to Born's rule in the only world you know for sure, THIS-WORLD. AG

Brent Meeker

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Oct 21, 2025, 4:35:05 PMOct 21
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The no collapse hypothesis gives wrong results in some worlds and not in others.  The problem is how you assign probabilities to these worlds.  MWI advocates use the Born rule to assign probabilities to the different branches and so produce an interpretation empirically identical to the neo-Copenhagen interpretation.  I think it fails in the sense that it can produce many observers, even a majority, existing in low probability branches who cannot know they are in low probability branches and so are deceived by their observations into falsifying QM.  MWI dismisses them as low probability even though they are numerous.  Copenhagen says "low probability" means they likely don't exist.  So it is a philosophical disagreement about the meaning of applied probability.

Brent
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Alan Grayson

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Oct 22, 2025, 3:56:28 AMOct 22
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Since you're a master of plots, how difficult would it be to produce three plots of the double slit experiment, with as many single events as you deem suitable? First plot would simulate the result of the experiment; the second would demonstrate the prediction using the collapse model; and the third would simulate the no-collapse model. Before we allow the many-worlders to confuse the issue, let's see if the no collapse model make the predictive cut in THIS WORLD. AG

Brent Meeker

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Oct 22, 2025, 7:54:40 PMOct 22
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Did you miss the part about MWI advocates using the Born rule in their interpretation?  Without it, the MWI is the same as the Born rule when p=0.5, no matter what the Schroedinger equation says p is.  It's what MWI advocates dismiss as "branch counting".

Brent
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Alan Grayson

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Oct 25, 2025, 10:04:47 AMOct 25
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I'm not sure I understand your comment. You seem to be claiming the Many Worlders get the same result as the collapse model for a special case of p=0.5. But do they get the same result in THIS-WORLD for the double slit, which collapses to a huge number of outcomes? If not, then the MWI does not satisfy Born's rule. AG 

Brent Meeker

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Oct 25, 2025, 7:25:07 PMOct 25
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When p=0.5 branch counting is the same as the Born rule.  In a double slit experiment the probability of each slit is 0.5 and all you get is an interference pattern, no counts of this vs. that.

Brent
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Alan Grayson

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Oct 26, 2025, 12:17:57 AMOct 26
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CMIIAW, but Born's rule is used on a wf, and the wf I have in mind exists after the test particle goes through the slits. I don't see that the probability of 0.5 of going through a slit has anything to do with a wf. We can apply basic logic to get that result. AG 

Alan Grayson

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Oct 26, 2025, 12:21:00 AMOct 26
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And 0.5 depends on modeling the test entity as particle, which it is not if which-way isn't being observed. AG  

smitra

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Oct 27, 2025, 8:28:35 AMOct 27
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On 13-10-2025 03:56, Alan Grayson wrote:
> Correct me if I'm mistaken, but as far as I know the wf has never been
> observed; only the observations of the system it represents. This
> being the case, in a large number of trials. Born's rulle will be
> satisfied regardless of which interpretation an observer affirms;
> either the MWI with no collapse of the wf, or Copenhagen with collapse
> of the wf. That is, since we can only observe the statistical results
> of an experiment from a this-world perspective, and we see that Born's
> rule is satisfied, so I don't see how it can be argued that the rule
> fails to be satisfied if the MWI is assumed. I think the same can be
> said about the other worlds assumed by the MWI, namely, that IF we
> could measure their results, the rule would likewise be satisfied.AG
>

I've scanned through part of this thread, so perhaps someone has already
said what I'm going to say here. A special case of the Born rule is when
measuring an observable A when the state is in an eigenstate of A. In
that case you find the eigenvalue of A for that eigenstate the system is
already in with 100% certainty. We can then exploit this fact to try to
set up an experiment to verify the Born rule as a quantum experiment
where the outcome is either 0 corresponding to the Born rule being
falsified and 1 corresponding to the Born rule being verified. If we can
then come up with such an observable for testring the Born rule, then if
the Born rule is true, any quantum state should be an eigenstate with
eigenvalue 1.

The problem here is that we can only ever do a finite number of
measurements and we don't have a sharp rejection or verification of the
Born rule. And there are then always exceptional states where the
statistics will start to agree with the Born rule arbitrarily late. We
can at most write down for any arbitrary integer an observable for
repeating an experiment N times in a quantum coherent way where each
individual measurement doesn't collapse the wavefunction (e.g.
measurement conducted by a quantum computer and stored in that quantum
computer), and only at the end we perform a measurement on the entire
statistics stored in the quantum computer.

Suppose we set things up such that there is qubit that will take the
value 1 if the hypothesis that the Born rule is false is ruled out at
99% confidence level. Then that qubit will be in a superposition of 1
and 0 and with N very large, the amplitude for 0 will be extremely
small. But the amplitude for 0 will only become zero in the limit of N
to infinity.

So, there is still a problem here, but at heart this is actually a
problem with the very concept of probability, which is not well defined
in physical contexts where one can only ever do a finite number of
measurements:

https://www.youtube.com/watch?v=wfzSE4Hoxbc&t=1036s

Saibal

Brent Meeker

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Oct 29, 2025, 12:58:48 PMOct 29
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But that's a completely general problem with all of science.  No theory is ever proven true.  It may be proven false or restricted in its domain of application, but however much evidence is found in its favor, it is not proven true.  That's why there's interest in using Bayes's theorem to quantify the degree of confidence we should ascribe to a theory.  It uses probability as "degree of justified belief".  It has nothing to do with frequencies and limits at infinity.

The problem with MWI is that, in 1 or 0 test sequence, the branching structure produces numbers of observers with a binomial distribution of p=0.5 no matter the probability in the Born rule, P.  The Born rule is satisfied for the QM predicted probability, but not by the branch counting.  This is mathematically and empirically consistent.  But to me it is seems philosophically inconsistent.  

In an experiment with P=0.9 the branches consistent with this will have the most weight, but the branches consistent with p=0.5 will be much more numerous but of low weight.  In Copenhagen they simply have a low probability of being the one existent branch, but what are we to make of their existence in MWI?  Almost all physicists will be getting the wrong answer...but that's OK because they have "low weight"!  Their existence is thin in some sense?  What does having "low weight"  mean.  If it's a probability, as assumed in it's calculation, what is it a probability of?  If is a probability of happening, then a low value means it probably didn't happen...not that it happened thinly.

Brent

Alan Grayson

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Oct 30, 2025, 12:57:21 AMOct 30
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To clarify: since the MWI is a no-collapes model, I though we could compare a collapse model, Copenhagen, with a no-collapse model, and look for discrepencies in what they predict. I also thought we could use the double slit experiment to do this comparison. But now I think this is impossible because the wf is never observed, so there doesn't seem any way to distinguish the cases I'd like to compare. Further, to get interference in the double slit experiiment, we must NOT do the which-way measurement. THEREFORE, we can NOT assume a probability of 0.5 of the test particle going through each slit, since this model is strictly reserved for the model of doing a which-way measurement. Maybe you can explain again how the MWI incorporates Born's rule. TY, AG
 

Brent Meeker

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Oct 30, 2025, 1:37:00 AMOct 30
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If the probability is not the same at each slit, then where the waves are 180deg out of phase, where we expect dark bands, the probability amplitudes won't cancel and the bands won't be perfectly dark.

Brent
 
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Alan Grayson

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Oct 30, 2025, 2:17:52 AMOct 30
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On Wednesday, October 29, 2025 at 11:37:00 PM UTC-6 Brent Meeker wrote:


On 10/29/2025 9:57 PM, Alan Grayson wrote:


On Saturday, October 25, 2025 at 10:21:00 PM UTC-6 Alan Grayson wrote:
On Saturday, October 25, 2025 at 10:17:57 PM UTC-6 Alan Grayson wrote:
On Saturday, October 25, 2025 at 5:25:07 PM UTC-6 Brent Meeker wrote:


Did you miss the part about MWI advocates using the Born rule in their interpretation?  Without it, the MWI is the same as the Born rule when p=0.5, no matter what the Schroedinger equation says p is.  It's what MWI advocates dismiss as "branch counting".

Brent

I'm not sure I understand your comment. You seem to be claiming the Many Worlders get the same result as the collapse model for a special case of p=0.5. But do they get the same result in THIS-WORLD for the double slit, which collapses to a huge number of outcomes? If not, then the MWI does not satisfy Born's rule. AG 
When p=0.5 branch counting is the same as the Born rule.  In a double slit experiment the probability of each slit is 0.5 and all you get is an interference pattern, no counts of this vs. that.

Brent

CMIIAW, but Born's rule is used on a wf, and the wf I have in mind exists after the test particle goes through the slits. I don't see that the probability of 0.5 of going through a slit has anything to do with a wf. We can apply basic logic to get that result. AG 

And 0.5 depends on modeling the test entity as particle, which it is not if which-way isn't being observed. AG

To clarify: since the MWI is a no-collapes model, I though we could compare a collapse model, Copenhagen, with a no-collapse model, and look for discrepencies in what they predict. I also thought we could use the double slit experiment to do this comparison. But now I think this is impossible because the wf is never observed, so there doesn't seem any way to distinguish the cases I'd like to compare. Further, to get interference in the double slit experiiment, we must NOT do the which-way measurement. THEREFORE, we can NOT assume a probability of 0.5 of the test particle going through each slit, since this model is strictly reserved for the model of doing a which-way measurement. Maybe you can explain again how the MWI incorporates Born's rule. TY, AG
If the probability is not the same at each slit, then where the waves are 180deg out of phase, where we expect dark bands, the probability amplitudes won't cancel and the bands won't be perfectly dark.

Brent

Can you elaborate the exact relationship of your response, to my previous post? Sorry, but it's over my head. AG

Alan Grayson

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Oct 30, 2025, 7:56:24 AMOct 30
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On Thursday, October 30, 2025 at 12:17:52 AM UTC-6 Alan Grayson wrote:
On Wednesday, October 29, 2025 at 11:37:00 PM UTC-6 Brent Meeker wrote:


On 10/29/2025 9:57 PM, Alan Grayson wrote:
On Saturday, October 25, 2025 at 10:21:00 PM UTC-6 Alan Grayson wrote:
On Saturday, October 25, 2025 at 10:17:57 PM UTC-6 Alan Grayson wrote:
On Saturday, October 25, 2025 at 5:25:07 PM UTC-6 Brent Meeker wrote:


Did you miss the part about MWI advocates using the Born rule in their interpretation?  Without it, the MWI is the same as the Born rule when p=0.5, no matter what the Schroedinger equation says p is.  It's what MWI advocates dismiss as "branch counting".

Brent

I'm not sure I understand your comment. You seem to be claiming the Many Worlders get the same result as the collapse model for a special case of p=0.5. But do they get the same result in THIS-WORLD for the double slit, which collapses to a huge number of outcomes? If not, then the MWI does not satisfy Born's rule. AG 
When p=0.5 branch counting is the same as the Born rule.  In a double slit experiment the probability of each slit is 0.5 and all you get is an interference pattern, no counts of this vs. that.

Brent

CMIIAW, but Born's rule is used on a wf, and the wf I have in mind exists after the test particle goes through the slits. I don't see that the probability of 0.5 of going through a slit has anything to do with a wf. We can apply basic logic to get that result. AG 

And 0.5 depends on modeling the test entity as particle, which it is not if which-way isn't being observed. AG

To clarify: since the MWI is a no-collapes model, I thoughT we could compare a collapse model, Copenhagen, with a no-collapse model, and look for discrepencies in what they predict. I also thought we could use the double slit experiment to do this comparison. But now I think this is impossible because the wf is never observed, so there doesn't seem any way to distinguish the cases I'd like to compare. Further, to get interference in the double slit experiiment, we must NOT do the which-way measurement. THEREFORE, we can NOT assume a probability of 0.5 of the test particle going through each slit, since this model is strictly reserved for the model of doing a which-way measurement. Maybe you can explain again how the MWI incorporates Born's rule. TY, AG
If the probability is not the same at each slit, then where the waves are 180deg out of phase, where we expect dark bands, the probability amplitudes won't cancel and the bands won't be perfectly dark.

Brent
Can you elaborate the exact relationship of your response, to my previous post? Sorry, but it's over my head. AG

When there's no which-way measurement, the electron is a wave and that wave goes through both slits equally. Is this what you mean by a probability of 0.5 through each slit? If so, I think it's a non-standard usage. AG 

Alan Grayson

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Oct 31, 2025, 4:51:51 AM (14 days ago) Oct 31
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If there's no which-way measurement, the electron is a wave and goes through both slits simultaneously. In this case one cannot assign a probability of 0.5 for each slit. AG 
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