Physicists disagree wildly on what quantum mechanics says about reality

[John Clark]

In today's issue of the journal Nature is a report of a survey of physicists that Nature conducted about the meaning of Quantum Mechanics, it's interesting that even after a century there is not something even close to a consensus. Even two physicists who won a Nobel prize for performing an experiment that explores the qualities of quantum reality disagree with each other over what their experiment is trying to tell us.  Another physicist says "The implication [of the survey] is that many quantum researchers simply use quantum theory without engaging deeply with what it means — the ‘shut up and calculate’ approach."

Physicists disagree wildly on what quantum mechanics says about reality

John K Clark    See what's on my new list at  Extropolis

vas

[smitra]

See here:

https://www.youtube.com/watch?v=bux0SjaUCY0&t=885s

Of course, you can never get to 100% rigorous proof in physics like in
mathematics. You can never rule out waking up tomorrow in some alien
world and aliens telling you that your life here on Earth was a
simulation and that everything you thought you knew about the laws of
physics is false.

The nice thing about the argument by Deutsch is that it doesn't depend
on QM being correct, it is based on interpreting the interference
experiment. So, QM could be wrong, or it could be that some of the
claims of the MWI proponents are wrong, and yet this argument by Deutsch
will still stand.

Saibal

[Bruce Kellett]

Sounds like the usual sort of nonsense from Deutsch.

Bruce

[Brent Meeker]

Whatever goes thru the other slit to create the cancellation is doing it in the same universe; and incidentally it also increases the incidence at other points.  So I don't see that it implies a parallel universe.

Brent

[smitra]

But then one is defining "this universe" post hoc after the photon lands
on some point on the screen. One can do that, but this means that there
are two possibilities that really exist. Whether that's considered to be
in a single universe or two universes is just semantics. There are two
classes of paths for the photon, one class is the set of paths through
one slit, the other are the paths through the other slit.

Given that's the case, we can then do another measurement where we
simply measure through which slit the photon goes. We then don't bother
to let the photon move through toward the screen anymore, we just detect
the outcome of measuring whether the photon after moving past either
slits ends up being detected immediately after the left or the right
slit.

If I then perform one such measurement, and I decide to go on vacation
destination X if the photon is detected behind the left slit and I go to
vacation destination Y if the result is the right slit, and I end up
going to X, the question is if there exists a parallel world where I go
to Y.

The question for people who would say that only one world where I go to
X exists, is then to explain why both possibilities for the photon going
to the left or right slit objectively exists when we detect the photon
only at the screen, but only one possibility exists when we detect the
photon directly after passing the slits.

Saibal

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[Brent Meeker]

On 8/25/2025 12:04 AM, smitra wrote:

But then one is defining "this universe" post hoc after the photon lands on some point on the screen. One can do that, but this means that there are two possibilities that really exist. Whether that's considered to be in a single universe or two universes is just semantics. There are two classes of paths for the photon, one class is the set of paths through one slit, the other are the paths through the other slit.

Given that's the case, we can then do another measurement where we simply measure through which slit the photon goes. We then don't bother to let the photon move through toward the screen anymore, we just detect the outcome of measuring whether the photon after moving past either slits ends up being detected immediately after the left or the right slit.

If I then perform one such measurement, and I decide to go on vacation destination X if the photon is detected behind the left slit and I go to vacation destination Y if the result is the right slit, and I end up going to X, the question is if there exists a parallel world where I go to Y.

The question for people who would say that only one world where I go to X exists, is then to explain why both possibilities for the photon going to the left or right slit objectively exists when we detect the photon only at the screen, but only one possibility exists when we detect the photon directly after passing the slits. 

The answer is, Don't confound possibility with reality.  Possibilities "objectively exist" doesn't mean the corresponding realities exist.  When you measure at one slit there is a different reality than when measuring at the screen.

Brent

[Quentin Anciaux]

Brent,

Saying that a possibility "objectively exists" but does not correspond to any reality is paradoxical. Either it exists, in which case it is part of reality, or it doesn’t, in which case it’s not an objective possibility.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Brent Meeker]

"Correspond" is a weak word. An expectation corresponds to the thing expected.  And the expectation is real.  But it doesn't follow that the thing expected is realized.  Possibility is not equivalent to reality. 

Brent

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[Quentin Anciaux]

If you claim possibilities “objectively exist” but are never realized, you’re redefining “existence” into something empty. An "objective" possibility with no ontological counterpart is indistinguishable from non-existence.

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Brent Meeker]

Where did I use the word "never"?

Is it your contention that every possibility must be realized, in which case "possibility" loses all meaning and might as well be replaced by reality.

Brent

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[Quentin Anciaux]

Not all possibilities are realized, but those with non-zero amplitude are realized somewhere in the superposition. In MWI, "possibility" refers to branches with different measures, not to mere logical abstractions. A "possible" event with zero measure is equivalent to non-existence.

Quentin 

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Tue, Aug 26, 2025 at 2:03 PM Quentin Anciaux <allc...@gmail.com> wrote:

Not all possibilities are realized, but those with non-zero amplitude are realized somewhere in the superposition. In MWI, "possibility" refers to branches with different measures, not to mere logical abstractions. A "possible" event with zero measure is equivalent to non-existence.

In quantum mechanics, the Born rule is confirmed in experiments by counting the number of times a particular result is obtained in repeated trials. The proportion of 'successes' gives an approximate measure of the Born rule probability of that result. In other words, the observed proportion is a reliable estimate of the absolute value squared of the amplitude of that eigenfunction in the wave function (the Born rule).

In any model (such as MWI) in which every possibility is realized on every trial, the Born rule is not confirmed in the majority of cases. This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence. The proportion of ones takes on any value between zero and unity, whereas these sequences are all generated from the same wave function with a particular value for the amplitude of a |0> outcome. In most cases, therefore, the observers who obtain each particular proportion of zeros will fail to confirm the Born rule. This is more apparent as the amplitude of the |0> component can be changed to any value between zero and one, while the number of successes in each string remains unchanged.

Bruce

[Brent Meeker]

On 8/25/2025 9:03 PM, Quentin Anciaux wrote:

Not all possibilities are realized, but those with non-zero amplitude are realized somewhere in the superposition. 

An article of faith among Everretians, with no empirical support.  Have you studied Barandes' MMI?

Brent

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[Quentin Anciaux]

You’re conflating mathematical possibilities with experiential frequencies. Yes, all 2^N sequences exist in the wavefunction, but their measures are not equal. The Born rule emerges because observers are not uniformly distributed across these sequences: almost all self-locating observers end up in high-measure branches. If you assume equal weighting, you’re rejecting the very structure of the wavefunction and replacing quantum mechanics with branch counting.

Quentin 

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Tue, Aug 26, 2025 at 4:21 PM Quentin Anciaux <allc...@gmail.com> wrote:

You’re conflating mathematical possibilities with experiential frequencies. Yes, all 2^N sequences exist in the wavefunction, but their measures are not equal. The Born rule emerges because observers are not uniformly distributed across these sequences: almost all self-locating observers end up in high-measure branches. If you assume equal weighting, you’re rejecting the very structure of the wavefunction and replacing quantum mechanics with branch counting.

I do not assume equal weighting. This argument has been tried before and still fails because all 2^N sequences necessarily exist when every possibility is realized on every trial. There is no probability associated with the sequences, and all have equal weight simply by construction. The "measure" you talk about is your own fantasy -- there is no unequal "measure" between the sequences. You cannot give any reasonable account of any such "measure" -- it does not follow from the Schroedinger equation or any other postulate of quantum mechanics.

Bruce

[Quentin Anciaux]

MMI simply postulates a single realized outcome and takes the Born rule as given, so it doesn't actually address the derivation problem. In MWI, amplitudes naturally define relative measures, which MMI leaves unexplained.

Quentin 

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Quentin Anciaux]

Bruce,

The measure isn’t something I’m inventing, it’s implicit in the squared amplitudes of the wavefunction. The Schrödinger equation preserves the L² norm, and decoherence ensures that branches with extremely low amplitudes contribute negligibly to observer statistics. Ignoring that structure and treating all branches as equally weighted is not quantum mechanics, it’s just branch counting under a flat prior.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Tue, Aug 26, 2025 at 4:56 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

The measure isn’t something I’m inventing, it’s implicit in the squared amplitudes of the wavefunction. The Schrödinger equation preserves the L² norm, and decoherence ensures that branches with extremely low amplitudes contribute negligibly to observer statistics. Ignoring that structure and treating all branches as equally weighted is not quantum mechanics, it’s just branch counting under a flat prior.

The trouble with that idea is that the Schrodinger equation is insensitive to the amplitudes. You can change the amplitudes on the wavefunction with binary outcomes and you get exactly the same set of 2^N sequences. So actually all the sequences have the same weight -- construction from repeated trials with both outcomes realized on every trial ensures that all 2^N sequences occur with equal weight. So there is no way in which there are low weight or low probability sequences. Anyway, such an argument fails because you can change the Born probability of a zero at will, by changing the amplitudes in the original wave function. And the proportion of zeros in any sequence is an estimate of the probability of obtaining a zero. These proportions change over sequences, so most will give a probability estimate that disagrees with the Born probability.

Bruce

[Quentin Anciaux]

Bruce,

You’re conflating two different levels: the combinatorial existence of 2^N sequences and their statistical weight. Yes, the Schrodinger equation produces all 2^N sequences regardless of amplitudes, but the amplitudes control the measure over those sequences. Changing |a|² and |b|² doesn’t alter which sequences exist, it alters how observers are distributed among them. That’s exactly what the Born rule captures: almost all self-locating observers end up in high-measure branches. Treating all sequences as equally weighted is equivalent to replacing quantum mechanics with flat branch counting, which is not what the formalism prescribes.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Tue, Aug 26, 2025 at 6:15 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

You’re conflating two different levels: the combinatorial existence of 2^N sequences and their statistical weight. Yes, the Schrodinger equation produces all 2^N sequences regardless of amplitudes, but the amplitudes control the measure over those sequences. Changing |a|² and |b|² doesn’t alter which sequences exist, it alters how observers are distributed among them. That’s exactly what the Born rule captures: almost all self-locating observers end up in high-measure branches. Treating all sequences as equally weighted is equivalent to replacing quantum mechanics with flat branch counting, which is not what the formalism prescribes.

It seems like you are using the Born rule to assign your weights, or measure, to the sequences. This is illegitimate, because the the Born rule is exactly what is in question in my analysis. I have shown that the Born rule is disconfirmed by the data obtained in the vast majority of sequences. The test of the Born rule is that the proportion of successes in the sequence approximates the Born probability of that success. In the binary case under discussion, the majority of sequences will have approximately equal numbers of zeros and ones as the number of trials, N, becomes large. The Born probability of obtaining a zero for these 50/50 sequences can be any value between zero and one, depending on the original amplitudes. Your model takes no account of this.

Bruce

[Quentin Anciaux]

Bruce,

You’re assuming that each sequence has equal weight by construction, but that’s exactly what quantum mechanics denies. The Schrödinger equation doesn’t just produce sequences, it produces them with amplitudes, and those amplitudes determine the statistical weight via |a|² and |b|². The fact that sequences exist mathematically doesn’t mean they are sampled uniformly by observers. Without introducing measure, your argument effectively replaces quantum mechanics with flat branch counting, which is inconsistent with the formalism itself.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Tue, Aug 26, 2025 at 8:25 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

You’re assuming that each sequence has equal weight by construction, but that’s exactly what quantum mechanics denies. The Schrödinger equation doesn’t just produce sequences, it produces them with amplitudes, and those amplitudes determine the statistical weight via |a|² and |b|². The fact that sequences exist mathematically doesn’t mean they are sampled uniformly by observers. Without introducing measure, your argument effectively replaces quantum mechanics with flat branch counting, which is inconsistent with the formalism itself.

I see. You are actually using the Born rule probabilities to provide your weights. That is putting the cart before the horse: you have not actually demonstrated that the Born rule even works in this model. It has nothing to do with branch counting as the origin of the Born probabilities. I do not use branch counting at any stage, and neither do I ever have to assume that the branches have equal weights or equal probabilities. That is all a canard of your own making. I simply count the number of zeros and ones in each sequence, and compare the proportions to the predicted Born probabilities. In most cases, one finds that the estimated probabilities disagree with the Born rule predictions. You have not countered this simple argument.

In fact, the model produces a single observer for every sequence, so your claim that the sequences are not sampled uniformly by observers is contradicted by the simple mathematics of unitary evolution.

Bruce

[Quentin Anciaux]

Bruce,

You keep assuming one observer per sequence and uniform sampling, but that assumption is yours, not Everett’s. In Everett’s framework, the relative weights aren’t arbitrary — they’re determined by the amplitudes in the wavefunction. By rejecting that, you’re refuting a simplified model of your own making, not MWI itself. If your argument truly applied to Everett’s theory, you should be able to show how it addresses the role of measure instead of ignoring it.

Quentin

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[Bruce Kellett]

On Tue, Aug 26, 2025 at 10:30 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

You keep assuming one observer per sequence and uniform sampling, but that assumption is yours, not Everett’s. In Everett’s framework, the relative weights aren’t arbitrary — they’re determined by the amplitudes in the wavefunction. By rejecting that, you’re refuting a simplified model of your own making, not MWI itself. If your argument truly applied to Everett’s theory, you should be able to show how it addresses the role of measure instead of ignoring it.

I think you need to actually work through the calculation/construction. One observer per sequence is implied by Everett's theory. If you have an initial state such as

  a|0> + b|1>

and every possible result is realized in every trial: the first trial gives you two branches, with results |0> and |1> respectively. The observer splits so that there is a single observer on each branch. Then, the next trial for the observer who saw |0> again causes a split, so we get two further branches, 00 and 01, with a single observer on each branch. Continuing this, we get branches 0000.., 0100..., 011,00... and so on, so there are 2^N sequences (branches) after N repetitions. On each repetition, the branch, and the associated observer again splits, so you get two new branches, with a single observer on each. Consequently, a straightforward Everettian construction proves that there is only one observer on each branch, and there is no branch without a single observer.

If you disagree with this result, then it is up to you to give the calculation that produces multiple observers on some branches.

Of course, each sequence in this process has an amplitude, coming from the amplitudes in the initial state. For example, the sequence with u zeros has an amplitude

   a^u*b^{N-u}. But these are just amplitudes carried through. They do not represent weights or probabilities until you impose Born's rule, and we are not doing that in this calculation. The question of measure is answered by the above construction: each sequence has the same measure.

It is important to realize that the same 2^N sequences are obtained whatever the initial amplitude a, and b might be. Thus the set of sequences does not come from the binomial theorem with a probability of success given by |a|^2, because you get the same sequence for all values of a and b, which is not the binomial theorem result. Equal measure over the sequences is a result of the construction, not of some mathematical theorem related to the binomial distribution.

Bruce

[Bruce Kellett]

I can expand a little on the argument for equal weight for each sequence. From the above, the sequence with u zeros has an amplitude a^ub^{N-u}. If we take the case with equal probabilities, a = b = 1/sqrt(2), this amplitude becomes 1/2^N. That is, in this case, the amplitude is independent of u, the number of zeros in the sequence. In other words, all sequences have the same amplitude, and hence the same weight or measure. Now we know that the same set of sequences is obtained for all values of a and b. So the result for a = b = 1/sqrt(2) is typical, and we can conclude that all sequences have the same weight for all cases. Note again that these sequences cover all possible binary sequences of length N, so we must get the same set of sequences for any a and b. The sequences are obtained by the construction, and, except in special cases like a = b, they are not the same as sequences from the binomial distribution for different probabilities of success.

Bruce

[ilsa]

Awesome, and the less dense inner knowledge of thought to come together with some material sciences always fascinating

Ilsa Bartlett

Institute for Rewiring the System
http://ilsabartlett.wordpress.com

http://www.google.com/profiles/ilsa.bartlett
www.hotlux.com/angel

"Don't ever get so big or important that you can not hear and listen to every other person."
-John Coltrane

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[Quentin Anciaux]

Bruce,

You keep repeating the same circular reasoning: you assume one observer per branch, then conclude uniform sampling by observers, then use that to claim equal measure. But Everett’s relative-state formulation does not require discrete worlds or a uniform observer distribution — that is your interpretation, not a derivation.

Carrying amplitudes through unitarity without giving them any role is precisely the point of contention. Ignoring them does not make their influence vanish; it only means you are not engaging with the core question. If you assert that your construction proves all branches have equal measure, then you are assuming the conclusion you're trying to establish.

If you are certain this invalidates any amplitude-based measure and thus the Born rule, the proper way forward is still the same: publish the derivation and let it stand under peer review. Repeating it here without addressing counterarguments doesn't make it more correct, just more dogmatic.

Quentin

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[Quentin Anciaux]

Bruce,

Your reasoning is flawed because you conflate the existence of sequences with their measure. Yes, the set of 2^N sequences is the same regardless of a and b, but the amplitudes attached to them are not irrelevant bookkeeping. In Everett's framework, the squared amplitudes define the structure of the wavefunction and thus the density of observer-instances across sequences.

Setting a = b to argue that all sequences have equal weight and then generalizing that conclusion to arbitrary a and b is invalid. It ignores the very feature that makes different amplitudes significant: they change the relative contribution of each branch to future correlations and observed frequencies. You cannot dismiss this and still claim to be deriving anything about measure.

If you want to argue that amplitudes have no bearing on measure, you need an independent justification for why quantum mechanics' core mathematical structure and its experimental validation via the Born rule should be discarded. Otherwise, you are assuming your conclusion.

Quentin

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[Bruce Kellett]

On Wed, Aug 27, 2025 at 2:53 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

You keep repeating the same circular reasoning: you assume one observer per branch, then conclude uniform sampling by observers, then use that to claim equal measure. But Everett’s relative-state formulation does not require discrete worlds or a uniform observer distribution — that is your interpretation, not a derivation.

No. The result is certainly derived from the fact that Everett assumes that every outcome is realized in every interaction. You have not shown this result to be invalid. You have merely asserted that it is so.

If you think there is more than one observer per branch, prove that this is so from the mathematics!

Carrying amplitudes through unitarity without giving them any role is precisely the point of contention. Ignoring them does not make their influence vanish; it only means you are not engaging with the core question. If you assert that your construction proves all branches have equal measure, then you are assuming the conclusion you're trying to establish.

The simple fact is that the amplitudes play no role in the argument I am making.

If you are certain this invalidates any amplitude-based measure and thus the Born rule, the proper way forward is still the same: publish the derivation and let it stand under peer review. Repeating it here without addressing counterarguments doesn't make it more correct, just more dogmatic.

If I can't convince you, how am I to convince a sceptical referee?

Bruce

[Quentin Anciaux]

Bruce,

If your derivation is as solid as you claim, then a skeptical referee is exactly who you should want to convince. Repeating the same argument here without engaging with the role of amplitudes will not make it any stronger. You cannot dismiss amplitudes entirely and then claim to have explained why measure must be uniform, that is circular.

If you truly believe your reasoning refutes the Born rule within Everett’s framework, then publishing it is the only way to settle the matter. Otherwise, endlessly asserting it here looks less like confidence and more like avoidance.

Your entire argument hinges on assuming uniform observer sampling by postulating one observer per branch. But that is precisely the point under debate, not a derived result. If you ignore the role of amplitudes in defining the structure of the wavefunction, you're not engaging with Everett's formulation, only with your own simplified model.

Until you demonstrate why amplitudes should be irrelevant within unitary evolution, claiming equal weights is just assuming your conclusion.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Wed, Aug 27, 2025 at 2:56 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

Your reasoning is flawed because you conflate the existence of sequences with their measure. Yes, the set of 2^N sequences is the same regardless of a and b, but the amplitudes attached to them are not irrelevant bookkeeping. In Everett's framework, the squared amplitudes define the structure of the wavefunction and thus the density of observer-instances across sequences.

The amplitudes play no role in the argument. Again, if you think that the squared amplitudes give multiple observers per branch, then it is up to you to demonstrate this mathematically.

Setting a = b to argue that all sequences have equal weight and then generalizing that conclusion to arbitrary a and b is invalid.

Prove that it is invalid! In the case of a = b, all sequences certainly have equal weight. And the same sequences occur for any values of a and b.

It ignores the very feature that makes different amplitudes significant: they change the relative contribution of each branch to future correlations and observed frequencies. You cannot dismiss this and still claim to be deriving anything about measure.

You still seem to be thinking in terms of binomial distributions from N trials with specific probabilities for success.

 You have missed the significance of the fact that all outcomes occur on each trial.

If you want to argue that amplitudes have no bearing on measure, you need an independent justification for why quantum mechanics' core mathematical structure and its experimental validation via the Born rule should be discarded. Otherwise, you are assuming your conclusion.

No, I am proving that the observed frequencies do not conform to the expected Born probabilities. How do you think that the Born rule is demonstrated in practice?

Look, the argument is very simple. Given the 2^N binary sequences from N trials, the number of zeros (ones) in each sequence varies, but the amplitude of zero in the original wavefunction, a, remains the same. Consequently, the expected Born probability is |a|^2 for every sequence. But the proportion of success (zeros) differs between sequences, leading to a contradiction. This argument is independent of the amplitudes for the sequences, and the number of observers per sequence. These, and other things, are just items that you have introduced as distractions. It is time you actually considered the argument being made.

Bruce

[Quentin Anciaux]

Bruce,

You keep treating the existence of all 2^N sequences as if it implied equal measure, but that conclusion is exactly what is under debate. Everett's framework does not equate existence with weight: amplitudes encode structural information about correlations and observer densities, and experiments confirm this via the Born rule.

Your argument sidesteps this entirely by assuming amplitudes are irrelevant, which is precisely the point you need to demonstrate, not assume.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Bruce Kellett]

On Wed, Aug 27, 2025 at 3:26 PM Quentin Anciaux <allc...@gmail.com> wrote:

Bruce,

If your derivation is as solid as you claim, then a skeptical referee is exactly who you should want to convince. Repeating the same argument here without engaging with the role of amplitudes will not make it any stronger. You cannot dismiss amplitudes entirely and then claim to have explained why measure must be uniform, that is circular.

If you truly believe your reasoning refutes the Born rule within Everett’s framework, then publishing it is the only way to settle the matter. Otherwise, endlessly asserting it here looks less like confidence and more like avoidance.

Your entire argument hinges on assuming uniform observer sampling by postulating one observer per branch.

The argument does not depend on this. This shows nothing more than that you have not understood the argument.

But that is precisely the point under debate, not a derived result. If you ignore the role of amplitudes in defining the structure of the wavefunction, you're not engaging with Everett's formulation, only with your own simplified model.

Until you demonstrate why amplitudes should be irrelevant within unitary evolution, claiming equal weights is just assuming your conclusion.

I think, rather, that you should show how the argument I have made depends on amplitudes when it clearly does not. It depends merely on the proportion of zero outcomes in each sequence. And that does not depend on the amplitudes.

Bruce

[Quentin Anciaux]

Bruce,

If your argument truly refutes the Born rule while staying within Everett’s framework, that would be groundbreaking. Where’s the Nobel prize then? Until it’s published and survives peer review, repeating it here won’t make it true.

Quentin

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Quentin Anciaux]

Bruce,

Everett’s original formulation describes a universal wavefunction evolving unitarily, not discrete worlds with one observer per branch. Your argument assumes this mapping, but it is an interpretative choice, not a result derived from the Schrödinger equation.

Also, your claim that all 2^N sequences have equal measure only holds if amplitudes are treated as irrelevant. In standard quantum mechanics, amplitudes directly determine observed frequencies via the Born rule, which has strong experimental support. Ignoring amplitudes means you are no longer analyzing Everett’s framework but a different model where the Born rule indeed fails.

To refute Everett with Born included, you would need to show that even when squared amplitudes define a natural measure, the predicted observed frequencies still fail. Assuming uniform sampling over sequences does not establish that.

This is why your derivation is not accepted: it relies on a hidden premise, one observer per branch with uniform sampling, which is not part of Everettian quantum mechanics.

Quentin 

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Brent Meeker]

I think some specificity would help this debate.  Suppose N=6, so there are 64 different sequences in 64 different worlds.  The number of observers is irrelevants; we can suppose the results are recorded mechanically in each world.  Further suppose that a=b so there is no question of whether amplitudes are being respected.  Then in one of the worlds we have 011000.  Per the Born rule its probability is 0.2344.  In MWI it is 1/64=0.0156.  The difference arises because the observers applying the Born rule looks at it as an instance of 2 out of 6 successes.  

So why can't the MWI observer do the same calculation?  He certainly can.  He can apply the Born rule.  But when he does so, it can't be interpreted as a probability of his branch since such probabilities would add up to much more than 1.0 when summed over the 64 different worlds.  From the standpoint of statistics 011000 is the same as 001010 and their probabilities sum.  Their difference is just incidental, but they are different worlds in MWI and summing them makes no sense.  

Brent

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[Quentin Anciaux]

Brent,

In Everettian QM, the Born rule applies to coarse-grained outcomes, not to individual fine-grained sequences treated as equiprobable. The amplitudes are not just bookkeeping: their squared norm defines the measure, which determines how observer-instances are distributed.

Think of a lottery with one million tickets, but where 400,000 of them are identical copies of the same number. All tickets "exist," but they are not equally weighted when predicting what a typical observer will see. Similarly, in your N=6 example, 011000 and 001010 belong to the same coarse-grained class of "2 successes out of 6," and the combined measure of all such sequences follows the Born rule.

Assuming each branch has equal weight and uniform observer sampling creates the contradiction, Everett’s formulation does not require that assumption.

Quentin 

All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)

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[Jesse Mazer]

Also, even without invoking the Born rule there is a result that if you consider a "pointer state" that records the relative fractions of different possible measurement results in a *series* of N systems prepared in the same initial state, and consider the limit as N approaches infinity, in this limit all the amplitude gets concentrated on the pointer state with the fractions that correspond to the probabilities for individual measurements predicted by the Born rule--see David Z Albert's comments at https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238 and the paper discussing Mittelstaedt’s theorem at https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement

Jesse

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[Brent Meeker]

On 8/27/2025 1:13 PM, Quentin Anciaux wrote:

Brent,

In Everettian QM, the Born rule applies to coarse-grained outcomes, not to individual fine-grained sequences treated as equiprobable. 

Not what they told me in graduate school.

The amplitudes are not just bookkeeping: their squared norm defines the measure, which determines how observer-instances are distributed.

Which is why I chose a=b to illustrate that the difference was due to different weights.

Think of a lottery with one million tickets, but where 400,000 of them are identical copies of the same number. All tickets "exist," but they are not equally weighted when predicting what a typical observer will see. 

Only if you implicitly define "typical" as probable under the Born rule.

Similarly, in your N=6 example, 011000 and 001010 belong to the same coarse-grained class of "2 successes out of 6," and the combined measure of all such sequences follows the Born rule.

But there's no "coarse graining" the multiple worlds.  They are all separate and orthogonal.  There is no observer who can see more than one of them.

Assuming each branch has equal weight and uniform observer sampling creates the contradiction, Everett’s formulation does not require that assumption.

That's right.  Everett's formulation allows adding the Born rule as a axiom giving different weight to different worlds.  But it's not implicit in MWI.

Brent

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[Brent Meeker]

Yes, given a device that calculates the Born probability, almost all worlds will agree on the probability.

Brent

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[Bruce Kellett]

On Thu, Aug 28, 2025 at 5:16 AM Brent Meeker <meeke...@gmail.com> wrote:

I think some specificity would help this debate.  Suppose N=6, so there are 64 different sequences in 64 different worlds.  The number of observers is irrelevants; we can suppose the results are recorded mechanically in each world.  Further suppose that a=b so there is no question of whether amplitudes are being respected.  Then in one of the worlds we have 011000.  Per the Born rule its probability is 0.2344.  In MWI it is 1/64=0.0156.  The difference arises because the observers applying the Born rule looks at it as an instance of 2 out of 6 successes.

The trouble with this is that you are treating this as an instance of Bernoulli trials with probability p= 0.5. When every outcome occurs with every trial we no longer have a Binomial distribution The Binomial distribution assumes that you have x successes out of N trials. In the Everettian case you have one success on every trial.

So your probability above for 2 successes applies to Bernoulli trials with one as the 'success'. The thing is that the probability of getting a zero is also 1/2, so we also have four successes out of six trials in your example. The Binomial probability for this result is also 0.2344. Actually, if we regard this experiment as a test of the Born rule, we have four zeros in 6 trials, which gives an estimate of the probability as 4/6 = 0.667, or as two ones in 6 trials which gives an estimate of the probability as 2/6 = 0.333, neither estimate is equivalent to |a|^2 = 0.5. The difference becomes more pronounced as N increases. The problem with your analysis is that you are assuming a binomial distribution. and we do not have any such distribution.

Bruce

[Jesse Mazer]

As I understand it this result isn't based on Born probability calculations, it's just based on looking at a measurement operator whose eigenvectors/eigenvalues correspond to adding the results of records (pointer states) of multiple trials and dividing by the total number of trials to get a fraction. For example if you have a measuring device designed to measure the spin on the z-axis on five successive trails and record the results, then if a detailed measurement of the records (using the assumption of collapse and the Born rule) gave the results "up, up, down, up, down", then in that case after collapse the state vector of the records would be parallel to the eigenvector of this new measurement operator with the eigenvalue "3/5 up". But if the experiment you're repeating is supposed to give probabilities 1/3 up and 2/3 down on each trial using the Born rule, you can apparently show that if we consider the limit as the number of trials approaches infinity, and just use the deterministic Schrodinger equation for wavefunction evolution *without* ever directly invoking the Born rule or state vector collapse, then all the amplitude of the records becomes concentrated on state vectors parallel to the eigenvector with eigenvalue "1/3 up", and orthogonal to all the other eigenvectors of this measurement operator. But since the collapse assumption was never invoked, the result for every *individual* trial can in general still be in a superposition of the spin eigenvectors for that the record of that trial (eg the record of trial #3 is in some superposition of spin-up and spin-down), it's only the sum-divided-by-number-of-trials that approaches a definite answer.

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[Brent Meeker]

On 8/27/2025 5:07 PM, Bruce Kellett wrote:

On Thu, Aug 28, 2025 at 5:16 AM Brent Meeker <meeke...@gmail.com> wrote:

I think some specificity would help this debate.  Suppose N=6, so there are 64 different sequences in 64 different worlds.  The number of observers is irrelevants; we can suppose the results are recorded mechanically in each world.  Further suppose that a=b so there is no question of whether amplitudes are being respected.  Then in one of the worlds we have 011000.  Per the Born rule its probability is 0.2344.  In MWI it is 1/64=0.0156.  The difference arises because the observers applying the Born rule looks at it as an instance of 2 out of 6 successes.

The trouble with this is that you are treating this as an instance of Bernoulli trials with probability p= 0.5. When every outcome occurs with every trial we no longer have a Binomial distribution The Binomial distribution assumes that you have x successes out of N trials. In the Everettian case you have one success on every trial.

So your probability above for 2 successes applies to Bernoulli trials with one as the 'success'. The thing is that the probability of getting a zero is also 1/2, so we also have four successes out of six trials in your example. The Binomial probability for this result is also 0.2344. Actually, if we regard this experiment as a test of the Born rule, we have four zeros in 6 trials, which gives an estimate of the probability as 4/6 = 0.667, or as two ones in 6 trials which gives an estimate of the probability as 2/6 = 0.333, neither estimate is equivalent to |a|^2 = 0.5. The difference becomes more pronounced as N increases. The problem with your analysis is that you are assuming a binomial distribution. and we do not have any such distribution.

That's how an experimenter will compute the probability of 2 successes in 6 trials given p=0.5.   Yes, the probability estimate is 0.33 given 2 in 6.  The probability of the result given p is not generally the same as the estimate of p given the result.  The former is binomial distributed.  The estimates of p aren't part of a distribution since they don't add up to 1.  

Brent

Bruce

So why can't the MWI observer do the same calculation?  He certainly can.  He can apply the Born rule.  But when he does so, it can't be interpreted as a probability of his branch since such probabilities would add up to much more than 1.0 when summed over the 64 different worlds.  From the standpoint of statistics 011000 is the same as 001010 and their probabilities sum.  Their difference is just incidental, but they are different worlds in MWI and summing them makes no sense.  

Brent

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[Bruce Kellett]

On Thu, Aug 28, 2025 at 2:30 PM Brent Meeker <meeke...@gmail.com> wrote:

On 8/27/2025 5:07 PM, Bruce Kellett wrote:

On Thu, Aug 28, 2025 at 5:16 AM Brent Meeker <meeke...@gmail.com> wrote:

I think some specificity would help this debate.  Suppose N=6, so there are 64 different sequences in 64 different worlds.  The number of observers is irrelevants; we can suppose the results are recorded mechanically in each world.  Further suppose that a=b so there is no question of whether amplitudes are being respected.  Then in one of the worlds we have 011000.  Per the Born rule its probability is 0.2344.  In MWI it is 1/64=0.0156.  The difference arises because the observers applying the Born rule looks at it as an instance of 2 out of 6 successes.

The trouble with this is that you are treating this as an instance of Bernoulli trials with probability p= 0.5. When every outcome occurs with every trial we no longer have a Binomial distribution The Binomial distribution assumes that you have x successes out of N trials. In the Everettian case you have one success on every trial.

So your probability above for 2 successes applies to Bernoulli trials with one as the 'success'. The thing is that the probability of getting a zero is also 1/2, so we also have four successes out of six trials in your example. The Binomial probability for this result is also 0.2344. Actually, if we regard this experiment as a test of the Born rule, we have four zeros in 6 trials, which gives an estimate of the probability as 4/6 = 0.667, or as two ones in 6 trials which gives an estimate of the probability as 2/6 = 0.333, neither estimate is equivalent to |a|^2 = 0.5. The difference becomes more pronounced as N increases. The problem with your analysis is that you are assuming a binomial distribution. and we do not have any such distribution.

That's how an experimenter will compute the probability of 2 successes in 6 trials given p=0.5.   Yes, the probability estimate is 0.33 given 2 in 6.  The probability of the result given p is not generally the same as the estimate of p given the result.  The former is binomial distributed.  The estimates of p aren't part of a distribution since they don't add up to 1. 

You are still assuming that the results follow a binomial distribution. This is not the case. As I said, in the Everett situation, when every outcome is realized in every trial, you have one success in every trial, and the resulting distribution is not binomial. The normal approximation for large N, as Jesse seems to assume, simply does not hold, since the distribution is not binomial.

Bruce

[Brent Meeker]

I said exactly that above: "In MWI it is 1/64=0.0156."  But the distribution of results in our single world, per the Born rule is binomial.

Brent

The normal approximation for large N, as Jesse seems to assume, simply does not hold, since the distribution is not binomial.

Bruce

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[Quentin Anciaux]

Bruce,

Your conclusion hinges on rejecting the binomial distribution in the Everettian context, but this misrepresents how the Born rule is tested experimentally. In actual experiments, observers find themselves in a single branch and estimate probabilities from relative frequencies within that branch. The binomial model is exactly what captures those frequencies. Everett does not magically change the combinatorics.

Your reasoning assumes equal weighting of all fine-grained sequences and uniform observer sampling across them. But that assumption is yours, not Everett's. In Everett's framework, the squared amplitudes define the measure over branches, so typical observers overwhelmingly find frequencies matching the Born rule even though all sequences exist.

The lottery analogy makes this clearer: having 1,000,000 tickets but 400,000 duplicates of the same number does not make every ticket equally predictive of what a typical draw looks like. Existence ≠ equal weight.

Dismissing amplitudes as irrelevant is exactly where your derivation breaks down.

Quentin 

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[Jesse Mazer]

I don't assume anything myself, I point to some sources by physicists below indicating there is a mathematical derivation of the fact that if physical records ('pointer states') of a series of N trials are kept, and you define a measurement operator that only tells you the fraction of those recorded trials with some result, then one can show that the deterministic equation of wavefunction evolution (without collapse assumption/Born rule) implies that the quantum state of the records approaches the "correct" eigenstate of this measurement operator (the one that matches what would be predicted using the Born rule) in the limit as N approaches infinity.

https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238

https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement

 

Bruce

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[Bruce Kellett]

It is not clear how this applies to the case under discussion. Could you be more specific?

Bruce

[Jesse Mazer]

You were discussing a case of this form: "This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence."

If we are interested in statistics for N trials, let's define a "supertrial" as a sequence of N trials of the individual measurement, and say that we are repeating many supertrials and recording the results of all the individual trials in each supertrial using some kind of physical memory (persistent 'pointer states'). Each supertrial has 2^N possible outcomes, and for a given supertrial outcome O (like up, down, up, up, up, down for N=6) you can define a measurement operator on the pointer states whose eigenvalues correspond to what the records would tell you about the fraction of supertrials where the outcome was O. If I'm understanding the result in those references correctly, then if one models the interaction between quantum system, measuring apparatus, and records using only the deterministic Schrodinger equation, without any collapse assumption or Born rule, one can show that in the limit as the number of supertrials goes to infinity, all the amplitude for the whole system including the records becomes concentrated on state vectors that are parallel to the eigenvector of the measurement operator with the eigenvalue that exactly matches the frequency of outcome O that would have been predicted if you *had* used the collapse assumption and Born rule for individual measurements. And this should be true even if the probability for up vs. down on individual measurements was not 50/50 given the experimental setup.

Jesse

[Bruce Kellett]

On Fri, Aug 29, 2025 at 2:47 AM Jesse Mazer <laser...@gmail.com> wrote:

You were discussing a case of this form: "This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence."

If we are interested in statistics for N trials, let's define a "supertrial" as a sequence of N trials of the individual measurement, and say that we are repeating many supertrials and recording the results of all the individual trials in each supertrial using some kind of physical memory (persistent 'pointer states'). Each supertrial has 2^N possible outcomes, and for a given supertrial outcome O (like up, down, up, up, up, down for N=6) you can define a measurement operator on the pointer states whose eigenvalues correspond to what the records would tell you about the fraction of supertrials where the outcome was O. If I'm understanding the result in those references correctly, then if one models the interaction between quantum system, measuring apparatus, and records using only the deterministic Schrodinger equation, without any collapse assumption or Born rule, one can show that in the limit as the number of supertrials goes to infinity, all the amplitude for the whole system including the records becomes concentrated on state vectors that are parallel to the eigenvector of the measurement operator with the eigenvalue that exactly matches the frequency of outcome O that would have been predicted if you *had* used the collapse assumption and Born rule for individual measurements. And this should be true even if the probability for up vs. down on individual measurements was not 50/50 given the experimental setup.

I haven't looked into this in any detail, but it seems to be a recasting of an idea that has been around for a long time. This idea hasn't made it into the mainstream because the details failed to work out. There are all sorts of problems with the idea, and it doesn't appear to translate well to the argument I am making. The 2^N sequences that result from repeated measurements on the basic binary system do not form a measurement in themselves. There is no operator for this, and no eigenfunctions and there is no obvious outcome. The results that one wants are the counts of zeros, say, in each sequence, so there is no single well-defined outcome. So I don't think it is worth worrying too much about the sort of argument you refer to -- it is not going anywhere.

Bruce

[Jesse Mazer]

On Thu, Aug 28, 2025 at 8:05 PM Bruce Kellett <bhkel...@gmail.com> wrote:

On Fri, Aug 29, 2025 at 2:47 AM Jesse Mazer <laser...@gmail.com> wrote:

You were discussing a case of this form: "This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence."

If we are interested in statistics for N trials, let's define a "supertrial" as a sequence of N trials of the individual measurement, and say that we are repeating many supertrials and recording the results of all the individual trials in each supertrial using some kind of physical memory (persistent 'pointer states'). Each supertrial has 2^N possible outcomes, and for a given supertrial outcome O (like up, down, up, up, up, down for N=6) you can define a measurement operator on the pointer states whose eigenvalues correspond to what the records would tell you about the fraction of supertrials where the outcome was O. If I'm understanding the result in those references correctly, then if one models the interaction between quantum system, measuring apparatus, and records using only the deterministic Schrodinger equation, without any collapse assumption or Born rule, one can show that in the limit as the number of supertrials goes to infinity, all the amplitude for the whole system including the records becomes concentrated on state vectors that are parallel to the eigenvector of the measurement operator with the eigenvalue that exactly matches the frequency of outcome O that would have been predicted if you *had* used the collapse assumption and Born rule for individual measurements. And this should be true even if the probability for up vs. down on individual measurements was not 50/50 given the experimental setup.

I haven't looked into this in any detail, but it seems to be a recasting of an idea that has been around for a long time. This idea hasn't made it into the mainstream because the details failed to work out.

Can you point to any sources that explain specific ways the details fail to work out? David Z Albert is very knowledgeable about results relevant to interpretation of QM so I'd be surprised if he missed any technical critique. Of course there is the philosophical argument that this doesn't resolve the measurement problem because it doesn't lead to definite results for individual trials (or supertrials) but that's not taking issue with the technical claim about measuring frequencies of results in the limit of infinite trials (and David Z Albert brings up this philosophical objection in the last paragraph before section VI at https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238 , and then in section VI he goes on to talk about why he thinks this objection means the fact about frequencies in the limit doesn't really resolve the measurement problem)

 

There are all sorts of problems with the idea, and it doesn't appear to translate well to the argument I am making. The 2^N sequences that result from repeated measurements on the basic binary system do not form a measurement in themselves. There is no operator for this, and no eigenfunctions and there is no obvious outcome.

I had thought that for any measurable quantity including coarse-grained statistical ones, it was possible to construct a measurement operator in QM--doing some googling, it may be that for some coarse-grained quantities one has to use a "positive operator valued measure", see answer at https://physics.stackexchange.com/a/791442/59406 , and according to https://quantumcomputing.stackexchange.com/a/29326 this is not itself an operator though it is a function defined in terms of a collection of positive operators. And the page at https://www.damtp.cam.ac.uk/user/hsr1000/stat_phys_lectures.pdf also mentions that in quantum statistical mechanics, macrostates can be defined in terms of the density operator which is used to describe mixed states (ones where we don't know the precise quantum microstate and just assign classical probabilities to different possible microstates). I don't know if either was used here, but p. 13 of the paper I mentioned at https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement indicates that some type of operator was used to derive the result about frequencies in the limit:

"The ingenious method of introducing a quantum-mechanical equivalent of probabilities that Mittelstaedt follows in his approach relies on a new operator F^N_k 

whose ‘intuitive’ role is to measure the relative frequency of the outcome a_k in a given sequence of N outcomes."

The full details would presumably be in Mittelstaedt's book The Interpretation of Quantum Mechanics and the Measurement Process in the paper's bibliography.

Jesse

[Bruce Kellett]

On Sat, Aug 30, 2025 at 2:04 AM Jesse Mazer <laser...@gmail.com> wrote:

On Thu, Aug 28, 2025 at 8:05 PM Bruce Kellett <bhkel...@gmail.com> wrote:

On Fri, Aug 29, 2025 at 2:47 AM Jesse Mazer <laser...@gmail.com> wrote:

You were discussing a case of this form: "This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence."

If we are interested in statistics for N trials, let's define a "supertrial" as a sequence of N trials of the individual measurement, and say that we are repeating many supertrials and recording the results of all the individual trials in each supertrial using some kind of physical memory (persistent 'pointer states'). Each supertrial has 2^N possible outcomes, and for a given supertrial outcome O (like up, down, up, up, up, down for N=6) you can define a measurement operator on the pointer states whose eigenvalues correspond to what the records would tell you about the fraction of supertrials where the outcome was O. If I'm understanding the result in those references correctly, then if one models the interaction between quantum system, measuring apparatus, and records using only the deterministic Schrodinger equation, without any collapse assumption or Born rule, one can show that in the limit as the number of supertrials goes to infinity, all the amplitude for the whole system including the records becomes concentrated on state vectors that are parallel to the eigenvector of the measurement operator with the eigenvalue that exactly matches the frequency of outcome O that would have been predicted if you *had* used the collapse assumption and Born rule for individual measurements. And this should be true even if the probability for up vs. down on individual measurements was not 50/50 given the experimental setup.

I haven't looked into this in any detail, but it seems to be a recasting of an idea that has been around for a long time. This idea hasn't made it into the mainstream because the details failed to work out.

Can you point to any sources that explain specific ways the details fail to work out? David Z Albert is very knowledgeable about results relevant to interpretation of QM so I'd be surprised if he missed any technical critique.

I quote David Albert from his contribution to the book "Many Worlds? Everett, Quantum Theory and Relativity" (Oxford,2010)

"But the business of parlaying this thought into a fully worked-out account of probability in the Everett picture quickly runs into very familiar and very discouraging sorts of trouble." I don't have any more detail about this, but it seems from the fact that this is not mainstream, that these difficulties proved insurmountable. For instance, it uses a frequentist definition of probability, and this is known to be full of problems.

Of course there is the philosophical argument that this doesn't resolve the measurement problem because it doesn't lead to definite results for individual trials (or supertrials) but that's not taking issue with the technical claim about measuring frequencies of results in the limit of infinite trials (and David Z Albert brings up this philosophical objection in the last paragraph before section VI at https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238 , and then in section VI he goes on to talk about why he thinks this objection means the fact about frequencies in the limit doesn't really resolve the measurement problem)

 

There are all sorts of problems with the idea, and it doesn't appear to translate well to the argument I am making. The 2^N sequences that result from repeated measurements on the basic binary system do not form a measurement in themselves. There is no operator for this, and no eigenfunctions and there is no obvious outcome.

I had thought that for any measurable quantity including coarse-grained statistical ones, it was possible to construct a measurement operator in QM--doing some googling, it may be that for some coarse-grained quantities one has to use a "positive operator valued measure", see answer at https://physics.stackexchange.com/a/791442/59406 , and according to https://quantumcomputing.stackexchange.com/a/29326 this is not itself an operator though it is a function defined in terms of a collection of positive operators. And the page at https://www.damtp.cam.ac.uk/user/hsr1000/stat_phys_lectures.pdf also mentions that in quantum statistical mechanics, macrostates can be defined in terms of the density operator which is used to describe mixed states (ones where we don't know the precise quantum microstate and just assign classical probabilities to different possible microstates). I don't know if either was used here, but p. 13 of the paper I mentioned at https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement indicates that some type of operator was used to derive the result about frequencies in the limit:

There is no single outcome from a repetition of the N trials and 2^N sequences. So it can't be an eigenvalue of some quantum operator.

"The ingenious method of introducing a quantum-mechanical equivalent of probabilities that Mittelstaedt follows in his approach relies on a new operator F^N_k 

whose ‘intuitive’ role is to measure the relative frequency of the outcome a_k in a given sequence of N outcomes."

The full details would presumably be in Mittelstaedt's book The Interpretation of Quantum Mechanics and the Measurement Process in the paper's bibliography.

I have no interest in looking into this further, since it clearly cannot work to give meaning to probability in an Everettian model. It was always a fringe idea that didn't work out.

Bruce

[Jesse Mazer]

On Fri, Aug 29, 2025 at 7:54 PM Bruce Kellett <bhkel...@gmail.com> wrote:

On Sat, Aug 30, 2025 at 2:04 AM Jesse Mazer <laser...@gmail.com> wrote:

On Thu, Aug 28, 2025 at 8:05 PM Bruce Kellett <bhkel...@gmail.com> wrote:

On Fri, Aug 29, 2025 at 2:47 AM Jesse Mazer <laser...@gmail.com> wrote:

You were discussing a case of this form: "This is easily seen if one considers a wave function with a binary outcome, |0> and |1> for example. After N repeated trials, one has 2^N strings of possible outcome sequences. One can count the number of, say, ones in each possible outcome sequence."

If we are interested in statistics for N trials, let's define a "supertrial" as a sequence of N trials of the individual measurement, and say that we are repeating many supertrials and recording the results of all the individual trials in each supertrial using some kind of physical memory (persistent 'pointer states'). Each supertrial has 2^N possible outcomes, and for a given supertrial outcome O (like up, down, up, up, up, down for N=6) you can define a measurement operator on the pointer states whose eigenvalues correspond to what the records would tell you about the fraction of supertrials where the outcome was O. If I'm understanding the result in those references correctly, then if one models the interaction between quantum system, measuring apparatus, and records using only the deterministic Schrodinger equation, without any collapse assumption or Born rule, one can show that in the limit as the number of supertrials goes to infinity, all the amplitude for the whole system including the records becomes concentrated on state vectors that are parallel to the eigenvector of the measurement operator with the eigenvalue that exactly matches the frequency of outcome O that would have been predicted if you *had* used the collapse assumption and Born rule for individual measurements. And this should be true even if the probability for up vs. down on individual measurements was not 50/50 given the experimental setup.

I haven't looked into this in any detail, but it seems to be a recasting of an idea that has been around for a long time. This idea hasn't made it into the mainstream because the details failed to work out.

Can you point to any sources that explain specific ways the details fail to work out? David Z Albert is very knowledgeable about results relevant to interpretation of QM so I'd be surprised if he missed any technical critique.

I quote David Albert from his contribution to the book "Many Worlds? Everett, Quantum Theory and Relativity" (Oxford,2010)

"But the business of parlaying this thought into a fully worked-out account of probability in the Everett picture quickly runs into very familiar and very discouraging sorts of trouble." I don't have any more detail about this, but it seems from the fact that this is not mainstream, that these difficulties proved insurmountable. For instance, it uses a frequentist definition of probability, and this is known to be full of problems.

But see my comment below on the distinction between (1) saying the result is wrong in a technical sense, vs. (2) raising the philosophical objection that it only tells us that overall frequencies in the infinite limit approach a definite answer, not that there can be a definite result to specific individual trials (or specific finite runs of trials), which may lead people to say that even though the result is valid, it doesn't resolve the measurement problem (the problem of why we only experience definite outcomes rather than superpositions when we make individual observations):

Of course there is the philosophical argument that this doesn't resolve the measurement problem because it doesn't lead to definite results for individual trials (or supertrials) but that's not taking issue with the technical claim about measuring frequencies of results in the limit of infinite trials (and David Z Albert brings up this philosophical objection in the last paragraph before section VI at https://books.google.com/books?id=_HgF3wfADJIC&lpg=PP1&pg=PA238 , and then in section VI he goes on to talk about why he thinks this objection means the fact about frequencies in the limit doesn't really resolve the measurement problem)

His essay in "Many Worlds? Everett, Quantum Theory and Relativity" states the technical result again on p. 357, saying it is one that can "easily be shown", and the part you quoted is from the second to last sentence of the paragraph where he again raises that philosophical objection to seeing this result (though valid) as a resolution to the measurement problem:

"Here’s an idea: Suppose that we measure the x-spin of each of an infinite ensemble of electrons, where each of the electrons in the ensemble is initially prepared in the state α|x-up⟩ + β|x-down⟩. Then it can easily be shown that in the limit as the number of measurements already performed goes to infinity, the state of the world approaches an eigenstate of the frequency of (say) up-results, with eigenvalue |α|^2. And note that the limit we are dealing with here is a perfectly concrete and flat-footed limit of a sequence of vectors in Hilbert space, not a limit of probabilities of the sort that we are used to dealing with in applications of the probabilistic law of large numbers. And the thought has occurred to a number of investigators over the years (Sidney Coleman, and myself, and others too) that perhaps all it means to say that the probability that the outcome a measurement of the x-spin of an electron in the state α|x-up⟩ + β|x-down⟩ up is |α|^2 is that if an infinite ensemble of such experiments were to be performed, the state of the world would with certainty approach an eigenstate of the frequency of (say) up-results, with eigenvalue |α|^2. And what is particularly beautiful and seductive about that thought is the intimation that perhaps the Everett picture will turn out, at the end of the day, to be the only picture of the world on which probabilities fully and flat-footedly and not-circularly make sense. But the business of parlaying this thought into a fully worked-out account of probability in the Everett picture quickly runs into very familiar and very discouraging sorts of trouble. One doesn’t know what to say (for example) about finite runs of experiments, and one doesn’t know what to say about the fact that the world is after all very unlikely ever to be in an eigenstate of my undertaking to carry out any particular measurement of anything."

 

 

There are all sorts of problems with the idea, and it doesn't appear to translate well to the argument I am making. The 2^N sequences that result from repeated measurements on the basic binary system do not form a measurement in themselves. There is no operator for this, and no eigenfunctions and there is no obvious outcome.

I had thought that for any measurable quantity including coarse-grained statistical ones, it was possible to construct a measurement operator in QM--doing some googling, it may be that for some coarse-grained quantities one has to use a "positive operator valued measure", see answer at https://physics.stackexchange.com/a/791442/59406 , and according to https://quantumcomputing.stackexchange.com/a/29326 this is not itself an operator though it is a function defined in terms of a collection of positive operators. And the page at https://www.damtp.cam.ac.uk/user/hsr1000/stat_phys_lectures.pdf also mentions that in quantum statistical mechanics, macrostates can be defined in terms of the density operator which is used to describe mixed states (ones where we don't know the precise quantum microstate and just assign classical probabilities to different possible microstates). I don't know if either was used here, but p. 13 of the paper I mentioned at https://www.academia.edu/6975159/Quantum_dispositions_and_the_notion_of_measurement indicates that some type of operator was used to derive the result about frequencies in the limit:

There is no single outcome from a repetition of the N trials and 2^N sequences. So it can't be an eigenvalue of some quantum operator.

Why can't there be an eigenvalue that tells you the frequency of a specific result, for example if N=6 and you want to know the frequency of the specific result "up, down, up, up, up, down" in a large collection of measurements of groups of 6 electrons? This frequency is just a type of coarse-grained macrostate that's a straightforward function of the spin microstate (the exact specification of the spin of every individual electron), so if there can be measurement operators for macrostates in QM I don't see why there'd be a problem with this one. But even assuming the result holds for such an operator when we consider the infinite limit, there would still be the same philosophical objection that this doesn't give us a definite outcome on any *specific* measurement of a group of 6 specific electrons in the infinite sequence.

Jesse

[smitra]

On 25-08-2025 20:40, Brent Meeker wrote:
> On 8/25/2025 12:04 AM, smitra wrote:
>
>> But then one is defining "this universe" post hoc after the photon
>> lands on some point on the screen. One can do that, but this means
>> that there are two possibilities that really exist. Whether that's
>> considered to be in a single universe or two universes is just
>> semantics. There are two classes of paths for the photon, one class
>> is the set of paths through one slit, the other are the paths
>> through the other slit.
>>
>> Given that's the case, we can then do another measurement where we
>> simply measure through which slit the photon goes. We then don't
>> bother to let the photon move through toward the screen anymore, we
>> just detect the outcome of measuring whether the photon after moving
>> past either slits ends up being detected immediately after the left
>> or the right slit.
>>
>> If I then perform one such measurement, and I decide to go on
>> vacation destination X if the photon is detected behind the left
>> slit and I go to vacation destination Y if the result is the right
>> slit, and I end up going to X, the question is if there exists a
>> parallel world where I go to Y.
>>
>> The question for people who would say that only one world where I go
>> to X exists, is then to explain why both possibilities for the
>> photon going to the left or right slit objectively exists when we
>> detect the photon only at the screen, but only one possibility
>> exists when we detect the photon directly after passing the slits.
> The answer is, Don't confound possibility with reality.
> Possibilities "objectively exist" doesn't mean the corresponding
> realities exist. When you measure at one slit there is a different
> reality than when measuring at the screen.


It's then not clear what physical mechanism would cause the two
realities to objectively exist in some cases and not in other cases. So,
they do exist when the photons are not observed (even when doing the
experiment with molecules), but when we observe the photons then we say
that the two possibilities are merely possibilities and only one reality
exists when there is nothing in the experimental results that compels
one to make that assumption.

Saibal

>
> Brent
>
>> Saibal
>>
>> On 18-08-2025 21:56, Brent Meeker wrote:
>> Whatever goes thru the other slit to create the cancellation is
>> doing
>> it in the same universe; and incidentally it also increases the
>> incidence at other points. So I don't see that it implies a
>> parallel
>> universe.
>>
>> Brent
>>
>> On 8/17/2025 10:31 AM, smitra wrote:
>>
>> See here:
>>
>> https://www.youtube.com/watch?v=bux0SjaUCY0&t=885s [1]
>>
>> Of course, you can never get to 100% rigorous proof in physics like
>> in mathematics. You can never rule out waking up tomorrow in some
>> alien world and aliens telling you that your life here on Earth was
>> a simulation and that everything you thought you knew about the laws
>>
>> of physics is false.
>>
>> The nice thing about the argument by Deutsch is that it doesn't
>> depend on QM being correct, it is based on interpreting the
>> interference experiment. So, QM could be wrong, or it could be that
>> some of the claims of the MWI proponents are wrong, and yet this
>> argument by Deutsch will still stand.
>>
>> Saibal

>>
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