In a recent talk, Sean Carroll explicitly claims that, in Everett, world branching occurs when two decoherence makes worlds "so different" that interference will no more become observable between them, or when they will no longer be able to interfere for some reason. I am not really sure what that could exactly mean, but I do not understand why two orthogonal eigenstates of the universal wavefunction would be forbidden to interfere because, in principle, one can always imagine an "huge" unitary transformation that could "rotate" such states towards a different basis, just like we could rotate $(1/\sqrt 2)|1\rangle + (1/\sqrt 2)|0\rangle$ into $|1\rangle$ using an $H$ gate. What am I missing?
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$\begingroup$ This may help - Parallel Worlds Probably Exist. Here’s Why $\endgroup$– mmesser314Commented Mar 28, 2023 at 2:44
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1$\begingroup$ @Luis Caires Aren't you just asking how decoherence works? This is quite a large subject and textbooks are written about it. $\endgroup$– SvenForkbeardCommented Mar 28, 2023 at 10:47
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1$\begingroup$ Some comments removed; comments are for criticizing and improving the post being commented upon, not for voicing your opinion on the general content of the question. If you want to answer the question, please write an answer. $\endgroup$– ACuriousMind ♦Commented Mar 28, 2023 at 20:25
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$\begingroup$ @ACuriousMind Pointing to a general and trivial critique of Everett is well within the range of the information the OP should get IMHO, especially since "state" in quantum mechanics always refers to an ensemble measurement. It can never be a property of the single quantum mechanical system. That there can be no universal wave function (despite continuous claims in the literature about it) is also fairly obvious. I don't know why you are criticizing this? My first comment clearly pointed out that the motivation of Sean Carrol is also beyond the reach of physics. Is it not? $\endgroup$– FlatterMannCommented Mar 28, 2023 at 20:47
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$\begingroup$ @FlatterMann I'm not making a judgement of whether or not you're right, I'm saying that comments are not the correct place for this critique of Everett. Perhaps you should review the commenting guidelines: I, for one, do not see how your comments guide OP in improving the post, nor do I consider them minor or transient. If you think this critique of Everett answers the question, leave it as an answer; in any case I don't see why it would belong in the comments here. If you disagree with this, I suggest you take it to Physics Meta. $\endgroup$– ACuriousMind ♦Commented Mar 28, 2023 at 21:00
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The decoherence argument is statistical. It tries to borrow the same reasoning by which the reversible laws of collisions between individual molecules lead to irreversible effects in accordance with the second law of thermodynamics. If you start with all the molecules of a gas in one side of the box, they will spread out to occupy both sides of the box equally. If you then reverse the velocity of every molecule, the reversibility of the Newtonian laws of collisions implies they would all spontaneously bounce back into the same side of the box. We never see this sort of thing happening in practice, not because the laws of molecular collisions forbid it, but because it is fantastically unlikely to happen that the velocities of $10^{23}$ particles should by chance be in one of the vanishingly rare configurations that unscrambled entropy.
The idea is that when entangling interactions start to involve astronomical numbers of particles, the probability of them ever spontaneously untangling back to pure uncorrelated states rapidly approaches zero. It's not theoretically impossible - as you say, some huge rotation in Hilbert space could do it - but it would be like trying to herd all the gas molecules into one side of the box.
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$\begingroup$ Decoherence doesn't require a large number of quanta. It happens already by coupling to a single additional quantum. That the world is uncertain doesn't require any of that, anyway. Relativity couples local physics to the volume of vacuum that's inside the local observer's forward light cone, including the surface of the light cone itself. Since the state on the forward light cone is not knowable (it travels away from us at the speed of light and we can't "catch" it ever again), there is necessarily a continuous loss of information about the local state going on in every relativistic system. $\endgroup$ Commented Mar 28, 2023 at 18:25
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$\begingroup$ @FlatterMann I don't think any of that is correct - but I don't think there is any point in me arguing with you about it. (By the way, if anyone does happen to read Everett's thesis (highly recommended!) they'll see that starting from the second sentence Everett is first presenting a summary of the standard interpretation found in textbooks before going on to explain why he thinks it is wrong and inconsistent. So you're not criticising MWI in your comment above, you're criticising the 'standard' theory.) $\endgroup$ Commented Mar 28, 2023 at 18:44
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$\begingroup$ How would the local observer know how much energy, momentum, angular momentum and charge has been transferred, already? That information is only available to observers who are covering increasingly larger distances to the original system and it can only be retrieved at the speed of light, hence it is never available "here and now". It can only be made available in the future. To me this is trivial. With regards to Everett's thesis... am I asking too much by pointing to a trivial mistake in the second sentence? I don't think that reaches any level of "extra work" by standards of science. $\endgroup$ Commented Mar 28, 2023 at 20:41
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$\begingroup$ @FlatterMann arxiv.org/abs/2008.02328 $\endgroup$– alanfCommented Mar 28, 2023 at 20:53
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$\begingroup$ The local observer doesn't need to know how much energy or momentum has been transferred elsewhere. The Everett Interpretation is an entirely local theory - there is no 'action at a distance' involved. All the information needed is available 'here and now' - that's one of the major selling points in its favour. Do you have any comment on my answer based on what I said, or what the actual MWI as Everett or Carroll presented it says, as opposed to your own made up version of it? $\endgroup$ Commented Mar 28, 2023 at 21:04