In the many world interpretation of quantum mechanics, after a superposition of states has interacted with some outside agent the two states smoothly continue to exist in two different worlds (so no collapse occurs).
But what if the weights of, say, two superimposed states are varied? To give an example, if in one superposition the weights are both $\sqrt2$ and in another (the same states) the weights are $\sqrt \frac{2}{3}$ and $\sqrt \frac{1}{3}$. After a measurement on both is made both states will continue to evolve in two different worlds. Why should they care about the weights?
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2Possible duplicate on Physics: physics.stackexchange.com/q/536522/195139– SandejoJun 12, 2021 at 20:19
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3 Answers
In the limit of infinitely many measurements on identically prepared systems, the amplitude of worlds in which the outcomes don't match the Born rule goes to zero. Worlds with an amplitude of exactly zero don't exist.
If you can only do finitely many trials, then that argument doesn't work: worlds corresponding to all sequences of outcomes exist, and the most you can say is that some have larger amplitudes than others, which is circular. Therefore I'd say that MWI can't explain the empirical fact that the Born rule works.
However, adding wavefunction collapse doesn't make this situation any better, because classical probability has the same problem. Any sequences of outcomes is possible in finitely many trials, and the most you can say is that some are more probable than others, which is circular. It's a general problem with any probabilistic, or amplitudinal, theory.
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In MWI, I'm pretty sure some branches will not see the born rule work, but they appear to be vanishingly small youtu.be/2R7elwozou4?t=1705 (~2min long)– J KusinJun 14, 2021 at 19:21
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+1! But in the case of a single "collapse" (in the MWI), what's the difference between a measurement of a 50/50 chance collapse and a 70/30 collapse? In both cases, a splitting of the branches appears. Can this difference only become clear in the larger context of more measurements, which will make the difference visible? Or is the 30/70 inherently no different from the 50/50 state? Of course, it is, but what id the difference for the splitting of entangled states on their own (the splitting on its own)?– user52804Jun 14, 2021 at 19:22
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@user52804 this is coming not out of expertise but just light hobby-reading, but I believe there are people who think that more than 2 worlds are created by such a split - probably not infinite, but many many worlds, and the distribution of these worlds follows the expected probability.– TKoLJul 20, 2021 at 16:32
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I would be very curious how we can approach the question without such an assumption though.– TKoLJul 20, 2021 at 16:32
"Can this difference only become clear in the larger context of more measurements, which will make the difference visible? Or is the 30/70 inherently no different from the 50/50 state?"
Each branch is only that branch, one of the outcomes, so in a sense yes, within a branch looking back there is only 'happened' or 'didn't happen' with no amplitude.
But, we don't just get magnitude of probability from repeated measurements, we get it from squaring the wavefunction (the Born rule). This wavefunction contains the energy of the quantum system in a way that captures it's dynamics (potential energy + kinetic energy, the Hamiltonian). When a measurement is taken, the wavefunction of the quantum system joins that of the measuring system or observer's wavefunction, merging some of tye information required to recover it's history with the wider system.
Similar question here: Probabilistic prediction (quantum mechanics) - what is the meaning of such a prediction and how do you falsify it?
answered Jun 14, 2021 at 21:30
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The developing wavefunction (into two branches or into one collapsed state) though doesn't "know" which state it's going to be in. Unless you assume hidden variables (which is not an interpretation, I think, like the MWI) the chances are really inherent to Nature. It's a bit difficult to imagine this being the case. What makes the wavefunction decide? Pure probability?– user52804Jun 14, 2021 at 21:39
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@Methadont: in MWI it doesn't decide between, it does all the things. The universe we are most likely to find ourselves in, closely follows the 2nd law of thermodynamics, because those universes are gigantically more common in the total phase-space of all possibilities. But in MWI, a version of us experiences all possible outcomes. Jun 14, 2021 at 21:45
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But don't the two states move away from each other? Each into its own branch? What decides where each is gonna end up? They have to end up somewhere, of course, but why not the other way round? By the way, does this site support MathJax?– user52804Jun 14, 2021 at 21:50
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They are in a superposition, they separate. There is no deciding between. All the possibilities happen. Cumulatively, more universes are going to have had chains of likely events, even though at each branching point there is only one world with each outcome Jun 14, 2021 at 21:55
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So the "middle" branches are the most, and the boundary ones the less?– user52804Jun 14, 2021 at 22:00
Differential weighting of "probabilities" for a QM event becomes meaningless for Everett's MW "interpretation" of QM. This is revealed by thinking through how Everett's MW deals with conservation laws, and the apparent "creation" of entire new universes as a result of trivial quantum events. Under Everett, the other universes were "already there", and just operating in sync -- "cohered" -- prior to the spit event. No new universes were created. The quantum split just causes two potential versions of the universe to decohere.
Thinking this through -- the universe initially had to have infinite cohered versions -- after all there is no limit to the number of splits possible so our universe had to have had infinite versions, and therefore infinite collective energy, momentum, etc.
However, there is ALSO, no limit to the future splits for any now detached universe. They can exist forever, with multitudes of quantum events causing further decoherences. EACH decohered universe therefore must have infinite versions within itself.
Since all options of a quantum event have infinite bandwidth, the 2/3:1/3 difference you ask about becomes a meaningless. The only way that MW can deal with probabilities is if one can say that the frequency of finding ourselves in the 1/3 world is only half that of fining ourselves in the 2/3 world, but this presumes an outside observer who can actually cross the boundaries between decohered universes and count up the frequency of the 1/3 vs 2/3 cases. Neglecting that Everett prohibits such an observer -- it also says that since, the number of both cases is infinite, the relative probability between them becomes meaningless.
