# How many worlds in the many-worlds interpretation of quantum mechanics?

How many branches are supposed to be created from each quantum event?

With the Copenhagen interpretation, this is cake. There’s only one. With the many-worlds interpretation, this is not so easy for my physics-ignorant mind to untangle, because many of these quantum choices don’t split fifty-fifty. You can’t have a 90% quantum chance split two ways, else we’d have experimental results giving us perfectly balanced coin tosses instead of the heavily weighted experimental outcomes. In order to get the many-world split to work at 90% odds consistently over many observations, you have to split it by some multiple of ten. Only if we imagine that ten/twenty/one hundred new branches are created can we assign the low probability outcome correctly to one of every ten. (In reality, these odds follow the Born rule and not my arbitrary choice of 90%, but you get the drift. It’s still not a coin flip.)

I’ve seen advocates of many-worlds argue that the math of decoherence is simpler than that of wavefunction collapse, and that therefore we should believe many-world decoherence instead. I’m not in a position to understand what they’re talking about, but if their simple math can’t answer my genuinely simple question about how many branches split off with the apparent collapse – all the while preserving the observed odds – then I’m not sure why I should believe them about this.

“Infinite branches” is an easy answer, I guess. Is there a finite answer about how many new branches are created every time we zing a photon off an electron? Bonus question: Is this an answer I’d be able to understand?

Every possible path is taken. That is different from an infinite number.

The branches do not have equal weights, in general.

Yes, but how does that occur?

I’ll extend the OP’s question a bit, because I think it’s relevant. Not all events are simple discrete this-or-that questions. A photon might interact with an atom, causing it to eject an electron. The momentum of the electron could have any of some continuum of values. How does many worlds handle that case?

All quantum events are discrete questions, by definition. The election will have one of a finite number of states–not a continuum of possibilities.

You’re looking at it wrong. It’s a misconception that “worlds” are “branching.”

The many-worlds interpretation is basically the observation that the linearity of wave functions implies that they are equivalent to the sum of infinitely many dirac-delta functions. If you don’t know what dirac-delta functions are – it basically means that you can look at a quantum mechanical wave function as an infinite number of different wave functions added together, where each wave function has a sharp peak – a definite eigenvalue, such as position. If you let the wave function evolve, then keeping track of each little delta function is informationally equivalent to an infinite number of individual world-lines existing simultaneously. There are no “added” worlds. No “branches.” It is just a mathematical re-statement of Schrodinger evolution of a single universal wave function.

So it’s a combination of state spaces that inevitably converges into a single state space? (Just making sure I understood what you said, I’m not sure I do)

No. See for example Compton scattering.

I’d like to see a cite for this.

This is not true in several ways: 1) discrete does not imply a finite range of values 2) electrons and more general systems can have an inifinite number of observed states 3) these observed states can be continious (i.e. operators representing observables can have continious eigenvalues) 4) Infact even in the most resticted quantum system the number of quantum states will always be infinite.

In some (notable) situations operators representing observables have a discrete (e.g. the energy of an electron in a potential well) or even finite (e.g. the spin of an electron in the z direction) range of eigenvalues which is where the association comes from, but this is not the general rule.

Okay, but what about the people who argue for a finite MWI, which has already appeared in this thread?

Tell me there are an infinite number of world-lines existing simultaneously, and I nod my head. Yes, that resolves the problem mathematically (even better than Copenhagen, given certain notions of parsimony). But the finite peeps are out there. This is genuine theoretical speculation. But I don’t get why they favor finite MWI, especially when there is an apparent contradiction between finite MWI and the observed weighted probabilities.

If we’re speculating about a finite universe, is there any way to make sense of MWI over Copenhagen? Or would you say it’s a bit crackpot to rest on any finite interpretations given the evidence we have?

I think in this way a world would correspond to a superselection sector, but I can’t say I know enough to answer your question confidently.

Well as any version of the MWI must fit with the mathematical postulates of quantum mechanics, any disagreement of the number of ‘worlds’ (if defined) would have to be due to a disagreement in the definition of ‘world’.

Can the number of ‘worlds’ be sensibly defined so they’re always finite? Not sure.

Ordinarily, I think Born’s rule is taken as axiomatic in the many worlds approach, just as in other interpretations. Deutsch has however argued, in a decision theoretic setting, that any rational agent who believes all the rules of quantum mechanics except Born’s, nevertheless acts as if the probabilities of outcomes (and hence, the weight of branches) are given by Born’s rule, making its axiomatic stipulation unnecessary. It’s also been argued that this derivation is only valid if one assumes the many worlds interpretation.

As for the number of worlds in many worlds, if* there truly exists a continuum of outcomes in some quantum processes, then I don’t see how a finite number of worlds could account for that.

*I say ‘if’ only because I’m not sure what effect a finite universe would have – naively, a finite volume would imply that ultimately all quantum numbers are discrete, it seems to me.

I strongly encourage you to read Everett’s dissertation on the subject, which can be found here:

The viewpoint I described is made obvious in many places throughout, for example here:

Let one regard an observer as a subsystem of the composite system:
observer + object-system. It is then an inescapable consequence that
after the interaction has taken place there will not, generally, exist a
single observer state. There will, however, be a superposition of the composite system states, each element of which contains a definite observer
state and a definite relative object-system state. Furthermore, as we shall
see, each of these relative object-system states will be, approximately,
the eigenstates of the observation corresponding to the value obtained by
the observer which is described by the same element of the superposition.
Thus, each element of the resulting superposition describes an observer
who perceived. a definite and generally different result, and to whom it
appears that the object-system state has been transformed into the corresponding eigenstate. In this sense the usual assertions of Process 1
appear to hold on a subjective level to each observer described by an element of the superposition.

Moreover, he over and over again makes explicit that this is a theory of one and only one universal wave function. The problem is that all of this is very clear in the technical literature, but popularly there is a lot of disinformation.

For a given scattering angle of measurement, there is exactly one possible outcome for the electron’s state.

1. Oops, you’re right, I meant countable, not finite.
2. Yes, but I wasn’t talking about observed states, I was talking about outcomes of quantum events.
3. Can you give an example of a quantum operator with a continuous set of eigenvalues?
4. Yes, I believe it’s countably infinite.

X operator, typically.

Okay, but it’s not really observable as a continuum. Nor does it represent a quantum interaction.

Momentum, position, free particle Hamiltonian… Generally, only bound quantum systems exhibit discrete quantum numbers, so for example a free particle can have any energy whatsoever, while a particle confined to a box has energy eigenvalues proportional to the square of some integer n.

Quoth Hellestal:

To an extent, it’s a bit crackpot to rest on any interpretation of quantum mechanics at all. Any of the standard interpretations you’ve heard of, and several more you haven’t, all give exactly the same predictions for the results of experiments, and therefore are scientifically exactly equivalent. It’s not the scientist’s place to choose between experimentally-indistinguishable models: That’s a job for the philosopher. Using one particular interpretation or another might make it a little more intuitive how to do some calculations, but any calculation can be done in any interpretation, and anyway, which interpretation is more intuitive depends on what calculation you’re doing (and on the intuition of the particular physicist, too).

Your statement makes no sense to me. “Observable as a continuum?” What does that mean? And why does the operator have to represent a “quantum interaction,” whatever you mean by that.