# MWI and quantum probability

One thing that I cannot wrap my head around, and which never seems to be addressed in the videos and articles I've read about the Many Worlds Interpretation, is how "branching" is supposed to account for the probabilistic outcomes we see in quantum experiments.

In the classical interpretation, the wavefunction of a particle exists in a superposition of possible outcomes until it is measured, after which decoherence kicks in and the particle settles into one if its possible outcomes. The "more likely” outcomes – corresponding to the square modulus of the wavefunction – occur more frequently. And yet any one particular outcome is essentially random.

In the MWI, we are in one fixed branch of the universal wavefunction. Whatever the measurement of the particle could be, will be, in some branch – just not necessarily our branch of reality.

So how is there to be any reconciling of the probability we see in experiments with this interpretation? If a quantum event is measured and outcome A is observed to have probability $$1/3$$, and probability B has probability $$2/3$$, then what does this mean in terms of the "branches" in the MWI?

The idea of "branching" or "branches" is fundamentally discrete. Does this mean there would be exactly 1 branch where outcome A happens, and exactly 2 branches where outcome B happens? What about for more complicated probabilities, like an outcome of $$12214/36537$$? And what about an irrational probably? Like $$1/\sqrt2$$ or $$1/\pi$$?

Are we to believe that there are no irrational quantum probabilities? That the density of the rational numbers in the reals is good enough? Or can branching somehow be made into a continuous phenomenon? And what would that even mean?

• Popular science sources that talk about "branches" are being inaccurate. Wavefunctions have a domain. This is $\mathbb{R}^3$ if you're talking about a single particle's position and something much larger if you're talking about the whole universe. If $x$ is the universe in which you perform the above experiment and see outcome A the first time, while $y$ is the universe in which you do the same and see outcome B the first time, it must be that $\psi(y) = 2\psi(x)$. But there are still no branches. Aug 17, 2021 at 19:11
• There's no problem with having an infinite number of branches. See my answer here: physics.stackexchange.com/a/536580/123208 Aug 17, 2021 at 21:40

I'm not clear on what videos and articles you have been reading but there is a large literature on probability in the MWI and on branching.

According to the MWI the universe you see around you is part of a much larger and more complex structure called the multiverse, which looks a bit like a collection of classical universes to some approximation. A universe is a structure within the multiverse in which information about a particular outcome of an event flows freely. For example, there is another universe in which I typed the letter 'a' at the end of this sentence: b. We're not in that universe but that universe exists and the people in that universe can see the letter 'a' at the end of the sentence and discuss it and check that I typed the letter 'a' and so on. In this universe I typed the letter 'b' instead and the people in the 'a' universe can't see the letter 'b' at the end of that sentence. So those two outcomes are in different universes.

For more detail see:

https://arxiv.org/abs/quant-ph/0104033

https://arxiv.org/abs/quant-ph/0107144

The parallel universe approximation breaks down in various kinds of experiments like single particle interference experiments and the EPR experiment. The approximation holds when information about a particular outcome has been copied from one system into another: this suppresses interference and that process goes by various names like decoherence and quantum darwinism:

https://arxiv.org/abs/1212.3245

The outcome of an experiment is discrete and is basically an eigenstate of some observable. Before the state of a system is decohered it will look a bit like this: $$\sum a_j|j\rangle$$.

The probabilities of doing a measurement on that state don't come from counting the number of branches, i.e. - the number of $$j$$ labels. Rather, the gist of the explanation goes like this. The probability of an outcome is a function of the state that satisfies particular properties required by decision theory, like if a person uses those probabilities you can't make him take a series of bets that will make him lower the expectation value of his winnings, the probability of an outcome can't be changed by a later measurement and some other assumptions and from this you can explain that the probability of the $$j$$th outcome will be $$|a_j|^2$$:

https://arxiv.org/abs/0906.2718

https://arxiv.org/abs/quant-ph/0303050

https://arxiv.org/abs/quant-ph/9906015

https://arxiv.org/abs/1508.02048

So if you are offered a bet such that if the outcome of the measurement if $$j$$ you will win £1 and otherwise you get nothing the probability of that outcome is $$1/\pi$$, then if somebody offers to let you place that bet for less then £$$1/\pi$$ you would be rational to take the bet, otherwise you wouldn't be rational to take it. Another way of thinking about it is that each branch has a certain thickness compared to the branches from which it is a descendant and the probability is the thickness of the branch that is relevant to betting that is rational in the decision theoretic sense.

There is another approach to deriving the probability rule called envariance that I think is less satisfactory but I'll give you a link for the sake of completeness:

https://arxiv.org/abs/quant-ph/0405161