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The many-worlds interpretation of quantum physics is built around a configuration space, where the position of a particle is three components of the position of that universe.

What happens with particle-antiparticle creation or annihilation? It can't just change the number of dimensions, can it?

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In the many-worlds interpretation of quantum mechanics, there is at all times just one state vector for the entire Universe. It is a vector in a certain (infinite-dimensional) vector space, and that vector space is always the same. So there's nothing in the theory whose dimension changes when the number of particles in the Universe changes.

To be a bit more precise, the space in which the state vector of the Universe lives is (something like) a Fock space. Vectors in that space include states with all possible different numbers of particles, as well as superpositions containing different numbers of particles. So if a particle-antiparticle pair is created, the state vector simply "wanders" from one part of that space to another; the space itself needn't get any bigger.

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  • $\begingroup$ That seems like it would violate locality, but I guess it's fine if you use the right distance metric. It still seems weird. I can't help noticing that most of the vectors are unused. For example, there's a vector corresponding to the sum of a one-particle system and a two-particle system, but that doesn't seem to make a lot of sense as a universe. $\endgroup$
    – user2898
    Commented Apr 4, 2011 at 3:35
  • $\begingroup$ I don't think there's any problem with locality, appropriately defined. As for your second comment, our Universe does consist of a superposition with different particle numbers, all the time. One way to see this is to imagine a single radioactive atom. Wait for an amount of time equal to the half-life. Either the atom has decayed or it hasn't. Those two possibilities correspond to states with different numbers of particles, and the state vector of the Universe is a superposition of them. $\endgroup$
    – Ted Bunn
    Commented Apr 4, 2011 at 14:18
  • $\begingroup$ That just means that, for example, <a,b> and <a,b,c> both have amplitudes. <a,b> + <a,b,c> still don't. I was thinking about this more. How do you take the laplace operator? $\endgroup$
    – user2898
    Commented Apr 5, 2011 at 0:05
  • $\begingroup$ Sorry, but I don't think I understand either sentence of that comment. $\endgroup$
    – Ted Bunn
    Commented Apr 5, 2011 at 13:50
  • $\begingroup$ First sentence: A vector in Fock space is a vector in one space plus a vector in two space, plus a vector in three space, etc. Our universe only has non-zero amplitudes in the vectors where all those vectors except one are zero. Second sentence: How do you take the Laplace operator in Fock space? It would have to give a different answer for amplitude to go from one number of dimensions to another (read: particle/antiparticle annihilation). $\endgroup$
    – user2898
    Commented Apr 6, 2011 at 23:14
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As I look upon the MWI: every event is a branch point of a particular universe (i.e. the universe before the event) to a set of universes after the event. Every universe in the set has another outcome of the event as it's start point. All universes are to be considered as objectively real, all outcomes of the event do occur, all in another universe and the different universes cannot interact.

You descrive an event with one particular outcome: the annihilation of a particle-antiparticle pair. The MWI is not related to this situation, your universe is the same before and after this event because you do not consider different outcomes of the event, you assume an event with only one result (annililation).

Some further remarks:

1) Your phrase "position of the universe" does not make sense, because position is measured ín some universe. For a universe to have a position we have to consider another universe that contains it.

2) The MWI is extremley absurd. When there are 10^80 paricles in the universe and each particle has 10^40 events per second, then after 10^17 seconds there are (10^80^40^17 = 10^54400) universes. MWI defies common sense also in other respects.

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  • $\begingroup$ The branches aren't distinct. For example, there could be a universe where an electron is in point A, one where it's in point B, and one for every point in between. They're also not non-interacting. In fact, momentum is an attribute of the relative amplitudes of the "universes" where the electron is really close to that spot. It's just that they don't interact noticeably at any real distance. I know the laws that govern how the universes interact. The way they interact in configuration space is the same as a single particle in "real" space. $\endgroup$
    – user2898
    Commented Apr 2, 2011 at 22:49
  • $\begingroup$ @Gerard You might argue that an instantaneous collapse of the wavefunction is absurd as well. :) $\endgroup$
    – Lagerbaer
    Commented Apr 3, 2011 at 0:26
  • $\begingroup$ More importantly, you might argue that 10^80 particles is absurd. I'm pretty sure Occam's razor is for laws, not entities. $\endgroup$
    – user2898
    Commented Apr 3, 2011 at 6:40
  • $\begingroup$ @Lagerbaer The collapse of the wavefunction is not understood at all on an ontological level, I wouldn't say it is absurd because we can demonstrate the problem in experiments, which is not the case for MWI: this theory is a speculative theory from its first principle, it can never be proved nor disproved by an experiment. $\endgroup$
    – Gerard
    Commented Apr 13, 2011 at 20:50
  • $\begingroup$ @user2898 Occam's razor is about not using unnecessary assumptions, MWI is full blown anti-Occam. 10^80 particles is a quite normal assumption when you believe the universe consists of particles. $\endgroup$
    – Gerard
    Commented Apr 13, 2011 at 20:53

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