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No expert by any means, but sometimes, in different contexts the term multiverse used. In quantum mechanics, some say that it is possible that there are actually many universes where all the possible states are manifested, cf. the many-worlds interpretation. I wondered what is the cardinality of the set of all possible universes? Does the answer to this question has or can have any theoretical consequences?

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Concerning many-worlds versus multiverse, see also physics.stackexchange.com/q/10140/2451 –  Qmechanic Mar 23 '13 at 20:03
Thanks @Qmechanic! I didn't know that distinction exists and that it is even oversimplified. I did found this thread though : link, which actually leads nowhere. Ending up with "this is not interesting". I find this assertion problematic as physics and in particular quantum mechanics is based on some non trivial mathematical tools. I mean, if the mathematics of the formulation will collapse or be problematic, wouldn't physics care? At the very least it seams to be worth exploring. –  Sonia Mar 23 '13 at 21:00
There are objects in mathematics which are not sets, for example class. It can be detrimental to the analysis. –  Sonia Mar 24 '13 at 6:47
There is not one 'interpretation' but several mutually incompatible 'interpretations'. Second, each one of them has been shown to be wrong. Check Against Many-Worlds Interpretations. A physics FAQ for general public is available here. –  juanrga Mar 24 '13 at 16:13
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2 Answers

One easy way of working out the cardinality of the "multiverse" (which does not have a unique, concrete model behind it) would be to think of how the "universes" within the multiverse are labeled. Suppose that your multiverse is based on different choices for some physical parameter $\Lambda$, and each universe would have its own setting for $\Lambda$. Then, the cardinality of the multiverse would be determined by the cardinality of the set of possible choices for $\Lambda$. Therefore, if you could only have $\Lambda \in \{1,2,7\}$, then there are clearly three possible universes; if you can only have $\Lambda \in \{1,2,3,\dots\}$, then there are countably infinite universes, etc.

Physical consequences of the cardinality of this set are of course determined by whether or not the universes interact. In both the many-worlds interpretation of QM and the string landscape as a multiverse pictures, there are no interactions between the universes so in those cases, there are no physical consequences.

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This line of reasoning can be expanded upon by assuming that the universe is finite. This implies that single-particle bases are countable and consequently so is a many-body basis, which implies there are countably many universes. I don't personally believe this has any important conceptual significance. –  Joshua Barr Mar 24 '13 at 4:10
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The many worlds interpretation assigns reality to every possible outcome of a measurement of the quantum state. So consider for example a Stern-Gerlack experiment on a single electron, measuring in some fixed axis, you get two outcomes to this experiment, so in the many worlds interpretation a complete description of reality would include two worlds. This example can be more concretely understood in terms of measurement operators and eigenstates of those measurement operators. Here our Stern-Gerlack experiment is a Pauli operator, and the Pauli operator has two eigenstates. So we can say that a measurement creates as many worlds as it has eigenstates. Since each measurement compounds the number of worlds, the cardinality of the multiverse must be at least as large as the cardinality of the largest possible set of eigenstates belonging to one possible measurement. I suppose the measurement with the largest number of eigenstates would be a measurement of a system with an infinite dimensional Hilbert space, such as measurement of position, which would have an infinite number of eigenstates. I can't claim to know what exactly the cardinality of that infinity must be, but if I were asked to guess I would say it is uncountably infinite.

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The position operator is pathological in that its eigenstates lie outside the space of physically realizable states (a "position eigenstate" is just a limiting case). For a finite universe, the spectrum of any operator with normalizable eigenstates will be countable. –  Joshua Barr Mar 24 '13 at 6:29
More on this, I think most Hilbert spaces we deal with in physics are separable, which means they have a countable basis. The Wightman axioms require that the state space be a separable Hilbert space. I think there is some discussion of this point in Streater and Wightman's quantum field theory textbook. I think if you start allowing for an infinite number of degrees of freedom (e.g. an infinite number of different particles) you might need to drop the assumption that the hilbert space is separable. –  asperanz Mar 24 '13 at 7:05
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