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This is obviously a follow on question to the Phys.SE post Hilbert space of harmonic oscillator: Countable vs uncountable?

So I thought that the Hilbert space of a bound electron is countable, but the Hilbert space of a free electron is uncountable. But the arguments about smoothness and delta functions in the answers to the previous question convince me otherwise. Why is the Hilbert space of a free particle not also countable?

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The dimension of the Hilbert space of a free particle is countable. To see this, simply note that

  1. The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$.

  2. The dimension theorem guarantees that any two bases of of a vector space have the same cardinality, which allows us to define the dimension of a vector space as the cardinality of any basis.

  3. The Hilbert space $L^2(\mathbb R^3)$ is separable; it admits a countable, orthonormal basis (e.g. hermite functions). Therefore, by the definition of the dimension of a vector space, it has countable dimension.

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So the total universe, or at least all of quantum mechanics, is countable? I was once told so by a very famous physicist, but I didn't think that was the majority position. –  Jim Graber May 22 '13 at 7:53
    
But then what are rigged Hilbert spaces good for? I thought their whole point was mixing countable and uncountable in a consistent way? A quick look at Wikipedia finds “They can bring together the 'bound state' (eigenvector) and 'continuous spectrum', in one place.” –  Jim Graber May 22 '13 at 7:53
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@Jim If you're using the rigged Hilbert space formalism, your state space is actually a subspace of a countable Hilbert space. See here: physics.stackexchange.com/questions/43515/… –  user1504 May 22 '13 at 11:39
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@user1504 : ok, but then it's not a Hilbert space. –  jjcale May 24 '13 at 18:02
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@jjcale Jim asked a second question (in his comment) about rigged Hilbert spaces. In this formalism, the state space is not actually a Hilbert space. Instead, it is a vector subspace of some Hilbert space. This is sufficient to answer Jim's question. For more details about the formalism, go see the answer to the question I linked to above. –  user1504 May 24 '13 at 19:19
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