Dimension of Hilbert Space in Quantum Mechanics

vinay uppal

We know that the Hilbert space of wavefunctions can be spanned by the |x> basis which is a non-countable set of infinite basis kets. Now consider the case of a particle in a box. We say that the space can be spanned by the energy eigenkets of the hamiltonian (each eigenket corresponds to an energy eigenvalue). Since the energy eigenvalues are discrete, therefore the set of corresponding eigenkets must form a countable set of infinite kets. But then isn't this a contradiction, since the cardinality of any set of basis vectors spanning the same space must be the same. In other words there must exist a bijective mapping from the one set of basis vectors to another set of basis vectors, whereas in this case we cannot define such a mapping from a countably infinite set to a non-countable infinite set.

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George Jones

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We know that the Hilbert space of wavefunctions can be spanned by the |x> basis which is a non-countable set of infinite basis kets.
This is not a basis for the Hilbert space, since the |x> do not actually live in the Hilbert space. See rigged Hilbert spaces or Gelfand triples.

George Jones

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For another paradox related to this, see

Standard quantum mechanic books that have short treatments of rigged Hilbert spaces/Gelfand triples include Nonrelativistic Quantum Mechanics by Anton Z. Capri, Quantum Mechanics: A Modern Development by Leslie Ballentine, and Quantum Mechanics: Foundations and Applications by Arno Bohm. See also the Chapter 14, Bras, kets, and all that sort of thing, in Mathematics for Physics and Physicists by Walter Appel.

A couple of heuristic not so careful posts by me explaining this sort of thing are

As far as I know, almost all (actual) Hilbert spaces used in quantum mechanics are separable, i.e., they have countable orthonormal bases (not to be confused with Hamel bases).

vinay uppal

That was an awesome paradox! Too good! And enlightening!

Anyway, returning to my question,I understood your point that |x> does not live in the Hilbert space.Thanks!

Let us not call the space of wavefunctions for a particle in a box as Hilbert space. Let us call it some space (say space X). This space can be spanned by |x> basis and the countably infinite energy basis kets (as in, you can express any wavefunction in |x> basis as well as the energy basis). How is this possible?

ice109

what are infinite basis kets |x>? why do they span the same subspace of the $L^2$ space as the eigenkets of the hamiltonian?

George Jones

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Let us not call the space of wavefunctions for a particle in a box as Hilbert space. Let us call it some space (say space X). This space can be spanned by |x> basis and the countably infinite energy basis kets (as in, you can express any wavefunction in |x> basis as well as the energy basis). How is this possible?
Think of wavefunction space as lying inside the larger space as a proper subset. Anything inside the larger space, including stuff in proper subsets, can be expanded in terms of the generalized eigenstates |x>. Find an orthonormal basis (e.g., energy eigenstates) for the proper subset, in terms of which anything inside the proper subset can be expanded. Stuff outside the proper subset, however, cannot be expanded in terms of a basis of the proper subset.

Count Iblis

Hmmm, why not use wave packets? The position representation is unphysical anyway. So, instead of attempting to definine a state corresponding to a particle being located at exactly a certain position using a mathematical tour de force, you can just as well work with Gaussian wave packets. You can cut them off to be exactly zero outside some range using infinitely differentiable functions if you like.

The Fourier transform of Gaussians are Gaussians, so you get wave packets for the momentum eigenstates in the same form.

George Jones

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Hmmm, why not use wave packets? The position representation is unphysical anyway. So, instead of attempting to definine a state corresponding to a particle being located at exactly a certain position using a mathematical tour de force, you can just as well work with Gaussian wave packets. You can cut them off to be exactly zero outside some range using infinitely differentiable functions if you like.

The Fourier transform of Gaussians are Gaussians, so you get wave packets for the momentum eigenstates in the same form.
Yes, I commented on this in the third reference that I gave in post #3,
George Jones said:
Nevertheless, quantum mechanics has a (somewhat) nice Hilbert space formulation (due to Von Neumann) that neither needs nor uses delta functions or any other non-square-integrable beasts (e.g., plane waves).
Two points: many physicists do not want to give up the elegance (and occasional ugliness) of full Dirac notation; quantum mechanics in Hilbert space, done in a mathematically honest way, is itself quite daunting (see, e.g., the books by Reed and Simon).

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