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Is the Hilbert space spanned by both bound and continuous hydrogen atom eigenfunctions?

+ 3 like - 0 dislike
577 views

As e.g. Griffiths says (p. 103, Introduction to Quantum Mechanics, 2nd ed.), if a spectrum of a linear operator is continuous, the eigenfunctions are not normalizable, therefore it has no eigenfunctions in the Hilbert space.

On the other hand, both bound and continuous eigenfunctions are required to have a complete set, to be able to expand an arbitrary wave function in terms of the eigenfunc­tions (Landau&Lifshitz, Quantum Mechanics, p.19). How are these results connected, how to explain the apparent contradiction?

Is a formulation that the Hilbert space is spanned by both bound and continuous hydrogen atom eigenfunctions correct?

Update: I just found Scattering states of Hydrogen atom in non-relativistic perturbation theory which is related (but only partially answers this question).

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user wondering
asked Feb 19, 2015 in Theoretical Physics by wondering (30 points) [ no revision ]
Related: physics.stackexchange.com/q/68639/2451 and links therein.

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user Qmechanic

2 Answers

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If a self-adjoint operator $H$ has a continuous spectrum, the closure of the space spanned by its eigenvectors is only a proper subspace of the Hilbert space. This holds since the orthogonal projection of $H$ to this subspace has discrete spectrum only. 

To get the full Hilbert space one needs to add the vectors from the complement. Vectors of this complement are described by an integral over a complete system of improper eigenvectors. The latter are not normalizable scattering states (for a particle in a potential solutions of the Lippman-Schwinger equations), hence do not belong themselves to the Hilbert space, but their integrals do. 

However, a simple example of this situation is the free particle with Hamiltonian $H=p^2/2m$. There is no point spectrum, the continuous spectrum is the halfline $E\ge 0$, and its improper eigenvectors are just the momentum kets $|k\rangle$ with $p|k\rangle=k|k\rangle$, with energy $E=k^2/2m$. The above complement is of course the whole Hilbert space, and the integral form of the eigenvector representation is just the familiar momentum representation of the wave function. 

For hydrogen, one gets a similar integral representation, but the details are somewhat messy to describe concretely as one needs the complete analysis of the scattering behavior.

answered May 1, 2015 by Arnold Neumaier (12,355 points) [ no revision ]
+ 1 like - 2 dislike

No, the Hilbert space is not spanned by continuous "eigenfunctions" because they are not eigenfunctions at all!

A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has a point spectrum, a continuous spectrum and a singular spectrum. The latter is physically irrelevant.

The point spectrum consists of the values $\lambda_i$ for which a true eigenvector $\psi_i\in L^2(\mathbb{R})$ with $T\psi_i = \lambda_i$ exists. The point spectrum is always discrete and countable because the Hilbert spaces of quantum mechanics are always separable, meaning they have countable bases, and so there are only countably many possible different eigenvectors.

The continuous spectrum is, as the name tells you, continuous, and apart from discrete intersections with the point spectrum, no elements of $L^2(\mathbb{R})$ are eigenvectors belonging to points of the continuous spectrum.

The quantum mechanical setting with bound and continuous states is described by a Gel'fand triple/rigged Hilbert space $\Phi \subset \mathcal{H} \subset \Phi^*$, where $\mathcal{H}$ is the physical Hilbert space and $\Phi^*$ is a larger space containing also eigenvectors for the elements of the continuous spectrum (e.g. for $L^2(\mathbb{R})$, this is frequently the space of tempered distributions containing the "eigenfunctions" $\delta(x-x_0)$ of the position operator). The large space of such a Gel'fand triple fails to be a Hilbert space - it has no proper inner product, for one. Therefore, there are indeed "eigenvectors" corresponding to the continuous spectrum in quantum mechanics, but they are not elements of the Hilbert space.

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user ACuriousMind
answered Feb 19, 2015 by ACuriousMind (730 points) [ no revision ]
I think that is important to note that we can represent a physical state (live in Hilbert space) by an integration over "free states"(sequences of physical states $|\psi_{\alpha}\rangle$ that satisfy $(A-a)|\psi_{\alpha}\rangle \rightarrow 0 $) for some $A$, with $a$ living in continuous spectrum of $A$.

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user Nogueira
So the complete basis of the Hilbert space is formed just from the "true eigenfunctions", i.e. bound state solutions. So as to be able to express an arbitrary (hydrogen atom) wavefunction we have to consider the complete basis, consisting of "eigenfunctions" of the continuum, in addition to the bound state (true) eigenfunctions. But such a (complete) space is bigger, than the Hilbert space. Correct?

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user wondering
@wondering: Essentially correct, though "But to be able to express an arbitrary (hydrogen atom) wavefunction we have to consider the complete basis, consisting of "eigenfunctions" of the continuum" is, strictly speaking, not true. Physical wavefunctions must be $L^2(\mathbb{R})$, and though the non-bound states are often conveniently obtained as "wavepackets" constructed out of the non-normalisable functions in the larger space, these wavepackets themselves are members of $L^2(\mathbb{R})$, and you could, in princple, express them in the bound basis (but that'd be ugly, and uninsightful).

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user ACuriousMind
So I can express wave packets created of hydrogen atom continuous wave functions in terms of the bound state basis? How to see it intuitively? Would you be able to give a reference to this?

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user wondering
@wondering: Proof: Wavepackets are in $L^2(\mathbb{R})$, and the bound states are, as eigenvectors of a self-adjoint operator, a full basis, hence wavepackets are expressible in terms of bound states. (This is why I said it is not particularly insightful, the proof is entirely non-constructive)

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user ACuriousMind

@wondering: The statement 

the bound states are, as eigenvectors of a self-adjoint operator, a full basis,

is valid only if there is no continuous spectrum; hence this is wrong in case of hydrogen. See my answer to the question. 

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