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Generally, the bound states (normalizable eigenvectors) of a Hamiltonian have discrete eigenvalues.

Is it possible for the eigenvalues to cover an interval? Say, $(a,b)$?

That is, for each $E \in (a,b)$, there is a corresponding bound state?

Qmechanic
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Jiang-min Zhang
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1 Answers1

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A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has in its spectrum three different kinds of subspectra: A discrete point spectrum, a continuous spectrum, and a singular spectrum. The latter is physically discarded.

The point spectrum consists of the eigenvalues of $T$, that is, the spectral values for which true eigenvectors in $L^2(\mathbb{R})$, and hence normalizable eigenstates, exist.

The continuous spectrum may intersect with the point spectrum, but except for these discrete intersections, the continuous spectral values do not have eigenvectors in $L^2(\mathbb{R})$, but only in a larger rigged Hilbert space, which are consequently non-normalizable (for example because the Hermitian product of $L^2(\mathbb{R})$ is not a proper inner product for them).

Thus, continuous bound states do not exist in the usual quantum mechanical setting.

A way to see that the true eigenvalues cannot form a continuum is to notice that they would have to be uncountably many, but the separable Hilbert spaces of quantum mechanics have only countably many basis vectors, and hence there are only countably many possible independent eigenvectors.

ACuriousMind
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  • A way to see that the true eigenvalues cannot form a continuum is to notice that they would have to be uncountably many, but the separable Hilbert spaces of quantum mechanics have only countably many basis vectors, and hence there are only countably many possible independent eigenvectors.------nice argument! – Jiang-min Zhang Feb 03 '15 at 15:19
  • Isn't it a bit too convenient to discard the pieces of the mathematical objects we don't like? Isn't it a signal that maybe the mathematical object does not correspond so well to the physical object? – Frank Mar 04 '15 at 21:33
  • @Frank: It indeed shows that not every mathematical object corresponds to a physical object. But there is no reason they should. – ACuriousMind Mar 04 '15 at 21:35
  • Yes, there is absolutely no reason they should - but the worry is, if it doesn't match perfectly, maybe there is a problem? Maybe it is NOT the mathematical object you use that's the right one? In which case you might get into serious trouble - of course nothing really new, just theories, that we will discard when they don't work anymore. I would prefer the mathematical object to perfectly match though, as it seems to remove a big source of possible problems... – Frank Mar 04 '15 at 21:44