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These facts are taken for granted in a QM text I read. The purportedly guaranteed non-normalizability of eigenfunctions which correspond to a continuous eigenvalue spectrum is only partly justified by the author, who merely states that the non-normalizability is linked to the fact that such eigenfunctions do not tend to zero at infinity.

Not a very satisfying answer. What I'm really after is an explanation based in functional analysis. I believe there is a generalized result about inner products being finite for discrete spectra but infinite for continuous spectra.

Can anyone shed some light on this?

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Maybe this will help you – nijankowski Jun 20 '13 at 15:56
Possible duplicate: – Qmechanic Jun 20 '13 at 15:56

2 Answers 2

This could be a math question. But if you think about the physical aspect of the question, it is interesting to look at the Schrodinger equation:

For free fields (without potential), you have (in units $\bar h = m = \omega = 1$):

$$i\frac{\partial \Psi( k, t)}{\partial t} = \frac{ k^2}{2}\Psi( k, t)$$ or

$$ E \tilde \Psi( k, E) = \frac{ k^2}{2} \tilde \Psi( k, E)$$

Whose solution is :

$$\Psi( k, t) \sim e^{- i \frac{ k^2}{2} t}$$


$$ \tilde \Psi( k, E) \sim \delta (E - \frac{ k^2}{2})$$

Here $ \tilde \Psi( k, E)$ is a Fourier transform of $\Psi( k, t)$.

It is clear, from the form of the equations, that there is no constraint about $E$. It is a continuous spectra, and this is clearly a non-normalizable solution.

However, with potentials, things appear differently, and you will have some differential equations, for instance, for the harmonic oscillator potential, you will have :

$$ E \Psi( k, E) = \frac{ k^2}{2} \Psi( k, E) - \frac{1}{2}\frac{\partial^2 \Psi( k, E)}{\partial k^2}$$

The solution for $\Psi$ involves a Hermite differential equation (multiply by some exponential $e^{-k^2}$).

If E is taking continuous, then the Hermite solution (with a real index) is not bounded at infinity, and so the solution is not normalizable.

If we want a normalizable solution, then we need a (positive) integer indexed solution $H_n$, whose name is Hermite polynomials. In this case, the spectrum of $E$ is discrete.

The choice of $E$ discrete (and so a normalizable solution) is then a physical choice. In the case of the harmonic oscillator, it is unphysical to suppose that the solution is not bounded at infinity.

The case of Hermite polynomials is a special case of orthogonal polynomials, which is very well suited to represent orthonormal states, corresponding to discrete eigenvalues of the hermitian operator Energy.

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This has come up gain in another context. While everything you say is correct you have ignored the usual next step for the unbound system, which is to form normalizable wave-packet solutions from the non-normalizable asymptotic solutions. – dmckee Sep 9 '13 at 2:35

The question is really one of definition. In the math literature on self adjoint opertors the "discrete spectrum" is by definition that part of the spectrum which consists of normalizable states, while the "continuous spectrum" is that part where they are non-normalizable. It is possible to have a physical system (a random potential on the entrire real line for example) in which all states are localized (and hence normalizable) but the eigen-energies form a continuous set. The physics example where only isolated energy levels are "discrete" only applies to simple models.

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No, the discrete and the continuous spectra are not by definition as such. They are defined as set of points and after that you may in principle verify whether or not their eigenstates are normalisable. – Gennaro Tedesco Jun 28 at 18:05
My bad... I tend to use "dicrete spectrum" as a synonym for "point spectrum" which is the set of eigenstates that are are elements of the Hilbert space. Anyway good place to look for the Physics<-> Math link is… – mike stone Jul 5 at 13:43
I see that I am not alone in identifying "discrete spectrum" with "point spectrum": – mike stone Jul 5 at 13:48

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