Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
Maybe this will help you physics.stackexchange.com/q/33668 –  Leonida Jun 20 '13 at 15:56
    
Possible duplicate: physics.stackexchange.com/q/33668/2451 –  Qmechanic Jun 20 '13 at 15:56
add comment

1 Answer

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 :

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

or

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

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.