3
$\begingroup$

Does a continuous spectrum of an observable always imply that the corresponding eigenvectors will not be normalizable? If yes, how to prove it?

$\endgroup$

marked as duplicate by Emilio Pisanty, user10851, Waffle's Crazy Peanut, Manishearth Sep 26 '13 at 13:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

5
$\begingroup$

If an observable – a Hermitian operator $L$ – has a continuous (or mixed) spectrum, it means that for any $\epsilon\gt 0$, there is an eigenstate of $L$ with eigenvalue $\ell+\epsilon$ aside from an eigenstate with the eigenvalue $\ell$.

The two eigenstates are bound to be orthogonal to each other for any $\ell,\epsilon$ due to the Hermiticity of the operator. So we have $$\langle \ell+\epsilon| \ell \rangle = 0$$ However, if we assume that the eigenstates are normalizable, then $$\langle \ell| \ell\rangle =1$$ but this means that the inner product $\langle \ell+\epsilon| \ell \rangle = 0$ is a discontinuous function (jumping between zero and one) of the bra's eigenvalue $\ell+\epsilon$ which contradicts the fact that the eigenstate itself should be a continuous function of the eigenvalue. Normalizable ket vectors inevitably go to zero at infinity and the inner products must be continuous functions of the eigenvalue if the wave functions themselves are continuous functions of the eigenvalue.

In the actual reality, this paradox is avoided because the inner product is proportional to the Dirac delta function. The dependence of the eigenvector on the eigenvalue is continuous but not "uniformly continuous" which allows distributions as the inner products.

$\endgroup$

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