Bras and kets of continuous spectrum Does anyone know why in quantum mechanics the second statement is always true?  

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space." 

 A: Because by definition the eigenvalues of an operator $A$ are part of the point spectrum $\sigma_p(A)$.  For self-adjoint operators, the continuous spectrum is the complement $\sigma_c(A)=\sigma(A) \setminus \sigma_p(A)$.
Therefore if, in any sense $Af=af$ for some $a \in \sigma_c(A)$, $f$ cannot be an eigenvector. For this reason it cannot belong to the Hilbert space.
As a matter of fact, the identity $Af=af$  where $a \in \sigma_c(A)$ holds in a different sense than the standard one, a distributional sense if the Hilbert space is $L^2(\mathbb R, d^nx)$.
It is worth noticing that the point spectrum, in spite of its name,  may be  a continuous set, all $\mathbb R$ for instance. In this  case, however, the Hilbert space would not be separable. A famous theorem by Stone and von Neumann proves that  the Hilbert space of a particle (irreducible representation of Weyl group) must be separable necessarily. For this reason Hilbert spaces of non-relativistic  elementary systems in QM are separable and point spectra are at most countable.
