Inverted Harmonic oscillator what are the energies of the inverted Harmonic oscillator?
$$ H=p^{2}-\omega^{2}x^{2} $$
since the eigenfunctions of this operator do not belong to any $ L^{2}(R)$ space I believe that the spectrum will be continuous, anyway in spite of the inverted oscillator having a continuum spectrum are there discrete 'gaps' inside it?
Also if I use the notation of 'complex frequency' the energies of this operator should be
$$ E_{n}= \hbar (n+1/2)i\omega $$ 
by analytic continuation of the frequency to imaginary values.
 A: The QHO does not permit analytic continuation, because its energies and wavefunctions depend not on $\omega$, but on $|\omega|$. Thus, their dependence on $\omega$ is not analytic and $\omega$ cannot be simply replaced by $i\omega$.
Moreover, this Hamiltonian is not Hermitian. Still, just like few other interesting cases ($ix^3$, $-x^4$), it has real spectrum.
Here you can find a short but comprehensive explanation:
http://arxiv.org/abs/quant-ph/0703234
A: The wave functions that are not $L^2$-integrable play no direct physical role. You may get such "mathematically nice" functions e.g. by the analytical continuation from the stable (non-inverted) harmonic oscillator but they won't have the same interpretation. That's easy to see: as you noticed, the analytic continuation gives you imaginary energies which can't be the eigenvalues of a Hermitian operator.
The actual eigenvalues are arbitrary real numbers (the energy may always be made higher positive, by the kinetic energy, as well as more negative, by the unbounded-from-below potential) and I am convinced that each of them has a degeneracy of two, one wave moving right and one moving left in some convention. There are not even "exceptional gaps" where the degeneracy would change.
The formal solutions with $E_n=\hbar(n+1/2) iw$ still exist as poles in the transition amplitudes for the unstable (inverted) potential but they don't directly affect physics at any particular real value of energy.
