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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.

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as an aside question, can we think of this in terms of Forces. normally a harmonic oscillator drives the particle back to where it was, whereas this force would drive the particle away from where it is faster and faster... seems like the energy would diverge very quickly, but still are there not quantized solutions?? -- oops i see lubos has given those explicitly is there any further information on this subject of interest? – Bozostein Oct 28 '11 at 17:40
In non relativistic QM the time of reaching infinity is finite, so the particle motion in such a repulsive potential is kind of periodic (assuming reflecting walls at infinity $\psi (x=\infty)=0$). This leads to quantization of energy levels (like quasi-classical quantization), which are real discrete numbers, not imaginary. Arbitrary can be only average energy in states being superpositions of the eigenstates. – Vladimir Kalitvianski Oct 28 '11 at 18:10

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.

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If $\psi(x=\pm\infty)=0$, then the discrete energy levels may split due to tunneling into "even" and "odd" eigenvalues corresponding to "even" and "odd" wave-functions, it seems to me. – Vladimir Kalitvianski Oct 28 '11 at 19:41
I vaguely recall that there is a PT way of dealing with this potential, which gives positive energies just like the ordinary HO. Bender used the inverted oscillator to give the "Dirac sea" construction for bosons! Formally this potential can be thought of as $1∗(ix)^\nu$ for $\nu=2$, which is a Bender potential, except perturbing the exponent starting at 1, not $x^2$, but I don't remember how the analytic continuation works exactly. – Ron Maimon Nov 6 '11 at 9:49

The QHO does not permit analytic continuation, because it's 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 hermitean. Still, just like few other interesting cases ($ix^3$, $-x^4$) it has real spectrum.

Here you can find a short but comprehensive explanation:

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