# Divergent Energies and Analytical Continuation - Two questions on the inverted harmonic oscillator and the inverted double well

I have two questions on the general topic of energy potentials that diverge at infinity.

First of all, the inverted harmonic oscillator. I found this post on Physics SE, Inverted Harmonic oscillator. The answer from fellow user Mazvolej states that

"<...> 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 $$i\omega$$.<...>".

I totally fail to see how the energies of the QHO depend on |$$\omega$$|. Will they not depend on $$\omega^2$$, which is analytic?

The paper Mavzolej links, Inverted Oscillator indeed shows that the naïve analytical continuation from ω to iω does not work, but I do not understand fully why.

Second question, I am trying to understand the reasoning in Anharmonic Oscillator. II A Study of Perturbation Theory in the Large Order

The authors consider a double well potential

$$\frac{x^2}{4}+\lambda\frac{x^4}{4}$$

When $$\lambda > 0$$ bound states exist and the QHO energy spectrum is just perturbed by the $$x^4$$ term dominating at infinity. When $$\lambda < 0$$ the energy at infinity diverges to $$-\infty$$. They approach the problem by considering the function $$E^k(\lambda)$$, where $$k$$ stand for the index of the energy eigenvalue, and considering its analytical continuation, as $$\lambda$$ is rotated from the positive to the negative real axis.

The first obstacle for me are the boundary conditions at infinity they set to solve Schroedinger's equation. I cite their reasoning on page 1623, which is totally delphic to me (they denote $$-\lambda = \epsilon$$):

*"<...> At $$x = +\infty$$ the boundary conditions are somewhat complicated <...>. It would appear that any linear combination of outgoing and incoming waves $$\exp(\pm \epsilon^{1/2} x^3/6)$$ would suffice. However, we recall that the analytical continuation of the energy levels into the complex plane is accomplished by simultaneously rotating $$x$$ into te complex $$x$$ plane. When $$\arg \lambda = \pi$$, the sector in which the boundary condition $$\lim _{\lvert x \rvert \to \infty} \psi(x) = 0$$ applies is given by $$-\frac{1}{3} \pi < \arg(\pm x)<0$$. Thus it is necessary to pick that asymptotic behaviour which vanishes exponentially if the argument of $$x$$ lies between $$0 ^\circ$$ and $$-60^\circ$$. Hence, $$\Psi(x)$$ must obey the boundary condition $$\Psi(x) \sim \frac{const}{x} \exp(-i \epsilon^{1/2} x^3/6)$$ as $$x \to +\infty$$ <...>".

I am not so sure I grasp this. Here are my thoughts.

For $$x \to +\infty$$, the quartic term will dominate and I understand the asymptotic behaviour $$\exp(\pm i\epsilon^{1/2} x^3/6)$$ is expected, when $$\lambda$$ is real and negative. The analytical continuation could be achieved by keeping $$\lambda$$ real and negative, and rotating $$x$$ (by why do they say, "simultaneously"? Is $$\lambda$$ also rotated? Why both?).

By writing $$x = \lvert x \rvert \exp(i \theta)$$ and substituing in the asymptotic expression (with the $$-$$ sign, the one the authors claim to be the right boundary condition $$\exp(- \epsilon^{1/2} \frac{x^3}{6})$$ I get $$\exp(- i\epsilon^{1/2} \frac{\lvert x \rvert^3}{6} (\cos 3\theta + i \sin 3\theta)) \sim \exp(- \epsilon^{1/2} \frac{\lvert x \rvert^3}{6} \sin 3\theta)$$ where in the last step the oscillatory component was discarded. The solution will hence decay to $$0$$ if $$0 > \theta > -\pi/3$$, and also this seem to match what they say.

Now, the wave function has to vanish for positive $$\lambda$$ and $$x \to +\infty$$, as in the QHO. This should be equivalent to keeping $$\lambda$$ negative and rotating $$x$$. But if $$x$$ is rotated by $$\pi$$, the asymptotic behaviour will not be exponentially decaying to $$0$$.

I would be grateful for any hint on my mistake.