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


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.


1 Answer 1


Scattering from an inverted parabolic potential is actually an exactly solvable problem in terms of the parabolic cylinder functions (it is better treated as a scattering rather than as eigenvalue problem). It produces Fermi-function-like transmission coefficient. Here is the reference where I have seen it, but it skims over the mathematical details.

While I do not have a complete answer to the question, here are a couple of tips:

  • the scattering boundary conditions are more appropriate here than the boundary conditions used for a normal oscillator, this is an important point;
  • once you have the solution on terms of the parabolic cylinder functions, you can check books on special functions about the analytical continuation relating these functions to Hermit polynomials
  • $\begingroup$ for some reason my browser says the address you linked is not valid, I would certainly like to check it. In any case, I am interested in the very specific topic I mention, on the analytical continuation methods used in the paper I reference and how to define the boundary conditions. Or is your answer referring to my first question? Thanks a lot $\endgroup$
    – Smerdjakov
    Commented Jul 21, 2020 at 9:59
  • $\begingroup$ @Smerdjakov I have corrected the link. I also added a couple of tips, based on your comment. $\endgroup$
    – Roger V.
    Commented Jul 21, 2020 at 10:09
  • 1
    $\begingroup$ thanks for the link and for the tips. I do understand that "the scattering boundary conditions are more appropriate here than the boundary conditions used for a normal oscillator", that is why I want to understand exactly how they are derived in the paper I reference. Thanks a lot. $\endgroup$
    – Smerdjakov
    Commented Jul 21, 2020 at 10:38

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