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

Physics Asked on October 28, 2021

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 $iomega$.<…>".

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

$$frac{x^2}{4}+lambdafrac{x^4}{4}$$

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 $$exp(pm iepsilon^{1/2} x^3/6)$$ 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
$$exp(- iepsilon^{1/2} frac{lvert x rvert^3}{6} (cos 3theta + i sin 3theta)) sim exp(- epsilon^{1/2} frac{lvert x rvert^3}{6} sin 3theta) $$
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.

One Answer

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.

Update
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

Answered by Roger Vadim on October 28, 2021

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