Complex scaling method for solving resonance states

Question

I am now reading about the complex scaling method for solving resonance states. As far as I understand, the procedure goes like this:

Let us take the 1d potential $V(x) = A e^{-x^2} x^2 $ as an example. Here $A > 0 $ .

The full single-particle Hamiltonian is

$$ H = - \frac{\partial^2}{\partial x^2} + A e^{-x^2} x^2 . $$

The equation for the resonance state is

$$ \left(- \frac{\partial^2}{\partial x^2} + A e^{-x^2} x^2 \right ) \psi(x) = E_{complex} \psi(x) . $$

Suppose that the function $\psi(x)$ can be analytically continued into the complex plane. We there can consider the variable $x$ as a complex variable. Now consider the equation on the line $ x = \rho e^{i\theta }$. Here $\rho, \theta \in \mathbb{R}$. The equation for the function $f(\rho) \equiv \psi(\rho e^{i \theta})$ is then

$$ \left(- \frac{1}{e^{2i\theta }} \frac{\partial^2}{\partial \rho^2} + A e^{-\rho^2 e^{2i \theta}} \rho^2 e^{2i \theta} \right ) f(\rho) = E_{complex} f(\rho) . $$

The point is that now $f(\rho)$ might be normalizable and we can use conventional method to diagonalize the new Hamiltonian on the left hand side.

The problem is that this procedure essentially relies on the fact that the potential here is an analytical one. What if we have the simple square well potential? Apparently, many potentials can support resonance states but cannot be analytically continued into the complex plane.

Answer 1

You are essentially correct. If you start on the real line with the Schrödinger equation $$ \left(-\frac12 \frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) = E\,\psi(x),$$ then at every point $x_0$ where $V(x)$ is analytic you are guaranteed that $\psi(x)$ will be analytic in a neighbourhood of that point.

However, if $V(x)$ is not analytic then you have no such guarantees, and indeed this will not generically be the case: you can write $V(x)=E+\tfrac12\psi''(x)/\psi(x)$, which is analytic if $\psi(x)\neq 0$ and $\psi(x)$ is analytic. In your particular example, if $V(x)$ has a discontinuity then $\psi''(x)$ will have one as well.

This does not, however, spell the doom of the method. The main idea is to use exterior complex scaling: solving the real-valued Schrödinger equation in some finite region $|x|\leq R$, which will typically contain the interesting parts of the potential including singularities and discontinuities, and then shooting off into the complex plane for large $x$ once you reach the analytic tail of the potential.

Thus, although it is true that

many potentials can support resonance states but cannot be analytically continued into the complex plane

because of discontinuities and whatnot, you will be very hard pressed to find a physically interesting potential that is not analytical in some ray $(R,\infty)$.

For resources on exterior complex scaling see for example arXiv:1002.2520 and references therein.