Why can an analytic continued Hamiltonian have squared integrable eigenfunctions? In 1D quantum mechanics, there are no bound states and there are resonant states for the following potentials:
$$
W(q)=\frac{1}{2}q^2-gq^3,\tag{1.3.2}
$$
$$
W(q)=\frac{1}{2}q^2+\frac{g}{4}q^4,\; g<0.\tag{1.3.3}
$$
Since we have to consider the decay of states, we consider complex energies of the following form:
$$
E=\mathrm{Re}\; E -\mathrm{i}\frac{\Gamma}{2},\;\Gamma>0.\tag{1.3.5}
$$
At this point, we consider the analytic continuation of the Hamiltonian to the complex domain.
$$
q\to \mathrm{e}^{\mathrm{i}\theta}q,\;\theta\in\mathbb{C}.
\tag{1.3.7}$$
The textbook I have says that when $|\theta|$ is larger than specific threshold, the rotated Hamiltonian has squared integrable eigenfunctions.
Question: How can this be proven?
References:

*

*Marcos Mariño (2015), "Instantons and Large N: An Introduction to Non-Perturbative Methods in Quantum Field Theory", p.11.

 A: I think that you have to analytically continue the boundary conditions if you want the energy eigenvalue to be analytic as well. I don't know your book, or the recent progress in this game, but we played with $\cos \beta x\to \cosh \beta x$ continuation of the potential in this way in our Late terms in the asymptotic expansion for the energy levels of a periodic potential. Physical Review D18 (1978) 4746, doi:10.1103/PhysRevD.18.4746.  This has some details.
A: *

*The Hamiltonian is
$$ H(q,p)~=~\frac{1}{2}p^2+W(q).\tag{1.2.1} $$
The momentum needs to be rotated oppositely
$$p\to \mathrm{e}^{-\mathrm{i}\theta}p$$
in order to preserve the CCR.


*For the unstable quartic potential (1.3.3) the rotated Hamiltonian
$$ \tilde{H}(q,p)~:=~H(\mathrm{e}^{\mathrm{i}\theta}q,\mathrm{e}^{-\mathrm{i}\theta}p)\mathrm{e}^{2\mathrm{i}\theta}~=~\frac{1}{2}p^2+ \tilde{W}(q),$$
is no longer self-adjoint as the rotated potential
$$\tilde{W}(q)~:=~ \frac{1}{2}\mathrm{e}^{4\mathrm{i}\theta}q^2+\frac{g}{4}\mathrm{e}^{6\mathrm{i}\theta}q^4, \qquad g~<~0, $$
is complex. The rotated TISE becomes
$$ -\frac{1}{2}\psi^{\prime\prime}(q)+\tilde{W}(q)\psi(q)~=~E\psi(q), \qquad E~\in~\mathbb{C}.$$


*If we choose $\frac{\pi}{12}<\theta<\frac{\pi}{4}$, then the coupling constant satisfy
$$ {\rm Re}(\mathrm{e}^{6\mathrm{i}\theta}g)~>~0, $$
so that ${\rm Re}\tilde{W}(q)$ is bounded from below.
For a fixed energy $E\in\mathbb{C}$
$$\exists k,K>0\forall |q|\geq K:~~ {\rm Re} (\tilde{W}(q)-E)   ~\geq~ \frac{k^2}{2}.$$


*The full solution $$\psi~=~C_1\psi_1+C_2\psi_2, \qquad C_1,C_2~\in~\mathbb{C},$$ to the homogeneous 2nd-order linear ODE is a linear combination of 2 independent solutions. Up to an irrelevant overall multiplicative constant, this means that the solutions effectively have 1 complex DOF.


*From my Phys.SE answer here, it follows that there exists a solution which is exponentially damped for $x\to\infty$, and a solution which is exponentially damped for $x\to-\infty$. If this is the same solution, it is normalizable. Such coincidence is expected to happen for discrete values of $E$ in the complex plane, although I offer no proof.
