Quantum Mechanics: Relate solutions for two dual hamiltonians? Consider a Hamiltonian in quantum mechanics:
$$H_x=-\frac{d^2}{dx^2}+V(x,c)$$
where $x\in\mathbb{R}$ and the potential $V(x,c)$ depends on position $x$ and a continuous parameter $c$. Furthermore, consider a dual Hamiltonian such that
$$\tilde H_y=-\frac{d^2}{dy^2}+V(y,\tilde c)$$
with $y\in\mathbb{R}$ and the same potential $V$, except that the parameters $c,\tilde c$ are not equal, but have a relation $\tilde c=\tilde c(c)$. The two Hamiltonians are considered to be dual since a function $F(x,y)$ is known, such that
$$H_xF(x,y)=\tilde H_yF(x,y)$$
If $\psi_c(x)$ and $\psi_{\tilde c}(y)$ are eigenfunctions of $H_x$ and $\tilde H_y$ with the same boundary conditions, how can one obtain i.e. $\psi_{\tilde c}(y)$ by using $\psi_c(x)$ and $F(x,y)$? I suspect that some integral kernel $K(x,y)$ should exist so that one could write
$$\psi_{\tilde c}(y)=\int dx~K(x,y)\psi_c(x)$$
where $K(x,y)$ and $F(x,y)$ should be related somehow. Any hint on how to make this more explicit?
 A: Let $\psi_\lambda(x, c)$, $\phi_\mu(y, {\tilde c})$ be the eigenfunctions of $H_x(c )$ and $H_y({\tilde c})$ respectively,
$$
H_x \psi_\lambda(x, c) = \lambda(c) \psi_\lambda(x, c)
$$
$$
H_y \phi_\mu(y, {\tilde c}) = \mu({\tilde c}) \phi_\mu(y, {\tilde c})
$$
Since both $\lbrace \psi_\lambda\rbrace_\lambda$ and $\{\phi_\mu\}_\mu$ must be basis sets, there exist expansion coefficients $a_{\lambda\mu} \in {\mathbb C}$ such that
$$
F(x, y) = \sum_{\lambda, \mu}{a_{\lambda\mu} \psi_\lambda(x, c) \phi_\mu(y, {\tilde c})}
$$
Substituting this in the duality condition for $H_x$ and $H_y$ gives
$$
H_x \sum_{\lambda, \mu}{a_{\lambda\mu} \psi_\lambda(x, c) \phi_\mu(y, {\tilde c})} = H_y \sum_{\lambda, \mu}{a_{\lambda\mu} \psi_\lambda(x, c) \phi_\mu(y, {\tilde c})}
$$
$$
 \sum_{\lambda, \mu}{\lambda a_{\lambda\mu} \psi_\lambda(x, c) \phi_\mu(y, {\tilde c})} = \sum_{\lambda, \mu}{\mu a_{\lambda\mu} \psi_\lambda(x, c) \phi_\mu(y, {\tilde c})}
$$
and so 
$$
\left(\lambda(c ) - \mu({\tilde c}) \right) a_{\lambda\mu} = 0 \;\;\Rightarrow a_{\lambda\mu} = 0 \;\;\;\text{or} \;\;\; \lambda(c ) = \mu({\tilde c})
$$
The latter result implies 
$$
a_{\lambda\mu} = \delta_{\lambda\mu} a_\lambda
$$
which in turn means
$$
F(x, y) = \sum_\lambda{a_\lambda \psi_\lambda(x, c) \phi_{\mu = \lambda}(y, {\tilde c})}
$$
But then
$$
\int{dx\;\psi^*_\lambda(x, c)F(x, y)} = \sum_{\lambda'}{a_{\lambda'} \left[\int{dx\;\psi^*_\lambda(x, c)\psi_{\lambda'}(x, c)}\right] \phi_{\mu = \lambda'}(y, {\tilde c})} =\\
= a_\lambda \phi_{\mu = \lambda}(y, {\tilde c})
$$
or 
$$
\phi_{\mu = \lambda}(y, {\tilde c}) = a^{-1}_\lambda \int{dx\;\psi^*_\lambda(x, c)F(x, y)}
$$
And similar for $\psi_\lambda$.
