= Solution
Multiply the expression for $\psi$ by $F^*(x)$ and integrate. Cancelling the nonzero $C$ gives
$$
1=\frac{\alpha m}{\kappa\hbar^2}\int\!\int F^*(x)F(y)e^{-\kappa|x-y|}\,dx\,dy.
$$
The kernel is real and symmetric, so relabelling the integration variables puts the expression in the printed order:
$$
\boxed{\kappa=\frac{\alpha m}{\hbar^2}\int\!\int F(x)F^*(y)e^{-\kappa|x-y|}\,dx\,dy,
\qquad E=-\frac{\hbar^2\kappa^2}{2m},\quad\kappa>0.}
$$
This is the scalar consistency condition for the rank-one potential, rather than an independent <normalization> equation. The double <integral> is real and nonnegative because the <Fourier transform> of $e^{-\kappa|x|}$ is $2\kappa/(q^2+\kappa^2)>0$. Thus attractive $\alpha>0$ is needed for a nontrivial solution. Under the usual integrability assumptions the corresponding free <resolvent> expectation decreases with $\kappa$, giving at most one <bound state> for this rank-one attraction.
Back to article page
