I have recently been reading a chapter on the Hartree-Fock system where the authors sought to find solutions to equations of the form $$ i \hbar \frac{\partial \psi}{\partial t} + G(-\hbar^2 \Delta) \psi = 0 $$ where $G: \mathbb{R}_{+} \to \mathbb{R}$ is a real valued function. The author then claims that $G(-\hbar^2 \Delta) \psi$ is a Fourier multiplier when $G(x) = -\frac{x}{2m}$ and $G(x) = - \sqrt{c^2 x + m^2 c^4}$.
Using the fact that $$\mathcal{F}(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2) \hat{\psi}(p)$$ the original PDE in the frequency space becomes $$ i \hbar \frac{\partial \hat{\psi}}{\partial t} + G (|p|^2) \hat{\psi} = 0 $$ with the initial conditions $$ \hat{\psi}(0, p) = \hat{\psi}_0 (p) $$ The solution of this equation then follows from an integrating factor by treating it as a first order ordinary differential equation.
$$ \hat{\psi}(p, t) = \hat{\psi}_0 (p) e^{-it G(|p|^2)} $$
As $G$ does not have an imaginary component, the $-itG(|p|^2)$ only modifies the phase of the solution s.t. $|\hat{\psi}(p, t)|=|\hat{\psi}_0 (p)|$.
Question: From the way the authors have stated their claims, I would believe that I can take any $G(x):\mathbb{R}_+ \to \mathbb{R}$ as a means of producing a solution.
Then, for the case $G(x) = (a + x)^{-1}$, would the solution be a simple as
$$\hat{\psi}(p, t) = \hat{\psi}_0 (p) \exp(-it(a + |p|^{2})^{-1})$$ when $a>0$?
Given that $\sqrt{c^2 x + m^2 c^4}$ (which I perceive to be an ugly function on account of its branch cut) is a Fourier multiplier, I don't see why $\frac{1}{x + a}$ could not be a multiplier as well. Am I mistaken somewhere in my reasoning? In fact, I believe that I have already computed a Fourier transform for $\frac{1}{x + a}|_{\mathbb{R}_+}$ for $a>0$. I wish to know if and how this result carries over to solving PDEs.
Note: I originally posted my question in StackExchange Physics; however, I was advised to delete my post and repose my question here as it was more concerned with mathematical viability and less so with the physical implications it would entail.
