$q$ one cannot infer how the particle displacement depends on time.

This result has a purely mathematical basis, namely that the field correlation function can always be represented via a Laplace transform as $$g^{(1s)}(q,t) = \int_{0}^{\infty} A(\Gamma) \exp(- \Gamma t).
    \label{eq:laplace}$$ Here $\Gamma$ is a relaxation rate and $A(\Gamma)$ is the contribution of relaxations having decay rate $\Gamma$ to $g^{(1s)}(q,t)$. So long as the system does not have relaxational modes that have negative amplitudes, $A(\Gamma)$ is everywhere positive or zero. In this case, there is always a system having the same $g^{(1s)}(q,t)$ as the system of interest, and in which $ \langle (\Delta x(t) )^{2} \rangle$ increases linearly in time. The system can be physically constructed as a solution of of polystyrene spheres of all different sizes. The composition of the mixture is determined by $A(\Gamma)$: One adds to the mixture just enough polystyrene spheres having diffusion coefficient $\Gamma/q^{2}$ so that their contribution to the scattering spectrum is $A(\Gamma)$. For each sphere, $\langle (\Delta x(t) )^{2} \rangle$ increases linearly in time, so therefore $\langle (\Delta x(t) )^{2} \rangle$ of the mixture also increases linearly in time. Thus, an arbitrary physically-acceptable (i.e., $A(\Gamma) > 0 \ \forall \ \Gamma$) form for 