$t q^{2}$, then eq \[eq:gaussianform\] and the Langevin model cannot not possibly describe how the scattering particles move. If eq \[eq:gaussianform\] described the particle motions, then the spectrum would be a simple exponential. In systems in which the spectrum is more complex than a simple exponential, eq \[eq:g1swrong\] is invalid. That is, if $g^{(1s)}(q,t)$ is not a simple exponential in $t$, $\log(g^{(1s)}(q,t))$ does not reveal the mean-square displacement of the particles.

Why does eq \[eq:g1sanddisplacements\] ever reduce to eq \[eq:g1swrong\]? If and only if $P(\Delta x, t)$ is a Gaussian in $\Delta x$, $P(\Delta x, t)$ is entirely characterized by its second moment $\langle \Delta x^{2} \rangle$. For a Gaussian displacement distribution function, the higher moments of $P(\Delta x, t)$ have values such that the coefficients of the higher-order terms ($q^{2n}$ for $n \geq 2$) of eq \[eq:g1sanddisplacements\] all vanish. For a Gaussian $P(\Delta x, t)$, the only non-zero part of eq \[eq:g1sanddisplacements\] is eq \[eq:g1swrong\]. This disappearance of the higher-order terms is unique to a Gaussian $P(\Delta x, t)$. For any other $P(\Delta x, t)$, the higher-order terms of eq \[eq:g1sanddisplacements\] do not vanish.

Experimental Findings \[sectionexperiment\]
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What do experiments say about $P(\Delta x, t)$ and $g^{(1s)}(q,t)$? There 