= Atomic displacement probability level
{title2=$p$}
An \b[atomic displacement probability level] is the fraction of modeled positional probability enclosed by an <atomic displacement ellipsoid>. It specifies the size of the drawn surface for a given <anisotropic atomic displacement parameter>; changing the level does not change the fitted displacement matrix.<Atomic displacement parameter nomenclature (IUCr report)>[Section2.3.1, constant-probability ellipsoids]{ref}
For a centered three-dimensional <normal distribution> with positive-definite Cartesian <covariance matrix> $U$, define
$$
Q=\mathbf u^TU^{-1}\mathbf u.
$$
The variable $Q$ has a <chi-squared distribution> with3 degrees of freedom. To enclose probability $p$, choose $c$ satisfying $\Pr(Q\leq c)=p$ and draw $Q=c$. If $\lambda_i$ are the <eigenvalues> of $U$, its semiaxis lengths are $\sqrt{c\lambda_i}$.
An explicit calculation uses the <cumulative distribution function>
$$
F_3(c)=\operatorname{erf}\!\sqrt{c/2}
-\sqrt{2c/\pi}\,e^{-c/2},\qquad c\geq0,
$$
where $\operatorname{erf}$ is the <error function>. Solving $F_3(c)=0.5$ gives $c\simeq2.36597$; for $p=0.9$, $c\simeq6.25139$.
For an original comparison, keep $U=\operatorname{diag}(0.01,0.04,0.09)\,\mathrm{\mathring A^2}$. The50% semiaxes are approximately0.154,0.308 and0.461angstroms. The90% semiaxes are approximately0.250,0.500 and0.750angstroms. Every semiaxis grows by the same factor, about1.625, although the underlying directional spread is unchanged.
This also explains why a surface drawn at one root-mean-square displacement in each principal direction, $Q=1$, encloses only about19.9% in three dimensions. The probability inside a three-dimensional ellipsoid differs from the probability inside a one-dimensional standard-deviation interval. Comparisons of displayed atomic sizes therefore need the stated probability levels as well as the displacement parameters.
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