 Atomic displacement probability level

An 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.^[ref]

For a centered three-dimensional normal distribution with positive-definite Cartesian covariance matrix $U$, defineThe variable $Q$ has a chi-squared distribution with3 degrees of freedom. To enclose probability $p$, choose $c$ satisfying $\mathrm{Pr}(Q\le c)=p$ 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 functionwhere $\mathrm{erf}$ is the error function. Solving $F_3(c)=0.5$ gives $c\simeq 2.36597$; for $p=0.9$, $c\simeq 6.25139$.

For an original comparison, keep $U=\mathrm{diag}(0.01,0.04,0.09){\mathring{\mathrm{A}}}^2$. 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.