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An atomic displacement ellipsoid visualizes the directional positional spread represented by an anisotropic atomic displacement parameter. For a centered three-dimensional normal distribution with positive-definite Cartesian covariance matrix $U$, its probability surfaces have the formThe eigenvectors of $U$ give the principal directions; if its eigenvalues are ${\lambda }_i$, the semiaxis lengths are $\sqrt{c{\lambda }_i}$. The atomic displacement probability level specifies how much modeled probability is enclosed.^[ref]

The squared Mahalanobis distance ${\mathbf{u}}^T{U}^{-1}\mathbf{u}$ has a chi-squared distribution with3 degrees of freedom in this Gaussian model. A50% ellipsoid uses $c\simeq 2.36597$. Taking $c=1$ instead encloses only about19.9%, so a three-dimensional probability ellipsoid cannot be read as a one-dimensional one-standard-deviation interval.

For an original example, $U=\mathrm{diag}(0.01,0.04,0.09){\mathring{\mathrm{A}}}^2$ gives50% semiaxes approximately0.154,0.308 and0.461 angstroms. An isotropic matrix with the same equivalent isotropic atomic displacement parameter, $U_{\mathrm{eq}}=0.14/3{\mathring{\mathrm{A}}}^2$, instead gives a sphere of radius about0.332 angstroms.

The ellipsoid represents modeled positional spread, including possible static disorder, rather than the atom's physical surface or a confidence region for its refined mean position. Its interpretation depends on the displacement model; the Gaussian probability construction requires a positive-definite covariance and does not turn a negative fitted eigenvalue into a physically meaningful radius.^[ref]

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.