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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 parameter describes the spread of an atom's modeled position about its mean in a crystal structure. An isotropic atomic displacement parameter represents equal spread in all directions; an anisotropic atomic displacement parameter uses a directional matrix. These parameters can reflect thermal motion, unresolved static variation and limitations of the refinement model, rather than temperature alone.

For isotropic Cartesian displacement, let $U_{\mathrm{iso}}$ be the mean-square displacement of one component, in squared angstroms. The conventional alternative parameter isThe IUCr dictionary gives both conventions and prefers reporting $U$.^[ref]

For an original numerical example, $U_{\mathrm{iso}}=0.010{\mathring{\mathrm{A}}}^2$ givesThe second value is the root-mean-square spread along one Cartesian direction. Under isotropy the three-dimensional mean-square displacement is $3U_{\mathrm{iso}}$, so the corresponding radial root-mean-square spread is about $0.173\mathring{\mathrm{A}}$.

This spread is not the same as uncertainty in the refined mean coordinate. Nor does doubling $B$ by itself prove that an atom's temperature doubled. Crystallographic disorder, imperfect data and a different structural model can affect the fitted parameter. Interpretation requires the experimental and modeling context.

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.

The equivalent isotropic atomic displacement parameter summarizes an anisotropic atomic displacement parameter by the average mean-square displacement along three perpendicular Cartesian directions:It has units of squared length. The corresponding equivalent $B$ parameter is $B_{\mathrm{eq}}=8{\pi }^2{U}_{\mathrm{eq}}$. The CIF field for isotropic displacement may contain either a genuinely isotropic atomic displacement parameter or this anisotropic summary.^[ref]

For an original example, take a Cartesian displacement covariance matrixThen $U_{\mathrm{eq}}=0.020{\mathring{\mathrm{A}}}^2$ and $B_{\mathrm{eq}}\simeq 1.579{\mathring{\mathrm{A}}}^2$. The principal root-mean-square displacements are0.1,0.1 and0.2 angstroms. A genuinely isotropic matrix $0.020I$ has the same equivalent parameter but a different directional distribution.

The trace is invariant under Cartesian rotations, so the average does not depend on the laboratory orientation. In an oblique crystallographic basis, however, the raw diagonal entries of the stored displacement matrix do not in general have this Cartesian interpretation. The cell geometry must enter the conversion before taking the physical trace.^[ref]

This scalar is convenient for comparisons and tabulation, but discards directional information displayed by an atomic displacement ellipsoid. Equality of equivalent parameters therefore neither establishes isotropy nor proves that two sites have the same displacement distribution.

An isotropic atomic displacement parameter describes the same mean-square positional displacement along every Cartesian direction in a crystal structure. Its Cartesian displacement covariance matrix isThus $U_{\mathrm{iso}}$ is a one-component variance, in squared angstroms, rather than a displacement length. The alternate conventional parameter is $B=8{\pi }^2{U}_{\mathrm{iso}}$.^[ref]

For an original example, $U_{\mathrm{iso}}=0.020{\mathring{\mathrm{A}}}^2$ givesThe radial root-mean-square displacement is instead $\sqrt{3U_{\mathrm{iso}}}\simeq 0.245\mathring{\mathrm{A}}$ because it includes three components.

For a centered isotropic normal distribution, the structure factor amplitude attenuation at reciprocal-vector magnitude $s$, measured in cycles per angstrom, isAt $s=1{\mathring{\mathrm{A}}}^{-1}$, the stated example gives $T\simeq 0.674$ and squared-amplitude attenuation $T^2\simeq 0.454$. Amplitude and intensity attenuation are therefore distinct, even in this simple Gaussian model.

An equivalent isotropic atomic displacement parameter can also summarize an anisotropic atomic displacement parameter by its Cartesian trace divided by3. Such an average can appear in the same CIF field as a genuinely isotropic parameter, but does not assert equal directional spread.^[ref]

A split-site disorder model represents alternative atomic positions explicitly within an average crystal structure. Each alternative has its own coordinate and crystallographic site occupancy. For one atom distributed between exactly two alternatives, with no vacancy, the occupancies sum to1. Correlated alternative arrangements can be collected into disorder groups: atoms within one group occur together, while different groups in the same assembly represent alternatives.^[ref]

For an original scattering example, put the alternatives at fractional coordinates $(0,0,0)$ and $(1/2,0,0)$ with occupancies0.7 and0.3. Give both the same real atomic scattering amplitude $f$, and omit all other atoms and displacement attenuation. The coherent average contribution to a reflection $(h00)$ isFor an even integer $h$, this equals $f$; for an odd integer $h$, it equals $0.4f$. The corresponding squared amplitudes are $f^2$ and $0.16f^2$. Treating both alternatives as fully occupied would predict a different structure factor. This calculation concerns the periodic average contribution, not a complete description of diffuse scattering from local fluctuations.

A split-site disorder model is useful when an electron density map supports distinct alternative locations instead of one broad peak. An anisotropic atomic displacement parameter can describe directional broadening, but does not by itself encode two mutually exclusive chemical arrangements.

Occupancy constraints depend on the physical model: alternatives for a fully present atom sum to1, whereas a model admitting vacancies need not. An average crystal structure alone also cannot establish whether the alternatives exchange during the measurement, occur in different cells, or have particular local correlations.