In single-crystal x-ray crystallography both isotropic and anisotropic displacement parameters $U_{ij}$ of thermal ellipsoids have dimension of square angstrom ($Å^2$) as follows from the definition of Debye–Waller factor ($T$ is dimensionless):

  • For isotropic approximation (conditions of Bragg's law): $$T_\text{iso} = 8\pi^2U\left(\frac{\sin \Theta}{\lambda}\right)^2 \to [U] = Å^2 \tag{1}$$

  • For anisotropic displacement parameters (ADPs): $$T_\text{ani} = 2\pi^2 \sum_{i=1\\j=1}^3{H[i]H[j]U_{ij}e^\star[i]e^\star[j]} \to [U] = Å^2 \tag{2}$$ where $H$ is reflex's index; $e^\star$ is length of the reciprocal lattice basis vector.

In both cases $U_1$, $U_2$ and $U_3$ ($U_{ij}$ in general) are basically showing contribution of oscillation along three principal (orthogonal) axes and are measured in $Å^2$. So, is there any physical meaning that can be attributed to square angstrom in this case? At first glance, it seems rather confusing that a parameter that seems to represent a linear quantity is measured in square units of length.

  • $\begingroup$ Take a square root and be done with it. $\endgroup$ Commented Aug 27, 2017 at 5:48
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    $\begingroup$ I'm referring to the first equation in that link. "However, as can be seen from the basic expression for the isotropic Debye-Waller factor, $T=\exp(-8\pi^2\langle u^2\rangle \sin^2(\theta)/\lambda^2)$, the larger the second moment, the more rapidly the scattering from the atomic center in question falls off with increase in the scattering angle." So it seems as though the factor to consider is $\langle u^2\rangle$, the variance, which should have units of angstroms squared. $\endgroup$
    – Tyberius
    Commented Aug 27, 2017 at 16:16
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    $\begingroup$ As given by @Tyberius the $\langle u^2 \rangle $ term is the mean square displacement of the lattice points perpendicular to the scattering plane. As the scattering also depends on $\sin^2(\theta)/\lambda^2$ for a given set of displacements the diminution in intensity becomes more important for large scattering angles. $\endgroup$
    – porphyrin
    Commented Aug 28, 2017 at 12:31
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    $\begingroup$ ..and some more points I forgot about. (a) Debye assumed that the lattice points move independently from one another, (b) the average is over a longer time than lattice vibrations, probably only important for fs duration experiments and (c) as the calculation producing the Debye-Waller factor involves averaging displacements from lattice positions the sharpness of peaks is not affected only their intensity . $\endgroup$
    – porphyrin
    Commented Aug 28, 2017 at 12:48
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    $\begingroup$ @NightWriter I figured a combination of mine and porphyrin's comments would answer the question. The main reason I didn't answer at the time was that I don't know much about thermal ellipsoids, so I wasn't certain it addressed his concern. $\endgroup$
    – Tyberius
    Commented May 23, 2019 at 13:42

1 Answer 1


Model of a crystal and deriving the expected diffraction image

In single-crystal X-ray crystallography, the diffraction image is due to scattering of electrons in a crystalline sample. A convenient way of describing the electron density is to first specify crystal symmetry and mean atomic positions (coordinates), and then to describe in more or less detail how atoms deviate from these mean positions (modeled by atomic displacement factors). Combining this with a spherical description of the electrons "belonging" to each atom (modeled by atomic form factors), you arrive at an electron density.

This description is useful because in the model, the electron density is the convolution of the crystal lattice points with the coordinates in the unit cell with the distribution of atoms around mean positions with the distribution of electrons around the atoms. The convolution theory states that the Fourier transform of a convolution is the product of the Fourier transforms. So in this model, you can separately Fourier-transform the different levels of the model and then multiply them to arrive at the expected diffraction image (via the structure factors).

Atomic displacement and Debye-Waller factor

The atomic displacement around a mean position may be described through a probability density function $p_k(\mathbf{u})$, where $\mathbf{u}$ is the position (3D-vector) with the mean position as origin. The Fourier transform of this function defines the Debye-Waller factor (source: equation 1.4.8):

$$T_k(\mathbf{h}) = \int p_k(\mathbf{u}) e^{2 \pi i \mathbf{h} \cdot \mathbf{u}}d^3 \mathbf{u}\tag{3}$$

Here, $\mathbf{h}$ represents coordinates in reciprocal space (just as Fourier transform of a time series gives you information in frequency space, the Fourier transform connects real space - coordinate $\mathbf{u}$ - with reciprocal space - coordinate $\mathbf{h}$).

If the displacement is modeled by a Gaussian, the integral can be determined as (source: equation 1.4.10)

$$T_k(\mathbf{h}) = e^{-2 \pi^2 \langle(\mathbf{h} \cdot \mathbf{u})^2\rangle}\tag{4}$$

This reflects that the Fourier transform of a Gaussian is a Gaussian again, and the width of one Gaussian is the inverse of the other (i.e. if the atoms are widely distributed, the signal will diminish a lot with increasing resolution). Again using the time series as an example, if a sound is played for a very short time, the pitch will be ill-defined (or if an NMR FID signal decays rapidly, the peaks will be broad).

Displacement parameters $\mathbf{U}$

The displacement parameters $\mathbf{U}$ or $\mathbf{U^{jl}}$ in the OP's equation (1) and (2) are derived from the $\langle(\mathbf{h} \cdot \mathbf{u})^2\rangle$ expression in equation (3), either direction-independent in the case of an isotropic approximation (spherical distribution around the mean) or separately along the three lattice axes (leading to six anisotropic displacement factors). I am not explicitly showing this - it gets a bit complicated because the derivation requires transformation into the sometimes non-orthogonal coordinate system based on the crystal lattice (see here). In any case, the expression is a mean square displacement (analogous to a variance of a one-dimensional function). For that reason, the dimensions are length squared and the typical units are $Å^2$. As stated in the IUCR nomenclature document:

the elements of the tensor U have dimension (length)2 and can be directly associated with the mean-square displacements of the atom considered in the corresponding directions


If the atomic pdf is assumed to be a trivariate Gaussian, the characteristic function corresponding to this pdf - by definition, its Fourier transform - can be described by the second moments of the pdf, which in the present context are called anisotropic mean-square displacements.

Second moments (variance) have dimensions that are the square of the dimension of the quantity described by the probability distribution function (pdf). In this case, we are looking at the distribution of the position (dimensions length), and the second moment will have dimensions of length squared.


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