Learning Scientific Programming with Python (2nd edition)

P6.X.4: Structure factors for perfect crystals

Question P6.X.4

The scattering of X-rays by the atoms in a crystal is described mathematically by the structure factor, $F_{hkl}$. In particular, the intensity of a diffracted beam scattered by a crystal plane with Miller indices $(hkl)$ with a vector $\boldsymbol{K} = h\boldsymbol{\hat{x}}^* + k\boldsymbol{\hat{y}}^* +l\boldsymbol{\hat{z}}^* = 2\pi\left(\frac{h}{a} \boldsymbol{\hat{x}} + \frac{k}{b} \boldsymbol{\hat{y}} + \frac{l}{c} \boldsymbol{\hat{z}}\right)$ is given by the square of the quantity:

$$ F_{hkl} = \sum_j f_j \exp\left( -2\pi i \boldsymbol{K}\cdot \boldsymbol{r_j} \right), $$ where $f_j$ is the form factor of atom $j$ and $\boldsymbol{r_j}$ is its position in the crystal lattice unit cell.

In this exercise, we will focus on crystals that are cubic (so $a=b=c$ and $\boldsymbol{K} = \frac{2\pi}{a}(\boldsymbol{\hat{x}} + \boldsymbol{\hat{y}} + \boldsymbol{\hat{z}})$) and monatomic (so that $f_j$ is the same for all atoms and we will set it equal to unity).

Write a function to calculate $F_{hkl}$ using the above formula for a given crystal structure. Your functions should take two arguments: a sequence of $m$ scattering vectors, hkl, provided as a NumPy array with dimensions (m,3) and an array of atom positions within the unit cell with dimensions (n,3).

Test your function against the following analytical results:

  1. Body-centred cubic crystals

    Atom positions: $$ \textstyle r_1 = (0,0,0) \quad r_2 = (\frac{a}{2},\frac{a}{2},\frac{a}{2}) $$

    Structure factor: $$ \begin{align*} F_{hkl} = \left\{ \begin{array}{ll} 2f, & h+k + l \; \mathrm{even}\\ 0, & h+k + l \; \mathrm{odd} \end{array}\right. \end{align*} $$

    For example, this function could be defined by the code:

    def F_bcc(hkl):
    """Structure factor for a perfect bcc crystal."""

    # F = 2 if h+k+l is even; otherwise F = 0.
    return 2 * (~np.sum(hkl, axis=1) % 2)

    Here, the (m,3) array hkl is summed over axis=1 (the three $h$, $k$, $l$ values) to give an array of shape (m,). Each $m$ value is taken modulo 2 to leave the remainder 0 for even values and 1 for odd values. The complement (~) of this is then 1 for even values and 0 for odd values. We return twice this value as the structure factor for a bcc crystal.

  2. Face-centred cubic crystals

    Atom positions: $$ \textstyle r_1 = (0,0,0) \quad r_2 = (\frac{a}{2},\frac{a}{2},0) \quad r_3 = (0, \frac{a}{2},\frac{a}{2}) \quad r_4 = (\frac{a}{2}, 0, \frac{a}{2}) $$

    Structure factor: $$ \begin{align*} F_{hkl} = \left\{ \begin{array}{ll} 4f, & h, k, l \; \mathrm{all\;even\;or\;all\;odd}\\ 0, & \mathrm{otherwise} \end{array}\right. \end{align*} $$

  3. Diamond crystal structure (Harder)

    Atom positions $$ \textstyle r_1 = (0,0,0) \quad r_2 = (\frac{a}{4},\frac{a}{4},\frac{a}{4}) \quad r_3 = (\frac{a}{2},\frac{a}{2},0) \quad r_4 = (\frac{3a}{4},\frac{3a}{4},\frac{a}{4}) $$ $$ \textstyle r_5 = (\frac{a}{2},0,\frac{a}{2}) \quad r_6 = (0,\frac{a}{2},\frac{a}{2}) \quad r_7 = (\frac{3a}{4},\frac{a}{4},\frac{3a}{4}) \quad r_8 = (\frac{a}{4},\frac{3a}{4},\frac{3a}{4}) $$

    Structure factor: $$ \begin{align*} F_{hkl} = \left\{ \begin{array}{ll} 4f [1 + (-i)^{h+k+l}], & h+k+l \; \mathrm{odd}\\ 8f, & h+k+l = 4n \; \mathrm{(i.e.\;divisible\;by\;4)}\\ 0 & \mathrm{otherwise} \end{array}\right. \end{align*} $$