10_xray.md

!!! summary "Learning goals"

After this lecture you will be able to:

- Define the reciprocal space, and explain its relevance
- Construct a reciprocal lattice from a given real space lattice
- Compute the intensity of X-ray diffraction of a given crystal

Reciprocal lattice

For every real-space lattice

{\bf R}=n_1{\bf a_1}+n_2{\bf a_2}+n_3{\bf a_3}

where n_1, n_2 and n_3 are integers, there exists a reciprocal lattice

{\bf G}=m_1{\bf b_1}+m_2{\bf b_2}+m_3{\bf b_3}

where m_1, m_2 and m_3 are also integers.

(Note that in previous years we used the notation ({\bf a},{\bf b},{\bf c}),({\bf a^*},{\bf b^*},{\bf c^*}) instead of ({\bf a_1},{\bf a_2},{\bf a_3}),({\bf b_1},{\bf b_2},{\bf b_3}).)

The two lattices are related as follows:

{\rm e}^{i{\bf G}\cdot{\bf R}}=1

The above relation holds if

{\bf a_i}\cdot{\bf b_i}=2\pi\delta_{ij}

giving

{\rm e}^{i{\bf G}\cdot{\bf R}}={\rm e}^{2\pi i(n_1 m_1 + n_2 m_2 + n_3 m_3)}

which is clearly 1.

Example of a two-dimensional reciprocal lattice

Compose reciprocal lattice vectors. Directions:

{\bf a_1}\cdot{\bf b_2}=0\rightarrow{\bf a_1}\perp{\bf b_2}

{\bf a_2}\cdot{\bf b_1}=0\rightarrow{\bf a_2}\perp{\bf b_1}

Lengths: {\bf a_1}\cdot{\bf b_1}={\bf a_2}\cdot{\bf b_2}=2\pi

|{\bf b_1}|=\frac{2\pi}{|{\bf a_1}|{\rm cos}30^\circ}=\frac{4\pi}{|{\bf a_1}|\sqrt{3}}, |{\bf b_2}|=\frac{2\pi}{|{\bf a_2}|{\rm cos}30^\circ}=\frac{4\pi}{|{\bf a_2}|\sqrt{3}}

Normalization is arbitrary in reciprocal space.

The reciprocal lattice vectors can be composed directly from their real-space counterparts:

{\bf b_1}=\frac{2\pi({\bf a_2}\times{\bf a_3})}{ {\bf a_1}\cdot({\bf a_2}\times{\bf a_3})}

{\bf b_2}=\frac{2\pi({\bf a_3}\times{\bf a_1})}{ {\bf a_1}\cdot({\bf a_2}\times{\bf a_3})}

{\bf b_3}=\frac{2\pi({\bf a_1}\times{\bf a_2})}{ {\bf a_1}\cdot({\bf a_2}\times{\bf a_3})}

Note that the denominator is the volume of the real-space unit cell. These definitions are cyclic, because {\bf a_1}\cdot({\bf a_2}\times{\bf a_3})={\bf a_2}\cdot({\bf a_3}\times{\bf a_1})={\bf a_3}\cdot({\bf a_1}\times{\bf a_2}).

The reciprocal lattice can also be described as a Fourier transform. We imagine the real-space lattice as a density function consisting of delta peaks, first in 1D:

\rho(r)=\sum_{n} \delta(r-na)

We take the Fourier transform of this function to find:

{\mathcal F}_{k}\left[\rho(r)\right]=\int {\rm d}r\ {\rm e}^{ikr} \rho(r)=\sum_{n} \int {\rm d}r\ {\rm e}^{ikr} \delta(r-na)=\sum_{n} {\rm e}^{ikna}

This sum is non-zero only if k=2\pi m/a, so when k is in a recoprocal lattice point. Therefore, it can be rewritten as:

{\mathcal F}_{k}\left[\rho(r)\right]=\frac{2\pi}{|a|}\sum_{m} \delta\left(k-\frac{2\pi m}{a}\right)

The above can be generalized to three dimensions:

{\mathcal F}_{\bf k}\left[\rho({\bf r})\right]=\int {\rm d}{\bf r}\ {\rm e}^{i{\bf k}\cdot{\bf r}} \rho({\bf r})

Periodicity of the reciprocal lattice

Any wave in a crystal with wave vector {\bf k} can also be described with wave vector {\bf k'}={\bf k}+{\bf G}_0, where {\bf G}_0=h{\bf b_1}+k{\bf b_2}+l{\bf b_3}.

A primitive unit cell of the reciprocal lattice contains a set of unique {\bf k} vectors. Convention: 1st Brillouin zone. All {\bf k} vectors outside the 1st BZ have a copy inside the 1st BZ. So any wave process with arbitrary {\bf k} can be described inside the 1st BZ.

The 1st Brillouin zone = the Wigner Seitz cell of the reciprocal lattice.

Diffraction

We will now discuss how incoming waves incident onto a crystal can be scattered by the lattice. These waves can be x-rays, neutrons or electrons.

We assume that the incoming wave has wave vector {\bf k} and the scattered wave has wave vector {\bf k'}. Furthermore, we assume only elastic scattering, meaning that |{\bf k'}|=|{\bf k}|. In this case, a scattering atom at position {\bf r} can be seen as a radiation sources emitting a wave {\rm e}^{i\left(({\bf k'}-{\bf k})\cdot{\bf r}-\omega t\right)}. A complete atomic lattice {\bf R} will then emit a total wave amplitude:

A\propto\sum_{\bf R}{\rm e}^{i\left(({\bf k'}-{\bf k})\cdot{\bf R}-\omega t\right)}

We find that this sum will only yield a finite value if:

{\bf k'}-{\bf k}={\bf G}

in other words, if the difference between the outgoing and incoming wave vectors coïncides with a reciprocal lattice point. We can then expect constructive interference due to the lattice points. This requirement is known as the Laue condition. Note that the total intensity measured by the detector goes like I\propto A^2.

In the above we have assumed that at each lattice point there is one single atom. But what if there are multiple atoms per unit cell? In that case each atom acquires a phase shift of its own and the total amplitude becomes:

A\propto\sum_{\bf R}{\rm e}^{i\left({\bf G}\cdot{\bf R}-\omega t\right)}\sum_j f_j\ {\rm e}^{i{\bf G}\cdot{\bf r}_j}

where f_j is a phenomenological form factor that may be different for atoms of different elements. The second sum is called the structure factor:

S({\bf G})=\sum_j f_j\ {\rm e}^{i{\bf G}\cdot{\bf r}_j}

The structure factor can cause destructive interference due to the contents of the unit cell, even though the Laue condition is met. Examples:

• Simplest case: 1 atom at each lattice point, {\bf r}_1=(0,0,0)\rightarrow S=f_1. In this case each reciprocal lattice point gives one interference peak, none of which is cancelled.
• Example 2: conventional cell of the fcc lattice.

Basis contains 4 identical atoms:

\begin{aligned} {\bf r}_1&=(0,0,0)\\ {\bf r}_2&=\frac{1}{2}({\bf a}_1+{\bf a}_2)\\ {\bf r}_3&=\frac{1}{2}({\bf a}_2+{\bf a}_3)\\ {\bf r}_4&=\frac{1}{2}({\bf a}_3+{\bf a}_1)\\ f_1&=f_2=f_3=f_4\equiv f \end{aligned}

\begin{aligned} {\bf G}&=h{\bf b}_1+k{\bf b}_2+l{\bf b}_3\\ S&=f\left({\rm e}^{i{\bf G}\cdot{\bf r}_1}+{\rm e}^{i{\bf G}\cdot{\bf r}_2}+{\rm e}^{i{\bf G}\cdot{\bf r}_3}+{\rm e}^{i{\bf G}\cdot{\bf r}_4}\right)\\ &=f\left(1+{\rm e}^{i/2(h{\bf b}_1\cdot{\bf a}_1+k{\bf b}_2\cdot{\bf a}_2)}+{\rm e}^{i/2(k{\bf b}_2\cdot{\bf a}_2+l{\bf b}_3\cdot{\bf a}_3)}+{\rm e}^{i/2(h{\bf b}_1\cdot{\bf a}_1+l{\bf b}_3\cdot{\bf a}_3)}\right)\\ &=f\left(1+{\rm e}^{-i\pi(h+k)}+{\rm e}^{-i\pi(k+l)}+{\rm e}^{-i\pi(h+l)}\right) \end{aligned}

Parameters h, k, l are integers \rightarrow all terms are either +1 or -1 \rightarrow puzzle. Solution:

S=4f if h, k, l are all even or all odd

S=0 in all other cases