Unit cell

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Example of the BCC unit cell.

The unit cell is the basic building block of a crystal lattice (whether an atomic crystal or a nanoscale superlattice). Crystalline materials have a periodic structure, with the unit cell being the minimal volume necessary to fully describe the repeating structure. There are a finite number of possible symmetries for the repeating unit cell.

A unit cell can be defined by three vectors that lie along the edges of the enclosing parallelepped. We denote the vectors as \mathbf{a}, \mathbf{b}, and \mathbf{c}; alternately the unit cell can be described by the lengths of these vectors (a, b, c), and the angles between them:

\alpha, the angle between b and c
\beta, the angle between a and c
\gamma, the angle between a and b


Mathematical description

Vectors

\begin{array}{l}
\mathbf{a} = \begin{bmatrix}
a \\
0 \\
0
\end{bmatrix} \\
\mathbf{b} = \begin{bmatrix}
b \cos{\gamma} \\
b \sin{\gamma} \\
0
\end{bmatrix} \\
\mathbf{c} = \begin{bmatrix}
c \sin{\theta_c} \cos{\phi_c} \\
c \sin{\theta_c} \sin{\phi_c} \\
c \cos{\theta_c}
\end{bmatrix}
= \begin{bmatrix}
c \cos{\beta} \\
c \frac{ \cos{\alpha} - \cos{\beta}\cos{\gamma} }{\sin{\gamma}} \\
c \sqrt{  1 - \cos^2{\beta} - \left( \frac{\cos{\alpha} - \cos{\beta}\cos{\gamma}}{\sin{\gamma}} \right)^2 }
\end{bmatrix}
\end{array}

Relations

\mathbf{a} \cdot \mathbf{b} = a b \cos{\gamma}
\mathbf{a} \cdot \mathbf{c} = a c \cos{\beta}
\mathbf{b} \cdot \mathbf{c} = b c \cos{\alpha}

Volume

V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = |\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})| = |\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})|

If a, b, and c are the parallelepiped edge lengths, and α, β, and γ are the internal angles between the edges, the volume is


V = a b c \sqrt{1+2\cos(\alpha)\cos(\beta)\cos(\gamma)-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)}.

The volume of a unit cell with all edge-length equal to unity is:

v =\sqrt{1-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)+2\cos(\alpha)\cos(\beta)\cos(\gamma)}

Angles

  • \gamma is the angle between \mathbf{a} and \mathbf{b}
  • \beta is the angle between \mathbf{a} and \mathbf{c}
  • \alpha is the angle between \mathbf{b} and \mathbf{c}
Unit cell definition using parallelepiped with lengths a, b, c and angles between the sides given by α,β,γ (from Wikipedia fractional coordinates).

Reciprocal vectors

The repeating structure of a unit cell creates peaks in reciprocal space. In particular, we observe maxima (constructive interference) when:


\begin{alignat}{2}
\mathbf{q} \cdot \mathbf{a} & = 2 \pi h \\
\mathbf{q} \cdot \mathbf{b} & = 2 \pi k \\
\mathbf{q} \cdot \mathbf{c} & = 2 \pi l \\
\end{alignat}

Where h, k, and l are integers. We define reciprocal-space vectors:


\begin{alignat}{2}
\mathbf{u} & = \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot (\mathbf{b}\times\mathbf{c}) } = \frac{1}{V} \mathbf{b}\times\mathbf{c} \\
\mathbf{v} & = \frac{\mathbf{c}\times\mathbf{a}}{\mathbf{a}\cdot (\mathbf{b}\times\mathbf{c}) }  =\frac{1}{V} \mathbf{c}\times\mathbf{a} \\
\mathbf{w} & = \frac{\mathbf{a}\times\mathbf{b}}{\mathbf{a}\cdot (\mathbf{b}\times\mathbf{c}) }  =\frac{1}{V} \mathbf{a}\times\mathbf{b} \\
\end{alignat}

And we can then express the momentum transfer (\mathbf{q}) in terms of these reciprocal vectors:


\begin{alignat}{2}
\mathbf{q} & = (\mathbf{q}\cdot\mathbf{a})\mathbf{u} + (\mathbf{q}\cdot\mathbf{b})\mathbf{v} + (\mathbf{q}\cdot\mathbf{c})\mathbf{w}
\end{alignat}

Combining with the three Laue equations yields:


\begin{alignat}{2}
\mathbf{q}_{hkl} & = (2 \pi h)\mathbf{u} + (2 \pi k)\mathbf{v} + (2 \pi l)\mathbf{w} \\
  & = 2 \pi(h\mathbf{u} + k \mathbf{v} + l \mathbf{w}) \\
  & = 2 \pi \mathbf{H}_{hkl}
\end{alignat}

Where \mathbf{H}_{hkl} is a vector that defines the position of Bragg reflection hkl for the reciprocal-lattice.

Examples

Cubic

Since \alpha=\beta=\gamma=90^{\circ}, V=abc, and:


\begin{alignat}{2}
\mathbf{a}
  & = \begin{bmatrix} a \\ 0 \\ 0  \end{bmatrix} \\
\mathbf{b}
  & = \begin{bmatrix} 0 \\ b \\ 0  \end{bmatrix} \\
\mathbf{c}
  & = \begin{bmatrix} 0 \\ 0 \\ c  \end{bmatrix} \\
\end{alignat}

And in reciprocal-space:


\begin{alignat}{2}
\mathbf{u}
 & =\frac{1}{V} \mathbf{b}\times\mathbf{c}
 & =\frac{1}{V} \begin{bmatrix} b c \\ 0 \\ 0 \end{bmatrix}
 & = \begin{bmatrix} \frac{1}{a} \\ 0 \\ 0 \end{bmatrix}\\
\mathbf{v}
 & =\frac{1}{V} \mathbf{c}\times\mathbf{a}
 & =\frac{1}{V} \begin{bmatrix} 0 \\ a c \\ 0 \end{bmatrix}
 & = \begin{bmatrix} 0 \\ \frac{1}{b} \\ 0 \end{bmatrix}\\
\mathbf{w} 
 & =\frac{1}{V} \mathbf{a}\times\mathbf{b}
 & =\frac{1}{V} \begin{bmatrix} 0 \\ 0 \\ a b \end{bmatrix}
 & = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{c} \end{bmatrix}\\
\end{alignat}

So:


\begin{alignat}{2}
\mathbf{q}_{hkl} 
  & = (2 \pi h)\mathbf{u} + (2 \pi k)\mathbf{v} + (2 \pi l)\mathbf{w} \\
  & = (2 \pi h)\begin{bmatrix} \frac{1}{a} \\ 0 \\ 0 \end{bmatrix} 
    + (2 \pi k)\begin{bmatrix} 0 \\ \frac{1}{b} \\ 0 \end{bmatrix} 
    + (2 \pi l)\begin{bmatrix} 0 \\ 0 \\ \frac{1}{c} \end{bmatrix} \\
  & = \begin{bmatrix} \frac{2 \pi h}{a} \\ \frac{2 \pi k}{b} \\ \frac{2 \pi l}{c} \end{bmatrix} 
\end{alignat}

And:


 q_{hkl} = 2\pi \sqrt{ \left( \frac{h}{a} \right)^2 + \left( \frac{k}{b} \right)^2 + \left( \frac{l}{c} \right)^2 }

Hexagonal

Since \alpha=\beta=90^{\circ} and \gamma=60^{\circ}, V=\frac{\sqrt{3}}{2}abc, and:


\begin{alignat}{2}
\mathbf{a}
  & = \begin{bmatrix} a \\ 0 \\ 0  \end{bmatrix} \\
\mathbf{b}
  & = \begin{bmatrix} \frac{1}{2}b \\ \frac{\sqrt{3}}{2} b \\ 0  \end{bmatrix} \\
\mathbf{c}
  & = \begin{bmatrix} 0 \\ 0 \\ c  \end{bmatrix} \\
\end{alignat}

And in reciprocal-space:


\begin{alignat}{2}
\mathbf{u}
 & =\frac{1}{V} \mathbf{b}\times\mathbf{c}
 & =\frac{1}{V} \begin{bmatrix} \frac{\sqrt{3}}{2} b c \\ -\frac{1}{2} b c \\ 0 \end{bmatrix}
 & = \begin{bmatrix} \frac{1}{a} \\ \frac{1}{\sqrt{3}a} \\ 0 \end{bmatrix}\\
\mathbf{v}
 & =\frac{1}{V} \mathbf{c}\times\mathbf{a}
 & =\frac{1}{V} \begin{bmatrix} 0 \\ a c \\ 0 \end{bmatrix}
 & = \begin{bmatrix} 0 \\ \frac{2}{\sqrt{3}b} \\ 0 \end{bmatrix}\\
\mathbf{w} 
 & =\frac{1}{V} \mathbf{a}\times\mathbf{b}
 & =\frac{1}{V} \begin{bmatrix} 0 \\ 0 \\ \frac{\sqrt{3}}{2} a b \end{bmatrix}
 & = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{c} \end{bmatrix}\\
\end{alignat}

So:


\begin{alignat}{2}
\mathbf{q}_{hkl} 
  & = (2 \pi h)\mathbf{u} + (2 \pi k)\mathbf{v} + (2 \pi l)\mathbf{w} \\
  & = (2 \pi h)\begin{bmatrix} \frac{1}{a} \\ \frac{1}{\sqrt{3}a} \\ 0 \end{bmatrix} 
    + (2 \pi k)\begin{bmatrix} 0 \\ \frac{2}{\sqrt{3}b} \\ 0 \end{bmatrix} 
    + (2 \pi l)\begin{bmatrix} 0 \\ 0 \\ \frac{1}{c} \end{bmatrix} \\
  & = \begin{bmatrix} \frac{2 \pi h}{a} \\ \frac{2 \pi h}{\sqrt{3}a} + \frac{4 \pi k}{\sqrt{3}b} \\ \frac{2 \pi l}{c} \end{bmatrix}  \\
  & = \begin{bmatrix} \frac{2 \pi h}{a} \\ \frac{2 \pi (h + 2 k)}{\sqrt{3}a} \\ \frac{2 \pi l}{c} \end{bmatrix} 
\end{alignat}

See Also