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

$\mathbf {a} \cdot \mathbf {b} =ab\cos {\gamma }$

$\mathbf {a} \cdot \mathbf {c} =ac\cos {\beta }$

$\mathbf {b} \cdot \mathbf {c} =bc\cos {\alpha }$

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

$V=abc{\sqrt {1+2\cos(\alpha )\cos(\beta )\cos(\gamma )-\cos ^{2}(\alpha )-\cos ^{2}(\beta )-\cos ^{2}(\gamma )}}.$

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:

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

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

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

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

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

Combining with the three Laue equations yields:

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

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:

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

And in reciprocal-space:

${\begin{alignedat}{2}\mathbf {u} &={\frac {1}{V}}\mathbf {b} \times \mathbf {c} &={\frac {1}{V}}{\begin{bmatrix}bc\\0\\0\end{bmatrix}}&={\begin{bmatrix}{\frac {1}{a}}\\0\\0\end{bmatrix}}\\\mathbf {v} &={\frac {1}{V}}\mathbf {c} \times \mathbf {a} &={\frac {1}{V}}{\begin{bmatrix}0\\ac\\0\end{bmatrix}}&={\begin{bmatrix}0\\{\frac {1}{b}}\\0\end{bmatrix}}\\\mathbf {w} &={\frac {1}{V}}\mathbf {a} \times \mathbf {b} &={\frac {1}{V}}{\begin{bmatrix}0\\0\\ab\end{bmatrix}}&={\begin{bmatrix}0\\0\\{\frac {1}{c}}\end{bmatrix}}\\\end{alignedat}}$

So:

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

Hexagonal

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

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

And in reciprocal-space:

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

So:

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