Unit cell

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:

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