課題: fccの逆行使を求めよ。
解
fccの基本並進ベクトルは$\displaystyle\v{a}_1=\frac{a}{2}(\hat{\v{x}}+\hat{\v{y}}), \v{a}_2=\frac{a}{2}(\hat{\v{y}}+\hat{\v{z}}), \v{a}_3=\frac{a}{2}(\hat{\v{z}}+\hat{\v{x}})$である。
逆格子ベクトルは \[\left\{ \begin{array}{rl} \v{b}_1 & = 2\pi\frac{a_2 \times a_3}{a_1\cdot(a_2 \times a_3)}\\ \v{b}_2 & = 2\pi\frac{a_3 \times a_1}{a_1\cdot(a_2 \times a_3)}\\ \v{b}_3 & = 2\pi\frac{a_1 \times a_2}{a_1\cdot(a_2 \times a_3)} \end{array}\right. \] で表されるので， \[ \begin{align} \v{b}_1 &= 2\pi\frac{\frac{1}{4}(\hat{\v{x}}+\hat{\v{y}}-\hat{\v{z}})}{\frac12(\hat{\v{x}}+\hat{\v{y}})\cdot(\hat{\v{x}}+\hat{\v{y}}-\hat{\v{z}})}\\ &= 2\pi\frac{\frac{1}{4}(\hat{\v{x}}+\hat{\v{y}}-\hat{\v{z}})}{\frac18\times2}\\ &= 2\pi(\hat{\v{x}}+\hat{\v{y}}-\hat{\v{z}}) \end{align} \] である。同様にして， \[\left\{ \begin{array}{rl} \v{b}_1 & = 2\pi(\hat{\v{x}}+\hat{\v{y}}-\hat{\v{z}})\\ \v{b}_2 & = 2\pi(-\hat{\v{x}}+\hat{\v{y}}+\hat{\v{z}})\\ \v{b}_3 & = 2\pi(\hat{\v{x}}-\hat{\v{y}}+\hat{\v{z}}) \end{array}\right. \]