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## Interpretation on Point Group of Dnd

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 基金资助: 国家自然科学基金.  21001095

 Fund supported: 国家自然科学基金.  21001095

Abstract

Point group in crystallography is one of the important subjects in structural chemistry.Some topics are very difficult to understand.To name a few, why does point group of D2d belong to tetragonal? Why are D4d and D6d not included in 32 kinds of crystallographic point groups? The two questions are easy to answer if we understand the following topic:for the Dnd point group, when n is odd, it contains an In symmetry axis; when n is even, it contains an I2n symmetry axis.In this work, graphic method and matrix method are adopted to clarify why the Dnd point group includes an S2n axis, and thus give explanations that D2d belongs to tetragonal as well as D4d and D6d are not included in 32 kinds of crystallographic point groups.

Keywords： Point group ; Dnd ; Symmetry operation ; Matrix

ZHU Yan-Yan, WEI Dong-Hui, ZHANG Wen-Jing, TANG Ming-Sheng. Interpretation on Point Group of Dnd. University Chemistry[J], 2016, 31(11): 76-80 doi:10.3866/PKU.DXHX201603005

## 1 图解法

### 图1

r3(x′′, y′′, -z)与r1(x, y, z)的坐标关系可表示为方程组：

## 2 矩阵法

### 图2

$x = r\cos \beta \;\;\;\;\;\;y = r\sin \beta$

$\left\{ \begin{array}{l}x' = {\rm{ }}r\cos (\alpha + {\rm{ }}\beta ) = {\rm{ }}r\cos \alpha \cos \beta-r\sin \alpha \sin \beta \\y' = {\rm{ }}r\sin (\alpha + {\rm{ }}\beta ) = {\rm{ }}r\sin \alpha \cos \beta {\rm{ }} + {\rm{ }}r\sin \beta \cos \alpha \\z' = {\rm{ }}z\end{array} \right.$

$\left\{ \begin{array}{l}x'{\rm{ }} = {\rm{ }}x\cos \alpha-y\sin \alpha \\y'{\rm{ }} = {\rm{ }}x\sin \alpha {\rm{ }} + {\rm{ }}y\cos \alpha \\z'{\rm{ }} = {\rm{ }}z\end{array} \right.$

$\left( {\begin{array}{*{20}{c}}{x'}\\{y'}\\{z'}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{-\sin \alpha }&0\\{\sin \alpha }&{\cos \alpha }&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = R\left( {\alpha, z} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right)$

$\begin{array}{l}{C_2}(z) = \left( {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - 1}&0\\0&0&1\end{array}} \right)\\{\sigma _h} = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{array}} \right)\\i = \left( {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&{ - 1}&0\\0&0&{ - 1}\end{array}} \right)\end{array}$

$\begin{array}{*{20}{c}}{{C_2}(z){\sigma _h} = i}&{{C_2}(z)i{\rm{ }} = {\rm{ }}{\sigma _h}}&{{\sigma _h}i = {C_2}(z)}\end{array}$

### 图3

$\begin{array}{*{20}{c}}{x = r\cos \alpha }&{y = r\sin \alpha }\end{array}$

r2(x′, y′, z′)的直角坐标x′、y′和z′可用r、α和β表示为：

$\left\{ \begin{array}{l}x'{\rm{ }} = {\rm{ }}r\cos (2\beta-\alpha ) = {\rm{ }}r\cos \alpha \cos {\rm{ }}2\beta {\rm{ }} + {\rm{ }}r\sin \alpha \sin {\rm{ }}2\beta \\y'{\rm{ }} = {\rm{ }}r\sin (2\beta-\alpha ) = {\rm{ }}r\sin {\rm{ }}2\beta \cos \alpha-r\sin \alpha \cos {\rm{ }}2\beta \\z'{\rm{ }} = {\rm{ }}z\end{array} \right.$

$\left\{ \begin{array}{l}x'{\rm{ }} = {\rm{ }}x\cos {\rm{ }}2\beta {\rm{ }} + {\rm{ }}y\sin {\rm{ }}2\beta \\y'{\rm{ }} = {\rm{ }}x\sin {\rm{ }}2\beta-y\cos {\rm{ }}2\beta \\z'{\rm{ }} = {\rm{ }}z\end{array} \right.$

${\sigma _d} = \left( {\begin{array}{*{20}{c}}{\cos \frac{{2{\rm{\pi }}}}{{2n}}}&{\sin \frac{{2{\rm{\pi }}}}{{2n}}}&0\\{\sin \frac{{2{\rm{\pi }}}}{{2n}}}&{-\cos \frac{{2{\rm{\pi }}}}{{2n}}}&0\\0&0&1\end{array}} \right)$

C2是绕X轴逆时针转动180°，它将(x, y, z)变换为(x, -y, -z)，对应的矩阵为：

${C_2} = \left( {\begin{array}{*{20}{c}}1&0&0\\0&{-1}&0\\0&0&{-1}\end{array}} \right)$

$\begin{array}{l}{\sigma _d}{C_2} = \left( {\begin{array}{*{20}{c}}{\cos \frac{{2{\rm{\pi }}}}{{2n}}}&{\sin \frac{{2{\rm{\pi }}}}{{2n}}}&0\\{\sin \frac{{2{\rm{\pi }}}}{{2n}}}&{-\cos \frac{{2{\rm{\pi }}}}{{2n}}}&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0&0\\0&{-1}&0\\0&0&{-1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\cos \frac{{2{\rm{\pi }}}}{{2n}}}&{ - \sin \frac{{2{\rm{\pi }}}}{{2n}}}&0\\{\sin \frac{{2{\rm{\pi }}}}{{2n}}}&{\cos \frac{{2{\rm{\pi }}}}{{2n}}}&0\\0&0&{ - 1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\cos \frac{{2{\rm{\pi }}}}{{2n}}}&{ - \sin \frac{{2{\rm{\pi }}}}{{2n}}}&0\\{\sin \frac{{2{\rm{\pi }}}}{{2n}}}&{\cos \frac{{2{\rm{\pi }}}}{{2n}}}&0\\0&0&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&0\\0&0&{ - 1}\end{array}} \right) = {C_{2n}}{\sigma _h}\end{array}$

## 3 旋转反映与旋转反演的关系

$\left\{ \begin{array}{l}\begin{array}{*{20}{c}}{{I_1} = S_2^-}&{{S_1} = I_2^-}\end{array}\\\begin{array}{*{20}{c}}{{I_2} = S_1^-}&{{S_2} = I_1^ - }\end{array}\\\begin{array}{*{20}{c}}{{I_3} = S_6^ - }&{{S_3} = I_6^ - }\end{array}\\\begin{array}{*{20}{c}}{{I_4} = S_4^ - }&{{S_4} = I_4^ - }\end{array}\\\begin{array}{*{20}{c}}{{I_5} = S_{10}^ - }&{{S_5} = I_{10}^ - }\end{array}\\\begin{array}{*{20}{c}}{{I_6} = S_3^ - }&{{S_6} = I_3^ - }\end{array}\end{array} \right.$

$\left\{ \begin{array}{l}\begin{array}{*{20}{c}}{{I_1} = {S_2}}&{{S_1} = {I_2}}\end{array}\\\begin{array}{*{20}{c}}{{I_2} = {S_1}}&{{S_2} = {I_1}}\end{array}\\\begin{array}{*{20}{c}}{{I_3} = {S_6}}&{{S_3} = {I_6}}\end{array}\\\begin{array}{*{20}{c}}{{I_4} = {S_4}}&{{S_4} = {I_4}}\end{array}\\\begin{array}{*{20}{c}}{{I_5} = {S_{10}}}&{{S_5} = {I_{10}}}\end{array}\\\begin{array}{*{20}{c}}{{I_6} = {S_3}}&{{S_6} = {I_3}}\end{array}\end{array} \right.$

$\left\{ \begin{array}{l}{S_{2m + 1}} = {I_{4m + 2}}\\{S_{4m + 2}} = {I_{2m + 1}}\\{S_{4m}} = {I_{4m}}\end{array} \right.$

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