Chapter 10, Problem 10C.8BE

Interpretation Introduction

Interpretation:

The irreducible representations spanned by the set of basis functions that span a reducible representation of the group $O_{\text{h}}$ have to be determined.

Concept introduction:

The systematic discussion of symmetry is known as group theory.  The table showing all the characters of the operations of a group are termed as character tables.  The character of an operation in a particular matrix is the sum of the diagonal elements of the representative of that operation.

Answer to Problem 10C.8BE

The irreducible representations spanned by the set of basis functions that span a reducible representation of the group $O_{\text{h}}$ is ${\text{A}}_{\text{1g}}+{\text{E}}_{\text{g}}+{\text{T}}_{\text{1u}}$.

Explanation of Solution

The character table of the $O_{\text{h}}$ point group is given below.

	$E$	$8C_3$	$6C_2$	$6C_4$	$3C_2$	$i$	$6S_4$	$8S_6$	$3{\sigma }_{\text{h}}$	$6{\sigma }_{\text{d}}$
${\text{A}}_{\text{1g}}$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$
${\text{A}}_{\text{2g}}$	$1$	$1$	$-1$	$-1$	$1$	$1$	$-1$	$1$	$1$	$-1$
${\text{E}}_{\text{g}}$	$2$	$-1$	$0$	$0$	$2$	$2$	$0$	$-1$	$2$	$0$
${\text{T}}_{\text{1g}}$	$3$	$0$	$-1$	$1$	$-1$	$3$	$1$	$0$	$-1$	$-1$
${\text{T}}_{\text{2g}}$	$3$	$0$	$1$	$-1$	$-1$	$3$	$-1$	$0$	$-1$	$1$
${\text{A}}_{\text{1u}}$	$1$	$1$	$1$	$1$	$1$	$-1$	$-1$	$-1$	$-1$	$-1$
${\text{A}}_{\text{2u}}$	$1$	$1$	$-1$	$-1$	$1$	$-1$	$1$	$-1$	$-1$	$1$
${\text{E}}_{\text{u}}$	$2$	$-1$	$0$	$0$	$2$	$-2$	$0$	$1$	$-2$	$0$
${\text{T}}_{\text{1u}}$	$3$	$0$	$-1$	$1$	$-1$	$-3$	$-1$	$0$	$1$	$1$
${\text{T}}_{\text{2u}}$	$3$	$0$	$1$	$-1$	$-1$	$-3$	$1$	$0$	$1$	$-1$

The characters of the reducible representation corresponding to the $O_{\text{h}}$ point group are given below.

$O_{\text{h}}$	$E$	$8C_3$	$6C_2$	$6C_4$	$3C_2$	$i$	$6S_4$	$8S_6$	$3{\sigma }_{\text{h}}$	$6{\sigma }_{\text{d}}$
	$6$	$0$	$0$	$2$	$2$	$0$	$0$	$0$	$4$	$2$

The number of times $\left(n\left(\Gamma \right)\right)$ an irreducible representation $\left(\Gamma \right)$ is present in a reducible representation is given below.

    $n\left(\Gamma \right)=\frac{1}{h}{\displaystyle \sum _{c}N\left(C\right){\chi }^{\Gamma }\left(C\right)\chi \left(C\right)}$        (1)

Where,

• $h$ is the order of the group.
• $N\left(C\right)$ is the number of operation in each class.
• ${\chi }^{\Gamma }\left(C\right)$ is the character of the irreducible representation.
• $\chi \left(C\right)$ is the character of the reducible representation.

The order of the point group, $h$ is $48$.

The number of times the irreducible representation ${\text{A}}_{\text{1g}}$ appears is given below.

    $\begin{array}{c}n\left({\text{A}}_{\text{1g}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{A}}_{\text{1g}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 1\times 6\right)+\left(8\times 1\times 0\right)+\left(6\times 1\times 0\right)+\left(6\times 1\times 2\right)+\left(3\times 1\times 2\right)\\ +\left(1\times 1\times 0\right)+\left(6\times 1\times 0\right)+\left(8\times 1\times 0\right)+\left(3\times 1\times 4\right)+\left(6\times 1\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(6+0+0+12+6+0+0+0+12+12\right)\\ =1\end{array}$

The number of times the irreducible representation ${\text{A}}_{\text{2g}}$ appears is given below.

    $\begin{array}{c}n\left({\text{A}}_{\text{2g}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{A}}_{\text{2g}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 1\times 6\right)+\left(8\times 1\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(6\times \left(-1\right)\times 2\right)+\left(3\times 1\times 2\right)\\ +\left(1\times 1\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(8\times 1\times 0\right)+\left(3\times 1\times 4\right)+\left(6\times \left(-1\right)\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(6+0+0-12+6+0+0+0+12-12\right)\\ =0\end{array}$

The number of times the irreducible representation ${\text{E}}_{\text{g}}$ appears is given below.

    $\begin{array}{c}n\left({\text{E}}_{\text{g}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{E}}_{\text{g}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 2\times 6\right)+\left(8\times \left(-1\right)\times 0\right)+\left(6\times 0\times 0\right)+\left(6\times 0\times 2\right)+\left(3\times 2\times 2\right)\\ +\left(1\times 2\times 0\right)+\left(6\times 0\times 0\right)+\left(8\times \left(-1\right)\times 0\right)+\left(3\times 2\times 4\right)+\left(6\times 0\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(12+0+0+0+12+0+0+0+24+0\right)\\ =1\end{array}$

The number of times the irreducible representation ${\text{T}}_{\text{1g}}$ appears is given below.

    $\begin{array}{c}n\left({\text{T}}_{\text{1g}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{T}}_{\text{1g}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 3\times 6\right)+\left(8\times 0\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(6\times 1\times 2\right)+\left(3\times \left(-1\right)\times 2\right)\\ +\left(1\times 3\times 0\right)+\left(6\times 1\times 0\right)+\left(8\times 0\times 0\right)+\left(3\times \left(-1\right)\times 4\right)+\left(6\times \left(-1\right)\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(18+0+0+12-6+0+0+0-12-12\right)\\ =0\end{array}$

The number of times the irreducible representation ${\text{T}}_{\text{2g}}$ appears is given below.

    $\begin{array}{c}n\left({\text{T}}_{\text{2g}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{T}}_{\text{2g}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 3\times 6\right)+\left(8\times 0\times 0\right)+\left(6\times 1\times 0\right)+\left(6\times \left(-1\right)\times 2\right)+\left(3\times \left(-1\right)\times 2\right)\\ +\left(1\times 3\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(8\times 0\times 0\right)+\left(3\times \left(-1\right)\times 4\right)+\left(6\times 1\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(18+0+0-12-6+0+0+0-12+12\right)\\ =0\end{array}$

The number of times the irreducible representation ${\text{A}}_{\text{1u}}$ appears is given below.

  $\begin{array}{c}n\left({\text{A}}_{\text{1u}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{A}}_{\text{1u}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 1\times 6\right)+\left(8\times 1\times 0\right)+\left(6\times 1\times 0\right)+\left(6\times 1\times 2\right)+\left(3\times 1\times 2\right)\\ +\left(1\times \left(-1\right)\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(8\times \left(-1\right)\times 0\right)+\left(3\times \left(-1\right)\times 4\right)+\left(6\times \left(-1\right)\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(6+0+0+12+6+0+0+0-12-12\right)\\ =0\end{array}$

The number of times the irreducible representation ${\text{A}}_{\text{2u}}$ appears is given below.

  $\begin{array}{c}n\left({\text{A}}_{\text{2u}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{A}}_{\text{2u}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 1\times 6\right)+\left(8\times 1\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(6\times \left(-1\right)\times 2\right)+\left(3\times 1\times 2\right)\\ +\left(1\times \left(-1\right)\times 0\right)+\left(6\times 1\times 0\right)+\left(8\times \left(-1\right)\times 0\right)+\left(3\times \left(-1\right)\times 4\right)+\left(6\times 1\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(6+0+0-12+6+0+0+0-12+12\right)\\ =0\end{array}$

The number of times the irreducible representation ${\text{E}}_{\text{u}}$ appears is given below.

  $\begin{array}{c}n\left({\text{E}}_{\text{u}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{E}}_{\text{u}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 2\times 6\right)+\left(8\times \left(-1\right)\times 0\right)+\left(6\times 0\times 0\right)+\left(6\times 0\times 2\right)+\left(3\times 2\times 2\right)\\ +\left(1\times \left(-2\right)\times 0\right)+\left(6\times 0\times 0\right)+\left(8\times 1\times 0\right)+\left(3\times \left(-2\right)\times 4\right)+\left(6\times 1\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(12+0+0+0+12+0+0+0-24+0\right)\\ =0\end{array}$

The number of times the irreducible representation ${\text{T}}_{\text{1u}}$ appears is given below.

  $\begin{array}{c}n\left({\text{T}}_{\text{1u}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{T}}_{\text{1u}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 3\times 6\right)+\left(8\times 0\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(6\times 1\times 2\right)+\left(3\times \left(-1\right)\times 2\right)\\ +\left(1\times \left(-3\right)\times 0\right)+\left(6\times \left(-1\right)\times 0\right)+\left(8\times 0\times 0\right)+\left(3\times 1\times 4\right)+\left(6\times 1\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(18+0+0+12-6+0+0+0+12+12\right)\\ =1\end{array}$

The number of times the irreducible representation ${\text{T}}_{\text{2u}}$ appears is given below.

    $\begin{array}{c}n\left({\text{T}}_{\text{2u}}\right)=\frac{1}{48}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\text{T}}_{\text{2u}}}\left(C\right)\chi \left(C\right)}\\ =\frac{1}{48}\left(\begin{array}{l}\left(1\times 3\times 6\right)+\left(8\times 0\times 0\right)+\left(6\times 1\times 0\right)+\left(6\times \left(-1\right)\times 2\right)+\left(3\times \left(-1\right)\times 2\right)\\ +\left(1\times \left(-3\right)\times 0\right)+\left(6\times 1\times 0\right)+\left(8\times 0\times 0\right)+\left(3\times 1\times 4\right)+\left(6\times \left(-1\right)\times 2\right)\end{array}\right)\\ =\frac{1}{48}\left(18+0+0-12-6+0+0+0+12-12\right)\\ =0\end{array}$

Therefore, the span is ${\text{A}}_{\text{1g}}+{\text{E}}_{\text{g}}+{\text{T}}_{\text{1u}}$.