Chapter 10, Problem 10B.4AE

Interpretation Introduction

Interpretation:

Whether all the irreducible representations are orthonormal for the point group $C_{\text{2h}}$ has to be confirmed.

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 known 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 10B.4AE

All the irreducible representations are orthonormal for the point group $C_{\text{2h}}$.

Explanation of Solution

The condition for orthonormality of irreducible representation is given below.

    $\frac{1}{h}{\displaystyle \sum _{c}N\left(C\right){\chi }^{{\Gamma }^{\left(i\right)}}\left(C\right){\chi }^{{\Gamma }^{\left(j\right)}}\left(C\right)}=\{\begin{array}{l}0\text{for}i\ne j\\ 1\text{for}i=j\end{array}$        (1)

Where,

• $h$ is the number of operations of the group.
• $N\left(C\right)$ is the number of classes.
• ${\chi }^{{\Gamma }^{\left(i\right)}}\left(C\right),{\chi }^{{\Gamma }^{\left(j\right)}}\left(C\right)$ are irreducible representations.

The character table for $C_{\text{2h}}$ point group is given below.

	$E$	$C_2$	$i$	${\sigma }_{\text{h}}$
${\text{A}}_{\text{g}}$	$1$	$1$	$1$	$1$	$R_z$	$x^2,y^2,z^2,xy$
${\text{B}}_{\text{g}}$	$1$	$-1$	$1$	$-1$	$R_x,R_y$	$xz,yz$
${\text{A}}_{\text{u}}$	$1$	$1$	$-1$	$-1$	$z$
${\text{B}}_{\text{u}}$	$1$	$-1$	$-1$	$1$	$x,y$

The number of operation, $h$ is $4$.

The number of classes, $N\left(C\right)$ is $1$.

The condition in equation (1) is applied to ${\text{A}}_{\text{g}}$ and ${\text{B}}_{\text{g}}$ to check orthogonality as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{{\text{B}}_{\text{g}}}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times \left(-1\right)\right)+\left(1\times 1\right)+\left(1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+\left(-1\right)+1+\left(-1\right)\right)\\ =0\end{array}$

The normality of ${\text{A}}_{\text{g}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{{\text{A}}_{\text{g}}}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

The normality of ${\text{B}}_{\text{g}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{{\text{B}}_{\text{g}}}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

Therefore, the irreducible representations, ${\text{A}}_{\text{g}}$ and ${\text{B}}_{\text{g}}$ are orthonormal.

The condition in equation (1) is applied to ${\text{A}}_{\text{g}}$ and ${\text{A}}_{\text{u}}$ to check orthogonality as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times \left(-1\right)\right)+\left(1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+\left(-1\right)+\left(-1\right)\right)\\ =0\end{array}$

The normality of ${\text{A}}_{\text{g}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

The normality of ${\text{A}}_{\text{u}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(-1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

Therefore, the irreducible representations, ${\text{A}}_{\text{g}}$ and ${\text{A}}_{\text{u}}$ are orthonormal.

The condition in equation (1) is applied to ${\text{A}}_{\text{g}}$ and ${\text{B}}_{\text{u}}$ to check orthogonality as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times \left(-1\right)\right)+\left(1\times \left(-1\right)\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+\left(-1\right)+\left(-1\right)+1\right)\\ =0\end{array}$

The normality of ${\text{A}}_{\text{g}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{\left({\text{A}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times 1\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

The normality of ${\text{B}}_{\text{u}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(-1\times \left(-1\right)\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

Therefore, the irreducible representations, ${\text{A}}_{\text{g}}$ and ${\text{B}}_{\text{u}}$ are orthonormal.

The condition in equation (1) is applied to ${\text{A}}_{\text{u}}$ and ${\text{B}}_{\text{g}}$ to check orthogonality as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times \left(-1\right)\right)+\left(\left(-1\right)\times 1\right)+\left(\left(-1\right)\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+\left(-1\right)+\left(-1\right)+1\right)\\ =0\end{array}$

The normality of ${\text{A}}_{\text{u}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(-1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

The normality of ${\text{B}}_{\text{g}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

Therefore, the irreducible representations, ${\text{A}}_{\text{u}}$ and ${\text{B}}_{\text{g}}$ are orthonormal.

The condition in equation (1) is applied to ${\text{B}}_{\text{u}}$ and ${\text{B}}_{\text{g}}$ to check orthogonality as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(\left(-1\right)\times \left(-1\right)\right)+\left(\left(-1\right)\times 1\right)+\left(1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+\left(-1\right)+\left(-1\right)\right)\\ =0\end{array}$

The normality of ${\text{B}}_{\text{u}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(-1\times \left(-1\right)\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

The normality of ${\text{B}}_{\text{g}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

Therefore, the irreducible representations, ${\text{B}}_{\text{u}}$ and ${\text{B}}_{\text{g}}$ are orthonormal.

The condition in equation (1) is applied to ${\text{A}}_{\text{u}}$ and ${\text{B}}_{\text{u}}$ to check orthogonality as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{g}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times \left(-1\right)\right)+\left(\left(-1\right)\times \left(-1\right)\right)+\left(\left(-1\right)\times 1\right)\right)\\ =\frac{1}{4}\left(1+\left(-1\right)+1+\left(-1\right)\right)\\ =0\end{array}$

The normality of ${\text{A}}_{\text{u}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{A}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(-1\times \left(-1\right)\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

The normality of ${\text{B}}_{\text{u}}$ is checked as shown below.

    $\begin{array}{c}\frac{1}{4}\left({\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}{\chi }^{{\Gamma }^{\left({\text{B}}_{\text{u}}\right)}}\right)=\frac{1}{4}\left(\left(1\times 1\right)+\left(-1\times \left(-1\right)\right)+\left(-1\times \left(-1\right)\right)+\left(1\times 1\right)\right)\\ =\frac{1}{4}\left(1+1+1+1\right)\\ =1\end{array}$

Therefore, the irreducible representations, ${\text{A}}_{\text{u}}$ and ${\text{B}}_{\text{u}}$ are orthonormal.