Chapter 10, Problem 10B.3P

(a)

Interpretation Introduction

Interpretation:

The relation $C_2{\sigma }_{\text{v}}={{\sigma }^{\prime }}_{\text{v}}$ and ${\sigma }_{\text{v}}{{\sigma }^{\prime }}_{\text{v}}=C_2$ has to be confirmed using an explicit matrix multiplication method.

Concept introduction:

A symmetry operation is defined as an action on an object to reproduce an arrangement using a symmetry argument. The spatial arrangement of the object remains identical after a symmetry operation. The point of reference through which a symmetry operation takes place is termed as a symmetry element.

Answer to Problem 10B.3P

The relation $C_2{\sigma }_{\text{v}}={{\sigma }^{\prime }}_{\text{v}}$ and ${\sigma }_{\text{v}}{{\sigma }^{\prime }}_{\text{v}}=C_2$ is confirmed from the explicit matrix multiplication method.

Explanation of Solution

The water $\left({\text{H}}_{\text{2}}\text{O}\right)$ molecule taking two $\text{H1s}$ orbital and the four valence orbitals of the $\text{O}$ atom are shown below.

Figure 1

The two $\text{H1s}$ orbital taken as a basis are ${\text{1S}}_{\left(\text{H1}\right)}$ and ${\text{1S}}_{\left(\text{H2}\right)}$.  The four valence orbitals of the $\text{O}$ atom taken as a basis are ${\text{2S}}_{\left(\text{O}\right)}$, ${\text{2p}}_{\text{x}\left(\text{O}\right)}$, ${\text{2p}}_{\text{y}\left(\text{O}\right)}$ and ${\text{2p}}_{\text{z}\left(\text{O}\right)}$.

The $\text{E}$ operator is applied on the orbitals as shown below.

Figure 2

The changes taking place in the above operation are shown below

  $\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)\leftarrow \left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)$

The above result is expressed as a matrix multiplication as shown below.

  $\begin{array}{c}\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)=\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)\\ =\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)D\left(E\right)\end{array}$

The identity operation $\left(E\right)$ does not affect the basis and is represented in a $6\times 6$ matrix as shown below.

  $D\left(E\right)=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)$

The $C_2$ operator is applied on the orbitals as shown below.

Figure 3

The changes taking place in the above operation are shown below

  $\left({\text{2S}}_{\left(\text{O}\right)}-{\text{2p}}_{\text{x}\left(\text{O}\right)}-{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H2}\right)}{\text{1S}}_{\left(\text{H1}\right)}\right)\leftarrow \left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)$

The above result is expressed as a matrix multiplication as shown below.

$\begin{array}{c}\left({\text{2S}}_{\left(\text{O}\right)}-{\text{2p}}_{\text{x}\left(\text{O}\right)}-{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H2}\right)}{\text{1S}}_{\left(\text{H1}\right)}\right)=\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)\\ =\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)D\left(C_2\right)\end{array}$

The $C_2$ operation is represented in a $6\times 6$ matrix as shown below.

  $D\left(C_2\right)=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)$

The ${\sigma }_{\text{v}}$ operator is applied on the orbitals as shown below.

Figure 4

The changes taking place in the above operation are shown below

  $\left({\text{2S}}_{\left(\text{O}\right)}-{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H2}\right)}{\text{1S}}_{\left(\text{H1}\right)}\right)\leftarrow \left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)$

The above result is expressed as a matrix multiplication as shown below.

$\begin{array}{c}\left({\text{2S}}_{\left(\text{O}\right)}-{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H2}\right)}{\text{1S}}_{\left(\text{H1}\right)}\right)=\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)\\ =\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)D\left({\sigma }_{\text{v}}\right)\end{array}$

The ${\sigma }_{\text{v}}$ operation is represented in a $6\times 6$ matrix as shown below.

  $D\left({\sigma }_{\text{v}}\right)=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)$

The ${{\sigma }^{\prime }}_{\text{v}}$ operator is applied on the orbitals as shown below.

Figure 5

The changes taking place in the above operation are shown below

  $\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}-{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H2}\right)}{\text{1S}}_{\left(\text{H1}\right)}\right)\leftarrow \left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)$

The above result is expressed as a matrix multiplication as shown below.

$\begin{array}{c}\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}-{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H2}\right)}{\text{1S}}_{\left(\text{H1}\right)}\right)=\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)\\ =\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)D\left({{\sigma }^{\prime }}_{\text{v}}\right)\end{array}$

The ${{\sigma }^{\prime }}_{\text{v}}$ operation is represented in a $6\times 6$ matrix as shown below.

  $D\left({{\sigma }^{\prime }}_{\text{v}}\right)=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)$

The explicit matrix multiplication $C_2{\sigma }_{\text{v}}$ is calculated as shown below.

  $C_2{\sigma }_{\text{v}}=D\left(C_2\right)\times D\left({\sigma }_{\text{v}}\right)$        (1)

Substitute the value of $D\left(C_2\right)$ and $D\left({\sigma }_{\text{v}}\right)$ in equation (1) as shown below.

  $\begin{array}{c}{C}_2{\sigma }_{\text{v}}=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)\\ =\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)\\ ={\sigma }^{\prime }\end{array}$

Therefore, the relation $C_2{\sigma }_{\text{v}}={{\sigma }^{\prime }}_{\text{v}}$ is confirmed.

The explicit matrix multiplication ${\sigma }_{\text{v}}{{\sigma }^{\prime }}_{\text{v}}$ is calculated as shown below.

  ${\sigma }_{\text{v}}{{\sigma }^{\prime }}_{\text{v}}=D\left({\sigma }_{\text{v}}\right)\times D\left({{\sigma }^{\prime }}_{\text{v}}\right)$        (2)

Substitute the value of $D\left({\sigma }_{\text{v}}\right)$ and $D\left({{\sigma }^{\prime }}_{\text{v}}\right)$ in equation (2) as shown below.

  $\begin{array}{c}{\sigma }_{\text{v}}{{\sigma }^{\prime }}_{\text{v}}=\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)\\ =\left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)\\ =C_2\end{array}$

Therefore, the relation ${\sigma }_{\text{v}}{{\sigma }^{\prime }}_{\text{v}}=C_2$ is confirmed.

(b)

Interpretation Introduction

Interpretation:

The corresponding representation is reducible and span ${\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}$ has to be proved.

Concept introduction:

As mentioned in the concept of introduction in part (a).

Answer to Problem 10B.3P

The corresponding representation is reducible because the characters are not similar to any irreducible representation.  The corresponding representation span ${\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}$ because the character for the sum ${\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}$ is the same as the basis $\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)$.

Explanation of Solution

The characters of a representative are the sum of their diagonal elements as shown below.

$\begin{array}{ccccc}R& E& C& {\sigma }_{\text{v}}& {{\sigma }^{\prime }}_{\text{v}}\\ D\left(R\right)& \left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)& \left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)& \left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 1& 0\end{array}\right)& \left(\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 1\end{array}\right)\\ \chi \left(R\right)& 6& 0& 2& 4\end{array}$

The characters shown above are not similar to any irreducible representation.  Therefore, the representation is reducible.

The character for the sum ${\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}$ is shown below.

  $\begin{array}{ccccc}R& E& C& {\sigma }_{\text{v}}& {{\sigma }^{\prime }}_{\text{v}}\\ 3{\text{A}}_{\text{1}}& 3& 3& 3& 3\\ {\text{B}}_1& 1& -1& 1& -1\\ 2{\text{B}}_2& 2& -2& -2& 2\\ {\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}& 6& 0& 2& 4\end{array}$

The character for the sum ${\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}$ is the same as the basis $\left({\text{2S}}_{\left(\text{O}\right)}{\text{2p}}_{\text{x}\left(\text{O}\right)}{\text{2p}}_{\text{y}\left(\text{O}\right)}{\text{2p}}_{\text{z}\left(\text{O}\right)}{\text{1S}}_{\left(\text{H1}\right)}{\text{1S}}_{\left(\text{H2}\right)}\right)$.  Therefore, the corresponding representation spans ${\text{3A}}_{\text{1}}{\text{+B}}_{\text{1}}{\text{+2B}}_{\text{2}}$.