Chapter 14, Problem 14.2E

Determine if the following integrals can be nonzero if the molecular or atomic system has the given local symmetry. Use the great orthogonality theorem if necessary.

(a) ${\displaystyle \int {\Psi }_{{\stackrel{\ast }{A}}_u}}{\stackrel{⌢}{O}}_{B_{2u}}{\Psi }_{A_u}d\tau$ in $D_{\text{2h}}$ symmetry

(b) ${\displaystyle \int {\Psi }_{{\stackrel{\ast }{A}}_1}}{\stackrel{⌢}{O}}_{A_1}{\Psi }_{A_2}d\tau$ in $C_{\text{3v}}$ symmetry

(c) ${\displaystyle \int {\Psi }_{\stackrel{\ast }{{\Sigma }_{\text{g}}^{+}}}}{\stackrel{⌢}{O}}_{{\Sigma }_{\text{g}}^{-}}{\Psi }_{{\Sigma }_{\text{g}}^{-}}d\tau$ in $D_{\infty \text{h}}$ symmetry

(d) ${\displaystyle \int {\Psi }_{\stackrel{\ast }{E}}}{\stackrel{⌢}{O}}_{A_2}{\Psi }_{T_1}d\tau$ in $T_{\text{d}}$ symmetry

Interpretation Introduction

(a)

Interpretation:

The integral $\left({\displaystyle \int {\Psi }_{{\stackrel{\ast }{A}}_u}}{\stackrel{⌢}{O}}_{B_{2u}}{\Psi }_{A_u}d\tau \right)$ can be nonzero or not when the molecular or atomic system has the $D_{\text{2h}}$ symmetry is to be predicated.

Concept introduction:

The group of symmetry operations of which at least one point is kept fixed is called point group. The symmetry operations can be identity, rotation, reflection, inversion and improper rotation.

Answer to Problem 14.2E

The integral is exactly zero when the molecular or atomic system has the $D_{\text{2h}}$ symmetry.

Explanation of Solution

The integral is shown below.

${\displaystyle \int {\Psi }_{{\stackrel{\ast }{A}}_u}}{\stackrel{⌢}{O}}_{B_{2u}}{\Psi }_{A_u}d\tau$

The integral will have a nonzero numerical value when the irreducible representations of the three components of the integrand must be contain the totally symmetrical irreducible representation of the point group $D_{\text{2h}}$ that is $A_{\text{g}}$. The relation for orthogonality for the above integral is shown below.

$A_{\text{g}}=A_{\text{u}}\otimes B_{\text{2u}}\otimes A_{\text{u}}$

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

$D_{\text{2h}}$	$E$	$C_2$	$C_2^{\text{'}}$	$C_2^{"}$	$i$	$\sigma \left(xy\right)$	$\sigma \text{'}\left(yz\right)$	$\sigma "\left(xz\right)$
$A_{\text{g}}$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$
$A_{\text{u}}$	$1$	$1$	$1$	$1$	$-1$	$-1$	$-1$	$-1$
$B_{\text{2u}}$	$1$	$-1$	$1$	$-1$	$-1$	$1$	$-1$	$1$

The representations of $A_{\text{u}}\otimes B_{\text{2u}}\otimes A_{\text{u}}$ is calculated as follows:

	$E$	$C_2$	$C_2^{\text{'}}$	$C_2^{"}$	$i$	$\sigma \left(xy\right)$	$\sigma \text{'}\left(yz\right)$	$\sigma "\left(xz\right)$
$A_{\text{u}}\otimes B_{\text{2u}}\otimes A_{\text{u}}$	$1$	$-1$	$1$	$-1$	$-1$	$-1$	$-1$	$1$

The representations of $A_{\text{u}}\otimes B_{\text{2u}}\otimes A_{\text{u}}$ is not same as $A_{\text{g}}$. Therefore, the integral is exactly zero when the molecular or atomic system has the $D_{\text{2h}}$ symmetry.

Conclusion

The integral is exactly zero when the molecular or atomic system has the $D_{\text{2h}}$ symmetry.

Interpretation Introduction

(b)

Interpretation:

The integral $\left({\displaystyle \int {\Psi }_{{\stackrel{\ast }{A}}_1}}{\stackrel{⌢}{O}}_{A_1}{\Psi }_{A_2}d\tau \right)$ can be nonzero or not when the molecular or atomic system has the $C_{\text{3v}}$ symmetry is to be predicated.

Concept introduction:

The group of symmetry operations of which at least one point is kept fixed is called point group. The symmetry operations can be identity, rotation, reflection, inversion and improper rotation.

Answer to Problem 14.2E

The integral is exactly zero when the molecular or atomic system has the $C_{\text{3v}}$ symmetry.

Explanation of Solution

The integral is shown below.

${\displaystyle \int {\Psi }_{{\stackrel{\ast }{A}}_1}}{\stackrel{⌢}{O}}_{A_1}{\Psi }_{A_2}d\tau$

The integral will have a nonzero numerical value when the irreducible representations of the three components of the integrand must be contain the totally symmetrical irreducible representation of the point group $C_{\text{3v}}$ that is $A_1$. The relation for orthogonality for the above integral is shown below.

$A_1=A_1\otimes A_1\otimes A_2$

The character table for point group $C_{\text{3v}}$ is shown below.

$C_{\text{3v}}$	$E$	$2C_{\text{3v}}$	$3{\sigma }_{\text{v}}$
$A_1$	$1$	$1$	$1$
$A_2$	$1$	$1$	$-1$

The representations of $A_1\otimes A_1\otimes A_2$ is calculated as follows:

$C_{\text{3v}}$	$E$	$2C_{\text{3v}}$	$3{\sigma }_{\text{v}}$
$A_1\otimes A_1\otimes A_2$	$1$	$1$	$-1$

The representations of $A_1\otimes A_1\otimes A_2$ is not same as $A_1$. Therefore, the integral is exactly zero when the molecular or atomic system has the $C_{\text{3v}}$ symmetry.

Conclusion

The integral is exactly zero when the molecular or atomic system has the $C_{\text{3v}}$ symmetry.

Interpretation Introduction

(c)

Interpretation:

The integral $\left({\displaystyle \int {\Psi }_{\stackrel{\ast }{{\Sigma }_{\text{g}}^{+}}}}{\stackrel{⌢}{O}}_{{\Sigma }_{\text{g}}^{-}}{\Psi }_{{\Sigma }_{\text{g}}^{-}}d\tau \right)$ can be nonzero or not when the molecular or atomic system has the $D_{\infty \text{h}}$ symmetry is to be predicated.

Concept introduction:

The group of symmetry operations of which at least one point is kept fixed is called point group. The symmetry operations can be identity, rotation, reflection, inversion and improper rotation.

Answer to Problem 14.2E

The integral is nonzero when the molecular or atomic system has the $D_{\infty \text{h}}$ symmetry.

Explanation of Solution

The integral is shown below.

${\displaystyle \int {\Psi }_{\stackrel{\ast }{{\Sigma }_{\text{g}}^{+}}}}{\stackrel{⌢}{O}}_{{\Sigma }_{\text{g}}^{-}}{\Psi }_{{\Sigma }_{\text{g}}^{-}}d\tau$

The integral will have a nonzero numerical value when the irreducible representations of the three components of the integrand must be contain the totally symmetrical irreducible representation of the point group $D_{\infty \text{h}}$ that is ${\Sigma }_{\text{g}}^{+}$. The relation for orthogonality for the above integral is shown below.

${\Sigma }_{\text{g}}^{+}={\Sigma }_{\text{g}}^{+}\otimes {\Sigma }_{\text{g}}^{-}\otimes {\Sigma }_{\text{g}}^{-}$

The character table for point group $D_{\infty \text{h}}$ is shown below.

$D_{\infty \text{h}}$	$E$	$2C_{\varphi }$	$\infty C_2$	$i$	$2S_{\left(-\varphi \right)}$	$\infty {\sigma }_{\text{v}}$
${\Sigma }_{\text{g}}^{+}$	$1$	$1$	$1$	$1$	$1$	$1$
${\Sigma }_{\text{g}}^{-}$	$1$	$-1$	$1$	$1$	$1$	$-1$

The representations of ${\Sigma }_{\text{g}}^{+}\otimes {\Sigma }_{\text{g}}^{-}\otimes {\Sigma }_{\text{g}}^{-}$ is calculated as follows:

$D_{\infty \text{h}}$	$E$	$2C_{\varphi }$	$\infty C_2$	$i$	$2S_{\left(-\varphi \right)}$	$\infty {\sigma }_{\text{v}}$
${\Sigma }_{\text{g}}^{+}\otimes {\Sigma }_{\text{g}}^{-}\otimes {\Sigma }_{\text{g}}^{-}$	$1$	$1$	$1$	$1$	$1$	$1$

The representations of ${\Sigma }_{\text{g}}^{+}\otimes {\Sigma }_{\text{g}}^{-}\otimes {\Sigma }_{\text{g}}^{-}$ is same as ${\Sigma }_{\text{g}}^{+}$. Therefore, the integral is nonzero when the molecular or atomic system has the $D_{\infty \text{h}}$ symmetry.

Conclusion

The integral is nonzero when the molecular or atomic system has the $D_{\infty \text{h}}$ symmetry.

Interpretation Introduction

(d)

Interpretation:

The integral $\left({\displaystyle \int {\Psi }_{\stackrel{\ast }{E}}}{\stackrel{⌢}{O}}_{A_2}{\Psi }_{T_1}d\tau \right)$ can be nonzero or not when the molecular or atomic system has the $T_{\text{d}}$ symmetry is to be predicated.

Concept introduction:

The group of symmetry operations of which at least one point is kept fixed is called point group. The symmetry operations can be identity, rotation, reflection, inversion and improper rotation.

Answer to Problem 14.2E

The integral is exactly zero when the molecular or atomic system has the $T_{\text{d}}$ symmetry.

Explanation of Solution

The integral is shown below.

${\displaystyle \int {\Psi }_{\stackrel{\ast }{E}}}{\stackrel{⌢}{O}}_{A_2}{\Psi }_{T_1}d\tau$

The integral will have a nonzero numerical value when the irreducible representations of the three components of the integrand must be contain the totally symmetrical irreducible representation of the point group $T_{\text{d}}$ that is $A_1$. The relation for orthogonality for the above integral is shown below.

$A_1=E\otimes A_2\otimes T_1$

The character table for point group $T_{\text{d}}$ is shown below.

$T_{\text{d}}$	$E$	$8C_3$	$3C_2$	$6S_4$	$6{\sigma }_{\text{d}}$
$A_1$	$1$	$1$	$1$	$1$	$1$
$A_2$	$1$	$1$	$1$	$-1$	$-1$
$E$	$2$	$-1$	$2$	$0$	$0$
$T_1$	$3$	$0$	$-1$	$1$	$-1$

The representations of $A_{\text{u}}\otimes B_{\text{2u}}\otimes A_{\text{u}}$ is calculated as follows:

$T_{\text{d}}$	$E$	$8C_3$	$3C_2$	$6S_4$	$6{\sigma }_{\text{d}}$
$E\otimes A_2\otimes T_1$	$6$	$0$	$-2$	$0$	$0$

The representations of $E\otimes A_2\otimes T_1$ is not same as $A_1$. Therefore, the integral is exactly zero when the molecular or atomic system has the $T_{\text{d}}$ symmetry.

Conclusion

The integral is exactly zero when the molecular or atomic system has the $T_{\text{d}}$ symmetry.