Chapter 10, Problem 10C.3P

(a)

Interpretation Introduction

Interpretation:

Whether the integral of the function $3x^2-1$ would vanish or not when integrated over a symmetrical range in a cube has to be predicted.

Concept introduction:

A symmetry operation is defined as an action on an object to reproduce an arrangement that is identical to its original spatial arrangement.  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 10C.3P

The integral of the function $3x^2-1$ would not vanish when integrated over a symmetrical range in a cube.

Explanation of Solution

The give function is $3x^2-1$.

The point group of the cube is $O_{\text{h}}$.

The integration of the given function is over a symmetrical range in a cube is shown below.

    $\begin{array}{c}I={\displaystyle \int \left(3x^2-1\right)}\text{d}x\\ ={\displaystyle \int \left(3x^2\right)}\text{d}x-{\displaystyle \int \left(1\right)}\text{d}x\\ =3{\displaystyle \int \left(x\times x\right)}\text{d}x-{\displaystyle \int \left(1\right)}\text{d}x\end{array}$

The second integral term $\left({\displaystyle \int \left(1\right)}\text{d}x\right)$ is totally symmetrical and present in the $O_{\text{h}}$ point group.

The first integral term $\left(3{\displaystyle \int \left(x\times x\right)}\text{d}x\right)$ has $x$ variable. The variable corresponds to the $T_{\text{1u}}$ representation of $O_{\text{h}}$ point group.

The integral will have a nonzero numerical value when the irreducible representations of the components of the integrand must contain the totally symmetrical irreducible representation of the point group $O_{\text{h}}$ that is $A_{\text{1g}}$.  The relation for orthogonality for the above integral is shown below.

  $A_{\text{1g}}=T_{\text{1u}}\otimes T_{\text{1u}}$

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

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

The representations of $T_{\text{1u}}\otimes T_{\text{1u}}$ is calculated as follows.

$O_{\text{h}}$	$E$	$8C_3$	$6C_2$	$6C_4$	$3C_2$	$i$	$6S_4$	$8S_6$	$3{\sigma }_{\text{h}}$	$6{\sigma }_{\text{d}}$
$T_{\text{1u}}\otimes T_{\text{1u}}$	$9$	$0$	$1$	$1$	$1$	$3$	$-1$	$0$	$1$	$1$

The representations of $T_{\text{1u}}\otimes T_{\text{1u}}$ is not the same as $A_{\text{1g}}$.  Therefore, the integral of the function $3x^2-1$ would not vanish when integrated over a symmetrical range in a cube.

(b)

Interpretation Introduction

Interpretation:

Whether the integral of the function $3x^2-1$ would vanish or not when integrated over a symmetrical range in a tetrahedron has to be predicted.

Concept introduction:

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

Answer to Problem 10C.3P

The integral of the function $3x^2-1$ would not vanish when integrated over a symmetrical range in a tetrahedron.

Explanation of Solution

The give function is $3x^2-1$.

The point group of tetrahedron is $T_{\text{d}}$.

The integration of the given function is over a symmetrical range in a tetrahedron is shown below.

    $\begin{array}{c}I={\displaystyle \int \left(3x^2-1\right)}\text{d}x\\ ={\displaystyle \int \left(3x^2\right)}\text{d}x-{\displaystyle \int \left(1\right)}\text{d}x\\ =3{\displaystyle \int \left(x\times x\right)}\text{d}x-{\displaystyle \int \left(1\right)}\text{d}x\end{array}$

The second integral term $\left({\displaystyle \int \left(1\right)}\text{d}x\right)$ is totally symmetrical and present in the $T_{\text{d}}$ point group.

the The first integral term $\left(3{\displaystyle \int \left(x\times x\right)}\text{d}x\right)$ has $x$ variable.  The variable corresponds to the $T_{\text{2}}$ representation of $T_{\text{d}}$ point group.

The integral will have a nonzero numerical value when the irreducible representations of the components of the integrand must contain the totally symmetrical irreducible representation of the point group $T_{\text{d}}$ that is $A_{\text{1}}$.  The relation for orthogonality for the above integral is shown below.

  $A_{\text{1}}=T_2\otimes T_2$

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

$T_{\text{d}}$	$E$	$8C_3$	$3C_2$	$6S_4$	$6{\sigma }_{\text{d}}$
$A_1$	$1$	$1$	$1$	$1$	$1$
$T_2$	$3$	$0$	$-1$	$1$	$-1$

The representations of $T_2\otimes T_2$ is calculated as follows.

$T_{\text{d}}$	$E$	$8C_3$	$3C_2$	$6S_4$	$6{\sigma }_{\text{d}}$
$T_2\otimes T_2$	$6$	$0$	$1$	$1$	$1$

The representations of $T_2\otimes T_2$ is not same as $A_{\text{1}}$.  Therefore, the integral of the function $3x^2-1$ would not vanish when integrated over a symmetrical range in a tetrahedron.

(c)

Interpretation Introduction

Interpretation:

Whether the integral of the function $3x^2-1$ would vanish or not when integrated over a symmetrical range in a hexagonal prism has to be predicted.

Concept introduction:

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

Answer to Problem 10C.3P

The integral of the function $3x^2-1$ would not vanish when integrated over a symmetrical range in a hexagonal prism.

Explanation of Solution

The give function is $3x^2-1$.

The point group of hexagonal prism is $D_{\text{6h}}$.

The integration of the given function is over a symmetrical range in a hexagonal prism is shown below.

    $\begin{array}{c}I={\displaystyle \int \left(3x^2-1\right)}\text{d}x\\ ={\displaystyle \int \left(3x^2\right)}\text{d}x-{\displaystyle \int \left(1\right)}\text{d}x\\ =3{\displaystyle \int \left(x\times x\right)}\text{d}x-{\displaystyle \int \left(1\right)}\text{d}x\end{array}$

The second integral term $\left({\displaystyle \int \left(1\right)}\text{d}x\right)$ is totally symmetrical and present in the $D_{\text{6h}}$ point group.

The first integral term $\left(3{\displaystyle \int \left(x\times x\right)}\text{d}x\right)$ has $x$ variable.  The variable corresponds to the ${\text{E}}_{\text{1u}}$ representation of $D_{\text{6h}}$ point group.

The integral will have a nonzero numerical value when the irreducible representations of the components of the integrand must contain the totally symmetrical irreducible representation of the point group $D_{\text{6h}}$ that is ${\text{A}}_{1\text{g}}$.  The relation for orthogonality for the above integral is shown below.

  ${\text{A}}_{1\text{g}}={\text{E}}_{\text{1u}}\otimes {\text{E}}_{\text{1u}}$

The character table for $D_{6\text{h}}$ point group is given below.

$D_{\text{6h}}$	$E$	$2C_6$	$2C_3$	$C_2$	$3C_2^{\text{'}}$	$3C_2^{"}$	$i$	$2S_3$	$2S_6$	${\sigma }_{\text{h}}$	$3{\sigma }_{\text{d}}$	$3{\sigma }_{\text{v}}$
${\text{A}}_{1\text{g}}$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$	$1$
${\text{E}}_{\text{1u}}$	$2$	$1$	$-1$	$-2$	$0$	$0$	$-2$	$-1$	$1$	$2$	$0$	$0$

The representations of $T_2\otimes T_2$ is calculated as follows.

$D_{\text{6h}}$	$E$	$2C_6$	$2C_3$	$C_2$	$3C_2^{\text{'}}$	$3C_2^{"}$	$i$	$2S_3$	$2S_6$	${\sigma }_{\text{h}}$	$3{\sigma }_{\text{d}}$	$3{\sigma }_{\text{v}}$
${\text{E}}_{\text{1u}}\otimes {\text{E}}_{\text{1u}}$	$4$	$1$	$1$	$4$	$0$	$0$	$4$	$1$	$1$	$4$	$0$	$0$

The representations of ${\text{E}}_{\text{1u}}\otimes {\text{E}}_{\text{1u}}$ is not same as ${\text{A}}_{1\text{g}}$.  Therefore, the integral of the function $3x^2-1$ would not vanish when integrated over a symmetrical range in a hexagonal prism.