(p.385) A12 Vanishing integrals
(p.385) A12 Vanishing integrals
A12.1 Introduction
If the sine function
is integrated, the result is zero for any limit ±p, because of the antisymmetry of the sine function across the origin. This type of function could not span the totally symmetric A_{1} irreducible representation. Consider next the integral
where ${\psi}_{1}\text{}\text{}\text{and}\text{}\text{}{\psi}_{2}$ are wave functions and the integration is over the space of τ. Since the value of the integral $\mathcal{I}.$ is independent of molecular orientation, any symmetry operation acting on $\mathcal{I}.$ is equivalent to the transformation $\mathcal{I}\to \text{}\mathcal{I}$, that is, a product ${\psi}_{1}{\psi}_{2}$ can be the basis for the A_{1} irreducible representation. For example, in point group C_{3v}, suppose that ${\psi}_{1}\text{}\text{}\text{and}\text{}\text{}{\psi}_{2}$span A_{1} and E respectively, then the table hereunder follows:
E 
2C_{3} 
3σ_{v} 


${\psi}_{1}({A}_{1})$ 
1 
1 
1 
${\psi}_{2}(E)$ 
2 
−1 
0 
The product ${\psi}_{1}{\psi}_{2}$ is 2 −1 0, which has the symmetry of E: A_{1} is not present (${A}_{1}\otimes \text{}\text{}E=E,$ Appendix A10.2.1) and the integral, Eq. (A12.2), vanishes because ${\psi}_{1}\text{}\text{}\text{and}\text{}\text{}{\psi}_{2}$ do not transform as the same irreducible representation of the point group. Had ${\psi}_{2}$ also spanned A_{1}, then ${\psi}_{1}{\psi}_{2}$ would have had the symmetry of A_{1}, and the integral would not necessarily have vanished; it could however, vanish for reasons unrelated to symmetry [1]. The arguments may be extended to products of more than two functions.
Example A12.1
Can the integrals, Eq. (A12.2), of the functions (a) ${x}^{2}{y}^{2},$ and (b) ${x}^{2}{y}^{2}+{z}^{2}$be nonvanishing when integrated over a square centred at the origin?
A square corresponds with point group D_{4h}. Hence, from the character table for this point group: (a) The function ${x}^{2}{y}^{2}$spans B_{1g} and the integral vanishes. (b) The function ${x}^{2}{y}^{2}+{z}^{2}$spans A_{1g} + B_{1g}, and the integral can be now nonvanishing.
(p.386) A12.2 Spectroscopic applications
In applying quantum mechanical principles to spectroscopic transitions, integrals of the form
are encountered, where i and f are initial and final states of the system, and μ is a transitional moment operator, the dipole moment operator, for example, which may be written as
where e_{i} is the charge of the i^{th} particle, and x_{i}, y_{i} and z_{i} are the coordinates of its position vector r_{i}, where i, j and k are unit vectors along x, y and z, respectively. Now, Eq. (A12.3) can be resolved into three integrals, that along the x axis, for example, being
The integral, Eq. (A12.3), will be nonzero if any one of its component integrals like Eq. (A12.5) is nonzero: for this condition to exist, the direct product of ${\psi}_{i},\text{}\text{}{\mu}_{x}\text{}\text{}\text{and}\text{}\text{}{\psi}_{f}$, or its counterpart in y or z, should contain the fully symmetric type A_{1} representation [2].
The ground state vibrational wave function ${\psi}_{i}$ of an atom has the same mathematical form as that of an s atomic orbital, so it is spherically (totally) symmetric, which means that there are no symmetry restrictions attached to it.
Consider the water molecule, point group C_{2v}. The character table shows that x, y and z have the symmetries ${B}_{1},\text{}{B}_{2}\text{}\text{and}\text{}{\text{A}}_{1}$, respectively. Following Section 8.3.3.2, it can be shown that this molecule has vibrational (final) states ${\psi}_{f}$ corresponding to A_{1} and B_{2} symmetries. Hence, the integral Eq. (A12.3) may be investigated by the direct products
With ${\Gamma}_{{\psi}_{1}}={A}_{1}$ and ${\Gamma}_{{\psi}_{f}}={A}_{1}$,
(p.387) and with ${\Gamma}_{{\psi}_{1}}={A}_{1}$ and ${\Gamma}_{{\psi}_{f}}={B}_{2}$,
Thus, Eq. (A12.3) can be nonzero in both A_{1} and B_{2} symmetries because the direct products contain the fully symmetric A_{1} representation in each case.