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Vanishing Integrals

The character tables provide a quick and convenient way of judging whether an overlap, or transition integral is necessary zero.

Let us consider the overlap integral

$\displaystyle I = \int f_1 f_2 \,d\tau,$     (25)

where $f_1$ and $f_2$ are two atomic, or molecular orbitals.

The integral $I$ is always a scalar value which means that it does not changes under any symmetry transformations of the molecule. The volume element $d\tau$ is also a scalar as it is invariant under any coordinate transformations. Therefore, the product $f_1 f_2$ must also remain unchanged by any symmetry operations of the molecular point group. If the integrand changes its sign under a symmetry operation, the integral $I$ is necessary zero, because its positive part will necessary cancel its positive part. As we know, the the irreducible representation which is equivalent in the molecular point group is totally symmetric representation $A_1$. Thus, the integral $I$ differs from zero only if the integrand $f_1 f_2$ spans the symmetry species $A_1$.

If the symmetry species of the functions $f_1$ and $f_2$ are known, the group theory provides a formal procedure which can be used for determination of the symmetry species of the product $f_1 f_2$. Particularly, the character table of the product $f_1 f_2$ can be obtained just by multiplication of the characters from the character tables of the functions $f_1$ and $f_2$ corresponding to a certain symmetry operator.

As an example we consider the product of the $f_1=s_N$ orbital of the $N$ atom and the linear combination of three hydrogen atom orbitals, $f_2=s_1$ in eq. (20) in $NH_3$ molecule, each of the orbitals spans $A_1$ species:

$\displaystyle f_1$ $\textstyle :$ $\displaystyle \:\:1\:\:1\:\:1$  
$\displaystyle f_2$ $\textstyle :$ $\displaystyle \:\:1\:\:1\:\:1$ (26)
$\displaystyle f_1f_2$ $\textstyle :$ $\displaystyle \:\:1\:\:1\:\:1$  

It is evident from eq. (26) and the table 5 , that the product $f_1f_1$ also spans $A_1$ and therefore, the in integral $I$ in eq. (25) in this case is not necessary equal to zero. Therefore, bonding and antibonding molecular orbitals can be formed from linear combinations of $s_N$ and $s_1$.

The procedure of finding the irreducible representation of the product of two representations $\Gamma_1$ and $\Gamma_2$ is written as direct product of irreducible representations $\Gamma_1 \times \Gamma_2$ and and for the example above can be written as $A_1 \times A_1 = A_1$.

As another example, we consider the product of the $f_1=s_N$ orbital of the $N$ atom in $NH_3$ and $f_2=s_3$, where $s_3 = s_B-s_C$ is the linear combination of the hydrogen atom wavefunctions from eq. (21). Now one function spans the $A_1$ species and another the E species. The product table of characters is

$\displaystyle f_1$ $\textstyle :$ $\displaystyle \:\:1\:\:\:\:\:\:\:1\:\:\:\:\:\:\:1$  
$\displaystyle f_2$ $\textstyle :$ $\displaystyle \:\:2\:\:-1\:\:\:\:\:\:\:0$ (27)
$\displaystyle f_1f_2$ $\textstyle :$ $\displaystyle \:\:2\:\:-1\:\:\:\:\:\:\:0$  

The product characters 2, $-1$, 0 are those of the E species alone and therefore, the integral must be zero. Therefore, bonding and antibonding molecular orbitals cannot be formed from linear combinations of $s_N$ and $s_3$. The direct product of the representations in this case is written as $A_1 \times$ E$\; =\; $E.

The general rule is that only orbitals of the same symmetry species may have nonzero overlap and therefore, form bonding and antibonding combinations. This result makes a direct link between the group theory and construction of molecular orbitals from atomic orbitals by the LCAO procedure we discussed in previous chapter. Indeed, the molecular orbitals can be formed only from a particular set of atomic orbitals with nonzero overlap. These molecular orbitals are usually labelled with a lower-case letter corresponding to the symmetry species. For instance, the ($s_N, s_1$) molecular orbitals are called $a_1$ if they are bonding and $a_1^*$ if they are antibonding.

Note, that the relationship between the symmetry species of the atomic orbitals and their product, in general, is not as simple as in eqs. (26) and (27). As an example, let us consider the linear combinations $s_2$ and $s_3$ in eq. (21) which both have symmetry species E. As we know the $N(2s)$ atomic orbital cannot be used together with each of them for building the bonding and antibonding molecular orbitals. However, the $N2p_x$ and $N2p_y$ atomic orbitals also belong to the E species in $C_{3v}$ (see Character Table 5) and thus are suitable because they may have a nonzero overlap with $s_2$ and $s_3$. This construction can be verified by multiplying the characters as

$\displaystyle f_1$ $\textstyle :$ $\displaystyle \:\:2\:\:-1\:\:\:\:\:\:\:0$  
$\displaystyle f_2$ $\textstyle :$ $\displaystyle \:\:2\:\:-1\:\:\:\:\:\:\:0$ (28)
$\displaystyle f_1f_2$ $\textstyle :$ $\displaystyle \:\:4\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\,0$  

It can be easily verified from eq. (28) by making summation of characters in Table 5 that E $\times$ E $=\; A_1\; +\; A_2\; + $ E. The product $f_1 f_2$ in eq. (28) contains the totally symmetric species $A_1$ and, therefore, the corresponding integral may have a nonzero value.

next up previous contents
Next: Vanishing Dipole Moment Integrals Up: Character Tables Previous: Character Tables   Contents
Markus Hiereth 2005-02-09

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