Symmetry-Adapted Linear Combinations

Now, how would we have come up with these vibrational normal modes if we hadn't had the program? Group theory isn't sufficient to give us normal modes in general, but in this case, it woudl almost get us there, because other than $a_g$, no irrep has more than one vibration. In such cases, the normal mode is symmetry-determined. In the case of irreps with more than one vibration, group theory can at least give us a symmetry-adapted set of vectors (basis); these vectors are mixed to form the normal modes. It is important to point out that we could say similar things about molecular orbitals. Some MO's may be symmetry-determined (in a sufficiently small basis set), and others may be linear combinations of the symmetry-adapted AO's belonging to some irrep. Thus, we need to know how to form symmetry-adapted linear combinations (SALC's) of basis functions like atomic orbitals or vibrational displacement vectors. To explore this, we will stick with our O$_4^+$ vibrational example for now.

Let us use displacement vectors in the $x$, $y$, and $z$ directions on each atom as our basis functions, and then we will form SALC's from these to see what the vibrations should look like (without the need for computations). We do this using the technique of projection operators.

Figure 3: Displacement Vectors for Atom 1 in O$_4^+$

Our job is somewhat easier for this example because each atom is symmetry-equivalent to all the others. Hence, it will suffice to apply projection operators to only the $x$, $y$, and $z$ displacements on one of the atoms. Let's label the displacements as $x_i$, $y_i$, or $z_i$, where $i$ is the number of the atom displaced. See Figure 3 for the displacements of atom 1. First, we must determine what each symmetry operator does to each of our basis vectors. Referring to Fig. 3, we can construct Table 5, which provides the results of each symmetry operation on $x_1$, $y_1$, and $z_1$.

Table 5: Result of Symmetry Operations on Atomic Displacements

	$E$	$C_2(z)$	$C_2(y)$	$C_2(x)$	$i$	$\sigma(xy)$	$\sigma(xz)$	$\sigma(yz)$
$x_1$	$x_1$	$-x_3$	$-x_4$	$x_2$	$-x_3$	$x_1$	$x_2$	$-x_4$
$y_1$	$y_1$	$-y_3$	$y_4$	$-y_2$	$-y_3$	$y_1$	$-y_2$	$y_4$
$z_1$	$z_1$	$z_3$	$-z_4$	$-z_2$	$-z_3$	$-z_1$	$z_2$	$z_4$

To apply a projection operator, we dot each of the rows in Table 5 with the rows of the character table -- there will be a separate projection operator for each irrep. Because the totally symmetric irrep contains all 1's in its row, the rows in the Table are already the result of applying ${\hat P}_{A_g}$ to $x_1$, $y_1$, and $z_1$. Hence, ${\hat P}_{A_g}(x_1) = 2x_1 + 2x_2 - 2x_3 - 2x_4$, or $x_1 + x_2 - x_3 - x_4$ (we aren't usually too concerned about normalization). Referring to Figure 2, this is the 126.71 cm$^{-1}$ $a_g$ normal mode! Similarly, ${\hat P}_{A_g}(y_1) = y_1 - y_2 - y_3 + y_4$. This is the 1849.60 cm$^{-1}$ normal mode. Finally, ${\hat P}_{A_g}(z_1) = 0$ (all the $z$ displacements cancel). There is no third $a_g$ mode.

We can apply projection operators to $x_1$, $y_1$, and $z_1$ for each irrep to build up a complete basis of SALC's. As another example,

$${\hat P}_{A_u}(x_1) = x_1 - x_3 - x_4 + x_2 + x_3 - x_1 - x_2 + x_4 = 0,$$

$${\hat P}_{A_u}(y_1) = y_1 - y_3 + y_4 - y_2 + y_3 - y_1 + y_2 - y_4 = 0,$$

$${\hat P}_{A_u}(z_1) = z_1 + z_3 - z_4 - z_2 + z_3 + z_1 - z_2 - z_4 = z_1 - z_2 + z_3 - z_4.$$

Consulting Fig. 2, we see that ${\hat P}_{A_u}(z_1)$ is the 131.73 cm$^{-1}$ normal mode.

Finally, let's try the $B_{1u}$ irrep:

$${\hat P}_{B_{1u}}(x_1) = x_1 - x_3 + x_4 - x_2 + x_3 - x_1 + x_2 - x_4 = 0,$$

$${\hat P}_{B_{1u}}(y_1) = y_1 - y_3 - y_4 + y_2 + y_3 - y_1 - y_2 + y_4 = 0,$$

$${\hat P}_{B_{1u}}(z_1) = z_1 + z_3 + z_4 + z_2 + z_3 + z_1 + z_2 + z_4 = z_1 + z_2 + z_3 + z_4.$$

Now note that there is not a $b_{1u}$ normal mode of vibration! Why is this? If we examine the result of ${\hat P}_{B_{1u}}(z_1)$, we see that it displaces all of the atoms in the $z$ direction. This is just a translation of the entire molecule in the $z$ direction. Creating SALC's out of Cartesian displacements will, in general, create not only vibrations but also translations and rotations. We already discussed above how to identify the irreps of translations and rotations.