# 12.6: Character Tables

- Page ID
- 210892

Now that we’ve learnt how to create a matrix representation of a point group within a given basis, we will move on to look at some of the properties that make these representations so powerful in the treatment of molecular symmetry.

## Similarity Transforms

Suppose we have a basis set \(\begin{pmatrix} x_1, x_2, x_3, ... x_n \end{pmatrix}\), and we have determined the matrix representatives for the basis in a given point group. There is nothing particularly special about the basis set we have chosen, and we could equally well have used any set of linear combinations of the original functions (provided the combinations were linearly independent). The matrix representatives for the two basis sets will certainly be different, but we would expect them to be related to each other in some way. As we shall show shortly, they are in fact related by a *similarity transform*. It will be far from obvious at this point why we would want to carry out such a transformation, but similarity transforms will become important later on when we use group theory to choose an optimal basis set with which to generate molecular orbitals.

Consider a basis set \(\begin{pmatrix} x_1', x_2', x_3', ... x_n' \end{pmatrix}\), in which each basis function \(x_i'\) is a linear combination of our original basis \(\begin{pmatrix} x_1, x_2, x_3, ... x_n \end{pmatrix}\).

\[x_j' = \Sigma_i x_ic_{ji} = x_1c_{j1} + x_2c_{j2} + ... \tag{11.1}\]

The \(c_{ji}\) appearing in the sum are coefficients; \(c_{ji}\) is the coefficient multiplying the original basis function \(x_i\) in the new linear combination basis function \(x_j'\). We could also represent this transformation in terms of a matrix equation \(\textbf{x'}\) = \(\textbf{x}C\):

\[\begin{pmatrix} x_1', x_2', ... x_n' \end{pmatrix} = \begin{pmatrix} x_1, x_2, ... x_n \end{pmatrix} \begin{pmatrix} c_{11} & c_{12} & ... & c_{1n} \\ c_{21} & c_{22} & ... & c_{2n} \\ ... & ... & ... & ... \\ c_{n1} & c_{n2} & ... & c_{nn} \end{pmatrix} \tag{11.2}\]

Now we look at what happens when we apply a symmetry operation g to our two basis sets. If \(\Gamma(g)\) and \(\Gamma'(g)\) are matrix representatives of the symmetry operation in the \(\textbf{x}\) and \(\textbf{x'}\) bases, then we have:

\[\begin{array}{rcll} g\textbf{x'} & = & \textbf{x'}\Gamma'(g) \\ g\textbf{x}C & = & \textbf{x}C\Gamma'(g) & \text{since} \: \textbf{x'} = \textbf{x}C \\ g\textbf{x} & = & \textbf{x}C\Gamma'(g)C^{-1} & \text{multiplying on the right by} \: C^{-1} \text{and using} \: CC^{-1} = I \\ & = & \textbf{x}\Gamma(g) \end{array} \tag{11.3}\]

We can therefore identify the similarity transform relating \(\Gamma(g)\), the matrix representative in our original basis, to \(\Gamma'(g)\), the representative in the transformed basis. The transform depends only on the matrix of coefficients used to transform the basis functions.

\[\Gamma(g) = C\Gamma'(g)C^{-1} \tag{11.4}\]

Also,

\[\Gamma'(g) = C^{-1}\Gamma(g)C \tag{11.5}\]

## Character Tables

A character table summarizes the behavior of all of the possible irreducible representations of a group under each of the symmetry operations of the group. The character table for \(C_{3v}\) is shown below.

\[\begin{array}{lllll} \hline C_{3v},3m & E & 2C_3 & 3\sigma_v & h=6 \\ \hline A_1 & 1 & 1 & 1 & z, z^2, x^2+y^2 \\ A_2 & 1 & 1 & -1 & R_z \\ E & 2 & -1 & 0 & \begin{pmatrix} x, y \end{pmatrix}, \begin{pmatrix} xy, x^2+y^2 \end{pmatrix}, \begin{pmatrix} xz, yz \end{pmatrix}, \begin{pmatrix} R_x, R_y \end{pmatrix} \\ \hline \end{array} \label{14.1}\]

The various sections of the table are as follows:

- The first element in the table gives the name of the point group, usually in both Schoenflies (\(C_{3v}\)) and Hermann-Mauguin (\(3m\)) notation.
- Along the first row are the symmetry operations of the group, \(E\), \(2C_3\) and \(3\sigma_v\), followed by the order of the group. Because operations in the same class have the same character, symmetry operations are grouped into classes in the character table and not listed separately.
- In the first column are the irreducible representations of the group. In \(C_{3v}\) the irreducible representations are \(A_1\), \(A_2\) and \(E\) (the representation we considered above spans \(2A_1\) + \(E\)).
- The characters of the irreducible representations under each symmetry operation are given in the bulk of the table.
- The final column of the table lists a number of functions that transform as the various irreducible representations of the group. These are the Cartesian axes \(\begin{pmatrix} x, y, z \end{pmatrix}\), the Cartesian products \(\begin{pmatrix} z^2, x^2 + y^2, xy, yz \end{pmatrix}\), and the rotations \(\begin{pmatrix} R_x, R_y, R_z \end{pmatrix}\).

The functions listed in the final column of the table are important in many chemical applications of group theory, particularly in spectroscopy. For example, by looking at the transformation properties of \(x\), \(y\) and \(z\) (sometimes given in character tables as \(T_x\), \(T_y\), \(T_z\)) we can discover the symmetry of translations along the \(x\), \(y\), and \(z\) axes. Similarly, \(R_x\), \(R_y\) and \(R_z\) represent rotations about the three Cartesian axes. As we shall see later, the transformation properties of \(x\), \(y\), and \(z\) can also be used to determine whether or not a molecule can absorb a photon of \(x\)-, \(y\)-, or \(z\)-polarized light and undergo a spectroscopic transition. The Cartesian products play a similar role in determining selection rules for Raman transitions, which involve two photons.

Character tables for common point groups are given in Appendix B.

## A simple way to determine the characters of a representation

In many applications of group theory, we only need to know the characters of the representative matrices, rather than the matrices themselves. Luckily, when each basis function transforms as a 1D irreducible representation (which is true in many cases of interest) there is a simple shortcut to determining the characters without having to construct the entire matrix representation. All we have to do is to look at the way the individual basis functions transform under each symmetry operation. For a given operation, step through the basis functions as follows:

- Add \(1\) to the character if the basis function is unchanged by the symmetry operation (i.e. the basis function is mapped onto itself);
- Add \(-1\) to the character if the basis function changes sign under the symmetry operation (i.e the basis function is mapped onto minus itself);
- Add \(0\) to the character if the basis function moves when the symmetry operation is applied (i.e the basis function is mapped onto something different from itself).

Try this for the \(s\) orbital basis we have been using for the \(C_{3v}\) group. You should find you get the same characters as we obtained from the traces of the matrix representatives.

We can also work out the characters fairly easily when two basis functions transform together as a 2D irreducible representation. For example, in the \(C_{3v}\) point group \(x\) and \(y\) axes transform together as \(E\). If we carry out a rotation about \(z\) by an angle \(\theta\), our \(x\) and \(y\) axes are transformed onto new axes \(x'\) and \(y'\). However, the new axes can each be written as a linear combination of our original \(x\) and \(y\) axes. Using the rotation matrices introduced in Section 9, we see that:

\[\begin{array}{ccc}x' & = & \cos\theta \: x + \sin\theta \: y \\ y' & = & -\sin\theta \: x + \cos\theta \: y \end{array} \label{14.2}\]

For one-dimensional irreducible representations we asked if a basis function/axis was mapped onto itself, minus itself, or something different. For two-dimensional irreducible representations we need to ask how much of the ‘old’ axis is contained in the new one. From the above we see that the \(x'\) axis contains a contribution \(\cos\)\(\theta\) from the \(x\) axis, and the \(y'\) axis contains a contribution \(\cos\)\(\theta\) from the \(y\) axis. The characters of the \(x\) and \(y\) axes under a rotation through \(\theta\) are therefore \(\cos\)\(\theta\), and the overall character of the \(E\) irreducible representation is therefore \(\cos\)\(\theta\) \(+\) \(\cos\)\(\theta\) \(= 2\) \(\cos\)\(\theta\). For a \(C_3\) rotation through 120 degrees, the character of the \(E\) irreducible representation is therefore \(2\cos120\)° \(=\) \(-1\).

In general, *when an axis is rotated by an angle **\(\theta\) by a symmetry operation, its contribution to the character for that operation is \(\cos\)\(\theta\)**.*

## Irreducible representations with complex characters

In many cases (see Appendix B), the characters for rotations \(C_n\) and improper rotations \(S_n\) are complex numbers, usually expressed in terms of the quantity \(\epsilon\) = exp(2\(\pi\)i/n). It is fairly straightforward to reconcile this with the fact that in chemistry we are generally using group theory to investigate physical problems in which all quantities are real. It turns out that whenever our basis spans an irreducible representation whose characters are complex, it will also span a second irreducible representation whose characters are the complex conjugates of the first irreducible representation i.e. complex irreducible representations occur in pairs. According to the strict mathematics of group theory, each irreducible representation in the pair should be considered as a separate representation. However, when applying such irreducible representations in physical problems, we add the characters for the two irreducible representations together to get a single irreducible representation whose characters are real.

As an example, the ‘correct’ character table for the group \(C_3\) takes the form:

\[\begin{array}{l|l} C_3 & E \: \: \: \: \: \: \: \: C_3 \: \: \: \: \: \: \: \: C_3^2 \\ \hline A & 1 \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: 1 \\ \hline E & \begin{Bmatrix} 1 & \epsilon & \epsilon* \\ 1 & \epsilon* & \epsilon \end{Bmatrix} \end{array} \label{14.3}\]

Where \(\epsilon\) = exp(2\(\pi\)i/3). However, as chemists we would usually combine the two parts of the \(E\) irreducible representation to give:

\[\begin{array}{l|lll} C_3 & E & C_3 & C_3^2 \\ \hline A & 1 & 1 & 1 \\ E & 2 & -1 & 1 \end{array} \label{14.4}\]

## Contributors

Claire Vallance (University of Oxford)