Group theory allows us to determine the symmetries of a molecule, enabling an efficient way to obtain relevant information in infrared (IR) spectroscopy.
The objective in this article is to use group theory to work out the symmetries of the normal modes of a molecule and then ascertain whether they are IRactive. As an example, let’s consider , which belongs to the point group. Here’s the corresponding character table:
Instead of using Cartesian displacement vectors to form a basis for generating representations of the point group, as we did in the previous article, we shall use massweighted Cartesian coordinate unit vectors to form a basis (see diagram above). Since a symmetry operation of transforms into an indistinguishable copy of itself, it transforms each of the elements of the basis set into a linear combination of elements of the set (see eq55 to eq61). Therefore, centred on every atom of generate a representation of the point group. To find the matrices of the representation, we arrange the vectors into a row vector and apply the symmetry operations of the point group on it to obtain the transformed vector. The matrices of the representation are then constructed by inspection. For example,
The traces of the matrices are
Using eq27a and the character table of the point group, we have
which implies that the decomposition of the reducible representation is .
Applying the projection operator on each element of the basis set for all irreducible representations, we have
Each equation in the above table is a basis of the irreducible representation it belongs to. Since any linear combination of bases that transform according to a onedimensional irreducible representation is also a basis of that irreducible representation, we can generate symmetryadapted linear combinations (SALC) from the above basis vectors that belong to their respective irreducible representations.
We have shown in an earlier article that

 the number of orthogonal or orthonormal basis functions of a representation corresponds to the dimension of the representation.
 if is a basis of a representation of a group, then any linear combination of is a basis of a representation that is equivalent to .
This implies that we can always select a set of nine orthonormal SALC for a block diagonal representation that is equivalent to . Since none of the nine SALCs can be expressed as a linear combination of the others, each SALC, known as normal coordinates , must describe one of nine independent motions of the molecule.
A quick method to assign the irreducible representations from the decomposition of to the nine degrees of freedom of is to refer to the character table, where the basis functions , and represent the three independent translational motion and , and represent the three independent rotational motion. Deducting the irreducible representations associated with these six basis functions from the direct sum of , we are left with , and . These remaining three irreducible representations correspond to the vibrational degrees of freedom.
Question
Does the basis function in the character table of the point group correspond to the normal coordinate for ?
Answer
Yes. In fact, there is a correspondence between the basis functions and six of the normal coordinates of . The three normal coordinates describing the vibrational motions of are known as normal modes.
To verify whether the three normal modes are IRactive, we refer to the Schrodinger equation for vibration motion (see eq93):
where the separation of variables technique allows us to approximate the total vibrational wavefunction as and hence, .
Each describes a normal mode of the molecule and has the formula (see eq94):
where is the normalisation constant for the Hermite polynomials .
The total vibrational energy (see eq96) is
A vibrational state of a polyatomic molecule is characterised by quantum numbers. Hence, the vibrational ground state of , where , is or
Many IRspectroscopy experiments are conducted at room temperature, where most molecules are in their vibrational ground state. According to the timedependent perturbation theory, the transition probability between orthogonal vibrational states within a given electronic state of a molecule is proportional to
where and are the initial and final states respectively and is the operator for the molecule’s electric dipole moment.
In other words, no transition between states occurs when . Since the objective is to ascertain the IR activity of each of the three normal modes of , it is suffice to study the fundamental transitions, where and or or .
The next step involves determining which irreducible representation of the point group , , , and belong to. Since the components , and represent the electric dipole moment along the , and directions respectively, they transform in the same way as the basis functions , and respectively (see character table above).
Question
Show that , where represents the symmetry operations of a point group and .
Answer
In general, if , then the function is invariant under the symmetry operation . Every value of is mapped into by and we can say that if . The converse is also true, i.e. if the variable of the function transforms according to , then . From this article, the potential term of the vibrational Hamiltonian for is . Since the function is invariant under any symmetry operation, or simply
If each is distinct, then or
If is degenerate, eq55 states that and eq105 becomes
where we have assumed that and are degenerate.
As described in this article, forms a set of orthonormal eigenvectors of the vibrational Hamiltonian. So, and are orthonormal vectors. The matrix formed by using these vectors as its columns is an orthogonal matrix , which is defined as . This is because the dot product of different columns will be zero, and the dot product of a column with itself will be 1. Since if is an orthogonal matrix (see property 11 of this link for proof), or
which implies that and .
The first two terms on LHS of eq107 then becomes . This implies that must also be equal to in eq105 if is degenerate. Therefore, regardless of whether is nondegenerate or degenerate.
As mentioned in the above Q&A, if the variable of the function transforms according to , then . It follows that if the variable of the function transforms according to , then , in which case is the reflection operator about the vertical axis of the graph of against . Therefore, the ground state vibrational wave function of transforms according to the totally symmetric irreducible representation of the point group because
Next, let’s analyse the symmetries of , and . These three wave functions (see this article) have the same form of:
Eq108 is a product of two functions and . Since is totally symmetric, the theory of direct product representation states that must transform according to the irreducible representation of the point group that belongs to.
With reference to the character table above, the theory of direct product representation again finds that the functions , and in eq104 transform according to the irreducible representations of , and respectively. Therefore, the theory of vanishing integrals states that
Zero  Not zero  
Zero  Zero  
Not zero  Zero 
Since in eq104 for each of the three normal modes, all three normal modes of are IRactive.