16.02.02 · inorgchem / symmetry-group-theory

Character tables and reducible representations: vibrational modes and spectroscopy

stub3 tiersLean: nonepending prereqs

Anchor (Master): Cotton — Chemical Applications of Group Theory, 3e; Bishop — Group Theory and Chemistry (1973)

Intuition Beginner

Every molecule belongs to a point group, and every point group has a character table that summarises how the molecule's properties behave under symmetry operations. The character table lists rows called irreducible representations (irreps) — the fundamental symmetry types. Every molecular property (an orbital, a vibration, a transition) transforms as one of these irreps or a combination of them.

A molecule with N atoms can vibrate in 3N minus 6 independent ways (3N minus 5 if it is linear). Each vibrational mode has a symmetry label from the character table. The label tells you whether the vibration is IR-active (absorbs infrared light) or Raman-active (scatters light with a frequency shift). This is powerful: from the character table alone, without solving the Schrodinger equation, you can predict how many IR bands and Raman peaks a molecule will produce.

The practical recipe has three steps. First, assign the point group. Second, build the reducible representation for the 3N atomic displacements and decompose it into irreps using the character table. Third, subtract translations and rotations; the remaining irreps are the vibrational modes. Match each mode against the symmetry of the dipole operator (x, y, z) for IR activity, and against the polarisability tensor (quadratic functions like $x^{2}$ , $x y$ , etc.) for Raman activity.

Visual Beginner

A character table for the C $_{2 v}$ point group (water, formaldehyde) arranged as a grid. The columns are the symmetry operations E, C $_{2}$ , $σ_{v}$ , $σ_{v}^{'}$ . Each row is an irrep (A $_{1}$ , A $_{2}$ , B $_{1}$ , B $_{2}$ ) with its characters and the functions that transform as that irrep — Cartesian coordinates for IR selection rules and quadratic functions for Raman selection rules.

Worked example Beginner

Determine the IR-active and Raman-active vibrations of water (H $_{2}$ O, C $_{2 v}$ ).

Water has 3 atoms, so it has 3(3) - 6 = 3 vibrational modes. The C $_{2 v}$ character table lists four irreps:

C $_{2 v}$	E	C $_{2}$	$σ_{v}$ (xz)	$σ_{v}^{'}$ (yz)
A $_{1}$	1	1	1	1	z	$x^{2}, y^{2}, z^{2}$
A $_{2}$	1	1	-1	-1	$R_{z}$	$x y$
B $_{1}$	1	-1	1	-1	x, $R_{y}$	$x z$
B $_{2}$	1	-1	-1	1	y, $R_{x}$	$y z$

The three vibrational modes of water transform as 2A $_{1}$ + B $_{1}$ . To check IR activity: A $_{1}$ matches z (so both A $_{1}$ modes are IR-active along z), and B $_{1}$ matches x (so the B $_{1}$ mode is IR-active along x). All three modes are IR-active.

For Raman activity: A $_{1}$ matches $x^{2}, y^{2}, z^{2}$ and B $_{1}$ matches $x z$ . All three modes are Raman-active. Since water has no inversion centre, there is no mutual exclusion between IR and Raman — both can observe the same modes.

Check your understanding Beginner

Exercise (easy, multiple choice).

A molecule with an inversion centre has a vibrational mode transforming as A $_{g}$ (gerade). This mode is:

A. IR-active only B. Raman-active only C. Both IR-active and Raman-active D. Neither IR-active nor Raman-active

Hint

The IR dipole operator transforms as ungerade (u). The Raman polarisability tensor transforms as gerade (g).

Answer

B. In a centrosymmetric molecule, the dipole operator (x, y, z) is ungerade. For an IR transition to be allowed, the direct product of the ground and excited state irreps must match a u irrep. Since A $_{g}$ $\otimes$ A $_{g}$ = A $_{g}$ (gerade), no dipole component matches — the mode is IR-inactive. The polarisability tensor components ( $x^{2}$ , $x y$ , etc.) are gerade, so A $_{g}$ modes are Raman-active. This is the mutual exclusion rule.

Formal definition Intermediate+

Character tables

For a point group $G$ with conjugacy classes $C_{1}, \dots, C_{k}$ and irreducible representations $Γ_{1}, \dots, Γ_{k}$ , the character table is the $k \times k$ matrix $χ_{i} (C_{j})$ where $χ_{i}$ is the character (trace of the representation matrix) of irrep $Γ_{i}$ on class $C_{j}$ . The rows satisfy the great orthogonality theorem:

g \in G \sum χ_{i} (g)^{*} χ_{j} (g) = ∣ G ∣ δ_{ij}

Equivalently, summing over classes:

C \sum \frac{∣ C ∣}{∣ G ∣} χ_{i} (C)^{*} χ_{j} (C) = δ_{ij}

The character table also lists the functions that transform as each irrep: Cartesian coordinates (x, y, z) for IR selection rules and rotations ( $R_{x}$ , $R_{y}$ , $R_{z}$ ), plus quadratic functions ( $x^{2}$ , $x y$ , $x z$ , etc.) for Raman selection rules.

Reducible representations and the reduction formula

A reducible representation $Γ_{red}$ is a representation that can be decomposed into a direct sum of irreps. The multiplicity of irrep $Γ_{i}$ in the decomposition is given by the reduction formula:

n_{i} = \frac{1}{∣ G ∣} C \sum ∣ C ∣ χ_{red} (C) \overline{χ_{i} (C)}

where $∣ C ∣$ is the number of elements in conjugacy class $C$ , $χ_{red} (C)$ is the character of the reducible representation on class $C$ , and $χ_{i} (C)$ is the character of irrep $Γ_{i}$ on class $C$ .

Vibrational mode analysis

For a molecule with N atoms, the 3N Cartesian displacement vectors form a basis for a reducible representation $Γ_{3 N}$ of the point group. The character of $Γ_{3 N}$ on each symmetry operation $R$ is computed as:

χ_{3 N} (R) = N_{unmoved} (R) \times χ_{cart} (R)

where $N_{unmoved} (R)$ is the number of atoms left in place by operation $R$ , and $χ_{cart} (R)$ is the character of the Cartesian transformation matrix for $R$ :

Proper rotation by angle $θ$ : $χ_{cart} = 1 + 2 cos θ$
Improper rotation (rotation by $θ$ plus reflection): $χ_{cart} = - 1 + 2 cos θ$
Identity: $χ_{cart} = 3$

The vibrational representation is obtained by subtracting translations and rotations:

Γ_{vib} = Γ_{3 N} - Γ_{trans} - Γ_{rot}

For a non-linear molecule, $Γ_{trans}$ accounts for 3 translational degrees of freedom and $Γ_{rot}$ accounts for 3 rotational degrees of freedom, giving $3 N - 6$ vibrational modes. For a linear molecule, only 2 rotational degrees of freedom exist, giving $3 N - 5$ modes.

IR and Raman selection rules

IR selection rule. A vibrational mode transforming as irrep $Γ_{k}$ is IR-active if $Γ_{k}$ appears among the irreps spanned by the Cartesian coordinates (x, y, z). Physically, the vibration must change the molecule's dipole moment.

Raman selection rule. A vibrational mode is Raman-active if $Γ_{k}$ appears among the irreps spanned by the polarisability tensor components ( $x^{2}$ , $y^{2}$ , $z^{2}$ , $x y$ , $x z$ , $y z$ , $x^{2} - y^{2}$ , etc.). Physically, the vibration must change the molecule's polarisability.

Mutual exclusion rule. For a centrosymmetric molecule (one possessing an inversion centre), no vibrational mode can be both IR-active and Raman-active. The IR dipole operator is ungerade (u) and the Raman polarisability tensor is gerade (g), so their activity conditions are mutually exclusive.

Counterexamples to common slips

Having the same irrep label does not guarantee both IR and Raman activity in non-centrosymmetric groups. In C $_{3 v}$ , both z (A $_{1}$ ) and $x^{2} + y^{2}, z^{2}$ (A $_{1}$ ) appear, so A $_{1}$ modes are both IR- and Raman-active. But this is because the specific functions match, not because of a general rule.
The mutual exclusion rule is not "all modes are either IR or Raman active." Some modes may be both (in non-centrosymmetric molecules) and some may be neither (silent modes). A vibration transforming as A $_{2}$ in C $_{2 v}$ is neither IR-active (A $_{2}$ does not match x, y, or z) nor Raman-active (A $_{2}$ matches only $x y$ , but the relevant quadratic functions are $x^{2}, y^{2}, z^{2}, x z, y z$ for the polarisability components — A $_{2}$ appears as $x y$ in the table and is Raman-active). Actually, A $_{2}$ does match $x y$ and is Raman-active. Silent modes are rare but do occur in higher-symmetry groups.
Character of 0 does not mean the atom is unmoved. A character of zero in the displacement representation means the total trace of the transformation matrix for all atoms is zero. Some atoms may move while contributing zero trace individually (e.g., a rotation that swaps atoms).

Key theorem with proof Intermediate+

Theorem (Reduction of the displacement representation). Let $G$ be the point group of a molecule with $N$ atoms. The $3 N$ -dimensional displacement representation $Γ_{3 N}$ decomposes as $Γ_{3 N} = Γ_{vib} \oplus Γ_{trans} \oplus Γ_{rot}$ , where $Γ_{trans}$ spans the same irreps as the Cartesian coordinates (x, y, z), $Γ_{rot}$ spans the same irreps as the rotation axes ( $R_{x}$ , $R_{y}$ , $R_{z}$ ), and $Γ_{vib}$ contains the symmetry species of the $3 N - 6$ vibrational modes (or $3 N - 5$ for linear molecules).

Proof. The 3N displacement coordinates form a basis for a representation of $G$ . By Maschke's theorem, this representation decomposes into a direct sum of irreps. The displacement space splits into three subspaces invariant under $G$ : translations (3 dimensions, moving all atoms uniformly along x, y, or z), rotations (3 dimensions for non-linear molecules, 2 for linear, rotating the whole molecule about $R_{x}$ , $R_{y}$ , $R_{z}$ ), and vibrations (the orthogonal complement).

Translations transform identically to the Cartesian coordinates because a uniform translation by a vector $v$ transforms under $R$ as $R v$ , the same as the position vector. Hence $Γ_{trans}$ spans the irreps listed for (x, y, z) in the character table.

Rotations transform as the axial vectors $R_{x}$ , $R_{y}$ , $R_{z}$ , which are also listed in the character table. For a linear molecule, rotation about the molecular axis carries no angular momentum ( $R_{z} = 0$ ), so only two rotational degrees of freedom exist.

The vibrational subspace is the orthogonal complement: $Γ_{vib} = Γ_{3 N} - Γ_{trans} - Γ_{rot}$ , computed irrep-by-irrep using the reduction formula. The dimension of the vibrational subspace is $3 N - 6$ (or $3 N - 5$ for linear molecules), matching the standard vibrational mode count. $□$

Worked example: vibrational modes of BF $_{3}$ (D $_{3 h}$ ).

BF $_{3}$ has 4 atoms, so $Γ_{3 N}$ has dimension 12. Characters of the displacement representation:

D $_{3 h}$	E	2C $_{3}$	3C $_{2}$	$σ_{h}$	2S $_{3}$	3 $σ_{v}$
$N_{unmoved}$	4	1	2	4	1	2
$χ_{cart}$	3	0	-1	1	-2	1
$χ_{3 N}$	12	0	-2	4	-2	2

Decompose using $∣ G ∣ = 12$ :

$n_{A_{1}^{'}} = \frac{1}{12} (12 - 0 - 2 + 4 - 2 + 2) = \frac{14}{12} \approx 1.17$ ... Let us recalculate carefully.

$n_{A_{1}^{'}} = \frac{1}{12} [1 (12) (1) + 2 (0) (1) + 3 (- 2) (1) + 1 (4) (1) + 2 (- 2) (1) + 3 (2) (1)]$ $= \frac{1}{12} [12 + 0 - 6 + 4 - 4 + 6] = \frac{12}{12} = 1$

$n_{A_{2}^{'}} = \frac{1}{12} [12 (1) + 0 + 3 (- 2) (1) + 4 (1) + 2 (- 2) (1) + 3 (2) (- 1)]$ $= \frac{1}{12} [12 + 0 - 6 + 4 - 4 - 6] = \frac{0}{12} = 0$

$n_{E^{'}} = \frac{1}{12} [12 (2) + 0 + 3 (- 2) (- 1) + 4 (2) + 2 (- 2) (- 1) + 3 (2) (1)]$ $= \frac{1}{12} [24 + 0 + 6 + 8 + 4 + 6] = \frac{48}{12} = 4$

$n_{A_{1}^{''}} = \frac{1}{12} [12 (1) + 0 + 3 (- 2) (1) + 4 (- 1) + 2 (- 2) (1) + 3 (2) (- 1)]$ $= \frac{1}{12} [12 + 0 - 6 - 4 - 4 - 6] = \frac{- 8}{12}$

This negative value indicates a computational error. Let us use the correct D $_{3 h}$ character table values and recompute. The D $_{3 h}$ character table:

D $_{3 h}$	E	2C $_{3}$	3C $_{2}$	$σ_{h}$	2S $_{3}$	3 $σ_{v}$
$A_{1}^{'}$	1	1	1	1	1	1
$A_{2}^{'}$	1	1	-1	1	1	-1
$E^{'}$	2	-1	0	2	-1	0
$A_{1}^{''}$	1	1	1	-1	-1	-1
$A_{2}^{''}$	1	1	-1	-1	-1	1
$E^{''}$	2	-1	0	-2	1	0

Now decomposing $χ_{3 N} = (12, 0, - 2, 4, - 2, 2)$ :

$n_{A_{1}^{'}} = \frac{1}{12} [1 (12) + 2 (0) + 3 (- 2) + 1 (4) + 2 (- 2) + 3 (2)] = \frac{12 + 0 - 6 + 4 - 4 + 6}{12} = 1$

$n_{A_{2}^{'}} = \frac{1}{12} [12 + 0 + 3 (- 2) (- 1) + 4 + 2 (- 2) + 3 (2) (- 1)] = \frac{12 + 0 + 6 + 4 - 4 - 6}{12} = 1$

$n_{E^{'}} = \frac{1}{12} [12 (2) + 0 + 0 + 4 (2) + 2 (- 2) (- 1) + 0] = \frac{24 + 0 + 0 + 8 + 4 + 0}{12} = 3$

$n_{A_{1}^{''}} = \frac{1}{12} [12 + 0 + 3 (- 2) + 4 (- 1) + 2 (- 2) (- 1) + 3 (2) (- 1)] = \frac{12 + 0 - 6 - 4 + 4 - 6}{12} = 0$

$n_{A_{2}^{''}} = \frac{1}{12} [12 + 0 + 3 (- 2) (- 1) + 4 (- 1) + 2 (- 2) (- 1) + 3 (2) (1)] = \frac{12 + 0 + 6 - 4 + 4 + 6}{12} = 2$

$n_{E^{''}} = \frac{1}{12} [12 (2) + 0 + 0 + 4 (- 2) + 2 (- 2) (1) + 0] = \frac{24 + 0 + 0 - 8 - 4 + 0}{12} = 1$

So $Γ_{3 N} = A_{1}^{'} + A_{2}^{'} + 3 E^{'} + 2 A_{2}^{''} + E^{''}$ . Subtracting translations ( $(x, y) \to E^{'}$ , $z \to A_{2}^{''}$ ) and rotations ( $R_{z} \to A_{2}^{'}$ , $(R_{x}, R_{y}) \to E^{''}$ ):

Γ_{vib} = A_{1}^{'} + 2 E^{'} + A_{2}^{''}

The 6 vibrational modes decompose as: one $A_{1}^{'}$ (symmetric stretch, Raman-active only since $(x^{2} + y^{2}, z^{2})$ are $A_{1}^{'}$ but no Cartesian coordinate is), two $E^{'}$ pairs (both IR- and Raman-active since $(x, y)$ and $(x^{2} - y^{2}, x y)$ are $E^{'}$ ), and one $A_{2}^{''}$ (IR-active only since $z$ is $A_{2}^{''}$ but no quadratic function is). This gives 4 IR-active modes and 5 Raman-active modes (the $A_{1}^{'}$ and two $E^{'}$ pairs), with mutual exclusion operating on the $A_{1}^{'}$ / $A_{2}^{''}$ pair.

Bridge. The decomposition of $Γ_{3 N}$ into $Γ_{vib} + Γ_{trans} + Γ_{rot}$ is the central computational technique that connects the abstract character table to experimental spectroscopy. This technique builds directly toward 16.02.03 pending, where the projection operator extracts explicit SALC basis functions from the same decomposition, and it underpins the crystal-field splitting analysis in 16.03.01 where the d-orbital reducible representation of O $_{h}$ is decomposed into $E_{g} + T_{2 g}$ .

Exercises Intermediate+

Exercise 2 (easy, numeric).

In C $_{2 v}$ , the reducible representation has characters $χ = (6, 0, 2, 2)$ under (E, C $_{2}$ , $σ_{v}$ , $σ_{v}^{'}$ ). Compute $n_{A_{1}}$ .

Hint

Use the reduction formula: $n_{A_{1}} = \frac{1}{∣ G ∣} \sum_{C} ∣ C ∣ χ_{red} (C) \overline{χ_{A_{1}} (C)}$ . In C $_{2 v}$ , every class has one element and A $_{1}$ has characters (1, 1, 1, 1).

Answer

2. $n_{A_{1}} = \frac{1}{4} [1 (6) (1) + 1 (0) (1) + 1 (2) (1) + 1 (2) (1)] = \frac{10}{4}$ ... Wait, let me recompute: $\frac{6 + 0 + 2 + 2}{4} = \frac{10}{4} = 2.5$ . Since multiplicities must be integers, the input characters are inconsistent with C $_{2 v}$ . If the characters were $χ = (6, 0, 0, 2)$ : $n_{A_{1}} = \frac{6 + 0 + 0 + 2}{4} = 2$ . The correct answer depends on the exact character vector. For $χ = (6, 0, 2, 2)$ : the decomposition gives $n_{A_{1}} = \frac{6 + 0 + 2 + 2}{4} = 2.5$ , which is non-integer, indicating this is not a valid representation of C $_{2 v}$ .

With corrected characters $χ = (6, 0, 0, 2)$ : $n_{A_{1}} = \frac{6 + 0 + 0 + 2}{4} = 2$ .

Exercise 3 (medium, short-answer).

For XeF $_{4}$ (square planar, D $_{4 h}$ , 5 atoms), determine the symmetry species of the vibrational modes and identify which are IR-active and which are Raman-active.

Hint

XeF $_{4}$ has 5 atoms, so 3(5) - 6 = 9 vibrational modes. Build $Γ_{3 N}$ by counting unmoved atoms and computing the Cartesian trace for each class of D $_{4 h}$ .

Answer

XeF $_{4}$ has 5 atoms, giving 9 vibrational modes. The displacement representation $Γ_{3 N}$ has characters computed from unmoved atoms and Cartesian traces:

E: 5 unmoved, $χ = 15$
2C $_{4}$ : 1 unmoved (Xe only), $χ = 1 (1) = 1$
C $_{2}$ : 1 unmoved, $χ = 1 (- 1) = - 1$
2C $_{2}^{'}$ : 3 unmoved (Xe + 2 opposite F on the axis), $χ = 3 (- 1) = - 3$
2C $_{2}^{''}$ : 1 unmoved (Xe only), $χ = 1 (- 1) = - 1$
i: 1 unmoved, $χ = 1 (- 3) = - 3$
2S $_{4}$ : 1 unmoved, $χ = 1 (- 1) = - 1$
$σ_{h}$ : 5 unmoved, $χ = 5 (1) = 5$
2 $σ_{v}$ : 3 unmoved, $χ = 3 (1) = 3$
2 $σ_{d}$ : 1 unmoved, $χ = 1 (1) = 1$

Decomposition gives $Γ_{3 N} = A_{1 g} + A_{2 g} + B_{1 g} + B_{2 g} + E_{g} + 2 A_{2 u} + B_{2 u} + 2 E_{u}$ .

Subtracting translations ( $z \to A_{2 u}$ , $(x, y) \to E_{u}$ ) and rotations ( $R_{z} \to A_{2 g}$ , $(R_{x}, R_{y}) \to E_{g}$ ):

Γ_{vib} = A_{1 g} + B_{1 g} + B_{2 g} + A_{2 u} + B_{2 u} + E_{u}

IR-active: $A_{2 u}$ (matches z) and $E_{u}$ (matches x, y). Two IR bands expected.

Raman-active: $A_{1 g}$ (matches $x^{2} + y^{2}, z^{2}$ ), $B_{1 g}$ (matches $x^{2} - y^{2}$ ), $B_{2 g}$ (matches $x y$ ). Three Raman bands expected.

$B_{2 u}$ is neither IR-active nor Raman-active — a silent mode. The mutual exclusion rule is obeyed: no mode appears in both lists.

Exercise 4 (medium, short-answer).

Prove the mutual exclusion rule: in a centrosymmetric point group, no vibrational mode can be both IR-active and Raman-active.

Hint

The dipole operator $r = (x, y, z)$ is ungerade (u). The polarisability tensor components $r_{i} r_{j}$ are gerade (g). What does this imply for the symmetry of the transition integrand?

Answer

In a centrosymmetric point group, every irrep carries a g or u label. The ground vibrational state is always totally symmetric ( $A_{g}$ ).

IR activity requires $⟨ A_{g} ∣ r_{α} ∣ Γ_{k} ⟩ \neq = 0$ . The integrand transforms as $A_{g} \otimes Γ_{u} \otimes Γ_{k} = Γ_{u} \otimes Γ_{k}$ . This contains $A_{g}$ only if $Γ_{k}$ is ungerade (u). So IR-active modes must be u.

Raman activity requires $⟨ A_{g} ∣ r_{α} r_{β} ∣ Γ_{k} ⟩ \neq = 0$ . The operator $r_{α} r_{β}$ is gerade (product of two u functions). The integrand transforms as $A_{g} \otimes Γ_{g} \otimes Γ_{k} = Γ_{g} \otimes Γ_{k}$ . This contains $A_{g}$ only if $Γ_{k}$ is gerade (g). So Raman-active modes must be g.

A mode is either g or u, never both. Hence IR and Raman activity are mutually exclusive.

Exercise 5 (hard, short-answer).

For CH $_{4}$ (T $_{d}$ , 5 atoms), the vibrational modes decompose as $Γ_{vib} = A_{1} + E + 2 T_{2}$ . Determine which modes are IR-active and which are Raman-active. Explain why the mutual exclusion rule does not apply.

Hint

T $_{d}$ has no inversion centre. Check which irreps match the Cartesian coordinates and the quadratic functions in the T $_{d}$ character table.

Answer

In T $_{d}$ , the Cartesian coordinates (x, y, z) transform as $T_{2}$ . The quadratic functions transform as: $A_{1}$ ( $x^{2} + y^{2} + z^{2}$ ), $E$ ( $2 z^{2} - x^{2} - y^{2}, x^{2} - y^{2}$ ), and $T_{2}$ ( $x y, x z, y z$ ).

IR-active modes: Both $T_{2}$ modes (matching x, y, z). Two IR absorption bands expected.

Raman-active modes: $A_{1}$ (matching $x^{2} + y^{2} + z^{2}$ ), $E$ (matching quadratic functions), and both $T_{2}$ modes (matching $x y, x z, y z$ ). Four Raman bands expected.

Both $T_{2}$ vibrational modes are simultaneously IR- and Raman-active. This does not violate any rule because T $_{d}$ lacks an inversion centre — the mutual exclusion rule does not apply. Methane has no g/u labels on its irreps.

Exercise 6 (hard, short-answer).

Derive the characters of $Γ_{3 N}$ for a general proper rotation $C_{n}$ by angle $θ$ and a general improper rotation $S_{n}$ . Show that the formula $χ_{3 N} (R) = N_{unmoved} (R) \times (1 + 2 cos θ)$ holds for proper rotations and $χ_{3 N} (R) = N_{unmoved} (R) \times (- 1 + 2 cos θ)$ holds for improper rotations.

Hint

Write the 3D transformation matrix for each type of operation and compute its trace. Only atoms that remain in place contribute.

Answer

A proper rotation $C_{n}$ by angle $θ$ about the z-axis has the 3D matrix:

R = cos θ sin θ 0 - sin θ cos θ 0 001, tr (R) = 1 + 2 cos θ

An improper rotation $S_{n}$ adds a reflection through the xy-plane (inversion of z):

S = cos θ sin θ 0 - sin θ cos θ 0 00 - 1, tr (S) = - 1 + 2 cos θ

For the displacement representation, each atom unmoved by the operation contributes the full Cartesian trace to $χ_{3 N}$ . Atoms that move to a different position contribute zero (their displacement vectors are mapped to another atom's position, giving zero diagonal entries). Therefore:

χ_{3 N} (C_{n}) = N_{unmoved} \times (1 + 2 cos θ)

χ_{3 N} (S_{n}) = N_{unmoved} \times (- 1 + 2 cos θ)

Special cases: identity E ( $θ = 0$ , proper): $χ = N_{unmoved} \times 3$ . Reflection $σ$ ( $θ = 0$ , improper): $χ = N_{unmoved} \times 1$ . Inversion i ( $θ = π$ , improper): $χ = N_{unmoved} \times (- 3)$ .

Correlation tables, double groups, and factor-group analysis Master

Correlation tables and descent in symmetry

When a molecule's symmetry is lowered from group $G$ to a subgroup $H \subset G$ , each irrep of $G$ branches into one or more irreps of $H$ . The correlation table lists these branchings. This is essential for analysing distortions, substituted derivatives, and adsorption on surfaces.

For example, when O $_{h}$ descends to D $_{4 h}$ (tetragonal distortion), the correlation table gives:

O $_{h}$	D $_{4 h}$
A $_{1 g}$	A $_{1 g}$
E $_{g}$	A $_{1 g}$ + B $_{1 g}$
T $_{2 g}$	B $_{2 g}$ + E $_{g}$
T $_{1 u}$	A $_{2 u}$ + E $_{u}$

The two-fold degenerate E $_{g}$ irrep splits into two non-degenerate irreps (A $_{1 g}$ + B $_{1 g}$ ) in the lower-symmetry group. The three-fold T $_{2 g}$ splits into one non-degenerate and one two-fold degenerate irrep (B $_{2 g}$ + E $_{g}$ ). This splitting of degeneracies upon symmetry lowering is the group-theoretic basis for the additional spectral lines observed in distorted complexes.

Correlation tables are constructed by comparing characters. An irrep $Γ_{i}$ of $G$ restricts to a representation of $H$ (by evaluating $χ_{i}$ only on elements of $H$ ), which is then decomposed into irreps of $H$ using the reduction formula. The restriction is well-defined because $H$ is a subgroup and the character is defined for all elements of $H$ .

Double groups for spin-orbit coupling

When spin-orbit coupling is significant (heavy atoms, 4d and 5d transition metals, lanthanides, actinides), the single-valued representations of the point group are insufficient. The electron spin introduces a two-component wavefunction that transforms under a double group $G^{'}$ , obtained by adjoining a rotation by $2 π$ (denoted $\overset{ˉ}{E}$ , distinct from the identity E which corresponds to rotation by $4 π$ ). The double group has additional irreducible representations called double-valued or spinor representations.

For example, the O $_{h}$ double group O $_{h}^{'}$ has 8 irreducible representations (compared to 10 for O $_{h}$ , because some classes merge). The double-valued irreps include E $_{1/2 g}$ (often written E $_{g}^{'}$ or $Γ_{6}$ ), G $_{g}$ (often $Γ_{8}$ ), and their ungerade counterparts. These labels appear in the term symbols of transition-metal ions with strong spin-orbit coupling and determine the splitting patterns of spin-orbit-coupled states.

The practical impact: in the presence of spin-orbit coupling, the selection rules for electronic transitions must use the double group irreps. A transition that is spin-forbidden in the single group may gain intensity in the double group if spin-orbit coupling mixes states of different spin multiplicity. This is why heavy-atom complexes show more intense "spin-forbidden" bands than light-atom complexes.

Site symmetry and factor-group analysis in crystals

In a molecular crystal, each molecule occupies a crystallographic site whose local symmetry is described by the site-symmetry group $S$ — the subgroup of the full space group that leaves the molecular position fixed (possibly mapping the molecule onto itself). The relationship between the molecular point group $G$ , the site-symmetry group $S$ , and the crystal factor group $F$ determines how the molecular vibrations split in the solid state.

Factor-group analysis proceeds in three steps:

Correlate the molecular point-group irreps to the site-symmetry group irreps (descent in symmetry: $G \to S$ ).
Induce the site-symmetry irreps to the factor-group irreps (ascent in symmetry: $S \to F$ ).
The resulting factor-group irreps give the symmetry species of the crystal vibrations, including both internal modes (molecular vibrations) and external modes (translations and librations of the molecular units).

The number of factor-group components for each molecular mode equals the number of molecules per primitive unit cell (the number of equivalent positions in the Wyckoff site), provided the molecular mode is nondegenerate. Degenerate modes may split further upon correlation.

For a crystal with two molecules per primitive cell related by inversion, each molecular vibration splits into two factor-group components: one Davydov component (in-phase between the two molecules, potentially Raman-active) and one anti-Davydov component (out-of-phase, potentially IR-active). The splitting energy — the Davydov splitting or correlation field splitting — depends on the intermolecular coupling strength and is typically 1–20 cm $^{- 1}$ for molecular crystals.

This analysis explains why solid-state IR and Raman spectra typically show more bands than solution-phase spectra. The crystal environment lifts degeneracies and activates modes that are silent in the isolated molecule. Factor-group analysis predicts the number and symmetry of these additional bands from the crystal structure alone.

Symmetry-adapted linear combinations and projection operators

The decomposition of $Γ_{vib}$ into irreps tells you how many SALCs of each symmetry type exist. The projection operator constructs them explicitly:

P^{Γ} = \frac{d _{Γ}}{∣ G ∣} g \in G \sum \overline{χ_{Γ} (g)} \hat{R}_{g}

where $d_{Γ}$ is the dimension of irrep $Γ$ and $\hat{R}_{g}$ is the symmetry operation applied to the basis function. Applying $P^{Γ}$ to any basis function produces either zero or a function of symmetry $Γ$ . For higher-dimensional irreps, a second projection (using a transfer operator or applying a non-commuting symmetry operation to the first result) generates the partner functions.

The SALCs are the symmetry-correct basis functions for constructing vibrational wavefunctions and molecular orbitals. Their symmetry labels match the irreps obtained from the decomposition, providing a constructive verification of the abstract character-table calculation.

Connections Master

To quantum mechanics. The character-table machinery is a finite-group shadow of the representation theory of Lie groups that pervades quantum mechanics. The selection rules derived here (IR activity requires the integrand to contain A $_{1}$ ) are exactly the same Wigner-Eckart machinery that governs angular momentum coupling, transition rates in atomic physics, and tensor operator matrix elements. The reduction formula is the character-theoretic version of Clebsch-Gordan decomposition.

To crystallography. The 32 crystallographic point groups are precisely the point groups compatible with translational periodicity (the crystallographic restriction, proved in 16.02.01). Factor-group analysis connects the molecular vibrational spectrum to the solid-state spectrum through the space group. The 230 space groups are the semidirect products of the 14 Bravais lattices with the 32 point groups, and the correlation between molecular and crystal spectra is mediated by the site-symmetry subgroup chain.

To molecular orbital theory. The SALCs generated by the projection operator are the ligand-group orbitals that combine with central-atom orbitals to form molecular orbitals. The symmetry matching between SALCs and metal orbitals (A $_{1}$ SALC with s, E SALCs with d $_{z^{2}}$ and d $_{x^{2} - y^{2}}$ , T $_{1 u}$ SALCs with p orbitals) is the basis for the molecular orbital diagrams of coordination complexes developed in 16.03.01 and 16.05.01.

To spectroscopy. The IR and Raman selection rules derived here are the foundation for interpreting vibrational spectra. The mutual exclusion rule provides a quick diagnostic for centrosymmetry: if a molecule's IR and Raman spectra show no common bands, the molecule has an inversion centre. The number of IR-active and Raman-active bands predicted by the character table should match the observed spectrum — discrepancies indicate structural errors or phase transitions.

Historical notes Master

The systematic use of character tables in chemistry originates with Bethe's 1929 paper on crystal field splitting ("Termaufspaltung in Kristallen," Ann. Phys. 3, 133–208, 1929), which applied point-group character tables to predict the splitting of atomic energy levels in a crystal environment. Bethe's work was purely theoretical — he was interested in the quantum mechanics of ions in crystals, not in molecular spectroscopy.

Wigner extended group-theoretic methods to molecular vibrations in the 1930s, showing that the symmetry of normal modes is determined by the decomposition of the displacement representation into irreps. Wigner's classification theorem for molecular vibrations established that the character table completely determines the number and symmetry of vibrational modes.

The mutual exclusion rule for centrosymmetric molecules was first stated explicitly by Placzek in 1934 in his comprehensive treatise on Raman spectroscopy (in Marx, Handbuch der Radiologie, vol. 6). Placzek recognised that the parity distinction between IR-active (ungerade) and Raman-active (gerade) modes is an immediate consequence of the inversion symmetry of the polarisability and dipole-moment operators.

Cotton's Chemical Applications of Group Theory (first edition 1963, third edition 1990) standardised the presentation of character tables and selection rules for chemists and remains the canonical textbook. Cotton introduced the systematic use of reducible representation decomposition and the reduction formula as routine tools for vibrational analysis.

Wilson, Decius, and Cross (Molecular Vibrations, 1955) developed the complete GF-matrix method for computing vibrational frequencies, with the symmetry adaptation provided by the character-table decomposition. Their work connected the group-theoretic classification to the numerical computation of normal-mode frequencies, completing the bridge from symmetry to spectroscopy.

Double groups were introduced by Bethe in the same 1929 paper to handle the half-integer angular momenta arising from electron spin. Opechowski (1940) and Griffith (The Theory of Transition-Metal Ions, 1961) extended the double-group formalism to transition-metal spectroscopy, enabling the treatment of spin-orbit coupling in the crystal field framework.

Factor-group analysis for molecular crystals was developed independently by Halford (1946) and Hornig (1949), who showed how to predict the vibrational spectra of crystalline solids from the space group symmetry and the number of molecules per unit cell. The method remains the standard tool for assigning solid-state vibrational spectra.

Bibliography Master

Bethe, H. — Termaufspaltung in Kristallen, Ann. Phys. 3 (1929), 133–208. The foundational paper applying point-group character tables to electronic structure.
Wigner, E. P. — Gruppentheorie und ihre Anwendung auf die Quantenmechanik der Atomspektren (Vieweg, 1931). Group-theoretic treatment of molecular symmetry and selection rules.
Placzek, G. — Rayleigh-Streuung und Raman-Effekt, in Marx, Handbuch der Radiologie, vol. 6 (1934). First systematic statement of the mutual exclusion rule.
Wilson, E. B., Decius, J. C. & Cross, P. C. — Molecular Vibrations (McGraw-Hill, 1955). The GF-matrix method connecting symmetry classification to frequency computation.
Cotton, F. A. — Chemical Applications of Group Theory, 3rd ed. (Wiley, 1990). The canonical textbook for chemical group theory.
Bishop, D. M. — Group Theory and Chemistry (Clarendon Press, 1973). Comprehensive reference for representations, vibrations, and selection rules.
Halford, R. S. — Motions of Molecules in Condensed Systems, J. Chem. Phys. 14 (1946), 8–18. Factor-group analysis for molecular crystals.
Hornig, D. F. — The Vibrational Spectra of Molecules and Complex Ions in Crystals, J. Chem. Phys. 16 (1949), 1063–1076. Independent development of factor-group analysis.
Griffith, J. S. — The Theory of Transition-Metal Ions (Cambridge UP, 1961). Double-group treatment of spin-orbit coupling in crystal fields.
Miessler, G. L., Fischer, P. J. & Tarr, D. A. — Inorganic Chemistry, 5th ed. (Pearson, 2014), Ch. 4. Modern introductory treatment of symmetry and vibrational analysis.