Inorganic Chemistry / Symmetry
Character tables and reducible representations
Reading character tables, reducing representations with the reduction formula, and projection operators for symmetry-adapted orbitals and vibrations.
UndergradPhDIn a nutshell
A character table is a compact grid of numbers that summarises everything a point group can do. Each row describes one basic way that an object, such as an orbital or a vibration, can behave under the group's symmetry operations: staying the same (+1), flipping sign (−1), or mixing with partners. Chemists use character tables like a lookup tool. Build a simple description of, say, the bond stretches in ammonia, then use a short formula to split it into these basic symmetry types. That tells you how many distinct vibrations there are, which are degenerate, and which appear in infrared or Raman spectra. The same method builds molecular orbital diagrams. Learning to read a character table and apply the reduction formula is the core practical skill of group theory in chemistry.
Reading a character table
For C_{3v} (ammonia):
- Columns are classes of operations; g_c is the number in each class, and the order is h = \sum g_c = 6.
- Rows are irreducible representations. The character under E is the degeneracy: 1 for A/B, 2 for E, 3 for T.
- Mulliken labels: A is symmetric and B antisymmetric under the principal rotation; subscripts g/u show behaviour under i; primes show behaviour under \sigma_h.
- The right-hand columns list which irreducible representation the Cartesian axes, rotations and quadratic functions belong to; this is what links tables to IR and Raman activity.
Building a reducible representation
Choose a basis (bond vectors, orbitals, atomic displacements) and, for one operation per class, ask how much of each basis member ends up on itself: +1 unchanged, −1 reversed, 0 moved elsewhere. Sum these to get the character \chi_\Gamma.
The reduction formula
Worked example: N–H stretches. The three bond vectors give \Gamma = (3, 0, 1): all stay under E, all move under C_3, one stays under each \sigma_v.
- n(A_1) = \frac{1}{6}(3\cdot1 + 2\cdot0\cdot1 + 3\cdot1\cdot1) = 1
- n(A_2) = \frac{1}{6}(3 + 0 - 3) = 0
- n(E) = \frac{1}{6}(3\cdot2 + 0 + 0) = 1
So \Gamma = A_1 + E: one symmetric stretch and a degenerate pair, all IR and Raman active.
For all motions of a molecule, use unmoved atoms multiplied by a contribution per atom (3 for E, −1 for C_2, 1 for \sigma, −3 for i) and subtract translations and rotations to leave the vibrations. For water this gives 2A_1 + B_2.
Projection operators
To find the actual combinations, apply
to one basis function, summing over every operation R. For ammonia, projecting r_1 gives the A_1 mode \tfrac{1}{\sqrt3}(r_1+r_2+r_3) and an E component \tfrac{1}{\sqrt6}(2r_1-r_2-r_3); its orthogonal partner is \tfrac{1}{\sqrt2}(r_2-r_3). The same procedure yields ligand SALCs for MO diagrams.
Behind the reduction formula lies the great orthogonality theorem: irreducible representations behave like orthogonal vectors in a space whose dimension is the number of classes. This is why decomposition is unique and why the sum of the squared dimensions equals the order of the group. Researchers rarely reduce representations by hand, but understanding the structure matters when software output must be interpreted or when a problem falls outside standard tables.
Direct products decide whether integrals vanish: \langle\psi_f|\hat{O}|\psi_i\rangle can be non-zero only if \Gamma_f\otimes\Gamma_O\otimes\Gamma_i contains the totally symmetric representation. This single rule generates electric-dipole, Raman, magnetic-dipole and two-photon selection rules and the symmetry conditions for vibronic coupling. The symmetric part of the direct product of a degenerate state with itself determines which distortions are Jahn–Teller active.
Correlation tables follow representations as symmetry is lowered, from O_h to D_{4h} on tetragonal elongation, or from a molecular group to a crystal site group. They are used to predict band splittings, to assign spectra of distorted complexes and to map orbital labels from computational subgroups back to the true point group.
For heavy-atom and open-shell systems, double groups extend character tables to spinor states, essential for spin–orbit-coupled excited states, lanthanide crystal-field levels and magnetic anisotropy. Magnetic (Shubnikov) groups and space-group representations, with labels at each wavevector in the Brillouin zone, play the equivalent role for phonons, band structures and topological materials, where symmetry indicators computed from characters can flag non-trivial band topology.
Key points
- Columns of a character table are classes of operations; rows are irreducible representations.
- The character under E gives the degeneracy of a representation.
- Characters of a reducible representation come from counting basis members left in place (+1), reversed (−1) or moved (0).
- Reduction formula: n_i = \frac{1}{h}\sum_c g_c\chi_\Gamma(c)\chi_i(c).
- Projection operators turn the irreducible representation count into explicit SALCs or normal modes.
Check yourself
5 questions
1In a character table, the character of an irreducible representation under E equals:
The identity leaves every component unchanged, so its character counts the dimension of the representation.
2What does the Mulliken label B_{2g} tell you?
B means one-dimensional and antisymmetric under the principal C_n; g means unchanged by inversion.
3For the three N–H bond vectors in NH_3, why is the character under C_3 zero?
Only vectors that end up on themselves contribute; under C_3 all three move, so the character is 0.
4In C_{2v} (h = 4), a reducible representation has characters (3, 1, 1, 3) under E, C_2, \sigma_v, \sigma_v'. How many times does A_1 (1, 1, 1, 1) occur?
n(A_1) = \frac{1}{4}(3\cdot1 + 1\cdot1 + 1\cdot1 + 3\cdot1) = 2. A non-integer result would signal an error in building \Gamma.
5Why must a vibrational analysis from the 3N displacement basis subtract the representations of x, y, z and R_x, R_y, R_z?
The 3N basis includes all motions; removing the three translations and three rotations (two if linear) leaves the 3N-6 vibrations.
Read more
Recent preprints on this topic
- Loading…
Matched by keyword from ChemRxiv via Crossref. Not peer-reviewed; read critically.