Water: Symmetry Without Solving Anything
The problem
How many vibrational modes does H₂O have, what does each one look like, and which absorb infrared?
The remarkable part is that you can answer all three without writing down a single equation of motion. No masses, no spring constants, no eigenvalue problem. Only the symmetry of the molecule.
Step 1: the group
Water is bent, with the O at the apex and the two H's symmetric about a vertical axis. Its symmetry operations are:
| Operation | Meaning |
|---|---|
| \(E\) | do nothing |
| \(C_2\) | rotate 180° about the bisector |
| \(\sigma_v(xz)\) | reflect in the molecular plane |
| \(\sigma_v'(yz)\) | reflect in the perpendicular plane |
Four operations, each its own inverse: this is the group \(C_{2v}\), isomorphic to the Klein four-group. It is abelian, so by the theory of representations all its irreducible representations are one-dimensional — there are four of them, and the character table is:
| \(C_{2v}\) | \(E\) | \(C_2\) | \(\sigma_v(xz)\) | \(\sigma_v'(yz)\) | |
|---|---|---|---|---|---|
| \(A_1\) | 1 | 1 | 1 | 1 | \(z\) |
| \(A_2\) | 1 | 1 | −1 | −1 | \(R_z\) |
| \(B_1\) | 1 | −1 | 1 | −1 | \(x\), \(R_y\) |
| \(B_2\) | 1 | −1 | −1 | 1 | \(y\), \(R_x\) |
Step 2: count the modes
Three atoms, three Cartesian displacements each: 9 degrees of freedom. Three are whole-molecule translation, three are rotation. So
(For a linear molecule it would be \(3N-5\), because rotation about the axis is not a degree of freedom — which is why CO₂ has four modes, not three.)
Step 3: find their symmetries
Build the reducible representation \(\Gamma_{3N}\) on the 9 displacement vectors. The character of each operation is, by a standard shortcut,
with contributions \(+3\) for \(E\), \(-1\) for \(C_2\), \(+1\) for \(\sigma\):
Take the standard Mulliken convention: the \(C_2\) axis is \(z\) and the molecule lies in the \(yz\) plane. So \(\sigma_v'(yz)\) is the molecular plane (all three atoms lie in it) and \(\sigma_v(xz)\) is perpendicular to it (only the oxygen lies in it).
| \(E\) | \(C_2\) | \(\sigma_v(xz)\) | \(\sigma_v'(yz)\) | |
|---|---|---|---|---|
| atoms unmoved | 3 | 1 | 1 | 3 |
| per-atom factor | 3 | −1 | 1 | 1 |
| \(\chi(\Gamma_{3N})\) | 9 | −1 | 1 | 3 |
The convention is not optional bookkeeping
Put the molecule in the \(xz\) plane instead and the two mirror characters swap, giving \(\Gamma_{\rm vib} = 2A_1 + B_1\). The physics is identical — the antisymmetric stretch is still antisymmetric under \(C_2\) — but the label changes. Textbooks quote \(B_2\) for water because they use the \(yz\) convention, so state yours before comparing with anyone.
Reduce with the standard formula \(n_i = \frac{1}{h}\sum_R\chi(R)\chi_i(R)\), with \(h=4\):
Now subtract translations (\(A_1 + B_1 + B_2\), read off the right-hand column as \(z\), \(x\), \(y\)) and rotations (\(A_2 + B_1 + B_2\), the \(R\) entries):
Three modes, and we now know their symmetry species — having solved nothing.
Step 4: what they look like, and where they are
| Mode | Symmetry | Motion | Observed (cm⁻¹) |
|---|---|---|---|
| \(\nu_1\) | \(A_1\) | symmetric stretch — both O–H lengthen together | 3657 |
| \(\nu_2\) | \(A_1\) | bend — the H–O–H angle opens and closes | 1595 |
| \(\nu_3\) | \(B_2\) | antisymmetric stretch — one bond lengthens as the other shortens | 3756 |
The two \(A_1\) modes preserve every symmetry operation; the \(B_2\) mode changes sign under \(C_2\), which is precisely what "antisymmetric" means, stated group-theoretically.
Step 5: which are infrared active?
A mode absorbs infrared if it changes the dipole moment — formally, if its representation contains the same species as \(x\), \(y\) or \(z\). From the character table those are \(A_1\) (\(z\)), \(B_1\) (\(x\)) and \(B_2\) (\(y\)).
\(\Gamma_{\rm vib} = 2A_1 + B_2\), so all three modes are IR active. (They are also all Raman active, since \(C_{2v}\) has no centre of inversion and hence no mutual-exclusion rule.)
That is why water is such a potent greenhouse gas: every one of its vibrations couples to infrared radiation. Contrast CO₂, which is linear and centrosymmetric — its symmetric stretch changes no dipole and is IR silent, which is why only three of its four modes absorb.
What this method actually is
Nothing here required knowing an O–H force constant. Symmetry alone determined:
- how many modes exist,
- which irreducible representation each belongs to,
- which are degenerate (none, here — but a linear or highly symmetric molecule would have some, forced by two-dimensional irreps),
- which are IR or Raman active.
What symmetry cannot give you is the frequencies — for those you need the dynamics. But it tells you the structure of the answer before you start, and it tells you which matrix elements vanish identically, which is often the whole difficulty.
This is the same logic that gives selection rules in atomic spectroscopy, forbids certain particle decays, and blocks terms in a Lagrangian: if a quantity is not invariant under the symmetry, its integral vanishes. Group theory is the machinery for knowing that in advance.
Common mistakes
- Using \(3N-5\) for a bent molecule. Water is not linear; it has three rotational degrees of freedom, so \(3N-6\).
- Forgetting to subtract rotations as well as translations. A common way to get four modes.
- Miscounting unmoved atoms. Only atoms not displaced by the operation contribute; for \(C_2\) in water that is the oxygen alone.
- Assuming IR-active implies Raman-inactive. They exclude each other only in centrosymmetric molecules. \(C_{2v}\) has no inversion centre, so both are allowed.
Related
Symmetry groups · Group representations · Cosets & Lagrange · Lie groups — the continuous analogue · Normal modes (PHY621) — the dynamics symmetry cannot supply · Normal modes lab (PHY621) — the CO₂ triatomic, solved numerically