Molecular Symmetry

Molecular symmetry turns a geometric observation into exact statements about a specified molecular model. Once the equilibrium geometry, isotopic composition, Hamiltonian, and coordinate convention are fixed, symmetry can:

• divide electronic, vibrational, and rotational problems into independent blocks;
• label states and orbitals without relying on a particular visualization;
• identify degeneracies protected by the model;
• prove that selected matrix elements vanish;
• organize normal modes and spectroscopic activity;
• expose when a numerical calculation has broken an intended symmetry.

The useful question is therefore not merely “what shape is the molecule?” It is:

Which transformations leave the relevant Hamiltonian unchanged, how do the states and operators transform, and what follows from those transformation laws?

A molecular point group is the practical fixed-geometry answer. It consists of rotations, reflections, inversion, and improper rotations that map the nuclear framework onto itself while leaving at least one point fixed. Point groups are powerful, but they are not the whole symmetry of a freely rotating, vibrating, permuting molecule.

Helpful background. Groups and representations explains invariant subspaces; selection rules proves the state–operator symmetry test. The character-table reference supplies the tables used below.

Molecular Point Groups

A group attached to a geometry

After removing overall translation, describe the nuclear framework by body-fixed positions

$$\mathbf R = \left( \mathbf R_1,\ldots,\mathbf R_N \right).$$

An orthogonal transformation $g\in O(3)$ is a point-group operation when it maps this framework onto itself, allowing exchange only among identical nuclear species. Schematically,

$$G_{\mathbf R} = \left\{ g\in O(3): g\mathbf R = P_g\mathbf R \right\},$$

where $P_g$ is a permutation that preserves nuclear species. For an isotopically specified rovibrational problem, operations that exchange different isotopes need not remain symmetries of the nuclear kinetic energy.

The point-group label therefore belongs to a particular geometry and model. It is commonly assigned at an equilibrium structure, a transition structure, or another stationary geometry. A generic displaced geometry usually has less symmetry than the stationary point even when the displacement is a normal coordinate classified by the higher-symmetry reference group.

Common families

Schoenflies notation is standard in molecular spectroscopy and quantum chemistry:

• $C_1$ contains only the identity.
• $C_s$ has one mirror plane, and $C_i$ has inversion.
• $C_n$, $C_{nv}$, and $C_{nh}$ have a principal $n$-fold axis, with the indicated vertical or horizontal mirror planes.
• $D_n$, $D_{nh}$, and $D_{nd}$ also have $n$ twofold axes perpendicular to the principal axis.
• $S_{2n}$ groups are generated by an improper rotation.
• $T_d$, $O_h$, and $I_h$ describe ideal tetrahedral, octahedral, and icosahedral frameworks.
• $C_{\infty v}$ and $D_{\infty h}$ are the linear-molecule groups without and with inversion, respectively.

Ideal examples include:

• bent water, $C_{2v}$;
• pyramidal ammonia, $C_{3v}$;
• trigonal-planar boron trifluoride, $D_{3h}$;
• methane, $T_d$;
• benzene, $D_{6h}$;
• hydrogen chloride, $C_{\infty v}$;
• carbon dioxide, $D_{\infty h}$.

These labels assume idealized isolated structures. Substitution, isotopic labeling, an external field, a crystal environment, or a distorted geometry can reduce the symmetry.

A reliable identification sequence

For a finite equilibrium geometry:

1. Check whether the molecule is linear.
2. Check for one of the high-symmetry polyhedral groups.
3. Find the highest-order proper rotation axis and call it the principal axis.
4. Look for $n$ twofold axes perpendicular to that principal axis.
5. Test for a horizontal mirror plane, vertical or dihedral mirror planes, an inversion center, and improper axes.
6. Verify closure by listing every operation, not only every visible symmetry element.

The last step matters. A symmetry element is a geometric object such as an axis or plane. A symmetry operation is a transformation such as $C_3$ or $C_3^2$. One threefold axis supplies two nonidentity rotations.

Point groups are stabilizers, not molecular trajectories

The point group of $\mathbf R_e$ contains operations that leave the reference framework invariant as a set. It does not include arbitrary internal motions that carry the molecule through other geometries. Internal rotation, inversion tunneling, pseudorotation, and exchange of identical nuclei through feasible paths may require a molecular symmetry or permutation–inversion group.

That distinction is essential in high-resolution spectroscopy. A rigid equilibrium point group is often the correct first tool, but it is not always the exact symmetry group of the full rovibronic Hamiltonian.

Symmetry Operations

The basic operations

The standard spatial operations are:

$$\begin{aligned} E &: \text{identity}, \\ C_n^k &: \text{rotation through } 2\pi k/n, \\ \sigma &: \text{reflection in a plane}, \\ i &: \mathbf r\mapsto-\mathbf r, \\ S_n &: \text{rotation followed by reflection}. \end{aligned}$$

For a chosen principal axis:

• $\sigma_h$ is perpendicular to the axis;
• $\sigma_v$ contains the axis;
• $\sigma_d$ contains the axis and bisects two perpendicular $C_2$ axes.

With the principal axis chosen as $z$, one may write

$$C_n = R_z\!\left(\frac{2\pi}{n}\right), \qquad S_n = \sigma_h C_n.$$

Powers $S_n^k$ contain the reflection only for odd $k$, since $\sigma_h^2=E$. The order of factors is harmless here because the rotation about $z$ commutes with reflection in the plane perpendicular to $z$. For more general products, operator order must be kept explicit.

How an operation acts on a wavefunction

At fixed nuclear geometry, a spatial operation acts on every electronic coordinate. For a spinless coordinate-space wavefunction,

$$\left( \widehat U_g\psi \right) \left( \mathbf r_1,\ldots,\mathbf r_{N_e} \right) = \psi \left( g^{-1}\mathbf r_1,\ldots, g^{-1}\mathbf r_{N_e} \right).$$

with the nuclei simultaneously mapped onto symmetry-equivalent positions. If the clamped-nuclei electronic Hamiltonian respects the operation,

$$\left[ \widehat H_e(\mathbf R), \widehat U_g \right] = 0 \qquad \text{for every }g\in G_{\mathbf R}.$$

The inverse in the wavefunction argument makes the operators compose in the same order as the geometric transformations.

Water and its axis convention

Take water in the $yz$ plane, with the $C_2$ axis along $z$. The two vertical planes are then:

• $\sigma_v(xz)$, which exchanges the two hydrogens;
• $\sigma_v'(yz)$, the molecular plane, which fixes all three nuclei.

The $C_2(z)$ rotation also exchanges the hydrogens. All four operations

$$\left\{ E,\, C_2(z),\, \sigma_v(xz),\, \sigma_v'(yz) \right\}$$

map the nuclear framework onto itself and form $C_{2v}$.

Axis convention for the worked $C_{2v}$ example. The molecular plane is $yz$; $C_2(z)$ and $\sigma_v(xz)$ exchange the hydrogens, while $\sigma_v'(yz)$ fixes each nucleus.

Axis conventions are not cosmetic. If the molecule is instead placed in the $xz$ plane, labels called $B_1$ and $B_2$ in one convention can be exchanged in another. The displacement pattern and measurable polarization rule remain the same.

Spin and external fields

Ordinary point groups act on spatial coordinates. If spin–orbit coupling is important, spinors transform under a double group, and a $2\pi$ rotation cannot simply be represented as the identity on the spin state. Magnetic fields also require care because axial vectors and antiunitary time reversal enter the symmetry analysis.

An external electric field restricts the group to operations that preserve the field direction and polarity. A calculation performed in a field must therefore use the symmetry of “molecule plus field,” not the field-free molecular point group.

Irreducible Representations

States transform in invariant subspaces

Let $\Gamma$ label an irreducible representation of $G$. A basis of one copy of that irrep transforms as

$$\widehat U_g \lvert\Gamma,a,\mu\rangle = \sum_{\nu=1}^{d_\Gamma} D_{\nu\mu}^{(\Gamma)}(g) \lvert\Gamma,a,\nu\rangle.$$

Here:

• $d_\Gamma$ is the irrep dimension;
• $\mu$ labels components inside one irrep;
• $a$ distinguishes repeated copies of the same irrep.

If the Hamiltonian commutes with the group, its matrix is block diagonal in irreducible sectors:

$$\mathcal H = \bigoplus_\Gamma \left( \mathbb C^{m_\Gamma} \otimes V_\Gamma \right).$$

States belonging to inequivalent irreps cannot mix under a symmetry-preserving Hamiltonian. Repeated copies of the same irrep can mix, so a symmetry label alone need not identify a unique state.

Symmetry-protected degeneracy

On each isotypic sector, Schur’s lemma gives $H_\Gamma=h_\Gamma\otimes I_{d_\Gamma}$: the Hamiltonian can mix repeated copies but acts identically on their irrep components. Diagonalizing $h_\Gamma$ produces energy multiplets with degeneracy at least $d_\Gamma$. Thus a multidimensional irrep enforces degeneracy:

$$\begin{gathered} d_\Gamma>1 \quad\Longrightarrow\quad \text{at least }d_\Gamma\text{ components} \\ \text{in one symmetry-protected multiplet}. \end{gathered}$$

The converse is false. Distinct one-dimensional irreps can cross, and unrelated levels can be accidentally degenerate. Conversely, numerical use of an Abelian subgroup can print the components of a parent multidimensional irrep under different labels even though the exact parent symmetry keeps them degenerate.

Mulliken labels

For common molecular point groups:

• $A$ and $B$ usually denote one-dimensional irreps;
• $E$ denotes a two-dimensional irrep;
• $T$ denotes a three-dimensional irrep;
• subscripts distinguish irreps with different behavior under selected operations;
• $g$ and $u$ denote even and odd parity under inversion;
• primes and double primes denote even and odd behavior under a designated horizontal reflection.

The precise subscript convention depends on the group and chosen axes. Read the table header before interpreting a label. The symbol $E$ can denote either the identity operation or a two-dimensional irrep; context distinguishes them.

Reducible representations

A chosen set of coordinates, atomic orbitals, or displacement vectors usually transforms reducibly. If

$$\Gamma_{\mathrm{red}} \cong \bigoplus_\alpha n_\alpha\Gamma_\alpha,$$

then $n_\alpha$ is the number of copies of irrep $\Gamma_\alpha$. This decomposition is the bridge from geometry to useful basis functions.

Character Tables

Characters compress the representation

The character of a representation matrix is its trace:

$$\chi^{(\Gamma)}(g) = \operatorname{Tr} D^{(\Gamma)}(g).$$

Characters are invariant under a basis change and are constant on conjugacy classes. A character table therefore lists:

• conjugacy classes in columns;
• irreducible representations in rows;
• irreducible characters in the central block;
• common coordinate, rotation, and quadratic functions as interpretive aids.

The functions printed beside an irrep are examples of objects with that transformation law. They are not an exhaustive basis and do not mean that the irrep “is” a coordinate.

Orthogonality and reduction

For finite groups, irreducible characters obey

$$\frac{1}{|G|} \sum_{g\in G} \chi^{(\alpha)}(g)^* \chi^{(\beta)}(g) = \delta_{\alpha\beta}.$$

Taking the inner product of a reducible character with an irreducible character gives the multiplicity:

$$n_\alpha = \frac{1}{|G|} \sum_C |C|\, \chi^{(\alpha)}(C)^* \chi^{(\mathrm{red})}(C).$$

The sum is over conjugacy classes $C$, and $|C|$ is the number of operations in the class. A valid reduction must give nonnegative integers and satisfy the dimension check

$$\dim\Gamma_{\mathrm{red}} = \sum_\alpha n_\alpha d_\alpha.$$

Failure of either check usually means that an atom-fixed count, vector trace, class size, or axis convention is wrong.

Projection operators

The projector onto the full isotypic sector of irrep $\Gamma$ is

$$\widehat{\mathcal P}^{(\Gamma)} = \frac{d_\Gamma}{|G|} \sum_{g\in G} \chi^{(\Gamma)}(g)^* \widehat U_g.$$

Applying it to trial orbitals or internal coordinates produces symmetry-adapted combinations. This character projector selects all copies of $\Gamma$. Resolving individual rows and repeated copies requires the fuller matrix-element projectors or an additional orthogonalization and diagonalization within the projected space.

Direct products

Products of states or operators transform in tensor-product representations. Their characters multiply:

$$\chi^{(\Gamma\otimes\Lambda)}(g) = \chi^{(\Gamma)}(g) \chi^{(\Lambda)}(g).$$

The resulting character can be reduced with the same formula. For one-dimensional real irreps, direct products are especially simple: multiply the signs operation by operation.

The compact $C_{2v}$ and $C_{3v}$ tables used below are collected at Character Tables.

Worked Example: Water Vibrations

The Cartesian displacement representation

Each of the three nuclei contributes three Cartesian displacement coordinates, so water begins with a nine-dimensional representation $\Gamma_{3N}$. Its character under operation $g$ is

$$\chi_{3N}(g) = \sum_{A:\,gA=A} \operatorname{Tr}R_g.$$

Only nuclei left at their own positions contribute to the trace. A nucleus moved to another site contributes zero to the diagonal trace, even when the destination contains an identical nucleus.

For the axis convention in the figure:

$$\begin{aligned} \chi_{3N}(E) &=3(3)=9, \\ \chi_{3N}\!\left(C_2\right) &=1(-1)=-1, \\ \chi_{3N}\!\left(\sigma_v(xz)\right) &=1(1)=1, \\ \chi_{3N}\!\left(\sigma_v'(yz)\right) &=3(1)=3. \end{aligned}$$

The factors $3$, $-1$, and $1$ are traces of the corresponding three-dimensional vector transformations. Reducing the character

$$\chi_{3N} = \left( 9,-1,1,3 \right)$$

against the $C_{2v}$ table gives

$$\Gamma_{3N} = 3A_1 \oplus A_2 \oplus 2B_1 \oplus 3B_2.$$

The dimensions add to $9$, as required.

Remove translations and rotations

With $z$ along the $C_2$ axis and the molecule in the $yz$ plane,

$$\begin{aligned} \Gamma_{\mathrm{trans}} &= \Gamma(x,y,z) = A_1\oplus B_1\oplus B_2, \\ \Gamma_{\mathrm{rot}} &= \Gamma(R_x,R_y,R_z) = B_2\oplus B_1\oplus A_2. \end{aligned}$$

Therefore

$$\begin{aligned} \Gamma_{\mathrm{vib}} &= \Gamma_{3N} - \Gamma_{\mathrm{trans}} - \Gamma_{\mathrm{rot}} \\ &= 2A_1 \oplus B_2. \end{aligned}$$

The two $A_1$ modes are conventionally the symmetric stretch and bend; the $B_2$ mode is the asymmetric stretch in this axis convention. All irreps are one dimensional, so $C_{2v}$ does not enforce a vibrational degeneracy.

The subtraction is subtraction of irrep multiplicities after decomposition. It is not subtraction of sets or individual displacement arrows.

Selection Rules

The totally symmetric test

For a transition operator $\widehat O$, consider

$$M_{fi} = \langle\psi_f| \widehat O |\psi_i\rangle.$$

The matrix element can be nonzero only if

$$\Gamma_f^* \otimes \Gamma_O \otimes \Gamma_i \supset \Gamma_{\mathrm{ts}},$$

where $\Gamma_{\mathrm{ts}}$ is the totally symmetric irrep. Equivalently, the multiplicity of the totally symmetric irrep is

$$n_{\mathrm{ts}} = \frac{1}{|G|} \sum_{g\in G} \chi_f(g)^* \chi_O(g) \chi_i(g).$$

If $n_{\mathrm{ts}}=0$, symmetry forces the matrix element to vanish within the stated model. If $n_{\mathrm{ts}}>0$, the transition is symmetry allowed, but its intensity can still be extremely small or zero for dynamical reasons.

Electric-dipole transitions

The electric-dipole components transform like $x$, $y$, and $z$. In the water convention:

$$\begin{aligned} \Gamma(x)&=B_1, & \Gamma(y)&=B_2, \\ \Gamma(z)&=A_1. \end{aligned}$$

From an $A_1$ initial state, an $x$-, $y$-, or $z$-polarized transition can reach $B_1$, $B_2$, or $A_1$, respectively, provided the other quantum numbers and approximations also permit the transition.

For a fundamental vibration from a totally symmetric vibrational ground state, the normal coordinate must transform like at least one dipole component. Water’s $2A_1+B_2$ fundamentals are therefore all infrared active in the electric-dipole and harmonic approximations.

Raman transitions

The leading nonresonant Raman operator is the symmetric polarizability tensor. Its components transform like

$$x^2,\ y^2,\ z^2,\ xy,\ xz,\ yz.$$

A normal mode is first-order Raman active when its irrep occurs among those quadratic functions. This is a different condition from infrared activity.

For a centrosymmetric molecule, dipole components are ungerade and polarizability components are gerade. A harmonic fundamental from a gerade ground state therefore cannot be both electric-dipole infrared active and first-order Raman active. This mutual-exclusion rule depends on inversion symmetry and on the stated leading operators.

Which forbidden lines can gain intensity?

Mixing can relax an approximate label such as pure spin, electronic symmetry, or a harmonic mode assignment. A vibronic state, for example, transforms as the combined electronic and nuclear state rather than either factor separately. External fields or isotope substitution can also reduce the exact symmetry, and a magnetic-dipole or electric-quadrupole operator has a different transformation law from the electric dipole.

However, symmetry-preserving mixing cannot defeat an exact selection rule for the full states and the specified operator. If inversion remains exact, electric-dipole transitions still require opposite total parities, while electric-dipole–electric-dipole Raman transitions require equal parities. Anharmonicity or resonant intensity borrowing alone does not remove that distinction.

Normal Modes and Spectroscopy

Why the Hessian respects symmetry

At a symmetry-preserving stationary geometry, the mass-weighted Hessian $\mathbf F$ commutes with the displacement representation:

$$\mathbf D(g)^\dagger \mathbf F \mathbf D(g) = \mathbf F.$$

It can therefore be block diagonalized by irreducible representation. This gives three immediate consequences:

1. modes from inequivalent irreps do not mix while the symmetry is preserved;
2. multidimensional irreps enforce degenerate harmonic frequencies;
3. transition-property derivatives can be screened by symmetry before an intensity calculation.

Modes of the same irrep can mix freely. If several modes are labeled $A_1$, the label does not uniquely mean “the symmetric stretch.” Frequencies, displacement vectors, and internal-coordinate projections are needed for a physical assignment.

Degenerate modes

For ammonia in $C_{3v}$, an $E$ vibrational mode has two components. Any orthonormal basis spanning that two-dimensional eigenspace is valid. Numerical eigenvectors can rotate within the subspace as the method, basis, or geometry changes while the observable subspace and common harmonic frequency remain unchanged.

A small computed splitting of an expected $E$ pair can signal:

• a geometry that is not exactly symmetry adapted;
• loose optimization or integration thresholds;
• unequal numerical treatment of the components in a lower-symmetry computational subgroup;
• a genuinely symmetry-breaking perturbation;
• an electronically unstable high-symmetry reference.

Using a subgroup alone does not physically split an exact parent-symmetry multiplet. An unstable high-symmetry geometry can have a degenerate pair of negative Hessian eigenvalues; instability does not by itself split that pair.

Correlation with lower symmetry

When a molecule distorts from group $G$ to subgroup $H$, an irrep of $G$ may remain irreducible or split when restricted to $H$:

$$\Gamma^{(G)} \downarrow H = \bigoplus_\alpha n_\alpha \Gamma_\alpha^{(H)}.$$

Correlation tables record this restriction. They are indispensable for tracking orbital and vibrational labels along reaction paths, symmetry-breaking coordinates, and Jahn–Teller distortions. Equal subgroup labels permit mixing; they do not guarantee that mixing is large.

The detailed normal-coordinate construction and intensity derivatives are developed in Normal Modes of Polyatomics. Raman Spectroscopy specializes the quadratic-function test to measured Raman tensors, depolarization ratios, and oriented samples.

Orbital Classification

Symmetry-adapted linear combinations

Suppose the two hydrogen $1s$ functions in water are $|H_L\rangle$ and $|H_R\rangle$. Their two-dimensional permutation representation has character

$$\chi_H = \left( 2,0,0,2 \right),$$

because $C_2(z)$ and $\sigma_v(xz)$ exchange the functions, whereas $E$ and $\sigma_v'(yz)$ fix them. Character reduction gives

$$\Gamma_H = A_1 \oplus B_2.$$

For normalized real $1s$ functions with overlap $S_H=\langle H_L|H_R\rangle$, normalized symmetry-adapted combinations are

$$\begin{aligned} |H_{A_1}\rangle &= \frac{ |H_L\rangle+|H_R\rangle }{\sqrt{2(1+S_H)}}, \\ |H_{B_2}\rangle &= \frac{ |H_L\rangle-|H_R\rangle }{\sqrt{2(1-S_H)}}. \end{aligned}$$

The $1/\sqrt2$ coefficients apply only to an orthonormal site basis, or after neglecting overlap. Symmetry determines the relative signs, while the inner product determines normalization.

On oxygen, the $2s$ and $2p_z$ orbitals transform as $A_1$, $2p_y$ as $B_2$, and $2p_x$ as $B_1$ in this convention. A totally symmetric one-electron Hamiltonian couples only matching symmetry sectors:

$$\langle\phi_\alpha| \widehat h |\phi_\beta\rangle = 0 \qquad \text{if } \Gamma_\alpha\not\cong\Gamma_\beta.$$

Thus the $A_1$ hydrogen combination can mix with oxygen $A_1$ functions, and the $B_2$ combination can mix with oxygen $2p_y$. The $B_1$ oxygen $2p_x$ function has no partner in this minimal hydrogen $\sigma$ basis.

Symmetry proves the zeros. It does not determine the nonzero coupling strengths, orbital ordering, occupation, or degree of bonding.

Orbitals, configurations, and states

A molecular orbital has an irrep label, but a many-electron state is classified by the direct product of all occupied spin-orbital transformation laws, reduced with the required fermionic antisymmetry and spin coupling. The symmetry of a single frontier orbital is not automatically the symmetry of an excited state.

For open-shell systems or strong spin–orbit coupling, one may need:

• spin-adapted configuration state functions;
• double-group irreps;
• total angular-momentum projections for linear molecules;
• explicit treatment of near-degenerate electronic configurations.

Computational block structure

Quantum-chemistry programs often exploit only an Abelian subgroup even when the geometry has a non-Abelian point group. This keeps all working irreps one-dimensional and simplifies integral storage. Components of a parent $E$ or $T$ irrep then receive separate subgroup labels.

Compare the full point group and axis convention when matching labels from different programs: two different subgroup labels may be components of the same parent multiplet.

Beyond the Rigid Point Group

Isotopic substitution

Electronic Coulomb interactions depend on nuclear charges, not nuclear masses. Replacing one isotope can therefore leave the clamped-nuclei electronic potential symmetric at the same geometry while reducing the symmetry of the nuclear kinetic energy and the full rovibrational problem.

This is why an isotopic substitution can activate weak vibrational lines or split rovibrational structure without substantially changing the electronic potential-energy surface.

Large-amplitude and nonrigid motion

For tunneling between equivalent minima, internal rotation, umbrella inversion, or pseudorotation, the experimentally relevant states can extend over several geometries. A single equilibrium point group cannot represent every feasible permutation and inversion operation. Molecular symmetry groups and permutation–inversion groups provide the appropriate global classification.

Electronic degeneracy and distortion

A nonlinear molecule in a degenerate electronic state can be unstable to a symmetry-lowering distortion through the Jahn–Teller effect. The high-symmetry point group still organizes the parent electronic and vibrational spaces, but the distorted minima belong to subgroups.

At intersections of electronic potential-energy surfaces, point-group symmetry can permit or forbid couplings on special subspaces, while generic conical intersection topology depends on the local branching plane. See Conical Intersections for the branching-plane construction.

Approximate symmetry

Near symmetry is often chemically informative, but “almost an irrep” is not an exact group-theory concept. If a perturbation is small, write

$$\widehat H = \widehat H_0 + \lambda\widehat V,$$

where $\widehat H_0$ has the higher symmetry. The labels of $\widehat H_0$ then remain useful zeroth-order labels, and $\widehat V$ controls mixing and splitting. Quantifying $\lambda\widehat V$ is more informative than assigning a higher point group by visual tolerance alone.

Exercises

Exercise 1: Identify idealized point groups

Assign the equilibrium point group of idealized $\mathrm{HCl}$, $\mathrm{CO_2}$, $\mathrm{H_2O}$, $\mathrm{NH_3}$, and $\mathrm{CH_4}$. State the geometric feature that distinguishes each answer.

Solution

• $\mathrm{HCl}$ is linear and heteronuclear, so it is $C_{\infty v}$.
• $\mathrm{CO_2}$ is linear and centrosymmetric, so it is $D_{\infty h}$.
• Bent $\mathrm{H_2O}$ has one $C_2$ axis and two vertical mirror planes, so it is $C_{2v}$.
• Pyramidal $\mathrm{NH_3}$ has one $C_3$ axis and three vertical mirror planes, so it is $C_{3v}$.
• Ideal tetrahedral $\mathrm{CH_4}$ has tetrahedral symmetry, $T_d$.

These are equilibrium-geometry labels. Isotopic substitution, distortion, or an external environment can reduce them.

Exercise 2: Replace one hydrogen by deuterium

At the same fixed geometry, which point-group operations of water remain symmetries of the HDO nuclear kinetic energy? Does this change the clamped-nuclei Coulomb electronic potential?

Solution

Only $E$ and the molecular-plane reflection $\sigma_v'(yz)$ preserve the masses at every site, giving $C_s$. The operations that exchange H and D no longer preserve the kinetic energy. The clamped-nuclei Coulomb electronic potential still has $C_{2v}$ symmetry because H and D have the same nuclear charge. Thus the electronic potential and isotope-specific rovibrational Hamiltonian need not have the same point group.

Exercise 3: Construct hydrogen SALCs

Apply symmetry to the two water hydrogen $1s$ functions. Show that their representation is $A_1\oplus B_2$ and construct normalized symmetry-adapted linear combinations for normalized functions with real overlap $S_H=1/5$.

Solution

The identity and molecular-plane reflection fix both functions, while $C_2(z)$ and $\sigma_v(xz)$ exchange them. Thus

$$\chi_H = \left( 2,0,0,2 \right),$$

which reduces to $A_1\oplus B_2$. Their squared unnormalized norms are $2(1+S_H)$ and $2(1-S_H)$, respectively, giving

$$\begin{aligned} |H_{A_1}\rangle &= \frac{ |H_L\rangle+|H_R\rangle }{\sqrt{12/5}}, \\ |H_{B_2}\rangle &= \frac{ |H_L\rangle-|H_R\rangle }{\sqrt{8/5}}. \end{aligned}$$

Their relative signs describe transformation behavior. An overall sign change of either SALC has no physical effect.

Exercise 4: Polarization selection in water

Assume an $A_1$ initial state in $C_{2v}$. Which final-state irreps can be reached by $x$-, $y$-, and $z$-polarized electric-dipole transitions in the axis convention used on this page?

Solution

The coordinate components transform as

$$\begin{aligned} \Gamma(x)&=B_1, & \Gamma(y)&=B_2, \\ \Gamma(z)&=A_1. \end{aligned}$$

For one-dimensional real irreps, the product

$$\Gamma_f \otimes \Gamma(\mu_a) \otimes A_1$$

contains $A_1$ precisely when $\Gamma_f=\Gamma(\mu_a)$. Therefore

$$x:B_1, \qquad y:B_2, \qquad z:A_1.$$

Other quantum numbers and the actual transition moment can impose additional restrictions or make an allowed line weak.

Exercise 5: Prove vibrational mutual exclusion

For a centrosymmetric molecule with a gerade vibrational ground state, show that a one-quantum harmonic fundamental cannot be both electric-dipole infrared active and first-order Raman active.

Solution

The electric dipole is ungerade. A nonzero infrared matrix element from a gerade ground state therefore requires an ungerade final vibrational state:

$$g\otimes u\otimes u = g.$$

The polarizability tensor is gerade. A nonzero first-order Raman matrix element from the same ground state requires a gerade final state:

$$g\otimes g\otimes g = g.$$

A normal mode of a centrosymmetric molecule is either gerade or ungerade, not both. Hence the same fundamental cannot satisfy both conditions within these approximations. With exact inversion, the same pair of parity eigenstates cannot acquire both E1 and E1–E1 intensity merely through anharmonic or vibronic mixing. Breaking inversion or using a different multipole process changes the symmetry test.

Exercise 6: Prove the symmetry block zero

Let $|\phi_\alpha\rangle$ and $|\phi_\beta\rangle$ belong to inequivalent irreps of a finite point group, and let $\widehat h$ be totally symmetric. Why must

$$\langle\phi_\alpha| \widehat h |\phi_\beta\rangle =0?$$

Solution

The matrix element transforms in

$$\Gamma_\alpha^* \otimes \Gamma_{\mathrm{ts}} \otimes \Gamma_\beta \cong \Gamma_\alpha^* \otimes \Gamma_\beta.$$

By irreducible-character orthogonality, the totally symmetric irrep occurs in $\Gamma_\alpha^*\otimes\Gamma_\beta$ only when the two irreps are equivalent. For inequivalent irreps, group averaging projects the matrix element to zero. This is the finite-group form of the block-diagonal statement behind Schur’s lemma.

Exercise 7: Change the water axes

Move the water molecule from the $yz$ plane to the $xz$ plane while keeping $z$ as the $C_2$ axis. What happens to the $B_1$ and $B_2$ labels? Does the infrared polarization physics change?

Solution

Interchanging the roles of $x$ and $y$ interchanges the two vertical reflection planes. Under the standard $C_{2v}$ convention, labels attached to $x$ and $y$ therefore exchange:

$$B_1 \longleftrightarrow B_2.$$

The asymmetric stretch receives the other printed label. Its displacement pattern and the statement that it couples to dipole polarization in the molecular plane perpendicular to the $C_2$ axis do not change. The relabeling is conventional; the observable relation is geometric.

Exercise 8: Diagnose a split degenerate mode

A calculation on idealized $C_{3v}$ ammonia is run without imposed symmetry. Two modes expected to form an $E$ pair differ by a small frequency. Give a diagnostic sequence that distinguishes numerical symmetry breaking from a physical instability.

Solution

First inspect the optimized Cartesian geometry and quantify deviations from $C_{3v}$ under a stated atom-matching tolerance. Then:

1. tighten geometry, integration-grid, self-consistent-field, and Hessian thresholds;
2. symmetrize the geometry and recompute the analytic or finite-difference Hessian;
3. project the two eigenvectors onto the expected $E$ displacement subspace;
4. test whether the splitting shrinks with numerical convergence;
5. inspect the electronic and vibrational stability of the high-symmetry stationary point;
6. repeat with symmetry explicitly imposed and with it disabled.

At an exactly $C_{3v}$ geometry with an invariant Hamiltonian, the $E$ pair must remain degenerate even if its Hessian eigenvalue is negative. Persistent splitting there means that the Hamiltonian or numerical calculation breaks the assumed symmetry. A lower-energy distorted geometry can have physically split modes because its group is smaller; compare those modes with the unstable but degenerate parent pair.

Term symbols connects these point-group labels to linear-molecule angular projections and total-parity conventions.

References

• P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 5th ed., Oxford University Press, 2011.
• D. M. Bishop, Group Theory and Chemistry, Dover, 1993.
• P. R. Bunker and P. Jensen, Molecular Symmetry and Spectroscopy, 2nd ed., NRC Research Press, 1998, ISBN 978-0-660-17519-5.
• P. R. Bunker and P. Jensen, Fundamentals of Molecular Symmetry, IOP Publishing, 2005, doi:10.1201/9781315273334.
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