Physical Applications of Group Theory

Group theory earns its place in physics the moment you notice that a Hamiltonian commutes with a set of symmetry operations. If $\hat{H}$ is invariant under a group $G$ of unitary operators $\{U(g)\}$ , then $[\hat{H},U(g)]=0$ for every $g$ , and each $U(g)$ carries an eigenspace of $\hat{H}$ into itself. Those eigenspaces therefore carry representations of $G$ , and — barring accidental degeneracy — the representations are irreducible. That single sentence converts the vague word "symmetry" into hard, countable predictions: which levels are degenerate, how a perturbation splits them, which spectral lines are allowed, and how many molecular vibrations are infrared or Raman active. No wavefunction is solved along the way; the answers are read off character tables. The engine that does the reading is the character — the trace of a representation matrix. Because traces are invariant under change of basis and constant on conjugacy classes, decomposing any reducible representation into its irreducible pieces reduces to a single weighted inner product of two short lists of numbers. This "magic formula" splits a molecule's $3N$ displacement representation into translations, rotations, and genuine vibrations, and splits an atomic multiplet dropped into a crystal into the irreps of the crystal site. Selection rules follow from the same arithmetic: a transition amplitude is an integral over all space, and by Schur's lemma it vanishes unless its integrand contains the totally symmetric representation. The framework is uniform across the discrete point groups of spectroscopy and the continuous rotation group SO(3) of angular momentum — the same characters, orthogonality, and decomposition — which is why crystal-field splittings are computed by subducing SO(3) irreps to a finite point group.

Key takeaways

A symmetry group $G$ of $\hat{H}$ organizes the Hilbert space into irreducible representations: energy levels are degenerate precisely because they span an irrep, and the degeneracy equals the irrep's dimension.
Any reducible representation is decomposed into irreps by one weighted character inner product — the magic formula $n_\mu=\frac{1}{|G|}\sum_C N_C\,\chi^{(\mu)}(C)^\ast\chi(C)$ — with no diagonalization required.
A matrix element $\langle f|\hat{O}|i\rangle$ is forced to zero unless the trivial representation appears in $\Gamma_f^\ast\otimes\Gamma_O\otimes\Gamma_i$ — the origin of every optical selection rule.
Molecular normal modes and their IR/Raman activity come from building $\Gamma_{3N}$ from unmoved-atom counts and subtracting the $3+3$ translations and rotations; H₂O gives $\Gamma_{\text{vib}}=2A_1\oplus B_1$ .
The same machinery handles continuous groups: SO(3) has irreps $D^{(\ell)}$ of dimension $2\ell+1$ , and subducing $D^{(2)}$ to the octahedral group gives the ligand-field splitting $d\to e_g\oplus t_{2g}$ .
Key idea: A symmetry $G$ of $\hat{H}$ block-diagonalizes the problem by irrep — degeneracies are irrep dimensions, splittings are decompositions of subduced or perturbed representations, and a matrix element survives only if its integrand contains the trivial representation. Everything reduces to character arithmetic.

Where this lives today

The character-table calculations in this article are run, in effect, every day. Vibrational spectroscopists still build $\Gamma_{\text{vib}}$ to assign infrared and Raman bands, and the mutual exclusion rule remains a first-line diagnostic for a centre of symmetry in an unknown compound. Ligand-field theory — the subduction of $D^{(\ell)}$ to a site point group — underlies the colours and magnetic moments of transition-metal and lanthanide complexes. In the solid state, the same little-group analysis at high-symmetry points of the Brillouin zone classifies electronic bands and phonons by their irreps; multidimensional irreps force the degeneracies and symmetry-protected crossings that define topological insulators and Dirac and Weyl semimetals, whose entire classification is a group-theory exercise. The Wigner–Eckart theorem is the backbone of atomic-clock and trapped-ion quantum-computing calculations, where transition strengths between sublevels reduce to Clebsch–Gordan coefficients. And in particle physics the very notion of a particle is an irreducible representation of a symmetry group — the idea, scaled from SO(3) to the Poincaré and gauge groups, that Wigner set in motion.

The physical picture

Picture the Hamiltonian as a big matrix that must look identical after any operation of the symmetry group. Change to a basis of functions that transform "cleanly" — each set spanning an irrep — and that matrix cannot connect functions belonging to different irreps, because a symmetry-invariant operator has no invariant way to link them. The matrix therefore falls apart into independent blocks, one per irrep, and the states inside a single block are automatically degenerate. Symmetry performs this block-diagonalization for free, before any dynamics: the block sizes are fixed by group theory alone, and only each block's energy is left to the physics. A doubly degenerate $E$ level or a triply degenerate $T$ level exists because the group offers no operation that singles out one partner state from the others.

Figure

A symmetry group block-diagonalizes the Hamiltonian: in an irrep-adapted basis a fully coupled 7×7 matrix decouples into blocks 2A₁ ⊕ E ⊕ T₂ of sizes 1, 1, 2 and 3, so seven states collapse to four distinct energy levels whose degeneracies are exactly the irrep dimensions

A matrix element tells the complementary story. It is an integral over all space, and if you rotate the entire picture by a symmetry of the system the integral cannot change its value. But the only piece of an integrand that survives averaging over the whole group is the piece that is itself invariant — the totally symmetric representation. So a transition is allowed exactly when the product of "how the bra, the operator, and the ket transform" contains that invariant component; otherwise the integral is not merely small, it is identically zero. Lower the symmetry with a perturbation, and a level splits by exactly the way its irrep decomposes when restricted to the smaller group — which is what the crystal-field diagram below computes for a $d$ electron placed at the centre of an octahedron.

Figure

A free-ion d level with ℓ = 2 is five-fold degenerate; an octahedral field splits it into a lower triplet t₂g and an upper doublet eg separated by 10Dq, and because the barycentre of the level is preserved t₂g drops by 4Dq while eg rises by 6Dq so that 3×(−4Dq) + 2×(+6Dq) = 0

The core equations

The starting point is the defining statement of a symmetry: the Hamiltonian is unchanged by every group operation, so it commutes with each unitary representative. This is what forces energy eigenspaces to be invariant subspaces, and hence to carry representations of $G$ .

[\hat{H},U(g)]=0\quad\forall\,g\in G \qquad (1)

where: $\hat{H}$ — the Hamiltonian; $U(g)$ — the unitary operator representing group element $g$ on the Hilbert space; $G$ — the symmetry group. Reading it: if $\hat{H}\psi=E\psi$ then $\hat{H}\,U(g)\psi=U(g)\hat{H}\psi=E\,U(g)\psi$ , so $U(g)\psi$ is another eigenstate at the same energy — the eigenspace is closed under $G$ . Everything computational rests on decomposing a reducible representation $\Gamma$ into irreps. Because characters are basis-independent and constant on conjugacy classes, the multiplicity of each irrep is a single inner product, best summed class-by-class with the class sizes as weights.

n_\mu=\frac{1}{|G|}\sum_{g\in G}\chi^{(\mu)}(g)^\ast\,\chi(g)=\frac{1}{|G|}\sum_{C}N_C\,\chi^{(\mu)}(C)^\ast\,\chi(C) \qquad (2)

where: $n_\mu$ — how many times irrep $\mu$ appears in $\Gamma$ ; $|G|$ — the group order; $\chi(g)$ — character of the reducible representation; $\chi^{(\mu)}(g)$ — character of irrep $\mu$ ; $N_C$ — number of elements in class $C$ . Reading it: dot the reducible character into each irreducible one, weight by class size, divide by $|G|$ ; the answer is always a non-negative integer — a built-in check. Selection rules are the second pillar and follow from the same orthogonality. A matrix element cannot depend on the choice of frame, so it must equal its own group average, and that average projects onto the trivial (totally symmetric) representation $\Gamma_1$ .

\langle f|\hat{O}|i\rangle=0\quad\text{unless}\quad \Gamma_1\subset\Gamma_f^\ast\otimes\Gamma_O\otimes\Gamma_i \qquad (3)

where: $\Gamma_1$ — the trivial representation; $\Gamma_i,\Gamma_f$ — irreps of the initial and final states; $\Gamma_O$ — irrep under which the operator transforms; $\otimes$ — direct product of representations. Reading it: form the triple product, decompose it with equation (2), and check whether $n_1\ge 1$ ; if the invariant is absent the amplitude is exactly zero. To use equation (3) you need the symmetry of the operator. The electric-dipole operator is proportional to the position vector, so it transforms as the coordinates $(x,y,z)$ ; the Raman (polarizability) operator transforms as the quadratic forms.

\Gamma_O^{\text{E1}}=\Gamma(x,y,z),\qquad \Gamma_O^{\text{Raman}}=\Gamma(x^2,y^2,z^2,xy,yz,zx) \qquad (4)

where: E1 — electric-dipole process, $\hat{O}\propto\mathbf{r}$ ; Raman — polarizability process, $\hat{O}$ quadratic in coordinates. Reading it: a mode or transition is infrared active if its irrep appears among $(x,y,z)$ in the character table's basis column, and Raman active if it appears among the quadratic forms. For molecular vibrations the reducible representation is the $3N$ -dimensional space of Cartesian displacements. Only atoms left fixed by an operation contribute to its character, and each contributes the trace of the $3\times3$ orthogonal matrix acting on its $(x,y,z)$ .

\chi_{3N}(g)=N_{\text{unmoved}}(g)\,\big(\pm 1+2\cos\theta_g\big) \qquad (5)

where: $N_{\text{unmoved}}(g)$ — atoms left in place by $g$ ; $\theta_g$ — its rotation angle; the sign is $+$ for proper operations ( $E,C_n$ ) and $-$ for improper ones ( $\sigma,S_n,i$ ). Reading it: a proper rotation contributes $1+2\cos\theta$ per fixed atom, an improper one $-1+2\cos\theta$ ; multiply by the count of fixed atoms and you have $\chi_{3N}$ class by class. The physical vibrations are what is left after removing the three rigid translations and three rigid rotations, whose irreps are read straight off the character table's basis columns.

\Gamma_{\text{vib}}=\Gamma_{3N}\ominus\Gamma_{\text{trans}}\ominus\Gamma_{\text{rot}} \qquad (6)

where: $\Gamma_{3N}$ — full displacement representation; $\Gamma_{\text{trans}}$ — irreps spanning $(x,y,z)$ ; $\Gamma_{\text{rot}}$ — irreps spanning $(R_x,R_y,R_z)$ ; $\ominus$ — subtraction of multiplicities. Reading it: what remains has dimension $3N-6$ (or $3N-5$ for a linear molecule), the correct count of normal modes. The discrete story carries over unchanged to the continuous rotation group SO(3), whose irreps are the angular-momentum multiplets $D^{(\ell)}$ . A rotation by $\theta$ is diagonal in the $|\ell m\rangle$ basis, and summing the diagonal phases gives a compact closed form.

\chi^{(\ell)}(\theta)=\sum_{m=-\ell}^{\ell}e^{im\theta}=\frac{\sin\!\big[(\ell+\tfrac12)\theta\big]}{\sin(\theta/2)} \qquad (7)

where: $\ell$ — angular-momentum quantum number, with $\dim D^{(\ell)}=2\ell+1$ ; $\theta$ — rotation angle; $m$ — magnetic quantum number. Reading it: evaluating this at the rotation angles of a finite point group is exactly what subduces an atomic multiplet to a crystal site. Finally, combining two angular momenta is the decomposition of a tensor product of SO(3) irreps — the Clebsch–Gordan series, the continuous-group version of equation (2).

D^{(j_1)}\otimes D^{(j_2)}=\bigoplus_{J=|j_1-j_2|}^{j_1+j_2}D^{(J)} \qquad (8)

where: $j_1,j_2$ — the angular momenta combined; $J$ — total angular momentum, stepping by 1. Reading it: the dimensions must balance, $(2j_1+1)(2j_2+1)=\sum_J(2J+1)$ , another automatic check.

Derivation ASelection rules from the vanishing-integral theorem

The goal is to prove, from symmetry alone, that a matrix element $I=\langle f|\hat{O}|i\rangle=\int\psi_f^\ast\,\hat{O}\,\psi_i\,\mathrm{d}\tau$ is forced to vanish unless the integrand contains the trivial representation, then to specialize to the electric dipole and recover the atomic rule $\Delta\ell=\pm1$ . The whole argument is that an integral over all space is a number, and a number cannot change under a relabelling of coordinates by a symmetry of the system, so $I$ must equal its own average over the group — a projection onto the invariant part of the integrand.

Step 1. Because the volume measure $\mathrm{d}\tau$ is invariant under every $U(g)$ , the integral is unchanged if we act with $g$ on all three factors and then average over the group: $I=\frac{1}{|G|}\sum_g\int[U(g)\psi_f]^\ast\,[U(g)\hat{O}U(g)^{-1}]\,[U(g)\psi_i]\,\mathrm{d}\tau$ . Why: this rewrites the single integral as the group average of itself, which is the trick that exposes what part of the integrand can survive.

Step 2. The product $\psi_f^\ast\,\hat{O}\,\psi_i$ is an object built from three pieces living in $\Gamma_f^\ast$ , $\Gamma_O$ and $\Gamma_i$ , so under $G$ it transforms in the tensor product $\Gamma_f^\ast\otimes\Gamma_O\otimes\Gamma_i$ . Why: products of representation-carrying quantities transform in the product representation, which is the only fact about tensor products we need.

Step 3. Averaging a representation over the group is the projector onto its trivial-irrep content, because $\frac{1}{|G|}\sum_g\chi^{(\mu)}(g)=\delta_{\mu,1}$ — every non-trivial irrep sums to zero by orthogonality with the constant character. Why: the group average annihilates everything that is not invariant, leaving only the totally symmetric part standing.

Step 4. Therefore $I\propto n_1\big(\Gamma_f^\ast\otimes\Gamma_O\otimes\Gamma_i\big)$ : if the trivial representation does not appear in the triple product, the projection is zero and $I=0$ identically, not merely small. Why: this is the selection rule in its most general form, valid for any group and any operator.

Step 5. Specialize to the electric dipole, $\hat{O}=\mathbf{r}$ , which transforms as $D^{(1)}$ for a free atom. Then $D^{(1)}\otimes D^{(\ell)}=D^{(\ell-1)}\oplus D^{(\ell)}\oplus D^{(\ell+1)}$ , so a non-zero amplitude to a state $D^{(\ell')}$ requires $\ell'\in\{\ell-1,\ell,\ell+1\}$ ; parity ( $\mathbf{r}$ is odd) then kills $\Delta\ell=0$ , leaving $\Delta\ell=\pm1$ . Why: the abstract rule collapses to the textbook atomic result the instant you feed it the rotation group and the dipole operator. *The abstract rule specializes cleanly: for atoms it returns $\Delta\ell=\pm1,\ \Delta m=0,\pm1$ , and for molecules it becomes the point-group rule that a mode is IR active if and only if its irrep lies in $\Gamma(x,y,z)$ .*

Go deeper · Derivation B · 2 minNormal modes and IR/Raman activity of water

Derivation BNormal modes and IR/Raman activity of water

The goal is to count and label every vibration of H₂O and decide which are infrared and which Raman active, using nothing but the $C_{2v}$ character table. Water has $N=3$ atoms and point group $C_{2v}=\{E,C_2,\sigma_v(xz),\sigma_v'(yz)\}$ with $|G|=4$ ; we place the molecule in the $xz$ plane with the $C_2$ axis along $z$ , the standard convention that makes the antisymmetric stretch come out as $B_1$ . The plan is to build the $3N=9$ displacement character from unmoved-atom counts, decompose it with the magic formula, and strip away the translations and rotations.

Step 1. Count atoms left in place by each operation: $E$ fixes all three; $C_2$ (the $z$ axis) fixes only the oxygen; the molecular plane $\sigma_v(xz)$ fixes all three; the perpendicular plane $\sigma_v'(yz)$ fixes only the oxygen. So $N_{\text{unmoved}}=(3,1,3,1)$ . Why: only atoms that stay put contribute to the character, so this count is the whole geometric input.

Step 2. Multiply by the per-atom contribution $\pm1+2\cos\theta$ : for $E$ it is $1+2\cos0=3$ , for $C_2$ it is $1+2\cos\pi=-1$ , and for each reflection it is $-1+2\cos0=1$ . Hence $\chi_{3N}=(3\cdot3,\ 1\cdot(-1),\ 3\cdot1,\ 1\cdot1)=(9,-1,3,1)$ . Why: this turns the geometry of Step 1 into the reducible character that equation (2) will digest.

Step 3. Apply the magic formula against the four $C_{2v}$ irreps $A_1(1,1,1,1)$ , $A_2(1,1,-1,-1)$ , $B_1(1,-1,1,-1)$ , $B_2(1,-1,-1,1)$ , each class of size 1: $n_{A_1}=\tfrac14(9-1+3+1)=3$ , $n_{A_2}=\tfrac14(9-1-3-1)=1$ , $n_{B_1}=\tfrac14(9+1+3-1)=3$ , $n_{B_2}=\tfrac14(9+1-3+1)=2$ , giving $\Gamma_{3N}=3A_1\oplus A_2\oplus 3B_1\oplus 2B_2$ . Why: the nine displacement coordinates are now sorted into symmetry species, and $3+1+3+2=9$ confirms the dimension.

Step 4. Remove the rigid motions read from the table: $\Gamma_{\text{trans}}=(z,x,y)=A_1\oplus B_1\oplus B_2$ and $\Gamma_{\text{rot}}=(R_z,R_y,R_x)=A_2\oplus B_1\oplus B_2$ . Subtracting, $\Gamma_{\text{vib}}=(3-1)A_1\oplus(1-1)A_2\oplus(3-1-1)B_1\oplus(2-1-1)B_2=2A_1\oplus B_1$ . Why: the three translations and three rotations carry no vibrational information, and removing them leaves exactly $3N-6=3$ genuine modes.

Step 5. Read off the activities: $A_1$ carries $z$ (IR) and $x^2,y^2,z^2$ (Raman); $B_1$ carries $x$ (IR) and $xz$ (Raman). Both species are therefore simultaneously infrared and Raman active — the symmetric stretch and bend ( $A_1$ ) and the antisymmetric stretch ( $B_1$ ) all appear in both spectra. Why: with no inversion centre nothing forbids mutual activity, and this is exactly what is measured for water. *As a check: $\chi_{3N}=(9,-1,3,1)$ decomposes to $3A_1\oplus A_2\oplus 3B_1\oplus 2B_2$ , and after removing translations and rotations $\Gamma_{\text{vib}}=2A_1\oplus B_1$ — three fundamentals, all IR and Raman active, in agreement with the observed bands at 3657, 1595 and 3756 cm⁻¹.*

Go deeper · Derivation C · 1 minSplitting of a d level in an octahedral crystal field

Derivation CSplitting of a d level in an octahedral crystal field

The goal is to compute how a five-fold degenerate free-ion $d$ level splits when the ion sits at the centre of an octahedron, by subducing the SO(3) irrep $D^{(2)}$ to the finite octahedral rotation group $O$ . This is the founding calculation of ligand-field theory: it turns the continuous rotational symmetry of a free atom into the discrete cubic symmetry of a crystal site and reads the level pattern off a character table. The group $O$ has order 24 and classes $E,\,8C_3,\,3C_2,\,6C_4,\,6C_2'$ with rotation angles $0,\,2\pi/3,\,\pi,\,\pi/2,\,\pi$ .

Step 1. Evaluate the SO(3) character $\chi^{(2)}(\theta)=\sin(5\theta/2)/\sin(\theta/2)$ at each class angle. At $\theta=0$ it is the dimension, $5$ ; at $2\pi/3$ , $\sin(5\pi/3)/\sin(\pi/3)=-1$ ; at $\pi$ , $\sin(5\pi/2)/\sin(\pi/2)=1$ ; at $\pi/2$ , $\sin(5\pi/4)/\sin(\pi/4)=-1$ . So over $(E,8C_3,3C_2,6C_4,6C_2')$ the characters are $\chi^{(2)}=(5,-1,1,-1,1)$ . Why: these five numbers are the reducible character of the $d$ level regarded as a representation of the smaller group $O$ .

Step 2. Decompose with the magic formula, using class sizes $(1,8,3,6,6)$ and $|G|=24$ . Against $E:(2,-1,2,0,0)$ one gets $n_E=\tfrac{1}{24}(1\cdot2\cdot5+8\cdot(-1)(-1)+3\cdot2\cdot1+0+0)=\tfrac{24}{24}=1$ . Why: the doublet $E$ appears once, and the class-weighted inner product gives the integer directly.

Step 3. Against $T_2:(3,0,-1,-1,1)$ one gets $n_{T_2}=\tfrac{1}{24}(1\cdot3\cdot5+0+3\cdot(-1)\cdot1+6\cdot(-1)(-1)+6\cdot1\cdot1)=\tfrac{1}{24}(15-3+6+6)=1$ ; the same arithmetic against $A_1,A_2,T_1$ returns zero. Why: only $E$ and $T_2$ survive, so the five-dimensional level splits into a doublet and a triplet with nothing left over.

Step 4. Assemble the result, $D^{(2)}\big\downarrow_{O}=E\oplus T_2$ , with dimensions $2+3=5$ . In the full group $O_h$ (adding inversion) both keep the even label from $\ell=2$ , giving $e_g\oplus t_{2g}$ . Why: the barycentre-preserving split of the $d$ orbitals into the doublet $\{d_{z^2},d_{x^2-y^2}\}$ and the triplet $\{d_{xy},d_{yz},d_{zx}\}$ is the celebrated $10\,Dq$ ligand-field splitting. *As a check: $\chi^{(2)}=(5,-1,1,-1,1)$ over the classes of $O$ decomposes to $E\oplus T_2$ ; in the full group $O_h$ the labels become $e_g$ (upper, $+6Dq$ ) and $t_{2g}$ (lower, $-4Dq$ ), separated by the measured $10Dq$ .*

Limiting cases

How the general formalism specializes at the extremes of symmetry:

Case	Result	Meaning
Full rotational symmetry	$D^{(\ell)},\ \dim=2\ell+1$	Free atom: SO(3) irreps give the familiar $(2\ell+1)$ -fold degeneracy of orbital angular momentum.
Cubic crystal field ( $O_h$ )	$D^{(2)}\downarrow E_g\oplus T_{2g}$	A $d$ level ( $\ell=2$ , five-fold) splits into a doublet $E_g$ and a triplet $T_{2g}$ — the ligand-field splitting $10Dq$ .
No symmetry	$G=\{e\}$	Only the trivial one-dimensional irrep exists: every level is non-degenerate and every transition is symmetry-allowed.
Centrosymmetric molecule	IR active $\Leftrightarrow$ Raman inactive	An inversion centre imposes the mutual exclusion rule — no fundamental is both IR and Raman active.

Common trap“Symmetry-allowed” means the transition is strong, and group theory hands you its rate.

It does neither. Group theory is a pure yes/no oracle: it tells you whether an amplitude is forced to vanish and whether a degeneracy is required, but it never supplies the non-zero magnitude — that comes from the actual wavefunctions and the dynamics. A "symmetry-allowed" line can still be vanishingly weak, and a "symmetry-forbidden" one is exactly zero only within the assumed approximation: electric-dipole-forbidden transitions routinely appear through vibronic coupling, magnetic-dipole or quadrupole channels, or spin–orbit mixing. Two further traps hide here. First, group theory cannot rule out accidental degeneracies from a hidden larger symmetry — the $n^2$ degeneracy of hydrogen comes from an SO(4) that the geometric group $SO(3)$ knows nothing about. Second, the label you assign a mode (say $B_1$ versus $B_2$ for water's antisymmetric stretch) depends on your axis convention; the physics — three modes, all IR and Raman active — does not.

Worked examples

1 · Vibrations of ammonia (fully worked, C₃ᵥ)

Ammonia NH₃ is a trigonal pyramid with $N=4$ atoms and point group $C_{3v}=\{E,2C_3,3\sigma_v\}$ , $|G|=6$ . Find the number, symmetry species, and IR/Raman activity of its normal modes.

Step 1. Count unmoved atoms. $E$ fixes all four; each $C_3$ (about the principal axis through N) fixes only the nitrogen; each mirror plane $\sigma_v$ contains N and one hydrogen, fixing two atoms. So $N_{\text{unmoved}}=(4,1,2)$ over $(E,2C_3,3\sigma_v)$ .

Step 2. Per-atom contributions: $E$ gives $1+2\cos0=3$ ; $C_3$ gives $1+2\cos(2\pi/3)=1-1=0$ ; $\sigma_v$ gives $-1+2\cos0=1$ . Multiply: $\chi_{3N}=(4\cdot3,\ 1\cdot0,\ 2\cdot1)=\mathbf{(12,0,2)}$ .

Step 3. Decompose with the magic formula, class sizes $(1,2,3)$ , against $A_1(1,1,1)$ , $A_2(1,1,-1)$ , $E(2,-1,0)$ : $n_{A_1}=\tfrac16(12+0+3\cdot2)=3$ , $n_{A_2}=\tfrac16(12+0-3\cdot2)=1$ , $n_{E}=\tfrac16(2\cdot12+0+0)=4$ . Thus $\Gamma_{3N}=3A_1\oplus A_2\oplus 4E$ , and the dimension checks: $3+1+8=12$ .

Step 4. Remove translations $\Gamma_{\text{trans}}=(z)+(x,y)=A_1\oplus E$ and rotations $\Gamma_{\text{rot}}=(R_z)+(R_x,R_y)=A_2\oplus E$ . Then $\Gamma_{\text{vib}}=(3-1)A_1\oplus(1-1)A_2\oplus(4-1-1)E=\mathbf{2A_1\oplus 2E}$ , of dimension $2+4=6=3N-6$ .

Step 5. Read activities: $A_1$ carries $z$ and $z^2,x^2+y^2$ — IR and Raman active; $E$ carries $(x,y)$ and $(x^2-y^2,xy),(xz,yz)$ — IR and Raman active. So ammonia has four distinct fundamental frequencies (2A₁ + 2E), every one of them both infrared and Raman active — the symmetric stretch and the umbrella bend ( $A_1$ ), and the degenerate stretch and degenerate bend ( $E$ ), observed near 3337, 950, 3444 and 1627 cm⁻¹.

2 · Mutual exclusion in carbon dioxide (linear, D∞ₕ)

CO₂ is linear and centrosymmetric, $N=3$ . Show that it has four vibrational modes and that no fundamental is both IR and Raman active.

Step 1. A linear molecule has $3N-5$ vibrations, so CO₂ has $9-5=\mathbf{4}$ modes. Analysing the displacement representation of the $D_{\infty h}$ group sorts them into a symmetric stretch $\Sigma_g^+$ , an antisymmetric stretch $\Sigma_u^+$ , and a doubly degenerate bend $\Pi_u$ — a total of $1+1+2=4$ , as required.

Step 2. Classify by parity. The subscript $g$ (gerade, even under inversion) or $u$ (ungerade, odd) is decisive: quadratic forms are always $g$ , and the coordinates $(x,y,z)$ are always $u$ . Hence Raman activity lives on $g$ modes and IR activity on $u$ modes.

Step 3. Apply it. The symmetric stretch $\Sigma_g^+$ is Raman active only (it appears among $x^2+y^2,z^2$ ) and IR silent — the two C=O bonds lengthen in phase, so the dipole never changes. The antisymmetric stretch $\Sigma_u^+$ (2349 cm⁻¹) and the bend $\Pi_u$ (667 cm⁻¹) are IR active only. No mode is both: this is the mutual exclusion rule, and observing a band in only one of the two spectra is a direct fingerprint of a centre of symmetry.

Physics of the portrait

Group theory did not walk into physics on its own; it was carried there, against considerable resistance, by a handful of people who saw that symmetry was not decoration but computation. The clearest single figure is Eugene Wigner, whose 1931 book made representation theory a working tool of the quantum physicist.

Origins · 1 minEugene Wigner and how this was discovered

Eugene Wigner

1902–1995 · group theory and symmetry in quantum mechanics

Eugene Paul Wigner, a Hungarian-born chemical engineer turned theoretical physicist, was the person who most decisively made group theory indispensable to physics — at a time when many dismissed it as the "Gruppenpest," the group-theory plague. Drawn into the new quantum mechanics in the late 1920s, Wigner realized that the degeneracies and selection rules of atomic spectra were not accidents of the Schrödinger equation but consequences of rotational symmetry, expressible entirely through the representation theory of the rotation group. His 1927 papers introduced the representation-theoretic analysis of atomic terms, and the resulting Wigner–Eckart theorem — that every matrix element of an irreducible tensor operator factorizes into a geometric Clebsch–Gordan coefficient times a single reduced matrix element — remains the most-used consequence of Schur's lemma in all of spectroscopy. His 1931 monograph Gruppentheorie und ihre Anwendung auf die Quantenmechanik der Atomspektren systematized the field and taught a generation to compute with characters. Wigner also proved that symmetries must be represented by unitary or antiunitary operators, pioneered time-reversal, and laid the groundwork for classifying relativistic particles by the irreducible representations of the Poincaré group. He shared the 1963 Nobel Prize in Physics "through the discovery and application of fundamental symmetry principles" — the citation that made "symmetry principle" a household phrase in physics.

History

1927

— Wigner applies representation theory to the structure of atomic spectra

1928

— Weyl's Gruppentheorie und Quantenmechanik brings the machinery to a broad audience

1929

— Bethe computes crystal-field term splittings in his Termaufspaltung in Kristallen

1931

— Wigner's monograph systematizes group theory for the quantum mechanics of atomic spectra

1963

— Wigner shares the Nobel Prize for fundamental symmetry principles

Going further

Texts. Tinkham, Group Theory and Quantum Mechanics (Dover) — the classic physicist's treatment of characters and selection rules · Hamermesh, Group Theory and Its Application to Physical Problems (Dover) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., ch. 16–17 · Stone & Goldbart, Mathematics for Physics (CUP 2009), ch. on group representations and their applications · Hassani, Mathematical Physics, ch. on group representation theory · Cotton, Chemical Applications of Group Theory, 3rd ed. (for the spectroscopy). The key numerical results are: (1) the H₂O displacement character $\chi_{3N}=(9,-1,3,1)$ in the $xz$ -plane convention decomposes to $3A_1\oplus A_2\oplus 3B_1\oplus 2B_2$ , giving $\Gamma_{\text{vib}}=2A_1\oplus B_1$ ; (2) the NH₃ character $(12,0,2)$ decomposes to $3A_1\oplus A_2\oplus 4E$ and $\Gamma_{\text{vib}}=2A_1\oplus 2E$ ; (3) the SO(3) character $\chi^{(2)}=(5,-1,1,-1,1)$ over the classes of $O$ gives $D^{(2)}\to E\oplus T_2$ ; (4) the Clebsch–Gordan dimensions satisfy $(2j_1+1)(2j_2+1)=\sum_J(2J+1)$ . Note the convention subtlety: with the molecule in the $yz$ plane the same physics labels the antisymmetric stretch $B_2$ — the mode count and activities are convention-independent, only the Mulliken label flips.

Prerequisites. Group Representations · Reducible and Irreducible Representations · Schur's Lemmas · Characters and Orthogonality Theorems · Character Tables and Decomposition of Representations · Tensor Products and Clebsch–Gordan Decomposition. Leads to. The Rotation Group SO(3) · Lie Algebras and the Exponential Map · Tensor Products and Clebsch–Gordan Decomposition.

Ask the tutor: Tell me a molecule or an ion in a crystal field and I will build its character, decompose it, and read off the degeneracies, normal modes, and IR/Raman selection rules step by step.

Why does a symmetry of the Hamiltonian create degeneracy?

If $\hat{H}$ commutes with every $U(g)$ , and $\psi$ is an eigenstate with energy $E$ , then $U(g)\psi$ is also an eigenstate with the same $E$ , because $\hat{H}(U(g)\psi)=U(g)(\hat{H}\psi)=E\,U(g)\psi$ . The whole set $\{U(g)\psi\}$ therefore lies in one energy level and forms an invariant subspace — a representation of $G$ . Generically that representation is irreducible (if it were reducible, an extra symmetry would be hiding), so the degeneracy of the level equals the dimension of an irrep. This is why the allowed degeneracies of a system are dictated in advance by its symmetry group: for a point group they are the irrep dimensions listed in the character table, and nothing else can occur without an accident.

What is the "magic formula" really doing?

It is an orthogonal projection. The irreducible characters form an orthonormal basis for class functions under $\langle\chi',\chi\rangle=\frac{1}{|G|}\sum_g\chi'(g)^\ast\chi(g)$ . Any reducible character is an integer combination $\chi=\sum_\mu n_\mu\chi^{(\mu)}$ , so the inner product with $\chi^{(\nu)}$ picks out its coefficient $n_\nu$ by orthogonality — exactly as projecting a vector onto a basis direction returns its component. Summing class-by-class with the class sizes $N_C$ as weights is a bookkeeping speed-up, since the character is constant on each class. That every $n_\mu$ comes out a non-negative integer is a strong check: a fractional or negative answer means an arithmetic slip.

How do I know which operator symmetry to use for a transition?

Match the operator to the physical process and read its transformation from the character table's basis columns. Electric-dipole (E1) transitions use $\hat{O}\propto\mathbf{r}$ , transforming as $(x,y,z)$ — so a mode or state is IR active when its irrep sits beside $x$ , $y$ , or $z$ . Raman scattering probes the polarizability, quadratic in the field, so its operator transforms as $x^2,y^2,z^2,xy,yz,zx$ , and Raman activity is read from the quadratic-form column. Magnetic-dipole transitions use the rotations $(R_x,R_y,R_z)$ , and electric-quadrupole transitions the quadratic forms again. The table lists all of these precisely so you can look up the operator's irrep and feed it into the triple-product selection rule.

Why place water in the xz plane rather than the yz plane?

Purely to match the most common textbook label. The physics is identical either way: H₂O has three modes, two of species $A_1$ and one of the $B$ type, all IR and Raman active. But the name of that $B$ mode depends on which in-plane axis you call $x$ . In the $xz$ -plane convention the antisymmetric stretch comes out $B_1$ ; in the $yz$ -plane convention the identical mode is called $B_2$ , because $B_1$ and $B_2$ are distinguished only by their behaviour under the two vertical mirrors, and swapping the planes swaps the labels. Mulliken's conventions do not fully fix the choice for $C_{2v}$ , so both appear in the literature. Always state your axis choice; never treat the $B_1$ -versus- $B_2$ label as physical.

Can group theory tell me how strong an allowed transition is?

No — and this is its single most important limitation. Group theory is a binary oracle: it certifies only that an amplitude is either forced to zero or not forced to zero. When a transition is allowed, the actual magnitude depends on the overlap of the real wavefunctions and the dynamics, which no character table contains. A symmetry-allowed line can still be extremely weak, and a formally forbidden line can appear once the assumptions relax — through vibronic coupling, higher multipoles, or spin–orbit mixing. The value of the method is in the zeros it guarantees and the degeneracies it enforces; for intensities you hand the surviving matrix elements to perturbation theory or a direct computation.

How does the same framework handle both molecules and atoms?

Through the identical toolkit — characters, orthogonality, decomposition — applied to different groups. For a molecule the group is a finite point group with a handful of tabulated irreps, and you decompose the $3N$ displacement representation. For a free atom the group is the continuous rotation group SO(3), whose irreps are the multiplets $D^{(\ell)}$ of dimension $2\ell+1$ , with the character formula $\chi^{(\ell)}(\theta)=\sin[(\ell+\tfrac12)\theta]/\sin(\theta/2)$ replacing a finite table. Crystal-field theory is the bridge: take the atomic $D^{(\ell)}$ , evaluate its SO(3) character only at the rotation angles present in the crystal's point group, and decompose with the same magic formula — the continuous problem reduced to a finite one.