Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / 49

For the ordinary spatial symmetries here, a symmetry operator is a unitary operator $U$ preserving the Hamiltonian:It maps an energy eigenstate to another state with the same energy and preserves the quantum dynamics. More general quantum symmetries can be antiunitary; time reversal is an important separate case.

An ordinary molecular point group consists of the identity, proper rotations $C_n$, and, when present, reflections $\sigma$, inversion $i$, and improper rotations $S_n$. Each operation leaves the nuclear geometry and chemical labels equivalent, so only identical nuclei can be interchanged. Translations are not elements of the point group of a finite isolated molecule. The molecular symmetry operations determine how electronic and vibrational coordinates transform.

The spin isomers of hydrogen illustrate how nuclear spin statistics constrain the product of electronic, vibrational, rotational and nuclear-spin exchange symmetries. In the ground electronic and vibrational states of these homonuclear molecules, the first two factors are symmetric. Exchanging the nuclei reverses the internuclear direction, so a rotational state has parity $(-1{)}^J$.

For ${}^1{\mathrm{H}}_2$, the protons are spin-one-half fermions and the total nuclear-exchange wavefunction must be antisymmetric. The antisymmetric spin singlet pairs with even $J$ (para-hydrogen, one spin state); the symmetric spin triplet pairs with odd $J$ (ortho-hydrogen, three spin states). For ${}^2{\mathrm{H}}_2$, spin-one deuterons are bosons, so the total must be symmetric. Nuclear spins $I=0,2$ supply six symmetric states paired with even $J$ (ortho-deuterium); $I=1$ supplies three antisymmetric states paired with odd $J$ (para-deuterium). The high-temperature ortho/para statistical ratios are $3:1$ for hydrogen and $2:1$ for deuterium, while low-temperature equilibrium populations also depend on rotational energies and whether spin conversion can occur.

For representation spaces $V,W$, the tensor product of group representations acts on product basis vectors throughIts character is the trace. Summing diagonal elements explicitly,No step assumes one-dimensional matrices, so the result applies to both degenerate and nondegenerate representations.

For a real-character irreducible molecular representation with character $\chi$, the multiplicity of the fully symmetric representation in its tensor square isby character orthogonality. The identity uses the real-character assumption appropriate to the usual real irreducible representations intended here. For a general complex irreducible representation, the trivial representation in a tensor square has multiplicity one precisely when it is self-dual, and zero otherwise, by Schur's lemma. The universally valid tensor product is $V\otimes V^{\ast }$.

In the class order $(E,8C_3,3C_2,6S_4,6{\sigma }_d)$, the 15 Cartesian nuclear displacements of methane have characters $(15,0,-1,-1,3)$. Only nuclei left fixed by an operation contribute to that operation's trace. Subtract translations $T_2$ and rotations $T_1$ to obtain the methane normal-mode representationIts dimension is $1+2+3+3=9=3(5)-6$. There are four frequency families: a nondegenerate symmetric stretch, a doubly degenerate bend, and triply degenerate stretch and bend modes.

Choose mass-weighted coordinates $Q_A$, ${\mathbf{Q}}_E=(Q_{E1},Q_{E2})$, and two triplets ${\mathbf{Q}}_3,{\mathbf{Q}}_4$ for the two $T_2$ copies. About a stationary equilibrium geometry, the allowed harmonic potential isThe kinetic energy is one-half the sum of the squared conjugate momenta. Cross terms between inequivalent irreducible representations are forbidden, but the two equivalent $T_2$ triplets permit the displayed scalar-product term. This is equivalent-mode harmonic coupling. An orthogonal rotation between the two triplets diagonalizes their $2\times 2$ force-constant matrix. In molecular normal coordinates the complete Hamiltonian becomesSymmetry enforces degeneracies within each multiplet; it does not require equal frequencies for the two distinct $T_2$ normal modes.

For a real nonmagnetic $T_1$ electronic triplet, the symmetric square of a group representation decomposes asThe $A_1$ coordinate shifts all components equally and does not remove the degeneracy. The linear Jahn-Teller selection rule therefore permits $E$ and $T_2$ vibrational modes to couple and split the triplet. Both methane $T_2$ vibrational triplets are symmetry-eligible. A nonzero allowed coupling generally produces a symmetry-lowering instability, although symmetry alone does not determine its strength or the preferred minimum.

The relevant decomposition is unchanged:Thus the same $E$ and $T_2$ vibrational symmetries are linearly Jahn-Teller active. Their coupling constants and the resulting energy surface can differ from those of a $T_1$ state. The ground-geometry methane normal-mode representation and the symmetry-allowed harmonic invariants are also unchanged; actual force constants in the excited state need not have the same values.