 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3/49/c/ii/solution

Q_{A}

Q_{E} = (Q_{E 1}, Q_{E 2})

, and two triplets

Q_{3}, Q_{4}

for the two

T_{2}

copies. About a stationary equilibrium geometry, the allowed harmonic potential is

V^{(2)} = \frac{1}{2} k_{A} Q_{A}^{2} + \frac{1}{2} k_{E} ∣ Q_{E} ∣^{2} + \frac{1}{2} k_{33} ∣ Q_{3} ∣^{2} + k_{34} Q_{3} \cdot Q_{4} + \frac{1}{2} k_{44} ∣ Q_{4} ∣^{2} .

The 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

H = E_{0} + \frac{1}{2} (P_{A}^{2} + ω_{1}^{2} Q_{A}^{2}) + \frac{1}{2} j = 1 \sum 2 (P_{E j}^{2} + ω_{2}^{2} Q_{E j}^{2}) + \frac{1}{2} j = 1 \sum 3 (P_{3 j}^{2} + ω_{3}^{2} Q_{3 j}^{2} + P_{4 j}^{2} + ω_{4}^{2} Q_{4 j}^{2}) .

Symmetry enforces degeneracies within each multiplet; it does not require equal frequencies for the two distinct

T_{2}

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