Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 34D / b / ii

Solution

F_{i} = \frac{1}{2} (p_{i}^{2} + W_{i}^{2} Q_{i}^{2}) - \frac{a _{i}^{2}}{8 W _{i}^{2}} .

Hence the positive oscillator energy is

E_{i} = F_{i} + \frac{a _{i}^{2}}{8 W _{i}^{2}} .

Using the convention

I_{i} = \frac{1}{2 π} \oint p_{i} d q_{i},

2 π

. With oscillator frequency

ω_{i} = ∣ W_{i} ∣

, that area is

2 π E_{i} / ω_{i}

. Therefore

I_{i} = \frac{1}{∣ W _{i} ∣} (F_{i} + \frac{a _{i}^{2}}{8 W _{i}^{2}}) .

Equivalently,

H = \sum_{i = 1}^{n} (∣ W_{i} ∣ I_{i} - \frac{a _{i}^{2}}{8 W _{i}^{2}}),

∣ W_{i} ∣

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