 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1/2/b/solution

Treat all transformation coefficients as

O (λ)

, so products of them with

λ

are second order. Substitution into the Hamiltonian gives its harmonic part plus

(m ω^{2} α_{1} + λ) X^{3} + (2 m ω^{2} α_{2} + β_{1} / m) X^{2} P + (m ω^{2} α_{3} + 2 β_{2} / m) X P^{2} + (β_{3} / m) P^{3} .

Set these four coefficients to zero and impose the canonical constraints. The result is

α_{1} = - \frac{λ}{m ω ^{2}}, α_{2} = 0, α_{3} = - \frac{2 λ}{m ^{3} ω ^{4}}, β_{1} = 0, β_{2} = \frac{λ}{m ω ^{2}}, β_{3} = 0.

x = X - \frac{λ}{m ω ^{2}} X^{2} - \frac{2 λ}{m ^{3} ω ^{4}} P^{2}, p = P + \frac{2 λ}{m ω ^{2}} XP,

O (λ^{2})

. It yields

K = P^{2} / (2 m) + m ω^{2} X^{2} /2 + O (λ^{2})

. The inverse map, if old initial data are specified, is

X = x + λ x^{2} / (m ω^{2}) + 2 λ p^{2} / (m^{3} ω^{4})

and

P = p - 2 λ x p / (m ω^{2})

to the same order. There is no time-dependent generating term because the map is autonomous.

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