Solution

A transformation of phase space is canonical when it preserves the symplectic form, or equivalently all Poisson brackets. For one degree of freedom this is

d Q \land d P = d q \land d p,

(131)

equivalently

{Q, P}_{q, p} = 1

Suppose a type-two generating function for a canonical transformation

F (q, P)

defines

p = F_{q}, Q = F_{P} .

(132)

Then

d q \land d p d Q \land d P = F_{qP} d q \land d P, = F_{Pq} d q \land d P .

(133)

Equality of the mixed partial derivatives gives

d q \land d p = d Q \land d P

, so the transformation is canonical wherever it is locally invertible.

For

F_{0} (q, P) = \int_{0}^{q} 2 P - u^{2} d u,

(134)

the fundamental theorem of calculus and differentiation under the integral sign give

p = 2 P - q^{2}, Q = \int_{0}^{q} \frac{d u}{2 P - u ^{2}} = arcsin \frac{q}{2 P} .

(135)

Thus

q = 2 P sin Q, p = 2 P cos Q,

(136)

and in particular

P = (p^{2} + q^{2}) /2

For the unit-frequency simple harmonic motion, the transformed Hamiltonian is simply

H = P

. Hamilton's equations become

\dot{Q} = \frac{\partial H}{\partial P} = 1, \dot{P} = - \frac{\partial H}{\partial Q} = 0.

(137)

Therefore

P (t) = E, Q (t) = t - t_{0},

(138)

and transformation back gives the familiar solution

q (t) = 2 E sin (t - t_{0}), p (t) = 2 E cos (t - t_{0}) .

(139)

Now consider the weak quartic oscillator

H (q, p) = \frac{1}{2} (p^{2} + q^{2}) + ϵ q^{4} .

(140)

Choose the modified generating function

F_{ϵ} (q, P) = \int_{0}^{q} 2 P - u^{2} - 2 ϵ u^{4} d u .

(141)

It gives

p = 2 P - q^{2} - 2 ϵ q^{4},

(142)

so the transformed Hamiltonian is again

H = P

. Hence

P = E

and

Q = t - t_{0}

. Expanding the square root with the Taylor series gives

F_{ϵ} (q, P) = F_{0} (q, P) - ϵ \int_{0}^{q} \frac{u ^{4}}{2 P - u ^{2}} d u + O (ϵ^{2}) .

(143)

Differentiation with respect to

P

therefore yields

Q = arcsin \frac{q}{2 P} + ϵ I (q, P) + O (ϵ^{2}),

(144)

where

I (x, y) = \int_{0}^{x} \frac{u ^{4}}{( 2 y - u ^{2} ) ^{3/2}} d u .

(145)

Set

θ = t - t_{0}

and

q_{0} = 2 E sin θ, p_{0} = 2 E cos θ .

(146)

Writing

q = q_{0} + ϵ q_{1} + O (ϵ^{2})

and expanding the inverse sine relation at

q_{0}

gives

0 = \frac{q _{1}}{p _{0}} + I (q_{0}, E), q_{1} = - p_{0} I (q_{0}, E) .

(147)

Expanding

p^{2} = 2 E - q^{2} - 2 ϵ q^{4}

similarly gives

p = p_{0} + ϵ p_{1} + O (ϵ^{2}), p_{1} = q_{0} I (q_{0}, E) - \frac{q _{0}^{4}}{p _{0}} .

(148)

Consequently

\frac{q}{p} = \frac{q _{0}}{p _{0}} + ϵ (\frac{q _{1}}{p _{0}} - \frac{q _{0} p _{1}}{p _{0}^{2}}) + O (ϵ^{2}) = tan θ - ϵ I (q_{0}, E) (1 + tan^{2} θ) + ϵ q_{0}^{2} tan^{3} θ + O (ϵ^{2}) .

(149)

Since

θ = t - t_{0}

, this is the required expression. The formula is understood away from turning points where

p_{0} = 0

and the ratio

q / p

itself is singular.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 15D / Solution

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