Past exam of the theoretical physics I course of the University of Cambridge / 2012 / Paper 1 / 2

A one-degree-of-freedom canonical transformation requires the Poisson bracket $\{x,p{\}}_{X,P}=1$. To first order,The product of the off-diagonal derivatives contributes only at second order. Consequentlywith ${\alpha }_3$ and ${\beta }_1$ unconstrained at this order. The self-brackets vanish identically. This is perturbative preservation of the symplectic form, not exact canonicity of the truncated polynomial map.

Treat all transformation coefficients as $O(\lambda )$, so products of them with $\lambda$ are second order. Substitution into the Hamiltonian gives its harmonic part plusSet these four coefficients to zero and impose the canonical constraints. The result isThus the first-order canonical removal of a cubic oscillator perturbation isup to $O({\lambda }^2)$. It yields $K=P^2/(2m)+m{\omega }^2{X}^2/2+O({\lambda }^2)$. The inverse map, if old initial data are specified, is $X=x+\lambda x^2/(m{\omega }^2)+2\lambda p^2/(m^3{\omega }^4)$ and $P=p-2\lambda xp/(m{\omega }^2)$ to the same order. There is no time-dependent generating term because the map is autonomous.

Hamilton's equations give $\dot{X}=P/m+O({\lambda }^2)$ and $\dot{P}=-m{\omega }^{2}X+O({\lambda }^2)$. Hence, with constants $A\ge 0$ and $\delta$ fixed by transformed initial data,An equivalent form uses $X(0)\cos \omega t+[P(0)/(m\omega )]\sin \omega t$ for $X(t)$. The indicated errors are a fixed-time perturbative statement; a higher-order frequency correction can accumulate phase on longer times.

Insert the harmonic variables into the forward map and reduce the squared trigonometric functions. This givesIndeed $p=m\dot{x}$ at this order. The second harmonic and mean displacement of a weakly cubic oscillator show asymmetry of the motion: for positive $\lambda$ the mean shifts towards negative $x$. There is no first-order frequency shift, although the waveform is no longer purely sinusoidal.

The small parameter is $|\lambda |A/(m{\omega }^2)\ll 1$. A purely cubic perturbation is unbounded in one direction, so these formulas describe small local oscillations around the stable origin, not globally confined motion at arbitrary energy. Amplitude $A$ labels the transformed orbit and must be inferred using the inverse map if physical initial position and momentum are supplied.