Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 4 / b / Solution

Put $g=k\hslash /(2m\omega )$, which has units of energy. In the original oscillator basis,Write $A=\tilde{a}$, $B=\tilde{b}$, $C_2=\cosh 2\theta$ and $S_2=\sinh 2\theta$. The transformed number and pair terms obeyHence the coefficient of $AB+A^{†}{B}^{†}$ is $g{C}_2-\hslash \omega S_2$, which vanishes for

This cancels the requested pair terms but does not diagonalize the printed Hamiltonian. Its number-conserving coupling also transforms:After the pair cancellation, the last creation terms have coefficient $-g{S}_2/2$, nonzero for $k\ne 0$. They create two quanta from the $A,B$ vacuum, so that vacuum is not an eigenstate of the literal Hamiltonian. This is an inconsistency in the ground-state assertion used by the next part, present in the printed PDF.

For the actual coordinate-coupled system, the normal coordinates $q_{\pm }=(x\pm y)/\sqrt{2}$ and their conjugate momenta diagonalize $H$ into independent oscillators withA normalizable ground state requires $|k|<m{\omega }^2$. Eliminating just the two indicated cross-pair terms is not sufficient to establish that ground state.