Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 2 / c / Solution

Set $X_j=i{H}_j$. The symplectic Lie condition is $X_j^{T}J+J{X}_j=0$, equivalently $H_j^{T}J+J{H}_j=0$. For the supplied diagonal $H_0$, the nonzero entries of $J$ pair levels $1,4$ and $2,3$; the corresponding energy sums are $-\alpha +\alpha =0$ and $-\beta +\beta =0$. Thus $H_{0}J+J{H}_0=0$ for all real $\alpha ,\beta$.

For the supplied real symmetric control matrix, direct multiplication givesBoth generators therefore satisfy the same alternating-form condition. That condition is closed under real linear combinations, and for two such generatorsas follows by substituting $X^{T}J=-JX$ and $Y^{T}J=-JY$ into $(XY{)}^{T}J-(YX{)}^{T}J$. Hence the whole Dynamical Lie algebra obeys it.

If $\dot{U}=X(t)U$ and $U(0)=I$, differentiating givesThus every reachable propagator satisfies $U^{T}JU=J$. Also $J^T=-J$, $J^{†}J=I$, so this is a symplectic dynamical symmetry in quantum control: the generated group lies in the corresponding compact symplectic group, unitarily equivalent to $Sp(2)$. It is not necessary to claim equality with $Sp(2)$, and special choices of $\alpha ,\beta$ can give a smaller group.