= Solution
Set $X_j=iH_j$. The symplectic Lie condition is $X_j^TJ+JX_j=0$, equivalently $H_j^TJ+JH_j=0$. 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 $H_0J+JH_0=0$ for all real $\alpha,\beta$.
For the supplied real symmetric control <matrix>, direct multiplication gives
$$
H_1J=\begin{pmatrix}0&0&1&0\\0&-1&0&1\\1&0&1&0\\0&1&0&0\end{pmatrix}=-JH_1.
$$
Both generators therefore satisfy the same alternating-form condition. That condition is closed under real linear combinations, and for two such generators
$$
[X,Y]^TJ=-J[X,Y],
$$
as follows by substituting $X^TJ=-JX$ and $Y^TJ=-JY$ into $(XY)^TJ-(YX)^TJ$. Hence the whole <Dynamical Lie algebra> obeys it.
If $\dot U=X(t)U$ and $U(0)=I$, differentiating gives
$$
\frac d{dt}(U^TJU)=U^T(X^TJ+JX)U=0.
$$
Thus \b[every reachable propagator satisfies $U^TJU=J$]. Also $J^T=-J$, $J^\dagger J=I$, 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.
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