In my exercise the goal is to find a numerical method to solve:
$$ \dot{q}=p, \quad \dot{p}=-\omega^2q $$
the final method (if I did everything correctly) is:
$$ \begin{pmatrix} q_{k+1} \\ p_{k+1} \end{pmatrix} = \begin{pmatrix} 1-\omega^2h^2/2 & \omega h \\ -\omega h + \omega^3 h^3 / 4 & 1-\omega^2 h^2 / 2 \end{pmatrix} \begin{pmatrix} q_{k} \\ p_{k} \end{pmatrix} $$
Now I‘m supposed to show that this method is symplectic and is of second order. For symplecticity I know that it suffices to show that the determinant of the matrix is $1$. But I dont know how to go about proving the order here. Normally we said that a one step method $x_{k+1}=x_k+h\Phi(t_k,x_k, h)$ had order $p$ if its truncation error:
$$ \frac{x(t_{k+1})-x(t_k)}{h}- \Phi(t_k,x(t_k), h) $$ is in $\mathcal{O}(h^p)$ as $h \rightarrow 0$. But what definition of order should I use in this two dimensional case?
It appears that the method application used the slightly different system $$ \dot q=\omega p,\\ \dot p=-\omega q, $$ as one can read off the terms constant and linear in $h$.
Using the velocity Verlet scheme, this gives $$\begin{align} p_{n+1/2}&=p_n-\frac h2 ωq_n\\ q_{n+1}&=q_n+hωp_{n+1/2}\\&=\left(1-\frac{h^2ω^2}2\right)q_n+hωp_n\\ p_{n+1}&=p_{n+1/2}-\frac h2 ωq_{n+1}\\ &=\left(1-\frac{h^2ω^2}2\right)p_n-hω\left(1-\frac{h^2ω^2}4\right)q_n \end{align}$$ which is exactly what the equation in vector form represents.
There exist numerous sources explaining how velocity Verlet is a second order method, starting with how the time symmetry would cancel any first-order error terms.