Modified Hamiltonian in symplectic Euler method

Question

Now I consider the harmonic oscillator problem.
The ordinal differential equation is
\begin{align*} \dot{q} &= p \\ \dot{p} &= -q \end{align*} In symplectic Euler method, where \begin{align*} \cfrac{p^{(m+1)} - p^{(m)}}{\Delta t} &= -q^{(m)} \\ \cfrac{q^{(m+1)} - q^{(m)}}{\Delta t} &= p^{(m+1)} \end{align*} the flow operator $\psi_{{\rm d}, \Delta} = ((1-\Delta t^2)q + \Delta t \cdot p, -\Delta t \cdot q + p)$ is symplectic.

Here, the textbook suggests that this flow operator stricktly integrates the modified Hamiltonian and this Hamiltonian is invariant. This modified hamiltonian is \begin{align*} \tilde{H} = \cfrac{q^2+p^2}{2} - \cfrac{qp}{2} \Delta t \end{align*}
I cannot understand how to derive $\tilde{H}$.

Answer 1

Compare the energy levels at the intermediate point $$ (p^{(m)})^2+(q^{(m)})^2=(p^{(m+1)}+q^{(m)}Δt)^2+(q^{(m)})^2\\ =(p^{(m+1)})^2+(q^{(m)})^2+2p^{(m+1)}q^{(m)}Δt+(q^{(m)})^2Δt^2 $$ and $$ (p^{(m+1)})^2+(q^{(m+1)})^2=(p^{(m+1)})^2+(q^{(m)}+p^{(m+1)}Δt)^2\\ =(p^{(m+1)})^2+(q^{(m)})^2+2p^{(m+1)}q^{(m)}Δt+(p^{(m+1)})^2Δt^2 $$ The terms of first order in $Δt$ are equal and can be compensated for by subtracting $2pqΔt$, as the additional terms that introduces are of the order $Δt^2$ and higher.

\begin{align} (p^{(m)})^2+(q^{(m)})^2-2p^{(m)}q^{(m)}Δt &=(p^{(m)}-q^{(m)}Δt)^2+(q^{(m)})^2-(q^{(m)})^2Δt^2 \\ &=(p^{(m+1)})^2+(q^{(m)})^2-(q^{(m)})^2Δt^2 \\[0.5em] (p^{(m+1)})^2+(q^{(m+1)})^2-2p^{(m+1)}q^{(m+1)}Δt &=(p^{(m+1)})^2+(q^{(m+1)}-p^{(m+1)}Δt)^2-(p^{(m+1)})^2Δt^2 \\ &=(p^{(m+1)})^2+(q^{(m)})^2-(p^{(m+1)})^2Δt^2 \end{align}

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In the more general case with a Hamiltonian $H=\frac12p^2+V(q)$ and thus $$ p^{(m+1)}=p^{(m)}-V'(q^{(m)})Δt $$ the same approach gives the modified Hamiltionian $$ \tilde H=\frac12p^2+V(q)-pV'(q)Δt, $$ as also $V(q)-pV'(q)Δt=V(q-pΔt)+O(Δt)^2$.