related to the Kepler problem in Sect. \[sec:kepler\].

Now that we have broken down the full Hamiltonian into individual Hamiltonians, to all of which we know the solution (or can easily calculate it), we can construct a symplectic integrator for the total Hamiltonian using an operator split method [e.g. @SahaTremaine1992]. Let us describe the evolution of particles under a Hamiltonian $\operatorname{\mathcal{H}}$ for a time $\mathit{dt}$ using the operator notation $\hat{\operatorname{\mathcal{H}}}(\mathit{dt})$. The notation $\hat{\operatorname{\mathcal{H}}}_2(\mathit{dt})\circ\hat{\operatorname{\mathcal{H}}}_1(\mathit{dt})$ means applying operator $\hat{\operatorname{\mathcal{H}}}_1$ first, then applying operator $\hat{\operatorname{\mathcal{H}}}_2$. It is easy to see that many of the operators commute with each other, i.e. $$\begin{aligned}
\left[\hat \operatorname{\mathcal{H}}_0, \hat \operatorname{\mathcal{H}}_{\rm Kepler} \right] &=& 0 \\
\left[\hat \operatorname{\mathcal{H}}_0, \hat \operatorname{\mathcal{H}}_{\rm Interaction} \right] &=& 0\\
\left[\left(\hat \operatorname{\mathcal{H}}_{\rm Kepler}\right)_i , \left(\hat \operatorname{\mathcal{H}}_{\rm Kepler}\right)_j \right] &=& 0\quad \forall i,j,\end{aligned}$$ where $[\hat{\operatorname{\mathcal{H}}}_1,\hat{\operatorname{\mathcal{H}}}_2] = \hat{\operatorname{\mathcal{H}}}_1\circ \hat{\operatorname{\mathcal{H}}}_2- \hat{\operatorname{\mathcal{H}}}_2\circ \hat{\operatorname{\mathcal{H}}}_1$. This leads to the following Drift-Kick-Drift (DKD) operator splitting scheme, which we refer to as the Wisdom-Holman map:

1.  Evolve the system under $\hat \operatorname{\mathcal{H}}_{\rm Kepler}(\mathit{dt}/2) \circ \hat\operatorname{\mathcal{H}}_0(\mathit{dt}/2)$.

2.  Evolve the system under $\hat \operatorname{\mathcal{H}}_{\rm Interaction}(\mathit{dt})$.

3.  Evolve the system under $\hat \operatorname{\mathcal{H}}_{\rm Kepler}(\mathit{dt}/2) \circ \hat\operatorname{\mathcal{H}}_0(\mathit{dt}/2)$.

The ordering of $\hat \operatorname{\mathcal{H}}_{\rm Kepler}$ and $\operatorname{\mathcal{H}}_0$ in the first and last step doesn’t matter as they commute. The 