This result may be obtained via ordinary perturbation theory with the Hamiltonian $\mathcal{H}(x)=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(0,\Delta )+U^{†}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}[V(x),0]U={\mathcal{H}}^0+{\rm Y}$, where ${\mathcal{H}}^0$ is the diagonalized Hamiltonian (MSW solution) at point ($x$). We stress that only that separation of the Hamiltonian into a nonperturbative (${\mathcal{H}}^0$) and a perturbative (${\rm Y}$) part leads to results where terms proportional to the full distance traveled by neutrinos in matter are absent; this is, in fact, guaranteed by the existence of the MSW solution.

It is important to recall that $ϵ$ enters both through $V$ and through the adiabatic phase ${\varphi }^m$.

Even more general would be an expansion around a suitable potential for which there is a closed analytic form.

When $V\ll {\Delta }_s\ll {\Delta }_a$, then ${\Delta }_a^m\simeq {\Delta }_a+O(V)$ and $$\begin{gathered} {\Delta }_s^m\simeq \\ {\Delta }_s\sqrt{[\cos 2\theta -(V{c}_{13}^2/{\Delta }_s){]}^2+{sin}^{2}2\theta }+O[s_{13}^2(V^2/{\Delta }_a)] \end{gathered}$$.