I am reading the paper "On splitting methods for Schrödinger-Poisson and cubic nonlinear Schrödinger equations" by Lubich. On page 2147 the author claims
$$[T,V](\psi) = T'(\psi) V(\psi) - V'(\psi) T(\psi) $$
where $T(\psi) = i\Delta \psi$ and $V(\psi) = -i\tilde{V}[\psi]\psi$ with $\tilde{V}([\psi])= \Delta^{-1} |\psi|^2$ where $\psi$ is a complex-valued $H^2$-function on $\mathbb{R}^3$ (so that $\Delta\psi$ is well-defined). It is then claimed that $$ \begin{split} T'(\psi) V(\psi) - V'(\psi) T(\psi) = &\; i\Delta(-i\Delta^{-1}(\psi\bar{\psi}) \psi) + i\Delta^{-1}(-i\Delta \psi \bar{\psi})\psi \\ &\;+ i\Delta^{-1}(\psi \overline{i\Delta\psi}) + i\Delta^{-1}(\psi\bar{\psi})i\Delta \psi \end{split}\label{1}\tag{1} $$ So the derivative in $T'(\psi)$ is probably just some sort of formal derivative with respect to $\psi$ but what would $(i\Delta \psi)'$ be then? The first term in $(1)$ is cleary coming from $T'(\psi)V(\psi)$ but it doesn't make sense to me from the definition of $T'(\psi)$. It would make sense as $T(V(\psi))$, though. So is $T'(\psi)=T(\psi)$ then?
The other terms are confusing as well. Naively, we would have $$(\tilde{V}[\psi]\psi)' = \tilde{V}[\psi]'\psi + \tilde{V}[\psi]$$ which, again, doesn't fit with what we have in \eqref{1}.
More formal than above, $T$ and $V$ are vector fields and their and the corresponding differential equation is $$\partial_t \psi = T(\psi) + V(\psi)$$ In that sense the commutator is the Lie bracket of vector fields.
Maybe someone can give me some clarification?
