Change of variables for obtaining a unitary group Consider the following NLS:
$$i u_t + \Delta u- 2 \operatorname{Re} u = F(u),$$
where $F(u):=(u + \bar{u} + |u|^2)u.$
In Scattering for the Gross–Pitaevskii equation, the authors S. Gustafson, K. Nakanishi, and T.-P. Tsai used a change of variables to get the free evolution as a unitary group:
$$u \mapsto v:=V^{-1}u:= U^{-1} \operatorname{Re} u + i \operatorname{Im}u,$$
where $U:=\sqrt{- \Delta (2-\Delta)^{-1}}$.
Then $v$ satisfies the equation
$$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v).$$
I have been trying to understand the motivation of this change of variable but no result! The main obstacles for me is the squire root, I can not see where the square root came from? Any help is appreciated.
Updates
I tried to write the equation as a dynamical system of two PDEs, real and imaginary parts, and tried to diagonalized the operator but I couldn't get the same result.
 A: The problem is to show that the equation
$$
iu_t+\Delta u-2\,{\rm Re}\,u=F(u),\label{1}\tag{1}
$$
with $u=Vv$, is equivalent to
$$i v_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v).\label{2}\tag{2}$$
Let me decompose $u=u_1+iu_2$ into real and imaginary parts.
By definition
$$v= [- \Delta (2-\Delta)^{-1}]^{-1/2} u_1 + i u_2,\label{3}\tag{3}$$
hence
$$\sqrt{- \Delta (2-\Delta)} v=(2-\Delta)u_1+i\sqrt{- \Delta (2-\Delta)}u_2.\label{4}\tag{4}$$
I now start from equation (2) and multiply both sides by $Vi$,
$$-u_t+V[i(\Delta-2)u_1+\sqrt{- \Delta (2-\Delta)}u_2]=iF(u).\label{5}\tag{5}$$
I note that
$$V^{-1}(i\Delta u-2iu_1)=i(\Delta-2)u_1+\sqrt{- \Delta (2-\Delta)}u_2.\label{6}\tag{6}$$
Substitution of equation \eqref{6} into equation \eqref{5} gives equation \eqref{1}, so indeed, equations \eqref{1} and \eqref{2} are equivalent.
A: Let $u=u_1+iu_2$, $v=v_1+iv_2$ with $u_1,u_2,v_1,v_2$ being real valued.
What one would like to do is find a $2\times 2$ matrix of real symmetric operators
$$
\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}
$$
such that if one lets
$$
\begin{pmatrix}v_1\\ v_2\end{pmatrix}:=\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}
\begin{pmatrix}u_1\\ u_2\end{pmatrix}\ ,
$$
the free equation $i\partial_t u=-\Delta u+2\Re u$ is equivalent to an evolution for $v$ which is of the form $i\partial_t v=H v$ where $H$ is a real symmetric operator, or rather its complexification.
One could start with general $A,B,C,D$ and see what happens, but with the hindsight provided by the authors of the mentioned article, we know we will be able to arrange for $B,C=0$, and $D={\rm Id}$. Let's go ahead and rename $A=U^{-1}$, for some $U$ TBA.
In terms of real and imaginary components, the equation for $u$ is equivalent to the system
$$
\left\{
\begin{array}{ccc}
\partial_t u_1 &=&   -\Delta u_2\\
\partial_t u_2 &=& \Delta u_1 -2u_1
\end{array}
\right.
$$
while for the components of $v$ we must have
$$
\left\{
\begin{array}{ccc}
\partial_t v_1 &=&   H u_2\\
\partial_t v_2 &=& -H u_1
\end{array}
\right.
$$
Since we assumed the relations $v_1=U^{-1}u_1$, $v_2=u_2$, we immediately get
$$
\partial_t v_1=-U^{-1}\Delta v_2
$$
which suggests $H=-U^{-1}\Delta$.
But we also get
$$
\partial_t v_2=\partial_t u_2=(\Delta-2)u_1=(\Delta-2)Uv_1
$$
which requires also $H=-(\Delta-2)U$. So consistency for the choice of $H$ needs the equation
$$
-U^{-1}\Delta=(2-\Delta)U
$$
to hold. Assuming $U$ commutes with $\Delta$, we obtain the desired square root formula.
