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](https://arxiv.org/abs/math/0510080), 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.



Change of variables for obtaining a unitary group
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.


