Nonlinear Schrodinger equation (NSL) is $$ i\phi_t + \Delta \phi + \phi^{2\sigma+1} $$ I want to show for any $\lambda , \gamma \in \mathbb R$ and $x_0\in \mathbb R^N$, $$ \psi=\lambda^{1/\sigma}R(\lambda(x+x_0))\exp{i(\lambda^2t +\gamma)} $$ is a ground state of NSL. The ground state is solution of $$ \Delta R -R + R^{2\sigma+1}=0 $$
What I try: $$ i\frac{\partial \psi}{\partial t}=- \lambda^{1/\sigma +2} \exp(i(\lambda^2 t + \gamma))R(\lambda(x+x_0)) \\ \Delta \psi =\lambda^{1/\sigma +2} \exp(i(\lambda^2 t + \gamma))\Delta R(\lambda(x+x_0)) \\ \psi^{2\sigma+1} = \lambda^{2+1/\sigma}\exp(i(\lambda^2 t + \gamma)(2\sigma+1)) R^{2\sigma+1}(\lambda(x+x_0)) $$ so \begin{align} i\frac{\partial \psi}{\partial t}+ \Delta \psi +\psi^{2\sigma+1} &= \lambda^{1/\sigma +2}\exp(i(\lambda^2 t + \gamma)) (-R+\Delta R + R^{2\sigma +1}\exp(i(\lambda^2t+\gamma))^{2\sigma}) \\ &\ne 0 \end{align} where is my mistake ? This question is from the 1.8 of 55th page of [1].
[1]Weinstein, Michael I., Lyapunov stability of ground states of nonlinear dispersive evolution equations, Commun. Pure Appl. Math. 39, 51-67 (1986). ZBL0594.35005.
The Schrödinger equation is actually $ i \phi_t + \Delta \phi+|\phi|^{2\sigma}\phi$. When you take the absolute value of $|\psi|^{2\sigma}$, the phase disappears, and so $ i \psi_t + \Delta \psi+|\psi|^{2\sigma}\psi = 0$.