The nonlinear Schrödinger equation (NSLE) is given by:
$$\frac{\partial A}{\partial z} + i\beta_2\frac{\partial^2A}{\partial t^2} = i\gamma|A|^2A$$
It has a (trivial in $t$) solution $A(t,z) =\sqrt P e^{i\gamma Pz}$. According to Wikipedia if we perturb this to $\tilde A =(\sqrt P+\epsilon(t,z)) e^{i\gamma Pz}$ and linearize in $\epsilon$, we obtain
$$\frac{\partial \varepsilon}{\partial z}+i\beta_2\frac{\partial^2\varepsilon}{\partial t^2}=i\gamma P \left(\varepsilon+\varepsilon^*\right)$$
My question is: shouldn't the RHS be $$i\gamma P(2\epsilon + \epsilon^*)\quad?$$
Indeed, $$i\gamma|\tilde A|^2\tilde A = i\gamma(\sqrt P+\epsilon)(\sqrt P+\epsilon^*)(\sqrt P+\epsilon)e^{i\gamma Pz}$$
So once we collect the linear terms in $\epsilon$, and cancel the $e^{i\gamma Pz}$, we should have $2\epsilon$.