I am currently trying to solve the non-linear Schrodinger equation(NLSE) in cylindrical coordinates, given by $i\hbar\frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\psi+g|\psi|^{2}\psi$, where $g$ is a constant.
When I write the laplacian in cylindrical coordinates and split $\psi$ as $\psi=f(r,t)e^{is\theta}\lambda(z)$, the NLSE gets converted to $i\hbar\lambda(z)\frac{\partial f(r,t)}{\partial t}=-\frac{\hbar^{2}}{2m}\Big(\lambda(z)\frac{\partial^{2}f(r,t)}{\partial r^{2}}+\lambda(z)\frac{1}{r}\frac{\partial f(r,t)}{\partial r}-\frac{s^{2}\lambda(z)f(r,t)}{r^{2}}+f(r,t)\frac{\partial^{2}\lambda(z)}{\partial z^{2}}\Big)+g\;f^{3}(r,t)\lambda^{3}(z)$.
My question is regarding the right hand side of the above equation. Now that I have a $\theta$ independence in the equation above, can I just replace $r\rightarrow x$ in the above equation to get the $x-z$ profile, given by $\lambda(z)f(x,y=0,t)$? If yes, why? If not, why not?