I'm trying to replicate a result from this paper: Physical Review A 76, 063614 (2007). It's for a class in classical mechanics, so we're only concerned with Lagrangian densities and such. I must marginalize a Lagrangian density, but I'm not sure how to do that.
Here's the problem
I must substitute the Ansatz: $$ \phi(r,z,t)=\frac{ r^s}{\sqrt{\pi s!\,}\sigma^{s+1}(z,t)}\exp\left[-\frac{r^2}{2\sigma^2(z,t)}\right]f(z,t) $$ into the lagrangian density: $$ \mathcal{L}(\Psi^{},\Psi^{\dagger},\partial_{\mu}\Psi,\partial_{\mu}\Psi^{\dagger})= \frac{i}{2}\left(\Psi^{\dagger}\frac{\partial \Psi}{\partial t}-\Psi\frac{\partial\Psi^{\dagger}}{\partial t}\right)- \frac{1}{2}|\nabla\Psi|^{2}-\frac{1}{2}(x^{2}+y^{2})|\Psi|^{2}-\pi g|\Psi|^{4}, $$ where $$ \Psi(\vec{r},t)=\phi(r,z,t)e^{is\theta}, $$ And from there marginalize the density so that it only depends on $z$. This is where I'm not entirely sure if I'm doing it right.
My attempt
In cylindrical coordinates, $dv=r\,dr\,d\theta\, d\psi\,dz$ for $r>0$ and $0 \leq \psi <\infty$. Then I only integrated with respect to $r$ and $\psi$; is that right?
After substituting and differentiating with respect to time, and focusing only on the first Lagrangian density term, (ignoring the $i/2$ term): $$ \left(f^{\dagger}\frac{\partial f}{\partial t}-f\frac{\partial f^{\dagger}}{\partial t}\right)\int_{0}^{2\pi} \,d\psi \int_{0}^{\infty} \frac{r^{s+1}}{\sigma^{s+1}(z,t)\sqrt{\pi s!}}\exp\left[-\frac{r^2}{2\sigma^2(z,t)} \right]\,dr $$
$$ =\frac{2\pi}{\sigma^{s+1}(z,t)\sqrt{\pi s!}}\left(f^{\dagger}\frac{\partial f}{\partial t}-f\frac{\partial f^{\dagger}}{\partial t}\right)\int_{0}^{\infty} r^{s+1}\exp\left[-\frac{r^2}{2\sigma^2(z,t)}\right]\,dr. $$ Is this how the marginalization is done? The idea is a density that only depends on $z$.
