Gross-Pitaevskii equation and healing length I am given the following GPE:
$$-\frac{\hbar^2}{2mr}\frac{d}{dr} \Big( r \frac{d\chi}{dr} \Big) + \frac{\hbar^2}{2mr^2} \chi + (U_0 \chi^2-\mu)\chi=0.$$
I am asked to derive the healing length (not in tbe usual way of saying that the kinetic part equals the potential part). I have to write $\chi(r)=\sqrt{n_0} + \omega(r)$ where $\omega(r)\ll \sqrt{n_0}$. For the limit $r \rightarrow \infty$ I could derive that $\chi(r \rightarrow \infty)=\sqrt{\mu/U_0}$ and thus $\mu \approx n_0 U_0$. Plugging this all into the previous result yields to leading order in $\omega$:
$$-\frac{\hbar^2}{2mr} \frac{d}{dr} \Big( r \frac{d\omega}{dr} \Big)+\frac{\hbar^2}{2mr^2} (\omega + \sqrt{n_0}) + U_0 (2\omega \sqrt{n_0}),$$
From the that $\omega \approx \sqrt{n_0}$ I should be able to derive the healing length of a vortex $\xi=\hbar/\sqrt{2mU_0 n_0}$. Any suggestions how to do this?
 A: Your GPE should be written as:
$$-\frac{\hbar^2}{2mr}\frac{d}{dr} \Big( r \frac{d\chi}{dr} \Big) + \frac{\hbar^2\color{red}{\ell^2}}{2mr^2} \chi + U_0 \color{red}{|\chi|^2}\chi = \mu\chi,$$
so that it is in the typical Hamiltonian form $H\chi = \text{sth}$. 
$\ell$ is the angular momentum quantum number, because the second term in your expression is the kinetic energy arising from the centrifugal barrier. In a BEC in its ground state, $\ell =0$.  Also, $U_0$ is the interaction constant, and $|\chi|^2$ is the number density $n_0$. 
The size of a vortex is the same as the characteristic lengthscale over which the condensate spatial extent changes significantly. This is when the kinetic part of the Hamiltonian $H_{\mathrm{kin}}$ is roughly the same as the interaction part $H_{\mathrm{int}}$:
$$ \frac{\hbar^2}{2mr}\frac{d}{dr} \Big( r \frac{d\chi}{dr} \Big) \approx  U_0 |\chi|^2\chi . $$
Then use dimensional analysis. 
So $r \sim \xi$ and $\mathrm{d}/\mathrm{d}r \sim 1/\xi$:
$$ \frac{\hbar^2}{2m\xi}\frac{1}{\xi} \Big( \xi \frac{\chi}{\xi} \Big) \approx  U_0 \chi^3,  $$
$$ \Rightarrow \frac{\hbar^2}{2m\xi^2} = U \chi^2,  $$
then plug in $\chi \approx \sqrt{n_0}$:
$$ \frac{\hbar^2}{2m\xi^2} = U n_0,  $$
$$ \Rightarrow \xi \approx \sqrt{\frac{\hbar^2}{2mU_0n_0}} = \frac{\hbar}{\sqrt{2mU_0n_0}}.  $$
