= Solution
For the displayed scaling, assume $\mu>0$ and $n_0>0$. Set
$$
\psi_0(\mathbf x)=\sqrt{n_0}\,\widetilde\psi(\widetilde{\mathbf x}),\qquad
\mathbf x=\ell\widetilde{\mathbf x},\qquad
\boxed{\ell=\frac{\hbar}{\sqrt{2m\mu}}}.
$$
Substituting in the stationary <cubic-quintic Gross–Pitaevskii equation> and dividing by $\mu\sqrt{n_0}$ gives
$$
\widetilde\nabla^2\widetilde\psi+
\left(1-\alpha|\widetilde\psi|^2-\beta|\widetilde\psi|^4\right)\widetilde\psi=0,
\qquad
\boxed{\alpha=\frac{V_0n_0}{\mu},\quad\beta=\frac{W_0n_0^2}{\mu}}.
$$
The uniform-state relation proves $\alpha+\beta=1$, so the real bulk amplitude is one; its constant <quantum phase> can be chosen to make $\widetilde\psi\to1$ in the zero-winding sector. The unit of distance $\ell$ is the kinetic-to-chemical-potential scale, reducing to the stated <healing length> convention for purely repulsive two-body coupling.
The positive-$\mu$ assumption is necessary for the printed sign convention. If $\mu<0$, the real length is $\hbar/\sqrt{2m|\mu|}$, and the dimensionless linear term is $-1$, with interaction coefficients divided by $|\mu|$; the displayed equation with linear term $+1$ does not follow. At $\mu=0$ this scaling is undefined. These are limitations of this particular nondimensionalization, not a claim that all cubic-quintic bulk states have positive <chemical potential>.
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