Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 83 / 2 / iv

Set $r={\ell }_0\tilde{r}$, $R=\sqrt{\mu /U}f(\tilde{r})$, using the healing length ${\ell }_0=\hslash /\sqrt{2m\mu }$ and $U,\mu >0$. For zero external potential, division by $\mu R$ producesPrimes now denote dimensionless-radius derivatives. Let $f=1-a/{\tilde{r}}^2+O({\tilde{r}}^{-4})$. Then $f^{\prime \prime }+f^{\prime }/\tilde{r}=O({\tilde{r}}^{-4})$, while the centrifugal term is $-{\mathcal{N}}^2/{\tilde{r}}^2+O({\tilde{r}}^{-4})$ and $(1-f^2)f=2a/{\tilde{r}}^2+O({\tilde{r}}^{-4})$. Their leading balance gives $2a={\mathcal{N}}^2$. Therefore the inverse-square tail of a Gross–Pitaevskii vortex yieldsFor a singly quantized vortex the answer is $1/2$. This is the amplitude deficit; the density deficit $1-f^2$ has limiting coefficient ${\mathcal{N}}^2$ instead.