Series expansion of the OBDM

The one-body density matrix of the pure system at zero temperature is given by (1.47). One can rewrite this integral as

$$\rho^{(1)}(r) = \frac{na}{\pi r} \int\limits_0^\infty F(\xi) \sin\left(\frac{\xi r}{\sqrt{2}r_0}\right) d\xi$$ (6.1)

with the function $F(\xi)$ defined as

$$F(\xi) = 2\left(1+\frac{\xi^2}{4}\right)^{1/2} -\left(1+\frac{\xi^2}{4}\right)^{-1/2}-\xi$$ (6.2)

and $r_0 = a/\sqrt{8\pi na^3}$ being the healing length. Let us integrate (B.1) by parts $2{\cal K}$ times. All terms which have the form $\sin(\xi r/\sqrt{2}r_0)F^{(k)}(\xi)\Bigl.\Bigr\vert^\infty_0$ disappear, because the sinus function is equal to zero in $\xi=0$ and $F^{(k)}(\infty) = 0$, terms with cosine contribute $\cos(\xi r/\sqrt{2}r_0)F^{(k)}(\xi)\Bigl.\Bigr\vert^\infty_0 = -F^{(k)}(0)$. The result of the integration is the following

$$\begin{array}{rcl} \rho^{(1)}(r) &=&\displaystyle \frac{na}{\pi r... ...2}r_0}\right) \left(\frac{\sqrt{2}r_0}{r}\right)^{2{\cal K}+1} d\xi \end{array}$$ (6.3)

The $k$-th derivative of the function $F$ can be calculated from the differentiation of the Taylor expansion of $F$ in zero

$$F^{(2k)}(0) = \frac{1+2k}{1-2k} \frac{\sqrt{\pi}}{\Gamma\Bigl(\frac{1}{2}-k\Bigr)} \frac{(2k)!}{2^{2k}\,k!}$$ (6.4)

As a result one has a representation of $\rho(r)$ in series of powers of $1/r^2$. By using the expansion (B.3) one can calculate the leading terms of $\rho^{(1)}(r)$ for $r\to\infty$

$$\rho^{(1)}(r) = \frac{\sqrt{na}}{2\pi\sqrt{\pi}}\cdot\frac{1}{r^2... ...qrt{na}\,r_0^2}{4\pi\sqrt{\pi}}\cdot\frac{1}{r^4} + O\left(\frac{1}{r^6}\right)$$ (6.5)

Which means that the asymptotic behavior $r\to\infty$ of $\rho(r)$ is one over distance squared^6.1. This behavior is correct at distances where the contribution from the second term in (B.5) can be neglected, i.e. $r\gg\sqrt{3/2}~r_0\approx r_0$ which means distances much larger than the healing length.