One-body density matrix

For a homogeneous system the one body density matrix is defined as the Fourier transform of the momentum distribution (1.42)

$$\rho({\bf r}) = \frac{N_0}{V}+\int N_{\bf p} e^{i{\bf pr}/\hbar} \frac{\bf dp}{(2\pi\hbar)^3},$$ (1.45)

where the contribution of the condensate has been extracted from the integral. After angular integration one gets

$$\rho(r) = \frac{N_0}{V} +\rho^{(1)}(r) = \frac{N_0}{V}+ \frac{1}{... ...\limits_{0}^{\infty} N_p \sin\left(\frac{pr}{\hbar}\right) \frac{p dp}{\hbar^2}$$ (1.46)

At $T = 0$ the momentum distribution $N_p$ is given by (1.42). By introducing the dimensionless variable $\xi = p / mc$, one obtains the following result for the coordinate dependent part of the one-body density matrix

$$\rho^{(1)}(r) = n\frac{4}{\pi x} \int\limits_0^\infty \frac{\sin(\sqrt{4\pi na^3}x\xi) d\xi} {\sqrt{4+\xi^2}(\xi\sqrt{4+\xi^2}+\xi^2+2)},$$ (1.47)

where $r = ax$. For $r = 0$, $\rho^{(1)}(0) = \displaystyle\frac{8}{3\sqrt{\pi}} (na^3)^{3/2}$ and the one-body density matrix coincides with the total density $\rho(0) = n$. For $r \gg r_0$ (where $r_0 = \hbar/mc = a/\sqrt{8\pi na^3}$ is the healing length) $\rho^{(1)}(r) = \sqrt{na}/(2\pi\sqrt{\pi}r^2)$ (see derivation and comments in Appendix B). Thus the asymptotic value of the one-body density matrix coincides with the condensate density $\lim\limits_{r\to\infty} \rho(r) = N_{0}/V$. For arbitrary values of $r$ the integral (1.47) can be calculated numerically. Results for different values of $na^3$ are shown in Fig. 1.1

Figure 1.1: One-body density matrix