 Solution

Write $G(r)=⟨{\psi }^{†}(0)\psi (\mathbf{r})⟩$, $r=|\mathbf{r}|$, and ${\ell }^2=\hslash /(m\omega )$. At zero temperature all $N$ particles occupy ${\varphi }_0(\mathbf{r})=(\ell \sqrt{\pi }{)}^{-1}{e}^{-r^2/(2{\ell }^2)}$. Expanding the bosonic annihilation operator in single-particle orbitals givesThis one-body density matrix is a Gaussian function of width $\ell$; it contains one wavefunction evaluated at $r$, so its exponent is half that of the spatial number density $N|{\varphi }_0(r){|}^2$. In particular,the total number density at the trap centre, including condensed and noncondensed particles.

At positive temperature, with occupations ${\overline{n}}_a$ of the trap energy eigenstates, the exact ideal-gas decomposition isFor $\hslash \omega \ll k_{B}T<k_B{T}_c$, the sketch has a broad Bose-Einstein condensate component on scale $\ell$ and a narrower thermal peak near the origin. On approaching $T_c$ the broad component loses its weight $N_0$, while the thermal part remains. Above $T_c$ there is no macroscopic broad Bose-Einstein condensate component; close to $T_c$, low energy levels can still give a shoulder, so a single narrow Gaussian function is only a high-temperature description. At $k_{B}T\lesssim \hslash \omega$ individual excited energy levels matter and the two-scale description need not apply.

For $T\gg T_c$ the Maxwell-Boltzmann distribution applies. The leading off-diagonal one-body density matrix isHere ${\sigma }_r$ is the Gaussian function width defined by $e^{-r^2/(2{\sigma }_r^2)}$, and ${\lambda }_T$ is the thermal de Broglie wavelength. The thermal cloud itself extends over the much larger spatial scale $\sqrt{k_{B}T/\kappa }$; this is a correlation width, not the cloud radius.