Quantum depletion of the condensate

It is easy to obtain the particle momentum distribution $\langle N_{\bf p}\rangle = \langle \hat a^\dagger_{\bf p} \hat a_{\bf p}\rangle$ from the operator transformation (2.16).

$$\langle N_{\bf p}\rangle = \frac{\langle n_{\bf p}\rangle + L_p^2... ...(\frac{p^2}{2m}\right)^2 \frac{\langle V_{\bf p}V_{\bf -p}\rangle }{E^4(p)}N_0,$$ (2.20)

where $\langle n_{\bf p}\rangle$ is the number of excitations with momentum ${\bf p}$. At zero temperature such excitations are absent and the distribution (2.20) takes the form

$$\langle N_{\bf p}\rangle = \frac{L_p^2}{1-L_p^2} + \left(\frac{p^... ...{\chi\left(\frac{b}{a}\right)^2}{\left(4+\left(\frac{p}{mc}\right)^2 \right)^2}$$ (2.21)

The first term of the above result corresponds to the momentum distribution in the absence of disorder. The second term gives the contribution due to disorder. The presence of disorder induces an extra quantum depletion of the condensate arising from the particle-impurity interaction. By integrating over momentum one obtains the following result for the condensate fraction

$$\frac{N_0}{N} = 1 - \frac{8}{3\sqrt{\pi}} (na^3)^{1/2} - \frac{\sqrt{\pi}}{2} (na^3)^{1/2} \chi \left(\frac{b}{a}\right)^2$$ (2.22)

It is interesting to note, that the effect of disorder again is described by the combination of parameters $R = \chi (b/a)^2$, as it was found for the energy (2.17).