Ground state energy

In equation (1.38) the discrete summation over momenta in a volume $V$ should be replaced by integration over $V{\bf dp}/(2\pi\hbar)^3$. The result is

$$\frac{E_0}{N} = \frac{mc^2}{2} \left(1+\frac{128}{15\sqrt{\pi}} (na^3)^{1/2}\right)$$ (1.39)

The first term of this expression gives the mean-field energy and coincides with the result obtained from the Gross-Pitaevskii equation (1.14). The second term gives the correction to the ground state energy arising from the zero-point motion of the quasiparticles

$$\frac{E_0}{N} = \frac{\hbar^2}{2ma^2}\left(4\pi na^3+\frac{512\sqrt{\pi}}{15}(na^3)^{3/2}\right)$$ (1.40)

The result is valid if the system is dilute, i.e. if the gas parameter is small $na^3 \ll 1$.