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Ground state energy

By inserting the results (2.6) and (2.7) for $\overline
V_0$ and $\langle V_{\bf p}V_{\bf -p}\rangle$respectively into the Hamiltonian (2.15) and after integration one gets the following result for the ground state energy
$\displaystyle \frac{E_0}{N} = \frac{\hbar^2}{2ma^2} \left\{
4\pi na^3 \left(1+\...
...\sqrt{\pi}}{15} + 16 \pi^{3/2}
\chi\biggl(\frac{b}{a}\biggr)^2 \right)
\right\}$     (2.17)

The first term in the above equation simply gives the mean field energy (see eq.(1.14)) with the correction due to the presence of the random external potential. This becomes clearer if we rewrite this term in terms of the coupling constant $E_{MF} = 1/2\,(gn+g_{imp}n_{imp})$. The beyond mean-field correction to the ground state energy is given by
$\displaystyle \frac{E_0}{N} - \frac{E_{MF}}{N}
=\frac{\hbar^2}{2ma^2} (na^3)^{3...
...(\frac{512\sqrt{\pi}}{15} + 16 \pi^{3/2}
\chi\left(\frac{b}{a}\right)^2 \right)$     (2.18)

The correction due to disorder in the above result is proportional to
$\displaystyle R = \chi \left(\frac{b}{a}\right)^2,$     (2.19)

which as we will see represents an important parameter to describe the effect of disorder.
next up previous contents
Next: Quantum depletion of the Up: Bogoliubov theory in the Previous: Diagonalization of the Hamiltonian   Contents