Diagonalization of the Hamiltonian

Let us rewrite Hamiltonian (1.1) in terms of the creation and annihilation operators $\hat a_{\bf p}$ and $\hat a_{\bf p}^\dagger$ in momentum representation

$$\hat H = \sum \frac{p^2}{2 m} \hat a_{\bf p}^\dagger\hat a_{\bf p... ...angle{\bf p}\vert V\vert{\bf p'}\rangle\,\hat a_{\bf p}^\dagger \hat a_{\bf p'}$$ (2.9)

where we have included the external potential $V$. We use the Bogoliubov prescription $\hat a^\dagger_0 = \hat a_0 = \sqrt{N_0}$ and we consider $\hat a_{\bf p}, {\bf p \ne 0}$ as small perturbations. To second order in $\hat a^\dagger_{\bf p}$ for $\hat a_{\bf p}$ the external potential term can be written as

$$\sum \langle{\bf p}\vert V\vert{\bf p'}\rangle\,\hat a_{\bf p}^\d... ...V_0} + \sqrt{N_0} \sum(\hat a_{\bf p}^\dagger V_{\bf p} + \hat a_p V_{\bf -p}),$$ (2.10)

The term $N_0 \overline V_0$ must be calculated in the second Born approximation in order to obtain an expression which is correct up to second order in the particle-impurity scattering amplitude.

$$\overline{V_0} = \frac{mc^2}{2}\chi\left(\frac{b}{a}\right) +\fra... ...\left(\frac{b}{a}\right)^2 \sum\limits_{\bf p\ne 0} \left(\frac{mc}{p}\right)^2$$ (2.11)

The part of the Hamiltonian which is independent of the external potential can be diagonalized by the Bogoliubov transformation (1.30). The Hamiltonian takes the form

$$\hat H= N \frac{mc^2}{2}\left(1 + \chi\frac{b}{a} \right)+ \frac{... ...{\hat b_{\bf p}^\dagger + L_p \hat b_{\bf -p}}{\sqrt{1-L_p^2}}V_{\bf -p}\right)$$ (2.12)

where $E_p$ and $L_p$ are defined by (1.36) and (1.35) respectively. The linear term in the quasiparticle operators can be eliminated by means of the following transformation (analogous transformation, but for a different model of the disorder was introduced in [16])

$$\left\{ \begin{array}{ccl} \hat b_{\bf p}&=& \hat c_{\bf p} + Z_p... ...{\bf p}^\dagger&=&\hat c_{\bf p}^\dagger + Z_p V_{\bf -p}\\ \end{array}\right.$$ (2.13)

with $Z_p$ defined by

$$Z_p = -\sqrt{\frac{1+L_p}{1-L_p}} \frac{\sqrt{N_0}}{E(p)} = -\sqrt{\frac{p^2}{2m E(p)}} \frac{\sqrt{N_0}}{E(p)}$$ (2.14)

The transformation (2.13) does not change the commutation rules and the new quasiparticle operators $\hat c_{\bf p}$, $\hat c_{\bf p}^\dagger$ satisfy the usual bosonic commutation relations. Finally, the Hamiltonian takes the form

$$\hat H= N\,\frac{mc^2}{2}\left(1 + \chi\frac{b}{a} \right)+ \sum\... ... -2N_0\frac{p^2}{2m} \frac{\langle V_{\bf p} V_{\bf -p}\rangle }{E(p)^2}\right]$$ (2.15)

The creation and annihilation particle operators $\hat a_{\bf p}^\dagger$, $\hat a_{\bf p}$ are obtained from the corresponding quisiparticle operators $\hat c_{\bf p}^\dagger$, $\hat c_{\bf p}$ in the following way

$$\left\{ \begin{array}{ccccc} \displaystyle\hat a_{\bf p}&=& \disp... ...p^2}} -\sqrt{N_0}\frac{p^2}{2m} \frac{V_{\bf -p}}{E^2(p)}\\ \end{array}\right.$$ (2.16)