Total energy

For a fixed value of the chemical potential $\mu$ the energy $E' = E-\mu N$ reaches minimum on the ground state function $\phi_0$. From this it follows that $E'$ does not have terms linear in $\delta\psi_{{\bf k}}$ and $\delta\psi_{{\bf k}}^*$, and

$$E'= E^{(0)}+E^{(2)}+{g_{i}}(\phi_0^*\delta\psi(0)+\phi_0\delta\psi^*(0))$$ (7.6)

Here $E^{(0)}=Ngn/2+{g_{i}}\phi_0^2$ is the energy of the system in absence of the perturbation plus the mean-field shift in the energy due to the impurity. The next term comes from the linear expansion of the energy $\int \vert\psi({\bf r})\vert^2 {g_{i}} \delta({\bf r}){\bf dr}$. The term $E^{(2)}$ being quadratic in $\delta\psi_{{\bf k}}$ and $\delta\psi_{{\bf k}}^*$ satisfies the Euler identity:

$$2 E^{(2)} = \int \left[ \delta \psi({\bf r}) \frac{\delta E^{(2)}... ...({\bf r}) \frac{\delta E^{(2)}}{\delta(\delta \psi^*({\bf r}))} \right]{\bf dr}$$ (7.7)

which using the variational equation

$$i \hbar \frac{\partial\delta(\delta \psi)}{\partial t} = \frac{\delta E^{(2)}}{\delta(\delta\psi^*)} + {g_{i}}\phi_0\delta({\bf r})$$ (7.8)

can be rewritten as

$$E^{(2)}= \displaystyle\frac{i\hbar}{2} \int \left[ \delta \psi^*(... ... \right]{\bf dr}-\frac{{g_{i}}}{2}(\phi_0^*\delta\psi(0)+\phi_0\delta\psi^*(0))$$ (7.9)

To start with, let us Fourier transform the first term. Exchanging time derivatives with gradients by the rule (7.3) one obtains

$$E^{(2)} = \int\hbar{\bf k\cdot V}\vert\delta\psi_{{\bf k}}\vert^2\frac{d^3k}{(2\pi)^3}+\frac{{g_{i}}\phi_0}2 (\delta\psi +\delta\psi^*)_{{\bf r}=0}$$ (7.10)

In the energy calculation we assume that the velocity $V$ is small and will make an expansion in powers of $V$ up to quadratic terms. It means that in the calculation of $\vert\delta \psi_{{\bf k}}\vert^2$ the term $(\hbar{\bf kV})^2$ in the denominator of (7.5) can be neglected and $\vert\delta \psi_{{\bf k}}\vert^2$ is written as

$$\vert\delta \psi_{{\bf k}}\vert^2 = \frac{{g_{i}}^2 \vert\phi_0\v... ...2 k^2}{2m} \left( \frac{\hbar^2 k^2}{2m}+2 \mu\right)\right]^2} + {\cal O}(V^4)$$ (7.11)

The energy does not have terms linear in ${\bf V}$, because all terms independent of ${\bf V}$ in (7.11) are even in ${\bf k}$, so once multiplied by ${\bf k}$ and integrated over momentum space they provide zero contribution to the energy. The only term that is left is the following

$$E^{(2)} = 2{g_{i}}^2\vert\phi_0\vert^2 \int \frac{(\hbar{\bf kV})... ... k^2}{2m} \left( \frac{\hbar^2 k^2}{2m}+2 \mu\right)^2} \frac{\bf dk}{(2\pi)^3}$$ (7.12)

For the calculation of $\delta\psi(0)$ in (7.6) and (7.9) one should consider $\delta\psi_{{\bf k}}$ taking into account that $\hbar{\bf kV}\ll\mu$ and then integrate it over the momentum space

$$\begin{array}{rcl} \delta \psi_{k}&=& \displaystyle -\frac{{g_{i}... ...t]^2} +\frac{1}{\frac{\hbar^2 k^2}{2m}+2\mu} \right\} {g_{i}}\phi_0 \end{array}$$ (7.13)

The integral of the second term over momentum space is equal to zero.

The third term suffers from large-$\k$ divergency and one should renormalize the scattering amplitude. It is sufficient to express the coupling constant in the $2^{nd}$ term of eq. (7.10) in terms of the scattering amplitude $b$ using the second order Born approximation:

$${g_{i}}=\frac{2\pi \hbar^2b}{m}\left(1+\frac{2\pi \hbar^2b}{... ...(\frac{\hbar^2k^2}{2m}\right)^{-1}\frac{d^3k}{(2\pi)^3}\right)$$ (7.14)

Now the third term in (7.10) is converging and can be calculated

$${g_{i}}^2n\int\frac{2\mu}{\frac{\hbar^2 k^2}{2m}\left(\frac{\hbar... ...)^3} =8\pi\sqrt{\pi} (na^3)^{3/2}\left(\frac{b}{a}\right)^2\frac{\hbar^2}{ma^2}$$ (7.15)

The energy shift quadratic in velocity $V$ is defined by the following integral

$$\begin{array}{rcl} \delta E&=& \displaystyle E^{(2)} {g_{i}}(\phi... ...left(\frac{\hbar^2 k^2}{2m}+2 \mu\right)^2} \frac{\bf dk}{(2\pi)^3} \end{array}$$ (7.16)

Here we make use of relation $\phi_0=\sqrt{n}$. In a three-dimensional case the term $(\bf {kV})^2 d{\bf k}$ in the integral (7.16) can be replaced by $1/3~k^2 V^2 4\pi k^2 dk$ due to the equivalence of different directions.

$$\delta E = \frac{{g_{i}}^2n \hbar^2 V^2}{6\pi^2\left(\frac{\hbar^... ...t(\frac{\hbar k}{\sqrt{2m}}\right)} {\left(\frac{\hbar^2k^2}{2m}+2 gn\right)^2}$$ (7.17)

This integral can be easily calculated if one recall the integral identity $\int \frac{x^2dx}{(x^2+a^2)^2} = -\frac{x}{2(x^2+a^2)^2} +\frac{1}{2a} \mathop{\rm arctg}\nolimits \frac{x}{a}$. Finally, by collecting everything together and considering $N_{imp}$ impurities with a concentration given by $\chi=N_{imp}/N$ we obtain the energy per particle

$$\frac{E}{N}=\left\{2\pi na^3\left(1+\chi \frac{b}{a}\right) +8\pi... ...2} +\frac{2\sqrt{\pi}}3(na^3)^{1/2}\chi\left(\frac{b}{a}\right)^2\frac{mV^2}{2}$$ (7.18)

If we set $V=0$ we recover Bogoliubov's corrections to the energy in the presence of quenched impurities [HM92,ABCG02]. Note that even if the ``mean-field'' energy obtained from the GP equation in the absence of impurities ($\chi =0$) leaves out terms of the order of $(na^3)^{3/2}$, the equations we obtain in the presence of impurities in a perturbative manner still correctly describe the effect of the disorder up to the terms of the order of $(na^3)^{3/2}$.