Find the fraction of bosons not in the concentrate in a weakly interacting bosonic superfluid


 * 1) Identify the general expression for the fraction of bosons not in the concentrate: \(f = N^{-1}\sum \limits_{\mathbf{k} \neq 0}\langle \hat{c}_{\mathbf{k}}^{\dagger} \hat{c}_{\mathbf{k}}\rangle\)
 * 2) Calculate \(\langle \hat{c}_{\mathbf{k}}^{\dagger} \hat{c}_{\mathbf{k}}\rangle\).
 * 3) Take the Bogoliubov transformation, as described here: \(\hat{c}_{\mathbf{k}}^{\dagger} = u_{\mathbf{k}} \alpha_{\mathbf{k}}^{\dagger} - v_{\mathbf{k}} \beta_{\mathbf{k}}\) \(\hat{c}_{\mathbf{k}} = u_{\mathbf{k}} \alpha_{\mathbf{k}} - v_{\mathbf{k}} \beta_{\mathbf{k}}^{\dagger}\)
 * 4) Substitute these into the expression for the boson fraction: \(f = N^{-1}\sum \limits_{{\mathbf{k}} \neq 0}\langle \left(u_{\mathbf{k}} \alpha_{\mathbf{k}}^{\dagger} - v_{\mathbf{k}} \beta_{\mathbf{k}} \right) \left(u_{\mathbf{k}} \alpha_{\mathbf{k}} - v_{\mathbf{k}} \beta_{\mathbf{k}}^{\dagger}\right)\rangle\)
 * 5) Simplify: \(f = N^{-1}\sum \limits_{{\mathbf{k}} \neq 0} \left(u_{\mathbf{k}}^2 \langle \alpha_{\mathbf{k}}^{\dagger}\alpha_{\mathbf{k}}\rangle + v_{\mathbf{k}}^2 \langle  \beta_{\mathbf{k}}^{\dagger}\beta_{\mathbf{k}} + 1\rangle \right)\)
 * 6) Use the general expression \(\langle \hat{a}_k^{\dagger} \hat{a}_q\rangle = \delta_{k,q} n_B E\left(k\right)\) to simplify: \(f = N^{-1}\sum \limits_{{\mathbf{k}} \neq 0} \left(u_{\mathbf{k}}^2 + v_{\mathbf{k}}^2 n_B \left(E\left({\mathbf{k}}\right) + v_{\mathbf{k}}^2\right) \right)\)
 * 7) Take the limit as temperature approaches absolute zero, so \(n_B \to 0\), and \(v_{\mathbf{k}} \to 0\) as \({\mathbf{k}} \to 0\): \(f = N^{-1}\sum \limits_ v_{\mathbf{k}}^2 \)
 * 8) Let \(N \to \infty\) and integrate over dimensions \(\mathbf{k}_1, \cdots, \mathbf{k}_d\): \(f = \int \frac{d^d {\mathbf{k}}}{\left(2\pi\right)^d} v_{\mathbf{k}}^2 \)
 * 9) Substitute \(v^2= \frac{1}{2}\left(\frac{F}{\sqrt{F^2 - G^2}} - 1\right)\): \(f = \int \frac{d^d {\mathbf{k}}}{2\left(2\pi\right)^d} \left(\frac{F}{\sqrt{F^2 - G^2}} - 1\right) \)
 * 10) Use the values for \(F\) and \(G\) for the bosonic superfluid: \(F=\left( \frac{|{\mathbf{k}}|^2}{2m} + 2gn_0\right)\) \(G=\left(2gn_0\right)\)
 * 11) Substitute these into the expression: \(f = \int \frac{d^d {\mathbf{k}}}{2\left(2\pi\right)^d} \left(\frac{\frac{|{\mathbf{k}}|^2}{2m} + 2gn_0}{\sqrt{\left( \frac{|{\mathbf{k}}|^2}{2m} + 2gn_0\right)^2 - \left(2gn_0\right)^2}} - 1\right) \)
 * 12) Simplify: \(f = \frac{1}{4gn_1\left(2\pi\right)^d} \int d^d {\mathbf{k}} \left(\frac{\frac{|{\mathbf{k}}|^2}{4mgn_0} + 1}{\sqrt{\left( \frac{|{\mathbf{k}}|^2}{2m} + 2gn_0\right)^2 - \left(2gn_0\right)^2}} - 1\right) \)
 * 13) Substitute \(k^2 = \frac{|{\mathbf{k}}|^2}{4mgn_0}\): \(f = \frac{1}{8gn_1\sqrt{mgn_0}\left(2\pi\right)^d} \int d^d k \left(\frac{k^2 + 1}{\sqrt{2gn_0k^2\left(2gn_0k^2 + 4gn_0\right)}} - 1\right) \)
 * 14) Simplify: \(f = \frac{1}{\left(2\pi\right)^d} \sqrt{\frac{m}{gn_0}}\frac{1}{2gn_0}\int d^d k \left(\frac{k^2 + 1}{\sqrt{k^4 + 2k^2}} - 1\right) \)