canonical Hamiltonian can be expressed in terms of the fluctuations (excitations) $\delta\hat{\Psi}({\bf r})=\hat{\Psi}({\bf r})-\sqrt{N_0}\phi_0({\bf r})$, and up-to quadratic order in $\delta \hat{\Psi}$, it is given by [@Biswas2010] $$\begin{aligned}
\label{eqn:62}
\hat{\mathcal H}&=&\Omega_0+\int d^3{\bf r} \left[\delta\hat{\Psi}^{\dagger}({\bf r})\bigg(-\frac{\hbar^2}{2M}\nabla^2\bigg)\delta\hat{\Psi}({\bf r})\text{d}^3{\bf r} + \frac{g\bar{n}}{2} \right. \nonumber \\ 
&\times&\left. \bigg(2\delta\hat{\Psi}^{\dagger}({\bf r})\delta\hat{\Psi}({\bf r})+\delta\hat{\Psi}^{\dagger}({\bf r})\delta\hat{\Psi}^{\dagger}({\bf r})
+ \delta\hat{\Psi}({\bf r})\delta\hat{\Psi}({\bf r})\bigg) \right],~~~\end{aligned}$$ where $\bar{n}\approx N_0\phi_0^2(\textbf{r})$ denotes the bulk particle density, $\mu\approx g\bar{n}$, and $\Omega_0$ is the grand potential for the condensate, which is given by [@Biswas2010] $$\begin{aligned}
\label{eqn:63}
\frac{\Omega_0}{N_0} = \int\bigg(\frac{\hbar^2}{2M}|\nabla\phi_0(\textbf{r})|^2-\mu|\phi_0(\textbf{r})|^2+\frac{gN_0}{2}|\phi_0(\textbf{r})|^4\bigg)\text{d}^3{\textbf{r}}.\end{aligned}$$ In a Bose-Einstein condensate low energetic excitations are probabilistically favored as temperature of the system decreases. We already have experienced in the previous section that, while the Casimir energy for the TM modes (which obey Dirichlet boundary condition) are positive, that for the TE modes (which obey Neumann boundary condition) are negative. Thus, the natural boundary condition for the elementary excitations in the condensate in an infinitely long cylinder, is expected to be of Neumann type. However, by the definition of the quantum vacuum energy, contribution of the elementary excitations of Dirichlet type is also unavoidable even for $T\rightarrow0$ unless any filtering of modes is engineered (e.g. like that of a resonant cavity) at the two ends of 