Bogoliubov) excitations in the condensate can be treated perturbatively. The elementary excitations of a BEC are generally described by expanding the Hamiltonian up-to the second order in quantum fluctuations over the BEC order parameter ($\sqrt{N_0}\phi_0(\bf r)$), via a standard perturbative approach [@Biswas2010]. We follow a similar approach here for obtaining the elementary excitations for the cylindrical geometry.

Let the position vector and mass of a single Bose particle be denoted as ${\bf r}$ and $M$ respectively. The grand canonical Hamiltonian operator for such a system of interacting Bose gas is given by [@Biswas2010] $$\begin{aligned}
\label{eqn:61}
\hat{\mathcal H}&=&\int\hat{\Psi}^{\dagger}({\bf r})\bigg(-\frac{\hbar^2}{2M}\nabla^2-\mu\bigg)\hat{\Psi}({\bf r})\text{d}^3{\bf r}\nonumber \\ &+&\frac{1}{2}\int\int\bigg(\hat{\Psi}^{\dagger}({\bf r})\hat{\Psi}^{\dagger}({\bf r'})V({\bf r-r'})\hat{\Psi}({\bf r})\hat{\Psi}({\bf r'})\bigg)\text{d}^3{\bf r}\text{d}^3{\bf r'},~~~\end{aligned}$$ where $\mu$ is the chemical potential, $V({\bf r-r'})$ is the inter-particle interaction potential, and $\hat{\Psi}({\bf r})$ is the field operator for Bose particles. For the simplest case, the interaction potential can be considered as $V({\bf r-r'})=g\delta^3({\bf r-r'})$, where $g=4\pi\hbar^2a_s/M$ is the coupling constant and $a_s$ is the s-wave scattering length [@Biswas2010]. The field operator can be expressed in terms of the single particle orthonormal wave functions $\{\phi_i({\bf r})\}$, as $\hat{\Psi}({\bf r})=\sum_{i=0}^{\infty}\phi_i({\bf r})\hat{a}_i$, where $\hat{a}_i$ ($\hat{a}_i^{\dagger}$) annihilates (creates) a Bose particle in the state $\phi_i({\bf r})$. Within a perturbative approach, the grand 