the cylinder. The quantum fluctuations (*i.e.* the elementary excitations) for the Neumann boundary on a cylindrical tube can be expressed as $\delta\hat{\Psi}(\textbf{r})= L (2 \pi \hbar)^{-1} \sum_{\sigma=-\infty,
n=1}^{\infty}\int_{-\infty}^\infty \phi_{p_z,\sigma,n} \hat{a}_{p_z,\sigma,n} {d}p_z$, where $\hat{a}_{p_z,\sigma,n}$ annihilates a Bose particle specified by the state $\phi_{p_z,\sigma,n}(z,\rho,\theta)=\sqrt{\frac{2}{R^2L}}\text{e}^{i\frac{p_zz}{\hbar}}\frac{e^{i\sigma\theta}}{\sqrt{2\pi}}\frac{J_\sigma(\beta_{\sigma,n}\rho/R)}{J_{\sigma+1}(\beta_{\sigma,n})}$ [@Cavity-Modes], whose momentum in the $z$-direction is $p_z$ and whose energy is $\xi=\frac{p_z^2}{2M}+\frac{\hbar^2\beta_{\sigma,n}^2}{2MR^2}$. The Hamiltonian in Eq.  can now be diagonalized in terms of the phononic operators through the usual Bogoliubov transformation: [@Roberts] $\hat{a}_{p_z,\sigma,n}=u_{p_z,\sigma,n}\hat{b}_{p_z,\sigma,n}+v_{p_z,\sigma,n}\hat{b}_{-p_z,\sigma,n}^{\dagger}$ , where $u_{p_z,\sigma,n}=\big(\frac{\xi^2+g\bar{n}}{2\epsilon_N(|p_z|,\sigma,n)}+\frac{1}{2}\big)^{1/2}$, $v_{p_z,\sigma,n}=-\big(\frac{\xi^2+g\bar{n}}{2\epsilon_N(|p_z|,\sigma,n)}-\frac{1}{2}\big)^{1/2}$, and $$\label{eqn:64}
\epsilon_N(|p_z|,\sigma,n)=\bigg[\frac{g\bar{n}}{M}\bigg(p_z^2+\frac{\hbar^2\beta_{\sigma,n}^2}{R^2}\bigg)\bigg(1+\frac{p_z^2+\frac{\hbar^2\beta_{\sigma,n}^2}{R^2}}{4Mg\bar{n}}\bigg)\bigg]^{\frac{1}{2}}. \\$$

For the TM modes (i.e. for the Dirichlet boundary condition), on the other hand, the Bogoliubov excitation energy would be $\epsilon_D(|p_z|,\sigma,n)$ which is almost like $\epsilon_N(|p_z|,\sigma,n)$ in Eq.  except $\beta_{\sigma,n}$ be replaced by $\alpha_{\sigma,n}$ everywhere. For the TM modes, we also call the phononic annihilation and creation operators by $\hat{d}_{p_z,\sigma,n}$ and $\hat{d}_{p_z,\sigma,n}^{\dagger}$ respectively instead of $\hat{b}_{p_z,\sigma,n}$ and $\hat{b}_{p_z,\sigma,n}^{\dagger}$ used for the TE modes.

Thus from Eqs.  and , and also from the subsequent discussion about the TM modes, we have the diagonalized Hamiltonian in terms of both the (Neumann and Dirichlet) types of the Bogoliubov excitations, as $$\begin{aligned}
\label{eqn:65}
\hat{{\mathcal H}}&=& \Omega_0+{\mathcal E}_0(R,\bar{n}) + \sum_{p_z,\sigma,n}\epsilon_N(|p_z|,\sigma,n)\hat{b}_{p_z,\sigma,n}^{\dagger}\hat{b}_{p_z,\sigma,n}\nonumber\\&&+ \sum_{p_z,\sigma,n}\epsilon_D(|p_z|,\sigma,n)\hat{d}_{p_z,\sigma,n}^{\dagger}\hat{d}_{p_z,\sigma,n}, \end{aligned}$$ where $$\begin{aligned}
{\mathcal E}_0(R,\bar{n})&=&\frac{1}{2}\sum_{p_z,\sigma,n}\bigg[ \epsilon_N(|p_z|,\sigma,n)-\bigg(\frac{p_z^2}{2M}+\frac{\hbar^2\beta_{\sigma,n}^2}{2MR^2}\bigg)-g\bar{n}\bigg]\nonumber\\&+&\frac{1}{2}\sum_{p_z,\sigma,n}\bigg[\epsilon_D(|p_z|,\sigma,n)-\bigg(\frac{p_z^2}{2M}+\frac{\hbar^2\alpha_{\sigma,n}^2}{2MR^2}\bigg)-g\bar{n}\bigg]\nonumber\\\end{aligned}$$ is the contribution to 