## 3 Field Quantization  
When the obtained Hamiltonian is moved to the realm of quantum mechanics, it is first needed to define the non-commutation rules \([\hat{q},\hat{p}]=i\hbar\) and \([\hat{Q}_{k},\hat{P}_{j}]=i\hbar\delta_{kj}\), with the commutation rules\(\left[\hat{q},\hat{Q}_{k}\right]=\left[\hat{Q}_{k},\hat{p}\right]=\left[\hat{q}, \hat{p}_{j}\right]=\left[\hat{P}_{j},\hat{p}\right]=0\). This allows us to introduce the field creation and annihilation operators according to  
\[\hat{Q}_{k}=\sqrt{\frac{\hbar}{2\omega_{k}(\hat{q})}}\left(a_{k}^{ \dagger}+\hat{a}_{k}\right)=\sqrt{\frac{\hbar}{\omega_{k}(\hat{q})}}\mathbb{Q} _{k}, \tag{30}\] \[\hat{P}_{k}=i\sqrt{\frac{\hbar\omega_{k}(\hat{q})}{2}}\left(\hat{a }_{k}^{\dagger}-\hat{a}_{k}\right)=\sqrt{\hbar\omega_{k}(\hat{q})}\mathbb{P} _{k},\]  
where \(\omega_{k}(\hat{q})=c\pi k/\hat{q}\) is defined according to (2). Also for a mechanical resonant frequency \(\Omega\) and a spring restoring potential  
\[V(\hat{q})=\frac{1}{2}m\Omega^{2}(\hat{q}-l)^{2}, \tag{31}\]  
the displacement operator may be defined as \(\hat{q}=l+\hat{x}\), where \(l\) is the reference position of mirror, and hence the phonon ladder operators as  
\[\hat{x}=\sqrt{\frac{\hbar}{2m\Omega}}\left(\hat{b}^{\dagger}+\hat{b}\right)= \sqrt{\frac{\hbar}{m\Omega}}\mathfrak{x},\qquad\hat{p}=i\sqrt{\frac{\hbar m \Omega}{2}}\left(\hat{b}^{\dagger}-\hat{b}\right)=\sqrt{\hbar m\Omega} \mathfrak{P}, \tag{32}\]  
with \(\left[\hat{b},\hat{b}^{\dagger}\right]=1\) and \(\left[\mathfrak{x},\mathfrak{P}\right]=i\). Now, it is necessary first to symmetrize [67] the classical Hamiltonian prior to insertion of operators, to ensure correct quantization of parameters. The process of symmetrization is done according to [71; 72; 73]  
\[\mathcal{S}\{\mathbb{AB}\}=\frac{1}{2}(\mathbb{AB}+\mathbb{BA}),\,\mathcal{S }\{\mathbb{ABC}\}=\frac{1}{3}(\mathbb{AB}\{\mathbb{BC}\}+\mathbb{B}\mathcal{S }\{\mathbb{AC}\}+\mathbb{C}\mathcal{S}\{\mathbb{AB}\}), \tag{33}\]  
etc. Therefore, after symmetrization the final form of the Hamiltonian is given by  
\[ \begin{array}{l}\mathbb{H}=\frac{1}{2m}\hat{p}^{2}+V(\hat{q})+\frac{1}{2} \sum\left(\hat{P}_{k}^{2}+\omega_{k}^{2}\hat{Q}_{k}^{2}\right)+\frac{1}{4m} \mathcal{S}\Big{\{}\frac{1}{\hat{q}^{2}}\left(\hat{p}\hat{q}+\frac{1}{2}\sum g _{kj}\hat{P}_{k}\hat{Q}_{j}\right)\sum g_{kj}\hat{P}_{k}\hat{Q}_{j}\Big{\}}\\ \quad-\frac{r}{8m^{2}}\mathcal{S}\Big{\{}\frac{1}{\hat{q}^{2}}\left(\hat{p} \hat{q}+\frac{1}{2}\sum g_{kj}\hat{P}_{k}\hat{Q}_{j}\right)^{2}\sum d_{kj}\hat {Q}_{k}\hat{Q}_{j}\Big{\}}.\end{array}  \tag{34}\]  
For a single-optical mode, (34) greatly simplifies and one gets  
\[\mathbb{H}=\frac{1}{2m}\hat{p}^{2}+V(\hat{q})+\frac{1}{2}\bigg{(}\hat{p}^{2}+ \frac{\pi^{2}}{\hat{q}^{2}}\hat{Q}^{2}\bigg{)}-\frac{r}{8m^{2}}\mathcal{S} \bigg{\{}\frac{1}{\hat{q}^{2}}\hat{p}^{2}\bigg{\}}\hat{Q}^{2}. \tag{35}\]  
This has to be applied to the last interacting term, which involves  
\[\mathcal{S}\bigg{\{}\frac{1}{\hat{q}^{2}}\hat{p}^{2}\bigg{\}}=\mathcal{S} \bigg{\{}\frac{\hat{p}^{2}}{\hat{q}^{2}}\bigg{\}}. \tag{36}\]  
However, symmetrization of a term which contains \(n\) non-commuting terms, results in \(n!\) terms, which for this case sum up to a total of \(4!=24\) different expressions. The direct way to get around this situation is to first make an estimate of which terms are the strongest in the limit of linearized interaction and ignore the rest.