## 2 Classical Hamiltonian
### Hamiltonian  
The definition of canonical momenta taken here is  
\[P_{k}=\dot{Q}_{k}-\frac{\dot{q}}{q}\sum A_{kj}Q_{j},\ p=m\dot{q}-\frac{1}{q} \sum B_{kj}P_{k}Q_{j}, \tag{21}\]  
with \(A_{kj}\) and \(B_{kj}\) being some transformation coefficients to be determined later. One here may take advantage of the degree of freedom in choice of \(A_{kj}\) and \(B_{kj}\) to get rid of unwanted summation terms in the Hamiltonian. It has to be here noticed again that the existence of the last term of (20), being completely new even under the single-mode operation, has nothing to do with the choice of canonical momenta. That implies the final resulting single-mode (and therefore multi-mode) Hamiltonian will be inevitably different, incorporating a few new terms. The Hamiltonian may be now derived from the Lagrangian by iterated use of (21) and through the relationship \(\mathcal{H}=p\dot{q}+\sum P_{k}\dot{Q}_{k}-\mathcal{L}\) as  
\[\mathcal{H}=\bigg{[}m\dot{q}-\frac{1}{q}\sum B_{kj}\bigg{(}\dot{Q}_{k}-\frac{ \dot{q}}{q}\sum A_{kl}Q_{l}\bigg{)}Q_{j}\bigg{]}\dot{q}+\sum\bigg{(}\dot{Q}_{k }-\frac{\dot{q}}{q}\sum A_{kj}Q_{j}\bigg{)}\dot{Q}_{k}-\ \mathcal{L}. \tag{22}\]  
Law [5] arbitrates the choice \(A_{kj}=B_{kj}=g_{kj}\). However, by going further with this choice for a single-mode optical field, it will be evident that \(P=\dot{Q}\) and \(p=m\dot{q}\), hence, still resulting in an extra nonlinear term proportional to \(p^{2}Q^{2}/q^{2}\) in the Hamiltonian, leading to a fourth-order momentum-field interaction. This will be discussed shortly in the subsection below. Hence, the Lagrangian \(\mathcal{L}\) found in the above yields the Hamiltonian below after some algebra  
\[ \begin{array}{l}\mathcal{H}=\frac{1}{2}m\dot{q}^{2}+V(q)+\frac{1}{2}\sum \bigg{(}\dot{Q}_{k}^{2}+\omega_{k}^{2}{Q_{k}}^{2}\bigg{)}-\frac{\dot{q}}{q} \sum\Big{(}A_{kj}+B_{kj}-g_{kj}\Big{)}\dot{Q}_{k}Q_{j}\\ +\frac{\dot{q}^{2}}{2q^{2}}\sum\Big{(}2\sum B_{kj}A_{kl}-d_{jl}\Big{)}Q_{l}Q_{ j}.\end{array}  \tag{23}\]  
Here is readily evident now by (19), that the last term can be made identically zero, only if \(d_{jl}=\sum g_{kj}g_{kl}=2\sum B_{kj}A_{kl}\). This not only cannot be satisfied by [PERSON]'s choice \(A_{kj}=B_{kj}=g_{kj}\), but also the second summation term nonlinear in \(\dot{q}\dot{Q}_{k}\) will also survive, further complicating the Hamiltonian formulation. Now, further elimination of \(\dot{q}\) and \(\dot{Q}_{k}\) from (21) gives  
\[ \begin{array}{l}\mathcal{H}=\frac{1}{2m}\left(p+\frac{1}{q}\sum B_{kj}P_{k}Q_{j }\right)^{2}+V(q)+\frac{1}{2}\sum\omega_{k}^{2}{Q_{k}}^{2}\\ \quad+\frac{1}{2}\sum\left[P_{k}+\frac{1}{mq}\left(p+\frac{1}{q}\sum B_{lm}P_{ l}Q_{m}\right)\sum A_{kj}Q_{j}\right]^{2}\\ \quad-\frac{1}{mq}\left(p+\frac{1}{q}\sum B_{kj}P_{k}Q_{j}\right)\sum C_{kj} \left[P_{k}+\frac{1}{mq}\left(p+\frac{1}{q}\sum B_{kj}P_{k}Q_{j}\right)\sum A_ {kj}Q_{j}\right]Q_{j}\\ \quad+\frac{1}{2m^{2}q^{2}}\left(p+\frac{1}{q}\sum B_{kj}P_{k}Q_{j}\right)^{2 }\sum D_{jl}Q_{l}Q_{j}.\end{array}  \tag{24}\]  
Here, \(C_{kj}=A_{kj}+B_{kj}-g_{kj}\) and \(D_{jl}=2\sum B_{kj}A_{kl}-d_{jl}\). Dealing directly with such an intractable and long expression is without doubt too tough. Instead, we may tweak [PERSON]'s choice slightly as \(A_{kj}=B_{kj}=\frac{1}{2}g_{kj}\), which allows (23) be greatly simplified as  
\[\mathcal{H}=\frac{1}{2}m\dot{q}^{2}+V(q)+\frac{1}{2}\sum\biggl{(}\dot{Q}_{k}^{ 2}+\omega_{k}^{2}{Q_{k}}^{2}\biggr{)}-\frac{\dot{q}^{2}}{4q^{2}}\sum d_{kj}Q_{ k}Q_{j}. \tag{25}\]  
This can be further eventually expanded and simplified as  
\[ \begin{array}{l}\mathcal{H}=\frac{1}{2m}\left(p+\frac{1}{2q}\sum g_{kj}P_{ k}Q_{j}\right)^{2}+V(q)+\frac{1}{2}\sum\left({P_{k}}^{2}+\omega_{k}^{2}{Q_{k}}^{ 2}\right)\\ \quad+\frac{1}{4 mq}\left(p+\frac{1}{2q}\sum g_{kj}P_{k}Q_{j}\right)\sum g_{kj} P_{k}Q_{j}-\frac{1}{8m^{2}q^{2}}\left(p+\frac{1}{2q}\sum g_{kj}P_{k}Q_{j} \right)^{2}\sum d_{kj}Q_{k}Q_{j}.\end{array}  \tag{26}\]  
The last two terms of this Hamiltonian can be expanded to obtain multiple orders of interactions.