## 2 Classical Hamiltonian
### The Equations of Motion  
The one-dimensional wave equation for transverse component of the magnetic potential \(A(x,t)\) in the dimensional form is expressed as [5]  
\[c^{2}A_{xx}(x,t)=A_{tt}(x,t), \tag{1}\]  
where \(x\) and \(t\) are respectively the position and time coordinates, and \(c\) is the speed of light in free space. Suppose the Fourier series relations for the magnetic vector potential are defined as [5]  
\[ \begin{split} Q_{k}(t)=\frac{1}{c}\sqrt{\frac{2S}{\mu_{0}q(t)}} \int\limits_{0}^{q(t)}A(x,t)\sin[x_{k}(t)x]dx,\\ A(x,t)=c\sqrt{\frac{2\mu_{0}}{Sq(t)}}\sum\limits_{k=1}^{\infty}Q_ {k}(t)\sin[x_{k}(t)x],\\ \omega_{k}(t)=c\pi k/q(t)=c\kappa_{k}(t),\end{split}  \tag{2}\]  
where \(S\) is the cross-sectional area, \(\mu_{0}\) is the permeability of vacuum. This arrangement ensures that the definition of canonical variables can be used later, so that \(\dot{Q_{k}}^{2}\) simply takes on the dimension of energy. One may furthermore define the functions \(f_{k}=\sqrt{2/q}\sin(\kappa_{k}x)\) and \(g_{k}=\sqrt{2/q}\cos(\kappa_{k}x)\), and hence \(A(x,t)=s\sum Q_{k}f_{k}\) where \(s=c\sqrt{\mu_{0}/S}\). Here, the inner product is also defined as \((a|b)=\int_{0}^{q}abdx\) such that the following relations may be found  
\[\big{(}f_{k}|f_{j}\big{)}=\delta_{kj},\ \big{(}f_{k}|\,\kappa_{j}x|g_{j}\big{)}= \alpha_{kj},\ \big{(}f_{k}\Big{|}\,\kappa_{j}^{2}x^{2}\Big{|}g_{j}\big{)}=\beta_{kj}. \tag{3}\]  
After straightforward calculations one obtains  
\[\alpha_{kj}=-\tfrac{1}{2}\delta_{kj}+g_{kj}, \tag{4}\] \[\beta_{kj}=\Big{(}k^{2}\tfrac{\pi^{2}}{3}-\tfrac{1}{2}\Big{)} \delta_{kj}+h_{kj}= \begin{cases}k^{2}\tfrac{\pi^{2}}{3}-\tfrac{1}{2},&k=j,\\ 8\tfrac{(-1)^{k+j}kk^{3}}{(k^{2}-\beta^{2})},&k\
eq j.\end{cases} \]  
Here, the anti-symmetric coefficients \(g_{kj}\) and \(h_{kj}\) are  
\[ \begin{split} g_{kj}&=-g_{jk}=\begin{cases}0,&k=j,\\ 2\tfrac{(-1)^{k+j}kj}{\beta^{2}-k^{2}},&k\
eq j,\end{cases}\\ h_{kj}&=\begin{cases}0,&k=j,\\ 8\tfrac{(-1)^{k+j}k^{3}}{(k^{2}-\beta^{2})},&k\
eq j.\end{cases}\end{split}  \tag{5}\]  
Differentiating \(f_{k}\) and \(g_{k}\) with respect to \(t\), noting that \(\dot{\kappa}_{k}=-\dot{q}\kappa_{k}/q\), gives  
\[ \begin{split}\dot{f}_{k}&=-\tfrac{\dot{q}}{q}\Big{(} \tfrac{1}{2}f_{k}+x\kappa_{k}g_{k}\Big{)},\\ \dot{\delta}_{k}&=-\tfrac{\dot{q}}{q}\Big{(}\tfrac{1}{2}g_{k}-x \kappa_{k}f_{k}\Big{)},\\ \ddot{f}_{k}&=-\tfrac{\ddot{q}q-\dot{q}^{2}}{q^{2}}\Big{(} \tfrac{1}{2}f_{k}+x\kappa_{k}g_{k}\Big{)}+\tfrac{\dot{q}^{2}}{q^{2}}\Big{(} \tfrac{1}{4}f_{k}+2 xx_{k}g_{k}-x^{2}\kappa_{k}{}^{2}f_{k}\Big{)}.\end{split}  \tag{6}\]  
It is possible now to differentiate \(A(x,t)\) with respect to position and time to obtain the relations  
\[ \begin{split}\tfrac{c^{2}}{s}A_{xx}(x,t)&=-\sum \omega_{k}{}^{2}Q_{k}f_{k},\\ \tfrac{1}{s}A_{t}(x,t)&=\sum\dot{Q}_{k}f_{k}-\tfrac{ \dot{q}}{q}\sum Q_{k}\Big{(}\tfrac{1}{2}f_{k}+xx_{k}g_{k}\Big{)},\\ \tfrac{1}{s}A_{tt}(x,t)&=\sum\ddot{Q}_{k}f_{k}-2 \tfrac{\dot{q}}{q}\sum\dot{Q}_{k}\Big{(}\tfrac{1}{2}f_{k}+x\kappa_{k}g_{k} \Big{)}-\tfrac{\ddot{q}q-\dot{q}^{2}}{q^{2}}\sum Q_{k}\Big{(}\tfrac{1}{2}f_{k} +xx_{k}g_{k}\Big{)}\\ &\qquad\qquad+\tfrac{\dot{q}^{2}}{q^{2}}\sum Q_{k}\Big{(} \tfrac{1}{4}f_{k}+2x\kappa_{k}g_{k}-x^{2}\kappa_{k}{}^{2}f_{k}\Big{)}.\end{split}  \tag{7}\]  
Therefore, the wave equation reads  
\[ \begin{split}-\sum\omega_{k}{}^{2}Q_{k}f_{k}&=\sum \ddot{Q}_{k}f_{k}-\tfrac{\dot{q}}{q}\sum\ddot{Q}_{k}(f_{k}+2x\kappa_{k}g_{k})- \tfrac{\ddot{q}}{q}\sum Q_{k}\Big{(}\tfrac{1}{2}f_{k}+x\kappa_{k}g_{k}\Big{)} \\ &\qquad\qquad+\tfrac{\dot{q}^{2}}{q^{2}}\sum Q_{k}\big{(}\tfrac {3}{4}f_{k}+3x\kappa_{k}g_{k}-x^{2}\kappa_{k}{}^{2}f_{k}\big{)}.\end{split}