We start with the Hill's equation of both transverse plane.
\begin{align} &x''\left(s\right) +\left(\frac{1}{\rho^2} + k_x\left(s\right) \right) x\left(s\right)=0\\ &y''\left(s\right) + k_y\left(s\right)y\left(s\right)=0 \end{align}In general, we need solve:
\begin{align} &x''\left(s\right) + K\left(s\right)x\left(s\right)=0 \label{eq:Hillseq} \end{align}For a ring accelerator or a linac with periodical structure, we have $K(s)=K(s+C)$. To have a stable motion, we try to express the solution of the Hill's equation as:
\begin{align} x\left(s\right)=aw\left(s\right)e^{i\psi\left(s\right)} \end{align}where $a$ is a constant, $w(s)$ is the envelope function and $\psi(s)$ is the phase function. Then the derivatives are:
\begin{align} x'\left(s\right)&=aw'e^{i\psi}+iawe^{i\psi}\psi'\\ x''\left(s\right)&=aw''e^{i\psi}+2iaw'e^{i\psi}\psi'-awe^{i\psi}\psi'^2+iawe^{i\psi}\psi'' \label{eq:x2prime} \end{align}Combine Equation $\ref{eq:x2prime}$ and $\ref{eq:Hillseq}$, then the real and imaginary part gives:
\begin{align} &2aw'\psi'+aw\psi''=a\frac{\left(w^2\psi'\right)'}{w}=0\label{eq:wpsi1}\\ &aw''+Kw-w\psi'^2=0 \end{align}Obviously, we are seeking for non-trivial solution ($w\ne 0$). The equation $\ref{eq:wpsi1}$ gives:
\begin{align} w^2\psi'=1 \end{align}Then by choose the constant a as 1, the equations for $w$ and $\psi$ become
\begin{align} &w''+Kw-1/w^3=0\\ &\psi'-1/w^2=0 \end{align}This shows that the phase function $\psi(s)$ is linked with the envelope function $w(s)$.
We define the beta function (represent the envelope of the oscillation) $\beta(s)=w^2(s)$ and alpha function (the slope of the envelope) $\alpha(s)=-\beta'(s)/2=-w(s)w'(s)$. The phase advance from location $s_0$ to $s$ becomes:
\begin{align} \psi(s,s_0)=\int_{s_0}^s\frac{1}{\beta(s')}ds' \end{align}The derivatives of beta function are:
\begin{align} \beta'&=2ww'\\ \beta''&=2w'^2+2ww'' \end{align}which satisfy:
\begin{align} \frac{1}{2}\beta''+K\beta-\frac{1}{\beta}\left[1+\frac{\beta'^2}{4}\right]=0 \end{align}Using the beta function, we can write the transverse motion as:
\begin{align} x(s)=a\sqrt{\beta(s)}\cos\left(\psi\left(s\right)\right)\label{eq:x_s_with_cos} \end{align}Its derivative becomes
\begin{align} x'(s)&=-a\sqrt{\beta}\psi'\sin\psi+a\frac{\beta'}{2\sqrt{\beta}}\cos\psi\\ &=-\frac{x}{\beta}\left(\tan\psi-\frac{\beta'}{2}\right)\\ &=-\frac{x}{\beta}\left(\tan\psi+\alpha\right) \end{align}An interesting relation can be revealed as:
\begin{align} \tan\psi(s)=-\frac{\beta(s)x'(s)+\alpha(s)x(s)}{x(s)}\label{eq:x_xp_with_tan} \end{align}Using the triganular relations $\cos^{-2}\alpha=1+\tan^2\alpha$ for any angle $\alpha$, we can combine equation $\ref{eq:x_s_with_cos}$ and $\ref{eq:x_xp_with_tan}$ and get:
\begin{align} \left(\frac{\beta(s)x'(s)+\alpha(s)x(s)}{x(s)}\right)^2+1&=\left(\frac{a\sqrt{\beta(s)}}{x(s)}\right)^2\\ \frac{\left(\beta(s)x'(s)+\alpha(s)x(s)\right)^2+x^2(s)}{\beta (s)}&=a^2 \end{align}Since the transverse Hamiltonian
\begin{align} H(x,p_x,y,p_y,s)=\frac{p_x^2}{2}+\frac{p_y^2}{2}+\frac{x^2}{2\rho(s)^2}+k(s)\left(x^2-y^2\right)/2 \end{align}is not a constant of motion,
\begin{align} \frac{dH}{ds}&=\frac{\partial H}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial H}{\partial p_x}\frac{\partial p_x}{\partial s} + \frac{\partial H}{\partial s}\\ &=\frac{\partial H}{\partial s}\ne 0 \end{align}We will follow the action-angle variable treatment in classical mechanics to simplify the transverse motion formula of one transverse direction using the Hamiltonian:
\begin{align} H(x,x',s)=\frac{x'^2}{2}+\frac{K(s)}{2}{x^2} \end{align}the transverse motion formula, by changing the coordinate from $(x,p_x)$ to $(J,\psi)$. We will use the generating function of the first kind
\begin{align} F_1(x,\psi)&=\int_0^x x'(\tilde x,\psi)d\tilde x \\ &= -\frac{x^2}{2\beta}\left(\tan\psi-\frac{\beta'}{2}\right) \end{align}The action is calculated by:
\begin{align} J&=-\frac{\partial F_1}{\partial\psi}=\frac{x^2}{2\beta}\sec^2\psi\\ &=\frac{x^2}{2\beta}\left(1+\left(\frac{\beta x'}{x}+\alpha\right)^2\right)\\ &=\frac{1}{2\beta}\left(x^2+\left(\beta x'+\alpha x\right)^2\right) \end{align}The new Hamiltonian is:
\begin{align} \tilde{H}&=H+\frac{\partial F_1}{\partial s} \\ &=\frac{1}{2}\frac{x^2}{\beta^2}\left(\tan\psi+\alpha\right)^2+\frac{1}{2}Kx^2 +\\ &\quad\frac{x^2}{2\beta^2}\beta'\left(\tan\psi-\frac{\beta'}{2}\right)-\frac{x^2}{2\beta}\left(\psi'\sec^2\psi-\frac{\beta''}{2}\right) \nonumber\\ &=\frac{J}{\beta} \end{align}Obviously, $J$ is constant since:
\begin{align} \frac{dJ}{ds}=-\frac{\partial \tilde{H}}{\partial \psi}=0 \end{align}Now, we can revisit the coordinate transformation $(x,x')$ to $(\psi, J)$
\begin{align} J&=\frac{1}{2}\left(\left(\frac{x}{\sqrt{\beta}}\right)^2+\left(\sqrt\beta x'+\frac{\alpha}{\sqrt{\beta}} x\right)^2\right) = \text{constant} \label{eq:action}\\ \psi&=-\arctan\frac{\sqrt\beta x'+\alpha x/\sqrt{\beta}}{x/\sqrt{\beta}} \end{align}Therefore we can define the normalized coordinate:
\begin{align} \mathcal{X}&=\frac{x}{\sqrt{\beta}} \\ \mathcal{P_x}&=\sqrt\beta x'+\frac{\alpha}{\sqrt{\beta}}x \end{align}In matrix form, we have
\begin{align} \left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)=\left(\begin{array}{cc} 1/\sqrt{\beta} & 0\\ \alpha/\sqrt{\beta} & \sqrt{\beta} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right) \end{align}or reversely:
\begin{align} \left(\begin{array}{c} x\\ x' \end{array}\right) =\left(\begin{array}{cc} \sqrt{\beta} & 0\\ -\alpha/\sqrt{\beta} & 1/\sqrt{\beta} \end{array}\right)\left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right) \end{align}For location 1 to location 2, the normalized coordinate follows:
\begin{align} \left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)_2=\left(\begin{array}{cc} \cos\psi_{12} & \sin\psi_{12}\\ -\sin\psi_{12} & \cos\psi_{12} \end{array}\right)\left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)_1 \end{align}The original coordintate has the map:
\begin{align} \left(\begin{array}{c} \mathcal{x}\\ x' \end{array}\right)_2= \left(\begin{array}{cc} \sqrt{\beta_2} & 0\\ -\alpha_2/\sqrt{\beta_2} & 1/\sqrt{\beta_2} \end{array}\right) \left(\begin{array}{cc} \cos\psi_{12} & \sin\psi_{12}\\ -\sin\psi_{12} & \cos\psi_{12} \end{array}\right)\left(\begin{array}{cc} 1/\sqrt{\beta_1} & 0\\ \alpha_1/\sqrt{\beta_1} & \sqrt{\beta_1} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right)_1 \end{align}Therefore we rederived the transfer map of two arbitrary location in usual form:
\begin{equation} M\left(s_{2}\mid s_{1}\right) =\left(\begin{array}{cc} \sqrt{{\frac{\beta_{2}}{\beta_{1}}}}\left(\cos\psi+\alpha_{1}\sin\psi\right) & \sqrt{\beta_{1}\beta_{2}}\sin\psi\\ -\frac{1+\alpha_{1}\alpha_{2}}{\sqrt{\beta_{1}\beta_{2}}}\sin\psi+\frac{\alpha_{1}-\alpha_{2}}{\sqrt{\beta_{1}\beta_{2}}}\cos\psi & \sqrt{{\frac{\beta_{1}}{\beta_{2}}}}\left(\cos\psi-\alpha_{2}\sin\psi\right) \end{array}\right) \label{eq:tmatrix_s1s2} \end{equation}When the two locations have same optical functions, the one turn map is rederived:
\begin{align} \left(\begin{array}{c} \mathcal{x}\\ x' \end{array}\right)_2&= \left(\begin{array}{cc} \sqrt{\beta} & 0\\ -\alpha/\sqrt{\beta} & 1/\sqrt{\beta} \end{array}\right) \left(\begin{array}{cc} \cos\psi & \sin\psi\\ -\sin\psi & \cos\psi \end{array}\right)\left(\begin{array}{cc} 1/\sqrt{\beta} & 0\\ \alpha/\sqrt{\beta} & \sqrt{\beta} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right)_1\\ &= \left(\begin{array}{cc} \cos\psi+\alpha\sin\psi & \beta \sin\psi\\ -\frac{1+\alpha^2}{\beta}\sin\psi & \cos\psi-\alpha\sin\psi \end{array}\right) \left(\begin{array}{c} x\\ x' \end{array}\right)_1 \end{align}Apparently, the solution of the Hills equation is unique except two initial conditions, one scales the envelope function $w(s)$ and the other one defines the initial phase of $\psi(s)$ when $s=0$.
However, is the parametrization a unique answer?
From the action's formula $\ref{eq:action}$, it is easy to see that the definition of the Parametrization can have infinite sets. In general, we can define the the normalized coordinate as:
\begin{align} \left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right) \end{align}The four independent parameter only need to satisfy three conditions:
\begin{align} a^2+c^2&=\gamma=\frac{1+\alpha}{\beta}\\ b^2+d^2&=\beta\\ ab+cd&=\alpha \end{align}The widely used parametrization version, takes the special case $b=0$. An 'equally good' choice can be found by forcing $c=0$. Then the normalized coordinate becomes:
\begin{align} \left(\begin{array}{c} \tilde{\mathcal{X}}\\ \tilde{\mathcal{P}_{x}} \end{array}\right)=\left(\begin{array}{cc} \sqrt{\gamma} & \alpha/\sqrt{\gamma}\\ 0 & 1/\sqrt{\gamma} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right) \end{align}The corresponding phase advance becomes:
\begin{align} \psi&=-\arctan\frac{x'/\sqrt\gamma }{\sqrt{\gamma}x+\alpha x'/\sqrt{\gamma}} \end{align}