Primer on symplecticity¶

Symplectic condition¶

Hamiltonian equation in matrix form¶

For a N-dimensional Hamiltonian system ($\mathbf{x}^N, \mathbf{p}^N$), the N sets of Hamiltonian equation reads:

\begin{align} \frac{dx_i}{dt}&=\frac{\partial H}{\partial p_i}\\ \frac{dp_i}{dt}&=-\frac{\partial H}{\partial x_i} \end{align}

Let's write a column vector $X=(x_1, p_1, x_2,p_2,\cdots,x_N,p_n)$, the Hamiltonian can be written as:

\begin{align} \frac{d}{dt}X & \equiv S_N \frac{\partial H}{\partial X} = \left(\begin{array}{cccc} S_{2} & 0 & \cdots & 0\\ 0 & S_{2} & \cdots & 0\\ \vdots & \vdots & \ddots & 0\\ 0 & 0 & 0 & S_{2} \end{array}\right) \frac{\partial H}{\partial X} \end{align}

where $S_1$ is the matrix for one dimension space:

\begin{align} S_2 = \left(\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right) \end{align}

Properties of S matrix¶

We found that the following facts about $S$ matrix

Determinant of $S$:

\begin{equation}\left|S\right|=1\end{equation}

Inverse of $S$: \begin{equation} S^{-1}=-S \text{ or } S^2 = -I\end{equation}
The transpose of $S$: \begin{equation}S^T=-S\end{equation}

For arbitrary 2 by 2 matrix $A$ we may prove that:

\begin{equation} A^TS_2A=AS_2A^T= \left(\begin{array}{cc} 0 & \det A\\ -\det A & 0 \end{array}\right)=S_2\det A \end{equation}

Or we may define a new matrix $\bar{A}=-S_2A^{T}S_2$ and yield:

\begin{equation} \bar{A}A=A\bar{A} = I\det{A} \end{equation}

Symplectic condition¶

For a map from location $s_0$ to $s_1=s_0+s$, we have:

\begin{align} X\left(s_1\right)=M X\left(s_0\right) \end{align}

M is a matrix and the components is defined as:

\begin{align} M_{ij}=\frac{\partial X_i\left(s_1\right)}{ \partial X_j\left(s_0\right)} \end{align}

If $M_{ij}$ are not functions of $X$, the map represents a linear map. Otherwise $M$ is a non-linear map. Regardless of the linearity, the condition that map $M$ need to follow due to the system is a Hamiltonian system is called symplectic condition. We just explored the condition in 1-D case. In N-D system, from the Map definition:

\begin{align} X\left(s_1\right)=M X\left(s_0\right) \end{align}

We take the derivative of the $i^{th}$ row of $X(s_1)$ with respect to $s$:

\begin{align} \frac{dX_i\left(s_1\right)}{ds} &=\frac{\partial X_i\left(s_1\right)}{\partial X_j\left(s_0\right)}\frac{dX_j\left(s_0\right)}{ds}\nonumber\\ &=M_{ij}S_{jk}\frac{\partial H}{\partial X_k\left(s_0\right)}\nonumber\\ &=M_{ij}S_{jk}\frac{\partial H}{\partial X_l\left(s_1\right)}\frac{\partial X_l\left(s_1\right)}{\partial X_k\left(s_0\right)}\nonumber\\ &=M_{ij}S_{jk}M_{kl}^T\frac{\partial H}{\partial X_l\left(s_1\right)} \end{align}

We know that

\begin{equation} \frac{dX_i\left(s_1\right)}{ds}=S_{il}\frac{\partial H}{\partial X_l\left(s_1\right)} \end{equation}

Therefore we have

\begin{align} MSM^T=S \end{align}

Using the property of $S$, we have the usual form:

\begin{align} M^TSM=S \label{eq:symplectic_form} \end{align}

The equation $\ref{eq:symplectic_form}$ is referred as symplectic condition of a transfer map (not limited to matrix).

Properties of symplectic map¶

The properties of the symplectic map:

If $M$ is symplectic, so does $M^T$ and $M^{-1}$.
If $M$ and $N$ are symplectic, so does $MN$.
The determinant of symplectic map is 1.

Properties of symplectic matrix¶

The symplectic matrix has the following properties from the symplectic condition:

If $\lambda$ is an eigenvalue of a symplectic matrix $M$, so does $1/\lambda$.
Every sympletic matrix has the inverse matrix: $M^{-1}=-SM^{T}S$

Block matrix for 2-D system¶

Consider a 4-D phase space transfer map:

\begin{equation} \left(\begin{array}{cc} A & B\\ C & D \end{array}\right) \end{equation}

The transfer map has to satisfy the symplectic condition, both $M^TSM=S$ and $MSM^T=S$. For the first condition:

\begin{align} \left(\begin{array}{cc} A^T & C^T\\ B^T & D^T \end{array}\right) \left(\begin{array}{cc} S_2 & 0\\ 0 & S_2 \end{array}\right) \left(\begin{array}{cc} A & B\\ C & D \end{array}\right)&= \left(\begin{array}{cc} S_2 & 0\\ 0 & S_2 \end{array}\right) \\ \left(\begin{matrix}A^{T} S_{2} A + C^{T} S_{2} C & A^{T} S_{2} B + C^{T} S_{2} D\\ B^{T} S_{2} A + D^{T} S_{2} C & B^{T} S_{2} B + D^{T} S_{2} D\end{matrix}\right)&= \left(\begin{array}{cc} S_2 & 0\\ 0 & S_2 \end{array}\right) \end{align}

Therefore the following relation satisfy:

\begin{align} A^{T} S_{2} A + C^{T} S_{2} C&=B^{T} S_{2} B + D^{T} S_{2} D=S_2 \label{Mmatdet}\\ A^{T} S_{2} B + C^{T} S_{2} D&=B^{T} S_{2} A + D^{T} S_{2} C=0 \label{Mmatzero} \end{align}

Change $M$ to $M^T$, we have

\begin{align} A S_{2} A^{T} + B S_{2} B^{T}&=C S_{2} C^{T} + D S_{2} D^{T}=S_2 \label{Mtmatdet}\\ A S_{2} C^{T} + B S_{2} D^{T}&=C S_{2} A^{T} + D S_{2} B^{T}=0 \label{Mtmatzero} \end{align}

import sympy 
sympy.init_printing(use_unicode=True)
a=sympy.MatrixSymbol('A', 2,2)
b=sympy.MatrixSymbol('B', 2,2)
c=sympy.MatrixSymbol('C', 2,2)
d=sympy.MatrixSymbol('D', 2,2)
s_2=sympy.MatrixSymbol('S_2', 2,2)
zero=sympy.ZeroMatrix(2,2)
M=sympy.BlockMatrix([[a,b],[c,d]])
ss=sympy.BlockMatrix([[s_2,zero],[zero,s_2]])
sympy.block_collapse(M*ss*M.T)

From the general property of 2-by-2 matrix $A$:

\begin{align} A^TS_2A=AS_2A^T= \left(\begin{array}{cc} 0 & \det A\\ -\det A & 0 \end{array}\right) \end{align}

From \ref{Mmatdet} and \ref{Mtmatdet}, we know that:

\begin{align} \det A + \det C=\det D + \det B &=1\\ \det D + \det C=\det A + \det B &=1 \end{align}

Therefore, we always have:

\begin{align} \det A = \det D &\equiv a\\ \det B = \det C &\equiv b\\ a+b&=1 \end{align}

And from \ref{Mmatzero} and \ref{Mtmatzero}, we have

\begin{align} \bar{A}B+\bar{C}D=\bar{B}A+\bar{D}C=0\\ A\bar{C}+B\bar{D}=C\bar{A} + D\bar{B}=0 \end{align}

The inverse of a blocked 4-by-4 matrix is simply:

\begin{align} M^{-1}&= \left(\begin{array}{cc} A & B\\ C & D \end{array}\right)^{-1}=-SM^TS \\ &=-\left(\begin{array}{cc} S_{1} & 0\\ 0 & S_{1} \end{array}\right) \left(\begin{array}{cc} A^T & C^T\\ B^T & D^T \end{array}\right) \left(\begin{array}{cc} S_{1} & 0\\ 0 & S_{1} \end{array}\right)\\ &=\left(\begin{array}{cc} \bar{A} & \bar{C}\\ \bar{B} & \bar{D} \end{array}\right) \end{align}

Here is a useful example. If we are exploring $x$-$z$ or $y$-$z$ phase space, the matrix $U$ usually has the form below for DC magnets:

\begin{align} B= \left(\begin{array}{cc} \begin{array}{c} 0\\ 0 \end{array} & d=\left(\begin{array}{c} d_{1}\\ d_{2} \end{array}\right)\end{array}\right) \end{align}

Then

\begin{align} \det A = \det D &=1\\ \det B = \det C &=0 \end{align}

And from

\begin{align} \bar{B}A+\bar{D}C=0 \end{align}

We have

\begin{align} C=-D\bar{B}A \end{align}

Block matrix for 3-D system¶

Consider a 6-D phase space transfer map:

\begin{equation} \left(\begin{array}{c} x\\ p_{x}\\ y\\ p_{y}\\ z\\ \delta \end{array}\right)_{2}=\left(\begin{array}{cc} M_{1} & C_1 & U_1\\ C_2 & M_{2} & U_2 \\ V_1 & V_{2} & M_3 \\ \end{array}\right)\left(\begin{array}{c} x\\ p_{x}\\ y\\ p_{y}\\ z\\ \delta \end{array}\right)_{1} \end{equation}

import sympy 
sympy.init_printing(use_unicode=True)
m1=sympy.MatrixSymbol('M_1', 2,2)
m2=sympy.MatrixSymbol('M_2', 2,2)
m3=sympy.MatrixSymbol('M_3', 2,2)
u1=sympy.MatrixSymbol('U_1', 2,2)
u2=sympy.MatrixSymbol('U_2', 2,2)
c1=sympy.MatrixSymbol('C_1', 2,2)
v1=sympy.MatrixSymbol('V_1', 2,2)
v2=sympy.MatrixSymbol('V_2', 2,2)
c2=sympy.MatrixSymbol('C_2', 2,2)

s_2=sympy.MatrixSymbol('S_2', 2,2)
#s_2=sympy.Matrix([[0,1],[-1,0]])
zero=sympy.ZeroMatrix(2,2)
M=sympy.BlockMatrix([[m1,c1,u1],[c2,m2,u2],[v1,v2,m3]])
ss=sympy.BlockMatrix([[s_2,zero,zero],[zero,s_2,zero],[zero,zero,s_2]])
sympy.block_collapse(M.T*ss*M)

Therefore, we can get the relation of the determinant of each block:

\begin{align} \det M_1 + \det C_1 + \det U_1 =1 \\ \det C_2 + \det M_2 + \det U_2 =1 \\ \det M_3 + \det V_1 + \det V_2 =1 \\ \end{align}

and

\begin{align} \det M_1 + \det C_2 + \det V_1 =1 \\ \det C_1 + \det M_2 + \det V_2 =1 \\ \det U_1 + \det U_2 + \det M_3 =1 \\ \end{align}