Finding the Hermitian generator of a representation of a Symplectic transformation Consider a set of $n$ position operators and $n$ momentum operator such that
$$\left[q_{i},p_{j}\right]=i\delta_{ij}.$$
Lets now perform a linear symplectic transformation
$$q'_{i}    =A_{ij}q_{j}+B_{ij}p_{j},$$
$$p'_{i}    =C_{ij}q_{j}+D_{ij}p_{j}.$$
such that the canonical commutation relations are maintained
$$\left[q'_{i},p'_{j}\right]=i\delta_{ij}.$$
Any such symplectic transformation should be unitarily implemented due to the Stone-von Neumann theorem (right?)
$$U^{-1}q_{i}U    =q'_{i},$$
$$U^{-1}p_{i}U    =p'_{i}.$$
The question is: Assuming the coefficients $A$,$B$,$C$ and $D$ are given, is there a systematic way to calculate the generator $G$ of the unitary transformation $U=e^{-iG}$?
 A: *

*Classically, the symplectic group $Sp(2n, \mathbb{R})$ of dimension $n(2n+1)$ is the group of all linear time-independent canonical transformations (CTs)
$$z^{\prime I}~=~\sum_{J=1}^{2n}M^{I}{}_Jz^J.\tag{1}$$
The corresponding symplectic Lie algebra $sp(2n,\mathbb{R})$ is the set of all linear time-independent infinitesimal CTs, which have time-independent quadratic generating functions $$F(z)~=~\frac{1}{2}\sum_{I,J=1}^{2n}a_{IJ}z^Iz^J, \qquad a_{IJ}~=~a_{JI}.\tag{1}$$
A finite linear CT is of the form
$$z^{\prime I}~=~e^{\{F(z), ~\cdot~ \}}z^I,\tag{2}$$
where $\{\cdot, \cdot \}$ denotes the Poisson bracket.


*Quantum mechanically, the Poisson bracket $\{\cdot, \cdot \}$ is replaced with the commutator $\frac{1}{i\hbar}[\cdot, \cdot ]$, so a finite linear CT becomes
$$\hat{z}^{\prime I}~=~e^{\frac{1}{i\hbar}[F(\hat{z}), ~\cdot~ ]}\hat{z}^I~=~\hat{U}\hat{z}^I\hat{U}^{-1},\tag{3}$$
where
$$ \hat{U}~=~e^{\frac{1}{i\hbar}F(\hat{z})} \tag{4}$$
is a unitary operator with Hermitian generator
$$F(\hat{z})~=~\frac{1}{2}\sum_{I,J=1}^{2n}a_{IJ}\hat{z}^I\hat{z}^J, \qquad a_{IJ}~=~a_{JI}.\tag{5}$$
