Finding the Hermitian generator of a representation of a Symplectic transformation

Question

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 $U=e^{-iG}$?

Answer 1

1. 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.

2. 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 $$F(\hat{z})~=~\frac{1}{2}\sum_{I,J=1}^{2n}a_{IJ}\hat{z}^I\hat{z}^J, \qquad a_{IJ}~=~a_{JI}.\tag{5}$$