Find the Bogoliubov transformation $b=SaS^\dagger$ induced by the squeezed operator A definition a bogoliubov transformation is defined as $$b=ua+va^\dagger~,~ b^\dagger=u^*a^\dagger+v^*a$$ 
But, using squeeze operator $$S=\exp{\left[\frac{1}{2}(z (a^\dagger)^2-z^*a^2)\right]}$$ we can claim that $$b=SaS^\dagger $$ is also a bogoliubov transform. Using S, how do we find the value of $u$ and $v$ corresponding to the first set of expressions? I have tried applying BCH formula to $b=SaS^\dagger$ but didn't get anything helpful. 
 A: The unitary squeezing operator
$$
S(z)\stackrel{\rm def}{=} \exp\left\{{\textstyle \frac12}(z {a^\dagger }^2 -z^* a^2)\right\},
$$
with  $z=|z|e^{i\theta}$, 
implements the symplectic transformation
$$
S^\dagger(z) \left[\matrix{ a \cr a^\dagger}\right] S(z)=   \left[\matrix{\cosh|z| &e^{i\theta} \sinh |z| \cr e^{-i\theta} \sinh|z| & \cosh |z|}\right] \left[\matrix{ a \cr a^\dagger}\right]. 
$$
The is best proved by using the 
 faithful non-unitary representation 
$$
a^2\simeq 2i\sigma_-,
$$
$$
{a^\dagger }^2 \simeq 2i\sigma_+,
$$
$$
(a^\dagger  a+\textstyle \frac12)\simeq \sigma_3,
$$
of the  $\mathfrak {su} (1,1)$ Lie algebra. 
Then  a Gauss decomposition 
$$
\left(\matrix{a&b\cr c&d}\right)= \left(\matrix{1&0\cr A  &1}\right)\left(\matrix{\lambda_1 &0\cr 0 &\lambda_2}\right)\left(\matrix{1&B\cr 0  &1}\right)
$$
of the corresponding 2-by-2 matrices gives the    disentangling identities 
$$
S(z)= \exp\left\{{\textstyle \frac12}(z {a^\dagger}^2 -z^* a^2)\right\}
$$
$$
=\exp\left\{e^{i\theta}{\textstyle \frac12}\tanh |z|\, {a^\dagger}^2\right\}\exp\left\{ -\ln\cosh |z| (a^\dagger  a+{\textstyle \frac12})\right\} \exp\left\{-e^{-i\theta}{\textstyle \frac12}\tanh |z| \, a^2\right\}
$$
$$
=\exp\left\{-e^{-i\theta}{\textstyle \frac12}\tanh |z| \,a^2\right\}\exp\left\{ +\ln\cosh |z| (a^\dagger a+\textstyle \frac12)\right\} \exp\left\{e^{i\theta}{\textstyle \frac12}\tanh |z|\, {a^\dagger}^2\right\}.\nonumber
$$
To make the identification we expand and Gauss decompose
$$
\exp\{iz \sigma_+ -  iz^* \sigma_-\},
$$
 note
  that $\lambda_2=\lambda_1^{-1}$ because of the ${\rm SU}(1,1)$ property, and then use 
$$
\left(\matrix{1&0\cr x  &1}\right)= \exp\{x \sigma_-\},\\
\left(\matrix{\lambda &0\cr 0 &\lambda^{-1}}\right)= \exp\{\ln(\lambda)\sigma_3\},\\
\left(\matrix{1& y\cr 0  &1}\right)=\exp\{ y \sigma_+\}.
$$
