# Squeezed State and Bogoliubov transformation

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.

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 $$S(z)= \exp\left\{{\textstyle \frac12}(z {a^\dagger}^2 -z^* a^2)\right\}$$ $$=\exp\left\{e^{i\theta}{\textstyle \frac12}\tanh |z|\, {a^\dagger}^2\right\}\exp\left\{ -\ln\cosh |z| (a^\dagger a+{\textstyle \frac12})\right\} \exp\left\{-e^{-i\theta}{\textstyle \frac12}\tanh |z| \, a^2\right\}$$ $$=\exp\left\{-e^{-i\theta}{\textstyle \frac12}\tanh |z| \,a^2\right\}\exp\left\{ +\ln\cosh |z| (a^\dagger a+\textstyle \frac12)\right\} \exp\left\{e^{i\theta}{\textstyle \frac12}\tanh |z|\, {a^\dagger}^2\right\}.\nonumber$$