[Physics] Find the Bogoliubov transformation $b=SaS^\dagger$ induced by the squeezed operator

Question

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.

Answer 1

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 $$ \left(\matrix{1&0\cr x &1}\right)= \exp\{x \sigma_-\},\\ \left(\matrix{\lambda &0\cr 0 &\lambda^{-1}}\right)= \exp\{\ln(\lambda)\sigma_3\},\\ \left(\matrix{1& y\cr 0 &1}\right)=\exp\{ y \sigma_+\}. $$