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.


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

  • $\begingroup$ how do you do the last step? Are you taking the matrix exponential of $z a^{\dagger 2}-z^* a^2\simeq 2i(z \sigma_--z^* \sigma_+)$, thought of as a $2\times 2$ matrix? $\endgroup$
    – glS
    May 16, 2020 at 20:53
  • 1
    $\begingroup$ I'll add a bit to my answer. $\endgroup$
    – mike stone
    May 16, 2020 at 20:56
  • $\begingroup$ thanks, that helped. There is still something I don't quite understand though. You reduced the task to computing $\exp\begin{pmatrix}0& iz\\ -i z^*&0\end{pmatrix}$. But your Gauss decomposition doesn't work for this matrix: you have $A=c/a$ in your notation, but for the matrix we are dealing with $a=0$ $\endgroup$
    – glS
    May 17, 2020 at 10:38
  • $\begingroup$ or you meant that you first compute this exponential via standard Taylor expansion method, and then you apply the Gauss decomposition on the resulting exponential in order to get matrices which can be identified with the original operators $a^2,a^{\dagger 2},\sigma_3$? $\endgroup$
    – glS
    May 17, 2020 at 10:43
  • $\begingroup$ Yes. I screwed up $\exp x\sigma_\pm$! I'll fix after breakast. And yes again: expand exponential to get sinh and cosh's then Gauss. $\endgroup$
    – mike stone
    May 17, 2020 at 12:20

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