# How to obtain a two mode Squeezed state from two mode squeezed vacuum?

The squeezing operator for two modes $\hat{a}$ and $\hat{b}$ is given by

\begin{equation} \hat{S}(r)=\exp[r(\hat{a}\hat{b}-\hat{a}^{\dagger}\hat{b}^{\dagger})/2]. \end{equation} I am stuck to find out the action of the squeezing operator on a two mode vacuum state $\arrowvert 0,0\rangle$, i.e., $\hat{S}(r)\arrowvert 0,0\rangle$. The general expression is given by

\begin{equation} \hat{S}(r)\arrowvert 0,0\rangle=\sqrt{1-\lambda^{2}}\sum_{_n=0}^{\infty}\lambda^{n} \arrowvert n,n\rangle, \end{equation} where $\lambda=\tanh r$ and $r$ is the squeezing parameter. The operators satisfy the usual commutation relation $[\hat{a},\hat{a}^{\dagger}]=1=[\hat{b},\hat{b}^{\dagger}]$ and $|m,n\rangle=|m\rangle\otimes |n\rangle$, and $\hat{a}^{\dagger}|n\rangle_{A}=\sqrt{n+1}|n+1\rangle_{A}$ and $\hat{a}|n\rangle_{A}=\sqrt{n}|n-1\rangle_{A}$. I tried to diagonalize the expression inside the exponential using Bogoliubov transformation but it seems that i'm missing some trick. Any suggestions is highly appreciated.

## 1 Answer

Define operators $${\hat K}_- = {\hat a} {\hat b}\;,\;\;\; {\hat K}_+ = {\hat a}^\dagger {\hat b}^\dagger\;, \;\;\; {\hat K}_0 = \frac{1}{2} \Big({\hat a}{\hat a}^\dagger + {\hat b}{\hat b}^\dagger - {\hat I}\Big)$$ and notice that they define a $SU(1,1)$ algebra $$[{\hat K}_-\;, \;{\hat K}_+] = 2 {\hat K}_0\;, \;\;\; [{\hat K}_0\;,\;{\hat K}_\pm] = \pm {\hat K}_\pm$$ Then the following decomposition holds: $$e^{ \alpha_0 {\hat K}_0 + \alpha_+ {\hat K}_+ + \alpha_- {\hat K}_-}= e^{\gamma_+ {\hat K}_+} e^{\ln\gamma_0 {\hat K}_0} e^{\gamma_- {\hat K}_-}$$ where $$\gamma_0 = \Big( \cosh \theta - \frac{\alpha_0}{2\theta} \sinh \theta \Big)^{-2}\;,\;\;\; \gamma_\pm = \frac{2\alpha_\pm \sinh\theta}{2\theta \cosh\theta - \alpha_0 \sinh\theta}$$ for $$\theta^2 = \frac{\alpha_0^2}{4} - \alpha_+\alpha_-$$

Source: Klimov & Chumakov, "A Group-Theoretical Approach To Quantum Optics", Appendix 11.3.3.

For the case at hand we obviously have $\alpha_0=0$, $\alpha_- = - \alpha_+ = r/2$ and so $$\theta = r/2\;,\;\;\; \gamma_0 = \Big( \cosh \frac{r}{2}\Big)^{-2}\;, \;\;\; \gamma_\pm = \mp\tanh\frac{r}{2}$$ hence $$e^{(r/2)( {\hat a} {\hat b} - {\hat a}^\dagger {\hat b}^\dagger )} = e^{-\tanh(r/2) {\hat a}^\dagger {\hat b}^\dagger } e^{ - \ln\cosh (r/2) ({\hat a}{\hat a}^\dagger + {\hat b}{\hat b}^\dagger - {\hat I}) } e^{ \tanh(r/2) {\hat a} {\hat b}}$$ The rest is up to you.

• Hi! I am confused with $\tanh r/2$ and $\tanh (r/2)$ in the first and last exponents respectively (see the formula at the last line). From the symmetry consideration I would guess both would have same factor. Are they same? – user123 Apr 10 '17 at 3:52
• Yeah, they are the same, my bad typing. Unfortunately it looks like I had another couple of typos too: there was a missing unity term in ${\hat K}_0$, and a missing power of $(-2)$ in $\gamma_0$. I rechecked everything and it should be ok now. Sorry about the mess. – udrv Apr 10 '17 at 5:09