# Displacement operator similarity transformation using the squeezed operator

I have been trying to get the analytical expression for the Wigner function of squeezed vacuum states. Using the characteristic function representation, the WF can be written as

$$W(\alpha)=\frac{1}{\pi}\int e^{\lambda^*\alpha-\lambda\alpha^*}\hat{C}(\lambda)d^2\lambda$$

Where $$\hat{C}(\lambda)$$ is the characteristic function defined as $$\hat{C}(\lambda)=\mathrm{Tr}\lbrace \rho\hat{D}(\lambda)\rbrace=\langle\hat{D}(\lambda)\rangle$$, with the displacement operator defined as $$\hat{D}(\lambda)=\exp{(-|\lambda|^2/2)}\exp{(\lambda\hat{a}^\dagger)}\exp{(-\lambda^*\hat{a})}$$.

Using the squeezed vacuum state, the characteristic function is

$$\hat{C}_{\xi}(\lambda)=\langle\xi|\hat{D}(\lambda)|\xi\rangle=\langle 0|\hat{S}^{\dagger}(\xi)\hat{D}(\lambda)\hat{S}(\xi)|0\rangle$$

Where $$\hat{S}(\xi)=\exp{(\frac{1}{2}\xi^*\hat{a}^2-\frac{1}{2}\xi\hat{a}^{\dagger 2})}$$ is the squeezed operator. I have seen that the transformation $$\hat{S}^{\dagger}(\xi)\hat{D}(\lambda)\hat{S}(\xi)$$ is equal to $$\hat{D}(\lambda\mu+\lambda^*\nu)$$, where $$\mu,\nu$$ are the elements of the similarity transformation of creation and annihilation operators using the squeezed operator, i.e., $$\hat{a}^{'}=\hat{S}(\xi)\hat{a}\hat{S}^\dagger(\xi)=\mu\hat{a}+\nu\hat{a}^\dagger$$ and $$\hat{a}^{\dagger'}=\hat{S}(\xi)\hat{a}^\dagger\hat{S}^\dagger(\xi)=\nu^*\hat{a}+\mu\hat{a}^\dagger$$.

Does anyone know how to proof this property?

$$\hat{S}^{\dagger}(\xi)\hat{D}(\lambda)\hat{S}(\xi)=\hat{D}(\lambda\mu+\lambda^*\nu)$$

Thanks in advance for any ideas!

To begin with, let's rederive the effect of the squeezing operator on the creation/annhilation operator. Let's define $$\hat{X}_\xi := \frac{1}{2}(\xi^*\hat{a}^2-\xi\hat{a}^{\dagger2})$$ and compute the following commutators : $$\left\{ \begin{array}{ccl} [\hat{X}_\xi,\hat{a}] &=& -\frac{1}{2}\xi[\hat{a}^{\dagger2},\hat{a}] &=& \xi\hat{a}^\dagger \\ [\hat{X}_\xi,\hat{a}^\dagger] &=& \;\frac{1}{2}\xi^*[\hat{a}^2,\hat{a}^\dagger] &=& \xi^*\hat{a} \end{array} \right.$$ Then, the "Baker-Campbell-Hausdorff formula" permits to write : $$\begin{array}{l} \displaystyle \hat{S}(\xi)\hat{a}\hat{S}(\xi)^\dagger = e^{\hat{X}_\xi}\hat{a}e^{-\hat{X}_\xi} = \sum_{n\ge0}\frac{[\hat{X}_\xi,\hat{a}]_n}{n!} = \sum_{k\ge0}\left(\frac{|\xi|^{2k}}{(2k)!}\hat{a}+\frac{\xi|\xi|^{2k}}{(2k+1)!}\hat{a}^\dagger\right) \\ \displaystyle \color{white}{\hat{S}(\xi)\hat{a}\hat{S}(\xi)^\dagger} = \cosh|\xi|\;\hat{a}+\frac{\xi}{|\xi|}\sinh|\xi|\;\hat{a}^\dagger \end{array}$$ where $$[\,\cdot\,,\cdot\,]_n$$ denotes $$n$$-time nested commutators, hence $$\begin{cases} \mu(\xi) = \cosh|\xi| \\ \nu(\xi) = \frac{\xi}{|\xi|}\sinh|\xi| \end{cases}$$
Now, it is actually easier to work with the "single-exponential" form of the displacement operator, i.e. $$\hat{D}(\lambda) = e^{\lambda\hat{a}^\dagger-\lambda^*\hat{a}}$$, which you can recover with the help of Glauber's formula. Thus we have : $$\begin{array} \hat{S}(\xi)\hat{D}(\lambda)\hat{S}(\xi)^\dagger &=& \displaystyle \hat{S}_\xi e^{\lambda\hat{a}^\dagger-\lambda^*\hat{a}} \hat{S}_\xi^\dagger \\ &=& \displaystyle \sum_{n=0}^\infty\frac{1}{n!}\hat{S}_\xi(\lambda\hat{a}^\dagger-\lambda^*\hat{a})^n\hat{S}_\xi^\dagger \\ &=& \displaystyle \sum_{n=0}^\infty\frac{1}{n!}\left(\lambda\hat{S}_\xi\hat{a}^\dagger\hat{S}_\xi^\dagger-\lambda^*\hat{S}_\xi\hat{a}\hat{S}_\xi^\dagger\right)^n \\ &=& \displaystyle \sum_{n=0}^\infty\frac{1}{n!}\left(\lambda(\mu\hat{a}^\dagger+\nu^*\hat{a})-\lambda^*(\nu\hat{a}^\dagger+\mu\hat{a}\right)^n \\ &=& \displaystyle \sum_{n=0}^\infty\frac{1}{n!}\left((\lambda\mu-\lambda^*\nu)\hat{a}^\dagger-(\lambda^*\mu-\lambda\nu^*)\hat{a})\right)^n \\ &=& \displaystyle \exp\left((\lambda\mu-\lambda^*\nu)\hat{a}^\dagger-(\lambda\mu^*-\lambda^*\nu)^*\hat{a})\right) \\ &=& \displaystyle \hat{D}(\lambda\mu-\lambda^*\nu) \end{array}$$ since $$\mu^* = \mu$$, and finally : $$\hat{S}(\xi)^\dagger\hat{D}(\lambda)\hat{S}(\xi) = \hat{S}(-\xi)\hat{D}(\lambda)\hat{S}(-\xi)^\dagger = \hat{D}(\lambda\mu+\lambda^*\nu),$$ since $$\mu(-\xi)=\mu(\xi)$$ and $$\nu(-\xi)=-\nu(\xi)$$.