How to calculate the Wigner function for an entangled state

[khfrekek92]

I'm trying to calculate a Wigner Function of an entangled state, and I'm not quite sure how to proceed. I have created this state by sending in vacuum and a squeezed state into a 50/50 BS, where the output state has a density operator:

$$\rho_{34}=S_{3}S_{4}S_{34}|00\rangle_{34}\langle 00|_{34}S_{34}^{\dagger}S_{4}^{\dagger}S_{3}^{\dagger}$$

Where $S_{3}$, $S_{4}$ and $S_{34}$ are the usual single and two mode squeezing operators.

Now I'm kind of stuck on how to calculate the Wigner function from here. I'm assuming I need to trace out one of the states as follows:

$$
\begin{aligned}
\rho_{3}&=Tr_{4}(\rho_{34})\\
&=S_{3}Tr_{4}[S_{4}S_{34}|00\rangle_{34}\langle 00|_{34}S_{34}^{\dagger}S_{4}^{\dagger}]S_{3}^{\dagger}\\
&=S_{3}Tr_{4}[S_{34}|00\rangle_{34}\langle 00|_{34}S_{34}^{\dagger}]S_{3}^{\dagger}\\
&=S_{3}Tr_{4}[|TMSV\rangle_{34}\langle TMSV|_{34}]S_{3}^{\dagger}\\
&=S_{3}\sum_{n=0}^{\infty}[_{4}\langle n|TMSV\rangle_{34}._{34}\langle TMSV|n\rangle_{4}]S_{3}^{\dagger}\\
&=\frac{S_{3}}{\cosh^{2}(r)}\sum_{m=0}^{\infty}[\tanh^{2m}(r)|m\rangle_{4}\langle m|_{4}]S_{3}^{\dagger}
\end{aligned}
$$Then to calculate the Wigner function, we use:

$$W_{3}=\frac{1}{2\pi{h}}\int_{-inf}^{inf}._{3}<q-\frac{y}{2}|\frac{S_{3}}{cosh^{2}(r)}\sum_{m=0}^{\infty}tanh^{2m}(r)|m>_{3}<m|S_{3}^{\dagger}|q+\frac{y}{2}>e^{\frac{ipy}{h}}dy$$
And then I get stuck. I have no idea how I would calculate this. Did I miss anything? Does anyone have anything that could maybe point me in the right direction? Thanks so much in advance!

 

[soarce]

I am not sure that the state \rho_{34} you wrote is obtained by passing a vacuum squeezed state through a beam spliter.
If you start with a two-mode ($a$ and $b$) vacuum state, apply the squeezing operator on mode $a$ and then send the two modes through a beam spliter should look like this:
\left|\left.0_a0_b\right&gt;\right.
Apply squeezing operator on mode $a$, $S_a(\eta)=\exp\left(\frac{\eta}{2}\left(a^\dagger\right)^2-\frac{\eta^\ast}{2}a^2\right)$:
S_a(\eta)\left|\left.0_a0_b\right&gt;\right.
Now apply the rotation operator on both modes, $R(\xi)=\exp\left(\xi a^\dagger b-\xi^\ast a b^\dagger\right)$:
R(\xi)S_a(\eta)\left|\left.0_a0_b\right&gt;\right.
and the density matrix of the final two-mode state should look like
\rho=R(\xi)S_a(\eta)\left|\left.0_a0_b\right&gt;\right.\left&lt;\left.0_a0_b\right|\right.S^\dagger_a(\eta)R^\dagger(\xi)

 

[khfrekek92]

Ah yes I see what you mean, I'll have to go back through and see what I missed. So now, with your $\rho$, do I just need to Trace over b to get the single-mode density operator for state a, and then plug that into the Winter integral as usual? Or do you need to do something different for entangled states?

 

[soarce]

It might be difficult to work with position representation, i.e. to calculate the integral over the continuous variable $y$. I did similar calculations long time ago and we used Fock representation in order to obtain quasi-distribution functions (including Wigner function). First we derived the characteristic function of the state:

\chi(\alpha,\beta;p)=\exp\left(p\frac{|\alpha|^2+|\beta|^2}{2}\right)\mbox{Tr}\left(\rho D_a(\alpha)D_b(\beta)\right)

$D$ is the displacement operator. Then the quasi-probability function is just the Fourier transform of the characteristic function. To obtain the Wigner function one sets $p=0$.

By using Fock representation one transform the integral into (infinite) sums. Give it a try to see if you can handle the calculation, probably you won't arrive to a closed analytical form but maybe you can put it in a nice form of a (infinte) sum.

 

[khfrekek92]

Alright thank you so much for pointing me in the right direction! I didn't know you couldn't calculate Wigner functions like that. I'll read up on that and give it a try. Thanks so much for your help!