# A How to calculate the Wigner function for an entangled state

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1. Dec 12, 2016

### 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!

Last edited: Dec 12, 2016
2. Dec 12, 2016

### 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:
$$\left|\left.0_a0_b\right>\right.$$
Apply squeezing operator on mode $a$, $S_a(\eta)=\exp\left(\frac{\eta}{2}\left(a^\dagger\right)^2-\frac{\eta^\ast}{2}a^2\right)$:
$$S_a(\eta)\left|\left.0_a0_b\right>\right.$$
Now apply the rotation operator on both modes, $R(\xi)=\exp\left(\xi a^\dagger b-\xi^\ast a b^\dagger\right)$:
$$R(\xi)S_a(\eta)\left|\left.0_a0_b\right>\right.$$
and the density matrix of the final two-mode state should look like
$$\rho=R(\xi)S_a(\eta)\left|\left.0_a0_b\right>\right.\left<\left.0_a0_b\right|\right.S^\dagger_a(\eta)R^\dagger(\xi)$$

3. Dec 12, 2016

### 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 in to the Winter integral as usual? Or do you need to do something different for entangled states?

4. Dec 12, 2016

### 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.

5. Dec 13, 2016

### 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!