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How is it possible to compute the Wigner function for a composite system that is prepared in an entangled state? In particular, consider the state $|ψ_{AB}\rangle=\frac{1}{\sqrt{2}}(|0_A\rangle|1_B\rangle + |1_A\rangle|0_B\rangle)$. How can we compute the following integral for the said entangled state?

$$W_{\psi}(x_{AB} \, ,p_{AB})=\frac{1}{\pi\hbar}\int\limits_{\infty}^{+\infty}dy\, \psi_{AB}^{*}(x+y)\,\psi_{AB}^{*}(x-y)\,e^{2ipy/\hbar}$$

Is there any way to express $W_{\psi_{AB}}(x_{AB} \, ,p_{AB})$ in terms of $W_{\psi_{A}}(x_{A} \, ,p_{A})$ and $W_{\psi_{B}}(x_{B} \, ,p_{B})$?

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Is there any way to express $W_{\psi_{AB}}(x_{AB} \, ,p_{AB})$ in terms of $W_{\psi_{A}}(x_{A} \, ,p_{A})$ and $W_{\psi_{B}}(x_{B} \, ,p_{B})$?

No.

This is in sharp contrast to pure states. Wigner functions behave in phase space the way density matrices comport themselves in Hilbert space.

For your entangled state, it is impossible to resolve $\rho_{AB}\neq \rho_A\otimes \rho_B$, and it should be obvious for the Wigner function, mutatis mutandis, except you wrote it in obscure language.

Let me choose a better language for you: Use x for A and x' for B. You might then write $$ \psi_{AB}(x,x')=\langle x| \langle x'| ~~|\psi_{AB}\rangle= \frac{1}{\sqrt {2}} (\langle x| 0\rangle \langle x'|1\rangle + \langle x| 1\rangle \langle x'|0\rangle ), $$ so your Wigner function reduces to just $$W_{AB} (x,x',p,p')=\frac{1}{(\pi\hbar)^2}\int\limits_{\infty}^{+\infty}dy ~dy'\, \psi_{AB}^*(x+y,x'+y')\,\psi_{AB}(x-y,x'-y')\,e^{2i(py+p'y') /\hbar} . $$

I won't plug the above in for you, but you immediately see that the sum of four integrals does not factorize in any way, the way the equivalent density matrices would not.

Now observe what happens when you "trace over B", that is, integrate over x',p'... You get a sum over two $W_A(x,p)$s, centered on 0 and 1 respectively... a mixed state--nonvanishing impurity.

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  • $\begingroup$ You are absolutely right that I was not clear enough. I meant $W_{\psi_{AB}} (x_A , x_{B}, p_A ,p_{B})$ and $\psi_{AB}(x_A,x_B)= \frac{1}{2}(\psi_{0} (x_A)\psi_{1} (x_B)+ \psi_{1} (x_A)\psi_{0} (x_B))$. Also, $x=x_{AB}=(x_{A}, x_{B})$. Nonetheless, I was aware that the integral appearing in this Wigner function is complicated, and I was curious whether it is possible to find a way for converting it to the Wigner function corresponding to the subsystems; that is $W_{\psi_A}(x_A, p_A)$ and $W_{\psi_B}(x_B, p_B)$. Now, I understand that this is not possible. Many thanks for your answer :) $\endgroup$ – Omid Charrakh Jul 19 '18 at 18:51

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