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Consider a state $\rho_{A}\otimes\rho_{B}$ where both $\rho_{A}$ and $\rho_{B}$ are qubit system, undergoes unitary evolution as follows \begin{equation} \rho_{AB}'= (\cos\theta\mathbb{I}_{A}\otimes\mathbb{I}_{B}-i\sin\theta\hat{n}.\hat{\sigma}_{A}\otimes\hat{m}.\hat{\sigma}_{B})\rho_{A}\otimes\rho_{B}(\cos\theta\mathbb{I}_{A}\otimes\mathbb{I}_{B}+i\sin\theta\hat{n}.\hat{\sigma}_{A}\otimes\hat{m}.\hat{\sigma}_{B}) \end{equation} Now I want to apply a projector $\mathbb{I}\otimes\Pi_{B}$ with $\Pi_{B}=\frac{1}{2}(\mathbb{I}_{B}+\hat{l}.\hat{\sigma}_{B})$ on $\rho_{AB}'$ and look for the signature of both the interactions on A \begin{equation} \rho_{A}''=\frac{Tr_{B}(\mathbb{I}\otimes\Pi_{B}\rho_{AB}')}{Tr(\mathbb{I}\otimes\Pi_{B}\rho_{AB}')} \end{equation} Now my question is can we write a POVM which captures the transformation $\rho_{A}\rightarrow\rho_{A}''$? What will be the form of that POVM?

I have checked when $\rho_{B}$ is pure state one can successfully write a POVM like that. However, when $\rho_{B}$ is mixed state I am unable to find such POVM.

Solution for pure state:

Solution is rather long, so I will give the essential steps.

Let us consider $\rho_{A}=\frac{1}{2}(\mathbb{I}_{A}+\hat{a}.\hat{\sigma}_{A})$ and $\rho_{B}=\frac{1}{2}(\mathbb{I}_{B}+\hat{b}.\hat{\sigma}_{B})$. Then \begin{align} \rho_{A}''& = K[\cos^{2}\theta \rho_{A}-i\sin\theta\cos\theta(\hat{m}.\hat{\sigma}_{B})_{l}\hat{n}. \hat{\sigma}_{A}\rho_{A}-\nonumber\\ \label{e10} & (\hat{m}.\hat{\sigma}_{B})_{l}\rho_{A}\hat{n}.\hat{\sigma}_{A})+\sin^{2}\theta(\hat{m}.\hat{\sigma}_{B})_{l}^{1,1}(\hat{n}.\sigma_{A}\rho_{A}\hat{n}.\sigma_{A})] \end{align} where $(\hat{m}.\hat{\sigma}_{B})_{l}^{k,r}=\frac{Tr[\Pi_{l}\hat{m}.\hat{\sigma}_{B}^{r}\rho_{B}\hat{m}.\hat{\sigma}_{B}^{r}]}{Tr[\Pi_{l}\rho_{B}]}$, $(\hat{m}.\hat{\sigma}_{B})_{l}=\frac{Tr[\hat{\Pi}_{l}\hat{m}.\hat{\sigma}_{B}\rho_{B}]}{Tr[\hat{\Pi}_{l}\rho_{B}]}=a_{m}^{l}+ib_{m}^{l}$ and $K$ is the normalization factor given by \begin{equation} K=\frac{1}{\cos^{2}\theta+\sin^{2}\theta(\hat{m}.\hat{\sigma}_{B})_{l}^{1,1}+b_{m}^{l}\sin 2\theta\langle \hat{n}.\hat{\sigma}_{A}\rangle} \end{equation} It can be shown that if $\rho_{B}$ is pure state then $(\hat{m}.\hat{\sigma}_{B})_{l}^{1,1}=|(\hat{m}.\hat{\sigma}_{B})_{l}|^{2}$. Then one can have the corresponding POVM
\begin{equation} \hat{E}_{l}=Tr[\Pi_{l}\rho_{B}](cos\theta\mathbb{I}+i\sin\theta(\hat{m}.\hat{\sigma}_{B})_{l}\hat{n}.\hat{\sigma}_{A}) \end{equation} It can be easily checked that $\Sigma_{l}E_{l}^{\dagger}E_{l}=1$

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  • $\begingroup$ You formula for rho prime prime seems to be incorrect. It is always one. Maybe the Tr_B in the nominator is wrong? Also, maybe you can show your derivation for the pure state case. $\endgroup$
    – lalala
    May 24, 2017 at 11:16
  • $\begingroup$ edited. Will upload the the solution for pure state $\endgroup$
    – WInterfell
    May 24, 2017 at 14:37

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