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The estimate expectation values is

$$\text{tr}((P_A\otimes P_B)\rho_\mathcal{E})=\text{tr}((P_A\otimes P_B)(\mathcal{E}\otimes I)(|\psi\rangle\langle\psi|))\,,$$

where $P_A$ and $P_B$ are Pauli matrices. $\rho_\mathcal{E}$ is the Jamiolkowski state, $$\rho_\mathcal{E}=(\mathcal{E}\otimes I)(|\psi\rangle\langle\psi|)\,,$$ and $$|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{j=1}^d|j\rangle\otimes|j\rangle\,.$$ How can I get the equivalent expression: $$\text{tr}((P_A\otimes P_B)\rho_\mathcal{E})=\frac{1}{d}\text{tr}(P_A\mathcal{E}(P_B^*))\,,$$ by using the Kraus decomposition? where the * denotes complex conjugation in the standard basis.

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    $\begingroup$ what is $\rho_{\cal E}$ here? If it's the Choi, you might be using $\mathcal E(X)=\operatorname{tr}_2[\rho_{\cal E}(I\otimes X^T)]$. $\endgroup$
    – glS
    Jan 9 at 10:37
  • $\begingroup$ $\rho_\mathcal{E}$ is the Jamiolkowski state, $\rho_\mathcal{E}=(\mathcal{E}\otimes I)(|\psi\rangle\langle\psi|)$, and $|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{j=1}^d|j\rangle\otimes|j\rangle$. $\endgroup$
    – karry
    Jan 9 at 10:41
  • $\begingroup$ $\mathcal{E}$ is a channel from some Hilbert space $A'$ in which $\vert\psi\rangle$ lives to the space $AB$. How exactly does $\mathcal{E}$ act on $P_B^\star$? $\endgroup$ Jan 10 at 12:43
  • $\begingroup$ @user1936752u I don‘t know, see arxiv.org/pdf/1205.2300.pdf $\endgroup$
    – karry
    Jan 10 at 13:09

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The key step in calculating $$ \text{tr}((P_A\otimes P_B)\rho_{\mathcal{E}}) $$ is to split taking the trace into two steps. First, take the trace over the second subsystem, and only then take the trace over the first. You'll want to verify that $$ \text{tr}_2((I\otimes P_B)|\psi\rangle\langle\psi|)=P_B^\star, $$ which is what lets you say that $$ \text{tr}_2((I\otimes P_B)\rho_{\mathcal{E}})=\mathcal{E}(P_B^\star). $$

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