# Derive $\text{tr}((P_A\otimes P_B)\rho_\mathcal{E})=\frac{1}{d}\text{tr}(P_A\mathcal{E}(P_B^*))$

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.

• what is $\rho_{\cal E}$ here? If it's the Choi, you might be using $\mathcal E(X)=\operatorname{tr}_2[\rho_{\cal E}(I\otimes X^T)]$.
– glS
Jan 9 at 10:37
• $\rho_\mathcal{E}$ is the Jamiolkowski state, $\rho_\mathcal{E}=(\mathcal{E}\otimes I)(|\psi\rangle\langle\psi|)$, and $|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{j=1}^d|j\rangle\otimes|j\rangle$. Jan 9 at 10:41
• $\mathcal{E}$ 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 $P_B^\star$? Jan 10 at 12:43
• @user1936752u I don‘t know, see arxiv.org/pdf/1205.2300.pdf Jan 10 at 13:09

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 $$\text{tr}_2((I\otimes P_B)\rho_{\mathcal{E}})=\mathcal{E}(P_B^\star).$$