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For some projector $\Pi_A$ and state $\rho_{AB}$, let

$$\sigma_{AB} = \frac{\Pi_A\otimes I_B}{\text{Tr}((\Pi_A\otimes I_B)\rho_{AB})}\rho_{AB}$$

Is it the case that $\sigma_B = \rho_B$? It seems intuitively true since the projector is acting only on the $A$ system but I'm not sure how to prove this.

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No, this is not the case. Consider the situation where $$ \rho_{AB}=\frac12(|0\rangle\langle 0|\otimes |0\rangle\langle 0|+|1\rangle\langle 1|\otimes |1\rangle\langle 1|). $$ So, we have that $\rho_B=\text{Tr}_A(\rho_{AB})=\frac12(|0\rangle\langle 0|+|1\rangle\langle 1|)$.

Now let $\Pi_A=|0\rangle\langle 0|$, which means that $\sigma_{AB}=|00\rangle\langle 00|$ and hence $\sigma_B=|0\rangle\langle 0|$. Clearly, $\sigma_B$ and $\rho_B$ are different.

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