# How is $I(\rho^{QC})=I_{CC}(\rho^{QC})$

On page 3 of this paper, for the proof of theorem 1, it states that, using Lemma 2 from the previous page, that if $$I(\Lambda_{A}\otimes\Gamma_{B})[\rho]=I(\rho))$$ then there exists $$\Lambda_{A}^{*}$$ and $$\Gamma_{B}^{*}$$ s.t $$(\Lambda_{A}^{*}\otimes\Gamma_{B}^{*})\circ(\Lambda_{A}\otimes\Gamma_{B})[\rho]=\rho$$Using this, they show that $$\rho^{QC}=(M^{*}\otimes I)[\rho^{CC}]$$ where $$\rho^{CC}=(M\otimes N)[\rho]$$ with the assumption that $$I(\rho^{CC})=I(\rho)$$, so $$M^{*}$$ and $$N^{*}$$ exist.

However, they state that $$I(\rho^{QC})=I_{CC}(\rho^{QC})=I_{CC}(\rho)=I(\rho)$$

I understand the last equality fine. But how are they getting the second equality. $$(M\otimes N)\rho^{QC}=(M\otimes N) \circ (M^{*}\otimes I)[\rho^{CC}]=\sum_{ij}p_{ij}|i\rangle\langle i|\otimes N(|j\rangle\langle j|)$$ and so

$$\rho_{CC}^{QC}=\sum_{ij}p_{ij}|i\rangle\langle i|\otimes N(|j\rangle\langle j|)=(M\otimes N)\rho^{QC}$$ so $$(M^{*}\otimes N^{*}) \circ (M\otimes N)[\rho^{QC}]=\rho^{QC}$$ and so $$I_{CC}(\rho^{QC})=I(\rho^{QC})$$ But how do they get $$I_{CC}(\rho^{QC})=I_{CC}(\rho)$$

The only thing I can think of is $$(I \otimes N^{*}) \circ (I \otimes N)[\rho_{CC}]=\rho_{CC}$$ which should mean that $$I(\rho_{CC}^{QC})=I_{CC}(\rho)$$, but I am not sure if this reasoning is correct.

Edit: $$\Lambda$$, $$\Gamma$$, $$M$$, $$M^{*}$$, $$N$$ and $$N^{*}$$ are all quantum maps. $$M$$ and $$N$$ are local quantum-to-classical measurement maps. CC and QC, per the paper, mean the classical-classical and quantum-classical states resulting from the application of the maps.

• Can you define the relevant symbols you're using in the question? Aug 12, 2021 at 12:47
• @Rammus does rhe edit work? Aug 12, 2021 at 12:53