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.

  • $\begingroup$ Can you define the relevant symbols you're using in the question? $\endgroup$
    – Rammus
    Aug 12 at 12:47
  • $\begingroup$ @Rammus does rhe edit work? $\endgroup$ Aug 12 at 12:53

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