# How can $\chi(\hat{A},\hat{B}:C) \le \chi(\hat{A},B:C)$ be true?

The holevo information of $$\rho_{ABC}$$ w.r.t to measurements on A and B (for the sake of this we'll assume local measurements suffice), is given by $$\chi(\hat{A},\hat{B}:C)$$ where $$\hat{A}$$ and $$\hat{B}$$ are a local measurement channel using projectors after the maximisation is taken over all possible rank 1 projectors on the two subsystems.

These projective measurements on the subsystems $$A$$ and $$B$$ would give a classica-classical-quantum state after their action of $$\rho_{ABC}$$

However, this is just the mutual information of the resultant classical-quantum state you obtain after you perform said measurements on $$\rho_{ABC}$$, which means it will be upperbound by the mutual information of the state before the action of the local measurement map has taken place: $$I(A,B:C)\ge\chi(\hat{A},\hat{B}:C)$$

However, this should also obey the monotonicity of the relative entropy function under the action of a local channel:

$$S(\rho_{ABC}||\rho_{AB}\otimes \rho_{C})\ge S((I\otimes\Phi_{B})\rho_{ABC}||(I\otimes\Phi_{B})\rho_{AB}\otimes \rho_{C}) \ge$$

$$S((\Phi_{A} \otimes\Phi_{B})\rho_{ABC}||(\Phi_{A} \otimes\Phi_{B})\rho_{AB}\otimes \rho_{C})$$

This implies that $$\chi(\hat{A},\hat{B}:C) \le \chi(\hat{A},B:C)$$, where in this case $$\Phi_{A}$$ and $$\Phi_{B}$$ are the local measurement channels. This seems right given everything I know regarding the relative entropy.

However, how does having a whole other system to measure decrease the holevo information? I believe my confusion is maybe as a result of the maximisation, ie the measurements that maximise it given a local operation on one system don't remain the same when you consider another system to maximise across as well. Or I am misinterpreting $$\chi(\hat{A},B:C)$$ as a holveo information function given only one of the systems has been measured.

Either that or I am misunderstanding a key property regarding the monotonicity of the relative entropy function. Any other map and I would say that $$I(\hat{A},\hat{B}:C) \le I(\hat{A},B:C)$$, where the hat denotes the action of a map. But I am having trouble with the interpretation behind this. ie how does measurement on one subsystem give more informatio than measuring on two?

• to clarify: when you say "$\hat A$ and $\hat B$ are local measurement channels", what you mean is that you're computing the Holevo/mutual information wrt an ensemble that is classical on the side of $A$ and $B$, yes? I find talking about these being "measurement channels" a bit confusing. If I understand, you're just saying $A,B$ are the registers, and the state on which you compute the QMI is $\rho_{ABC}=\sum_{a,b} p_{a,b} (|a\rangle\!\langle a|\otimes|b\rangle\!\langle b|\otimes \rho_{ab})$ for some choice of bases. And the QMI of this state equals the corresponding Holevo $\chi$.
– glS
Commented May 3, 2023 at 15:16
• Though I agree that you can also think of this as computing the QMI of $(\Psi_A\otimes\Psi_B\otimes I)\rho$ for some pair of measurement (i.e. entanglement-breaking) channels, which is just a way to obtain the above classical-classical-quantum state, if that's what you mean
– glS
Commented May 3, 2023 at 15:18
• Yes, due to the action of the projective measurements, which I chose to represent as channels, you would have a CCQ state, where the classical side is AB. Hold one I'll explicitly state the CCQ state in the question. Commented May 3, 2023 at 15:18
• ok, but then isn't this essentially the same as asking why, for the standard case of bipartite states, the Holevo $\chi$ is upper bounded by the QMI? Answer being that the QMI contains contributions which are due to the quantum discord, whereas the Holevo $\chi$ is "more accessible", and it's the QMI minus the discord contributions (or at least, minus the discord contributions from one side)
– glS
Commented May 3, 2023 at 15:20
• I don't know if I'd call $\chi(\hat A,B:C)$ a "Holevo information", it looks like something in between the Holevo and the QMI. If I understand the notation, it's the QMI between $AB$ and $C$ after dephasing on $A$. That is, $I(AB:C)_{(\Phi_A\otimes I_{BC})\rho_{ABC}}$, maybe maximized over measurement channels $\Phi_A$. It makes sense that this is larger than $\chi(\hat A,\hat B:C)$ because in the latter you're only considering classical (or accessible, if you want) correlations on the $AB$ side, whereas in the former you keep part of the non-accessible (discordant) correlations
– glS
Commented May 3, 2023 at 20:52