# Why coherence measure monotonicity under selective incoherent operation should be form $C\left( \rho \right) \ge \sum_n{p_nC\left( \rho _n \right)}$?

In this paper, the authors give four postulates if a function $$C$$ can be taken as a coherence measure:

(C1) $$C(\rho) \geqslant 0$$, and $$C(\rho)=0$$ if and only if $$\rho \in \mathcal{I}$$, where $$\mathcal{I}$$ stands for sets of incoherent states.

(C2a) Monotonicity under incoherent operations, $$C(\rho) \geqslant$$ $$C(\Lambda(\rho))$$ if $$\Lambda$$ is an incoherent operation.

(C2b) Monotonicity under selective incoherent operations, $$C(\rho) \geqslant \sum_{n} p_{n} C\left(\rho_{n}\right)$$, where $$p_{n}=\operatorname{Tr}\left(K_{n} \rho K_{n}^{\dagger}\right), \rho_{n}=$$ $$K_{n} \rho K_{n}^{\dagger} / p_{n}$$, and $$\Lambda(\rho)=\sum_{n} K_{n} \rho K_{n}^{\dagger}$$ is an incoherent operation.

(C3) Nonincreasing under mixing of quantum states, i.e., convexity, $$\sum_{n} p_{n} C\left(\rho_{n}\right) \geqslant C\left(\sum_{n} p_{n} \rho_{n}\right)$$ for any set of states $$\left\{\rho_{n}\right\}$$ and any probability distribution $$\left\{p_{n}\right\}$$.

My question is about (C2b). Since they require $$K_n \mathcal{I} K_n^\dagger\subset \mathcal{I},\forall n$$, so why should (C2b) be $$C\left( \rho \right) \ge \sum_n{p_nC\left( \rho _n \right)}$$ instead of $$C\left( \rho \right) \ge C\left( \rho _n \right) ,\forall n$$?

• Why not to ask the paper's authors? Aug 4 at 9:24
• Interestingly, I think that $C(\rho) \geq C(\rho_n), \forall n$ is satisfied by the $l_1$-monotone for qubits.
– R.W
Aug 5 at 8:22

The reason why the stronger condition $$C(\rho) \geq C(\rho_n)$$ is not imposed is simply that this would rule out the vast majority of coherence/entanglement measures – coherence or entanglement, as measured by standard monotones such as the entropy, can actually be increased with some probability even under an incoherent/LOCC operation.
[...] we note that the measurement of $$k$$ occasionally yields a residual state $$\Psi_k$$ with more entropy of entanglement than the original state $$\Psi$$. However, neither the measurement of $$k$$ nor any other local processing by one or both parties can increase the expected entropy of entanglement between Alice’s and Bob’s subsystems.