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 $C\left( \rho \right) \ge C\left( \rho _n \right) ,\forall n$?

  • $\begingroup$ Why not to ask the paper's authors? $\endgroup$
    – kludg
    Aug 4, 2022 at 9:24
  • 1
    $\begingroup$ Interestingly, I think that $C(\rho) \geq C(\rho_n), \forall n$ is satisfied by the $l_1$-monotone for qubits. $\endgroup$
    – R.W
    Aug 5, 2022 at 8:22

1 Answer 1


These conditions aren't really related to coherence theory per se, they are directly inspired by the requirements that entanglement measures are typically required to satisfy, see e.g. the old review on entanglement measures.

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.

An explicit example can be found in Bennett et al.'s Procrustean protocol for entanglement concentration, with the relevant quote being as follows:

[...] 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.