Triangle inequalities in coherence measures and entanglement concurrence

Abstract

We provide detailed proofs of triangle inequalities in coherence measures and entanglement concurrence. If a rank-2 state $\ensuremath{\varrho}$ can be expressed as a convex combination of two pure states, i.e., $\ensuremath{\varrho}={p}_{1}|{\ensuremath{\psi}}_{1}\ensuremath{\rangle}\ensuremath{\langle}{\ensuremath{\psi}}_{1}|+{p}_{2}|{\ensuremath{\psi}}_{2}\ensuremath{\rangle}\ensuremath{\langle}{\ensuremath{\psi}}_{2}|$, a triangle inequality can be established as $|E(|{\mathrm{\ensuremath{\Psi}}}_{1}\ensuremath{\rangle})\ensuremath{-}E(|{\mathrm{\ensuremath{\Psi}}}_{2}\ensuremath{\rangle})|\ensuremath{\le}E(\ensuremath{\varrho})\ensuremath{\le}E(|{\mathrm{\ensuremath{\Psi}}}_{1}\ensuremath{\rangle})+E(|{\mathrm{\ensuremath{\Psi}}}_{2}\ensuremath{\rangle})$, where $|{\mathrm{\ensuremath{\Psi}}}_{1}\ensuremath{\rangle}=\sqrt{{p}_{1}}|{\ensuremath{\psi}}_{1}\ensuremath{\rangle}$ and $|{\mathrm{\ensuremath{\Psi}}}_{2}\ensuremath{\rangle}=\sqrt{{p}_{2}}|{\ensuremath{\psi}}_{2}\ensuremath{\rangle}$; $E$ can be considered either coherence measures or entanglement concurrence. This inequality displays mathematical beauty for its similarity to the triangle inequality in plane geometry. An illustrative example is given after the proof.