Holevo information of an equiprobable binary pure-state channel

Put $\bar{\rho }=\frac{1}{2}\mathrm{\mid }{\psi }_0⟩⟨{\psi }_0\mathrm{\mid }+\frac{1}{2}\mathrm{\mid }{\psi }_1⟩⟨{\psi }_1\mathrm{\mid }$. Its trace is one. After changing the global phase of $\mathrm{\mid }{\psi }_1⟩$ and choosing an orthonormal basis, write $\mathrm{\mid }{\psi }_0⟩=(1,0)$ and $\mathrm{\mid }{\psi }_1⟩=(c,\sqrt{1-c^2})$. Then $\bar{\rho }=\frac{1}{2}\left(\begin{array}{cc}1+c^2& c\sqrt{1-c^2}\\ c\sqrt{1-c^2}& 1-c^2\end{array}\right),$ whose trace is $1$ and determinant is $\frac{1}{4}((1+c^2)(1-c^2)-c^2(1-c^2))=(1-c^2)\mathrm{/}4$. Thus its characteristic polynomial is $t^2-t+(1-c^2)\mathrm{/}4$, and [F3] gives eigenvalues $(1+c)\mathrm{/}2$ and $(1-c)\mathrm{/}2$.

Each pure output state has eigenvalues $1,0$, hence entropy zero by [F2]. By the symmetry $h(t)=h(1-t)$ in its displayed definition, the average entropy from step 1.1 is $h((1+c)\mathrm{/}2)$, so [F2] gives the displayed $\chi$. This ensemble is among those maximized in [F5], so $C_{\mathrm{c}\mathrm{q}}\ge \chi$; the rate conclusion follows from [F5] for every $0\le R<\chi$ when $\chi >0$. If $c=0$, the value is $\ln 2$; if $c=1$, both outputs coincide up to phase and the value is zero.