Sector entropy and relative entropy identities

Choose orthonormal eigenbases for each ${\rho }_{\alpha }$, ${\sigma }_{\alpha }$, and ${\chi }_{\alpha }$ using [F2]. Write their eigenvalues as $r_{\alpha i}$, $t_{\alpha i}$, and $c_{\alpha j}$, respectively. In each block, the tensor products of the factor eigenvectors form an orthonormal basis and have eigenvalues $p_{\alpha }{r}_{\alpha i}{c}_{\alpha j}$ for $p_{\alpha }{\rho }_{\alpha }\otimes {\chi }_{\alpha }$ and $q_{\alpha }{t}_{\alpha i}{c}_{\alpha j}$ for $q_{\alpha }{\sigma }_{\alpha }\otimes {\chi }_{\alpha }$. Thus every block is positive semidefinite, the finite orthogonal sum is positive semidefinite, and its trace is ${\sum }_{\alpha }{p}_{\alpha }({\sum }_i{r}_{\alpha i})({\sum }_j{c}_{\alpha j})={\sum }_{\alpha }{p}_{\alpha }=1$; the same calculation with $q_{\alpha },t_{\alpha i}$ proves $\mathrm{tr}{\sigma }_A=1$. Since $q_{\alpha },t_{\alpha i},c_{\alpha j}$ are all strictly positive, every eigenvalue of ${\sigma }_A$ is positive, so the reference is faithful.

The eigenvalues of ${\rho }_A$ are exactly $p_{\alpha }{r}_{\alpha i}{c}_{\alpha j}$ over its orthogonal block eigenbases. Terms with $r_{\alpha i}=0$ contribute $0$ by the entropy convention; all other terms satisfy $\log (p_{\alpha }{r}_{\alpha i}{c}_{\alpha j})=\log p_{\alpha }+\log r_{\alpha i}+\log c_{\alpha j}$ by [F4]. Substituting in $s({\rho }_A)$ and summing the finite products, using ${\sum }_i{r}_{\alpha i}={\sum }_j{c}_{\alpha j}=1$, gives $s({\rho }_A)=-{\sum }_{\alpha }{p}_{\alpha }\log p_{\alpha }+{\sum }_{\alpha }{p}_{\alpha }s({\rho }_{\alpha })+{\sum }_{\alpha }{p}_{\alpha }s({\chi }_{\alpha })$, which is the asserted entropy identity.

Write $u_{\alpha i}$ for the chosen eigenbasis of ${\sigma }_{\alpha }$ and $v_{\alpha j}$ for that of ${\chi }_{\alpha }$, and put $a_{\alpha i}:=⟨{\rho }_{\alpha }{u}_{\alpha i},u_{\alpha i}⟩$. By [F4], the block-$\alpha$ eigenvalues of $K_{{\sigma }_A}:=-\log {\sigma }_A$ on $u_{\alpha i}\otimes v_{\alpha j}$ are $-\log q_{\alpha }-\log t_{\alpha i}-\log c_{\alpha j}$; the diagonal entries of ${\rho }_{\alpha }\otimes {\chi }_{\alpha }$ in that basis are $a_{\alpha i}{c}_{\alpha j}$. Its block trace against $K_{{\sigma }_A}$ is therefore $p_{\alpha }{\sum }_{i,j}{a}_{\alpha i}{c}_{\alpha j}(-\log q_{\alpha }-\log t_{\alpha i}-\log c_{\alpha j})$. Factoring the finite sums and using ${\sum }_i{a}_{\alpha i}=\mathrm{tr}{\rho }_{\alpha }=1$, ${\sum }_j{c}_{\alpha j}=1$, ${\sum }_i{a}_{\alpha i}(-\log t_{\alpha i})=\mathrm{tr}({\rho }_{\alpha }(-\log {\sigma }_{\alpha }))$, and $-{\sum }_j{c}_{\alpha j}\log c_{\alpha j}=s({\chi }_{\alpha })$, summing over $\alpha$ gives $\mathrm{tr}({\rho }_A{K}_{{\sigma }_A})={\sum }_{\alpha }{p}_{\alpha }[-\log q_{\alpha }+\mathrm{tr}({\rho }_{\alpha }(-\log {\sigma }_{\alpha }))+s({\chi }_{\alpha })]$.

By [F1] and faithfulness from step 1.1, $D({\rho }_A\mathrm{\parallel }{\sigma }_A)=\mathrm{tr}({\rho }_A{K}_{{\sigma }_A})-s({\rho }_A)$, since $\mathrm{tr}({\sigma }_A{K}_{{\sigma }_A})=s({\sigma }_A)$ by the spectral entropy definition. Applying the same identity [F1] sectorwise gives $D({\rho }_{\alpha }\mathrm{\parallel }{\sigma }_{\alpha })=\mathrm{tr}({\rho }_{\alpha }(-\log {\sigma }_{\alpha }))-s({\rho }_{\alpha })$. Subtract the entropy expression in step 2.1 from the trace expression in step 2.2; the ancillary terms cancel and $-{\sum }_{\alpha }{p}_{\alpha }\log q_{\alpha }-H(\mathbf{p})={\sum }_{\alpha }{p}_{\alpha }\log (p_{\alpha }\mathrm{/}{q}_{\alpha })$. This is the claimed relative-entropy identity.