Compression dimension bounds entanglement fidelity

Choose Kraus families $(E_a)$ and $(D_b)$ for the encoder and decoder from [F2]. The composite channel has Kraus operators $K_{ba}=D_b{E}_a$, each of rank at most $M$ because it factors through the $M$-dimensional code space. By [F3], ${\sum }_{a,b}{K}_{ba}^{\ast }{K}_{ba}={\sum }_a{E}_a^{\ast }\left({\sum }_b{D}_b^{\ast }{D}_b\right)E_a={\sum }_a{E}_a^{\ast }{E}_a=I\mathrm{.}$

By [F4], write $\mathrm{\mid }\varphi ⟩={\sum }_{j=1}^r{s}_j\mathrm{\mid }{r}_j⟩\mathrm{\mid }{a}_j⟩$, ordering the $s_j$ decreasingly. Expanding $\mathrm{\mid }\varphi ⟩⟨\varphi \mathrm{\mid }$ and contracting either orthonormal factor family by [F7] gives ${\rho }_R:={\mathrm{Tr}}_{A^{\otimes n}}(\mathrm{\mid }\varphi ⟩⟨\varphi \mathrm{\mid })={\sum }_{j=1}^r{s}_j^2\mathrm{\mid }{r}_j⟩⟨r_j\mathrm{\mid },{\rho }^{\otimes n}={\sum }_{j=1}^r{s}_j^2\mathrm{\mid }{a}_j⟩⟨a_j\mathrm{\mid }\mathrm{.}$ Indeed each cross term with indices $j,k$ contracts with coefficient $⟨a_k\mathrm{\mid }{a}_j⟩={\delta }_{jk}$, or $⟨r_k\mathrm{\mid }{r}_j⟩={\delta }_{jk}$, respectively. Thus both reduced operators have exactly the positive eigenvalues $s_j^2$, with any remaining eigenvalues zero. Complete $(r_j{)}_{j=1}^r$ by an orthonormal basis of $\ker {\rho }_R$, available from [F5] applied to the zero operator there. On this full reference basis, ${\rho }_R$ has eigenvalues ${\lambda }_j^{(n)}$, with the Statement's zero padding; in particular ${\sum }_j{\lambda }_j^{(n)}={\sum }_{j=1}^r{s}_j^2=1$.

For each $K=K_{ba}$ let $\mathrm{\mid }{\eta }_K⟩=(I_R\otimes K)\mathrm{\mid }\varphi ⟩$, and let $P_K$ be the projection onto its reference-side Schmidt support (zero if ${\eta }_K=0$). Since $\mathrm{\mid }{\eta }_K⟩\in H_R\otimes \mathrm{ran}K$, apply [F4] on these two factors: there are at most $\dim \mathrm{ran}K\le M$ orthonormal second-factor vectors. Thus $\mathrm{rank}{P}_K\le M$ and $(P_K\otimes I)\mathrm{\mid }{\eta }_K⟩=\mathrm{\mid }{\eta }_K⟩$. Cauchy–Schwarz and the trace-pairing identity in [F7] give $\mathrm{\mid }⟨\varphi \mathrm{\mid }{\eta }_K⟩{\mathrm{\mid }}^2\le ⟨\varphi \mathrm{\mid }(P_K\otimes I)\mathrm{\mid }\varphi ⟩\mathrm{\parallel }{\eta }_K{\mathrm{\parallel }}^2=\mathrm{tr}({\rho }_R{P}_K)\mathrm{\parallel }{\eta }_K{\mathrm{\parallel }}^2\mathrm{.}$

Put $q_j=⟨r_j,P_K{r}_j⟩$. Then $0\le q_j\le 1$, and the full reference basis from step 1.2 gives ${\sum }_j{q}_j=\mathrm{tr}(P_K)=\mathrm{rank}{P}_K\le M$. If $M\ge \dim H_R$, then $\mathrm{tr}({\rho }_R{P}_K)\le 1={\sum }_{j\le M}{\lambda }_j^{(n)}$, since the source has at most $\dim H_R$ positive eigenvalues. Otherwise, ordering the eigenvalues gives $\mathrm{tr}({\rho }_R{P}_K)={\sum }_j{\lambda }_j^{(n)}{q}_j\le {\sum }_{j\le M}{\lambda }_j^{(n)}{q}_j+{\lambda }_M^{(n)}(M-{\sum }_{j\le M}{q}_j)\le {\sum }_{j\le M}{\lambda }_j^{(n)}\mathrm{.}$ Thus the overlap in step 2.1 is bounded by $\left({\sum }_{j\le M}{\lambda }_j^{(n)}\right)\mathrm{\parallel }{\eta }_K{\mathrm{\parallel }}^2$.

Expanding the recovered state in the composite Kraus family from step 1.1 gives $F_e^{(n)}={\sum }_{a,b}\mathrm{\mid }⟨\varphi \mathrm{\mid }{\eta }_{K_{ba}}⟩{\mathrm{\mid }}^2$. By step 3.1 this is at most $\left({\sum }_{j\le M}{\lambda }_j^{(n)}\right){\sum }_{a,b}\mathrm{\parallel }{\eta }_{K_{ba}}{\mathrm{\parallel }}^2$. Step 1.1 gives ${\sum }_{a,b}\mathrm{\parallel }{\eta }_{K_{ba}}{\mathrm{\parallel }}^2=⟨\varphi \mathrm{\mid }{I}_R\otimes ({\sum }_{a,b}{K}_{ba}^{\ast }{K}_{ba})\mathrm{\mid }\varphi ⟩=\mathrm{\parallel }\varphi {\mathrm{\parallel }}^2=1$, proving the bound.