Finite-dimensional Schumacher quantum source coding theorem

Fix $\delta >0$, choose an ordered eigenbasis of $\rho$ once, and take the spectral typical projector ${\mathrm{\Pi }}_n$ from [F2]. Order its finitely many product eigenvectors lexicographically; the resulting map to the first $r_n$ standard code basis vectors defines a partial isometry $W_n$ with $W_n^{\ast }{W}_n={\mathrm{\Pi }}_n$, where $r_n=\mathrm{rank}{\mathrm{\Pi }}_n$. Let the code space have dimension $M_n=r_n+1$ and let $\mathrm{\mid }\perp ⟩$ be its extra flag, orthogonal to $P_n:=W_n{W}_n^{\ast }$. Define ${\mathcal{E}}_n(X)=W_{n}X{W}_n^{\ast }+\mathrm{tr}((I-{\mathrm{\Pi }}_n)X)\mathrm{\mid }\perp ⟩⟨\perp \mathrm{\mid },{\mathcal{D}}_n(Y)=W_n^{\ast }Y{W}_n+⟨\perp \mathrm{\mid }Y\mathrm{\mid }\perp ⟩{\rho }^{\otimes n}\mathrm{.}$ The encoder Kraus family consists of $W_n$ and $\mathrm{\mid }\perp ⟩⟨e_j\mathrm{\mid }$ over an orthonormal basis of $\ker {\mathrm{\Pi }}_n$; its Kraus squares sum to ${\mathrm{\Pi }}_n+(I-{\mathrm{\Pi }}_n)=I$. The decoder Kraus family consists of $W_n^{\ast }$ and $\sqrt{{\lambda }_j}\mathrm{\mid }{f}_j⟩⟨\perp \mathrm{\mid }$ from a spectral decomposition ${\rho }^{\otimes n}={\sum }_j{\lambda }_j\mathrm{\mid }{f}_j⟩⟨f_j\mathrm{\mid }$; its Kraus squares sum to $P_n+\mathrm{\mid }\perp ⟩⟨\perp \mathrm{\mid }=I$. Thus both maps are channels by [F4]–[F6]. If $(K_{\mathrm{\ell }}{)}_{\mathrm{\ell }}$ is the finite composite Kraus family, substituting its action into [F1] gives $F_e^{(n)}={\sum }_{\mathrm{\ell }}\mathrm{\mid }⟨\varphi \mathrm{\mid }({\mathrm{id}}_R\otimes K_{\mathrm{\ell }})\mathrm{\mid }\varphi ⟩{\mathrm{\mid }}^2$. The branch $K_0=W_n^{\ast }{W}_n={\mathrm{\Pi }}_n$ contributes $\mathrm{\mid }⟨\varphi \mathrm{\mid }({\mathrm{id}}_R\otimes {\mathrm{\Pi }}_n)\mathrm{\mid }\varphi ⟩{\mathrm{\mid }}^2=(\mathrm{tr}({\rho }^{\otimes n}{\mathrm{\Pi }}_n){)}^2$, and every other branch contributes a nonnegative squared modulus. Therefore $F_e^{(n)}\ge (\mathrm{tr}({\rho }^{\otimes n}{\mathrm{\Pi }}_n){)}^2\to 1$. The basis order makes the sequence of codes explicit and uses no choice axiom.

Suppose ${\mathrm{lim\hspace{0.17em}sup}}_n{n}^{-1}\ln M_n<S(\rho )$. Choose $\delta >0$ and $N$ so that $M_n\le e^{n(S(\rho )-2\delta )}$ for $n\ge N$. Let ${\mathrm{\Pi }}_n$ be the typical projector from [F2], and let ${ϵ}_n=1-\mathrm{tr}({\rho }^{\otimes n}{\mathrm{\Pi }}_n)\to 0$. Each eigenvalue in its range is at most $e^{-n(S(\rho )-\delta )}$. Therefore the sum of any $M_n$ eigenvalues of ${\rho }^{\otimes n}$ is at most ${ϵ}_n+M_n{e}^{-n(S(\rho )-\delta )}\le {ϵ}_n+e^{-n\delta }\to 0$: the part outside the typical range contributes at most its total mass, and each selected eigenvalue inside contributes at most the displayed maximum.

Since $r_n\le e^{n(S(\rho )+\delta )}$ and $e^{n(S+\delta )}\ge 1$, $M_n=r_n+1\le 2e^{n(S(\rho )+\delta )}$. Hence ${\mathrm{lim\hspace{0.17em}sup}}_n{n}^{-1}\ln M_n\le S(\rho )+\delta$. Given $R>S(\rho )$, choose $\delta <R-S(\rho )$ to obtain the direct assertion.

Apply [F3] to each $n$-block code with intermediate dimension $M_n$. Step 1.2 shows that its fidelity is at most a quantity tending to zero, hence $F_e^{(n)}\to 0$. The argument includes zero eigenvalues and requires no full-rank source assumption.