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The Kraus formula for entanglement fidelity applies to any input, including mixed qubit states. For the amplitude damping channel, the diagonal Kraus trace is $[1+s_z+\sqrt{1-p}(1-s_z)]/2$ and the jump Kraus trace has squared modulus $p(s_x^2+s_y^2)/4$. Summing them gives the formula. Ground-state inputs have unit entanglement fidelity; excited-state inputs have fidelity $1-p$.

If encoding and decoding quantum channels factor through a $k$-dimensional space, their composite Kraus operators all have matrix rank at most $k$. The rank bound for a weighted operator trace and the Kraus formula for entanglement fidelity implyThus high operation fidelity requires the leading $k$ eigenvalues of the source density operator to carry almost all its probability. The bound applies to arbitrary completely positive trace-preserving encoders and decoders.

For Kraus operators $A_j$, the Kraus formula for entanglement fidelity isIndeed each Kraus contribution to the purified-state overlap is $|⟨\Psi |(I\otimes A_j)|\Psi ⟩{|}^2$, and the expectation in a purification is $\mathrm{Tr}(\rho A_j)$. In this case,The entanglement fidelity of amplitude damping is thusFor $p=0$, this is one. For a ground-state input it is one for every $p$, whereas for an excited-state input it is $1-p$. These checks agree with the physical relaxation mechanism.

Diagonalize the source average $\pi ={\sum }_j{\lambda }_j|j⟩⟨j|$, using its positive eigenvalues. For fixed $\delta >0$, let ${\Pi }_{n,\delta }$ project onto product eigenvectors whose eigenvalues obeyThe typical subspace theorem states that, for every $ϵ>0$ and all sufficiently large $n$,The probability statement is the weak law of large numbers applied to $-{\log }_2{\lambda }_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the quantum typical subspace retains almost all probability using about $nS(\pi )$ qubits.

For $R>S(\pi )$ choose $0<\delta <R-S(\pi )$. Measure $\{{\Pi }_{n,\delta },I-{\Pi }_{n,\delta }\}$. On success, encode the projected state isometrically into a space of dimension $\dim {\mathcal{T}}_{n,\delta }$; on failure, output a separate fixed flag. Decode the successful sector by the inverse isometry, and map the flag to a fixed state $|{\phi }_0⟩$. The compressed dimension is at most $2^{n(S(\pi )+\delta )}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving quantum channels, not merely successful postselected operations.

Writing $\Pi ={\Pi }_{n,\delta }$, the composite channel isFor a pure source signal $|{\Psi }_k⟩$, set $a_k=⟨{\Psi }_k|\Pi |{\Psi }_k⟩$. Its squared quantum fidelity after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is ${\pi }^{\otimes n}$. Convexity of the square givesThe Kraus formula for entanglement fidelity gives the same lower bound for $F_e({\pi }^{\otimes n},{\mathcal{N}}_n)$, because one Kraus operator is $\Pi$ and the other terms are nonnegative. Taking $ϵ\to 0$ proves reliable compression at every $R>S(\pi )$, in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is typical-subspace compression with a failure flag.

Choose Kraus representationsThe composite quantum channel has Kraus operators $T_{ba}=D_b{C}_a$. Each has matrix rank at most $k$, because it factors through the $k$-dimensional space $K$, andThe operation fidelity uses the unsquared quantum fidelity on a purification of a density operator, so its square is entanglement fidelity. The Kraus formula for entanglement fidelity givesIndeed each overlap $⟨{\Psi }_{\rho }|(I_R\otimes T_{ba})|{\Psi }_{\rho }⟩$ equals $\mathrm{Tr}({\rho }_Q{T}_{ba})$.

For any one of these Kraus operators $T$, let ${\Pi }_T$ be the orthogonal projection onto its image, of matrix rank $r_T\le k$. Since ${\Pi }_{T}T=T$, the Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives the rank bound for a weighted operator trace:The last step uses part (i) at $r_T$, followed by nonnegativity of the eigenvalues of the density operator ${\rho }_Q$. Summing over the Kraus operators and using their completeness relation proves the finite-dimensional quantum compression converse:No invertibility of ${\rho }_Q$, or restriction to an isometric decoder, was used.

Measure the quantum typical subspace projector $\Pi$. Encode its successful sector isometrically and reserve one orthogonal compressed flag for failure; decode the flag to a fixed state. The composite channel is ${\mathcal{N}}_n(\tau )=\Pi \tau \Pi +\mathrm{Tr}[(I-\Pi )\tau ]|{\phi }_0⟩⟨{\phi }_0|$. It is trace preserving. The Kraus formula for entanglement fidelity gives the displayed lower bound, while convexity of the square gives the same bound for average pure-source fidelity. One flag adds only a vanishing asymptotic rate overhead.