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Iterating the Bloch-vector map of amplitude damping givesThereforeThe repeated amplitude damping limit describes irreversible relaxation of the excited state $|1⟩$ into the ground state $|0⟩$, with energy transferred to the environment and no compensating thermal excitation. At $p=1$, the ground state is reached in one action. At the endpoint $p=0$, the quantum channel is the identity and the initial Bloch vector remains unchanged; the ground-state conclusion requires nonzero damping.

In the computational basis, writeThe Kraus representation of the amplitude damping channel givesReading off its Bloch vector gives the Bloch-vector map of amplitude damping:Here $0\le p\le 1$. The shift in $s_z$ shows that this quantum channel is not unital unless $p=0$; its fixed ground state lies at the north pole, rather than the center of the Bloch sphere.

The $+1$ measurement effect of the Pauli operator ${\sigma }_x$ is $P_{+}=(I+{\sigma }_x)/2$. The Born rule and the Bloch vector representation giveThe other two Bloch vector components do not enter this projective measurement probability.

Conjugation by a Pauli operator preserves its matching Bloch vector component and reverses the other two. Summing all three conjugations therefore sends $\mathbf{s}$ to $-\mathbf{s}$. The specified quantum depolarizing channel has retention factorThis Pauli-mixture parametrization of qubit depolarization differs from a convention in which $p$ itself is the retention factor. For $0\le p\le 1$, $-1/3\le \eta \le 1$. A pure input gives output eigenvalues $(1\pm |\eta |)/2$, and hence entropy $h_2((1+\eta )/2)$ by symmetry of binary entropy. Mixed inputs have a smaller output radius and at least this much entropy. Every average output has entropy at most one. ThusTwo equiprobable orthogonal pure inputs have antipodal output Bloch vectors, maximally mixed average output, and this same minimal individual output entropy. They attain the bound. Applying the coding theorem givesAt $p=1$ it is one; at $p=1/4$ it is zero; at $p=0$ it is $1-h_2(1/3)\simeq 0.0817$. Negative retention reverses the labels of the two signals but does not eliminate their distinguishability.