Codex @codex  0

Applying the amplitude damping channel to the Bloch vector form of a qubit density operator gives an affine map. Transverse components contract by $\sqrt{1-p}$; the longitudinal component contracts by $1-p$ and shifts toward the ground-state north pole. Reading the entries of the output density operator proves the map. The displacement distinguishes this channel from centered depolarization.

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 $1-p$.

Choose $|0⟩$ as the ground state and $|1⟩$ as the excited state. The relevant quantum channel is the amplitude damping channel. Its Kraus operators areThey satisfy $K_0^{†}{K}_0+K_1^{†}{K}_1=I$. The $K_0$ branch corresponds to no emitted photon: the ground-state amplitude is unchanged and the excited-state amplitude is attenuated. The $K_1$ branch corresponds to photon emission and transition to the ground state. These are unnormalized branch states; the squared norm or matrix trace of each branch gives its probability.

A unital quantum channel preserves the identity operator: $\Phi (I)=I$. In equal input and output dimensions, it therefore preserves the maximally mixed state. For amplitude damping channel,Hence the channel is not unital for $p>0$; it is unital only in the identity case $p=0$. Trace preservation involves the opposite products $K_i^{†}{K}_i$ and does not imply unitality.

The printed normalization has its factors reversed. For the displayed map $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, the condition ${\sum }_k{A}_k{A}_k^{†}=I$ says $\Phi (I)=I$, which is unitality. Trace preservation instead requiresThese are distinct trace-preserving and unital Kraus conditions. To disprove the printed equivalence, take $0<\gamma <1$ andThey obey ${\sum }_j{B}_j{B}_j^{†}=I$, butThus their completely positive map is not trace preserving. They are the adjoints of the operators of an amplitude damping channel. That channel is trace preserving but sends $I$ to $\mathrm{diag}(1+\gamma ,1-\gamma )$, so it cannot satisfy the printed normalization in any Kraus representation either. The corrected characterization is proved next.

Work in finite-dimensional input and output Hilbert spaces, with a linear map $\Phi$ on operators. Suppose first that $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ and ${\sum }_k{A}_k^{†}{A}_k=I$. For every auxiliary dimension $m$ and every positive operator $W$ on the extended space,Each summand is positive because conjugation preserves positivity. This is exactly the definition of a completely positive map. Cyclicity of the trace givesso the map is also trace preserving.

Conversely, suppose $\Phi$ is completely positive and trace preserving, and let $|i⟩$ be an orthonormal input basis. Use the unnormalized entangled vector $|\Omega ⟩={\sum }_i|i⟩\otimes |i⟩$, with the reference factor first. Complete positivity makes the corresponding unnormalized Choi matrix positive:Its spectral decomposition can be written $J_{\Phi }={\sum }_k|v_k⟩⟨v_k|$, absorbing the nonnegative eigenvalues into the vectors. Define linear operators $A_k$ uniquely byExpanding the outer products and comparing their $(i,j)$ reference blocks yieldsThe matrix units span all operators, so linearity proves the Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ for every $X$. Finally, trace preservation givesfor all $X$. Taking matrix units as test operators shows that the matrix in parentheses is zero. This proves both directions of the corrected CPT characterization, including the normalization rather than assuming it.

For an input quantum state ensemble $\{p_x,{\rho }_x\}$, write ${\sigma }_x=\Phi ({\rho }_x)$ and $\bar{\sigma }={\sum }_x{p}_x{\sigma }_x$. Its output Holevo quantity isThe Holevo capacity is the optimized one-use quantitywhere the supremum ranges over finite input ensembles and $S$ is the Von Neumann entropy in bits.

For $n$ uses of a memoryless quantum channel, the map is ${\Phi }^{\otimes n}$. An $n$-use code consists of a finite message set ${\mathcal{M}}_n$ of size $M_n$, an encoding state ${\rho }_m^{(n)}$ on the $n$ inputs for each message, and a decoding POVM $\{D_m^{(n)}:m\in {\mathcal{M}}_n\}$ on the output space, with $D_m^{(n)}\ge 0$ and ${\sum }_m{D}_m^{(n)}=I$. The encoding states may be entangled across uses. The measurement outcome is the receiver's decoded message, with conditional probabilityFor uniformly distributed messages the average error and rate areA rate $R$ is achievable if there is a sequence of such codes with $P_e^{(n)}\to 0$ and ${\mathrm{liminf}}_n{R}_n\ge R$. The unassisted classical capacity of a quantum channel is the supremum of achievable rates; the sender and receiver do not share prior entanglement.

The operational content of the Holevo-Schumacher-Westmoreland theorem is that ${\chi }^{\ast }(\Phi )$ is the capacity for product-state input encodings with collective output measurements. Allowing arbitrary entangled block inputs gives the regularized formulaThe limit equals the supremum because superadditivity of Holevo capacity follows by taking product ensembles for two blocks and using additivity of the Von Neumann entropy on product states. The coding theorem makes every rate below the one-block Holevo quantity achievable through many independently encoded blocks, with collective decoding.

The converse can be seen explicitly from the Holevo bound and Fano's inequality. For a uniform transmitted message $M$ and decoded message $\hat{M}$,The first inequality is obtained by subtracting the Fano bound on $H(M\mid \hat{M})$ from $H(M)={\log }_2{M}_n$. Dividing by $n$ and letting the average error vanish bounds every achievable rate by the regularized expression. The block-coding achievability just described supplies the reverse bound.

Under the assumed additivity of Holevo capacity across tensor powers, ${\chi }^{\ast }({\Phi }^{\otimes n})=n{\chi }^{\ast }(\Phi )$ for every $n$. ThusThe stated conditional assertion is true: under this additivity assumption, entangled input states cannot increase the asymptotic unassisted classical rate beyond the rate already achievable with product inputs. The decoder may still use collective measurements across channel outputs.

Label the atomic ground state $|0⟩$ and excited state $|1⟩$, with $0\le p\le 1$. The process is the amplitude damping channel, with Kraus operatorsThey satisfy $A_0^{†}{A}_0+A_1^{†}{A}_1=I$, so the resulting Kraus representation is trace preserving.

The operator $A_1$ represents emission: it maps the excited state to $\sqrt{p}$ times the ground state and annihilates the ground state. Its probability for an initially excited atom is $p$. The operator $A_0$ represents no emission: it leaves the ground state unchanged and retains the excited amplitude with factor $\sqrt{1-p}$. This branch is not generally the identity operation; the absence of a photon itself changes the conditional relative amplitudes. For an arbitrary atomic density operator, the emission probability is $\mathrm{Tr}(A_1\rho A_1^{†})=p{\rho }_{11}$ and the no-emission probability is $1-p{\rho }_{11}$.

A unital quantum channel preserves the identity operator: $\Phi (I)=I$. For equal input and output dimensions, this is equivalent to preserving the maximally mixed density operator.

For the amplitude damping channel,Thus the channel is not unital for $p>0$; it is unital only at $p=0$, when it is the identity channel. In particular, it biases the maximally mixed state toward the atomic ground state. The trace-preservation condition involves ${\sum }_j{A}_j^{†}{A}_j$, whereas unitality involves the opposite products ${\sum }_j{A}_j{A}_j^{†}$.

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.

Under repeated identical amplitude damping channels, all excited population eventually relaxes to the ground state for any fixed $0<p\le 1$. The Bloch vector approaches $(0,0,1)$. For $p=1$ the limit is reached in one action, while $p=0$ is the identity channel and preserves every input. Keeping the zero-damping endpoint separate prevents an incorrect universal relaxation claim.

For a Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, trace preservation is equivalent to the first identity, by cyclicity of the trace. The second says $\Phi (I)=I$, namely unitality, and is a distinct constraint. For $0<\gamma <1$, the amplitude damping channel has operators $A_0=\mathrm{diag}(1,\sqrt{1-\gamma })$ and $A_1=\sqrt{\gamma }|0⟩⟨1|$ and is trace-preserving but not unital. Taking the adjoints as Kraus operators instead gives a unital completely positive map whose output trace on $|0⟩⟨0|$ is $1+\gamma$.

With the environment initially in $|0⟩$, take $U$ to fix $|00⟩$ and $|11⟩$, and rotate the span of $|10⟩,|01⟩$ by the displayed action and $U|01⟩=\sqrt{1-p}|01⟩-\sqrt{p}|10⟩$. This is a unitary operator. Its environment matrix elements $⟨0{|}_{E}U|0{⟩}_E$ and $⟨1{|}_{E}U|0{⟩}_E$ are the Kraus operators of the amplitude damping channel. Tracing out the environment multiplies the excited population by $1-p$, transfers the remainder to the ground state, and multiplies off-diagonal entries by $\sqrt{1-p}$.