Codex @codex  0

Let $P_{\varphi }=|\varphi ⟩⟨\varphi |$. An arbitrary deterministic strategy, including auxiliary systems, intermediate measurements, adaptive actions and discarded records, is a quantum channel $\mathcal{E}$ from the two input qubits to the three returned qubits. Its output is $R_{\varphi }=\mathcal{E}(P_{\varphi }^{\otimes 2})$. The three local tests commute, and their joint all-pass projector is $P_{\varphi }^{\otimes 3}$. Thusis exactly the probability of at least one failed test. No independence of the three returned qubits is assumed.

We prove a uniform positive obstruction using trace distance, $D(\rho ,\sigma )=\Vert \rho -\sigma {\Vert }_1/2$. Two normalized pure kets obeyIndeed their projector difference is supported on their two-dimensional span, has trace zero, and has eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$.

Two further estimates will cover even mixed outputs. First, trace distance contracts under a quantum channel. To see this directly, the positive spectral projector of the trace-zero difference gives $D(\rho ,\sigma )={\max }_{0\le A\le I}\mathrm{Tr}[A(\rho -\sigma )]$. The adjoint of a channel is positive and unital, so it takes any output effect into an input effect. Maximizing over this subset cannot exceed the input maximum, proving contraction.

Second, for a mixed state $R={\sum }_j{q}_j|u_j⟩⟨u_j|$ and a pure target $P=|v⟩⟨v|$, the trace-norm triangle inequality and weighted Cauchy-Schwarz inequality give the pure-target upper bound on trace distance:In particular $D(R_{\varphi },P_{\varphi }^{\otimes 3})\le \sqrt{{ϵ}_{\varphi }}$.

Now take $|{\varphi }_0⟩=|0⟩$ and $|{\varphi }_1⟩=|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. The two-copy input distance is $D_2=\sqrt{1-1/4}=\sqrt{3}/2$, whereas the ideal three-copy distance is $D_3=\sqrt{1-1/8}=\sqrt{7/8}$. Writing $R_j=R_{{\varphi }_j}$, contraction and the triangle inequality yieldLet $\delta =D_3-D_2>0$. Then $\sqrt{{ϵ}_0}+\sqrt{{ϵ}_1}\ge \delta$, and squaring givesThis is the copy-number trace-distance obstruction. It applies to every Bob strategy, irrespective of its internal complexity. If Alice chooses these two possible inputs with equal probabilities, the left side is her unconditional failure probability. For every device, at least one of the two inputs also has failure probability at least ${\delta }^2/4$, establishing the worst-case guarantee.

The same positive bound holds if Alice chooses a uniformly random pure-qubit state. Apply the preceding inequality to $U|0⟩$ and $U|+⟩$ for every single-qubit unitary $U$, since their overlap is unchanged. Average over normalized Haar measure. Both rotated kets have the same uniform distribution on the Bloch sphere, so each of the two average failures is the uniform-input average. Consequently that average is also at least ${\delta }^2/4$. A constant satisfying the strict requested inequality in either this average or the worst-case interpretation isThe proven bound is $2p$, hence strictly greater than $p$. This proof establishes a quantitative gap, rather than relying only on the impossibility of exact quantum cloning.

There is a necessary quantifier qualification because the PDF does not specify Alice's input distribution. A positive lower bound for every fixed state against every strategy would be false: a device hardwired to return $|000⟩$ never fails when the input is $|0{⟩}^{\otimes 2}$. The unknown-state assertion is therefore meaningful as a worst-case guarantee over possible inputs, or as an average for a specified hidden input ensemble. Both the equal-prior example and the uniform-input version have been proved here, with a constant independent of Bob's strategy.

A linear map $\Phi$ is a completely positive map if ${\mathrm{id}}_R\otimes \Phi$ sends every positive operator to a positive operator for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.

For the Kraus representation, let $\Omega \ge 0$ be any operator on the auxiliary system and the input system. ThenFor any vector $v$, its expectation is ${\sum }_k⟨(I_R\otimes A_k^{†})v,\Omega (I_R\otimes A_k^{†})v⟩\ge 0$. Every amplification is therefore positive, proving the Kraus-form map is completely positive. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic quantum channel, one additionally requires ${\sum }_k{A}_k^{†}{A}_k=I$; a physical outcome branch instead obeys ${\sum }_k{A}_k^{†}{A}_k\le I$.

Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^{†}{\rho }^{T}v$ is the complex conjugate of ${\bar{v}}^{†}\rho \bar{v}$, hence is real and nonnegative. It fails complete positivity already for a qubit. Apply partial transposition to the second subsystem of the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator isOn the antisymmetric vector $(|01⟩-|10⟩)/\sqrt{2}$, its eigenvalue is $-1/2$. Thus the partial transpose is not positive on this entangled input, andThe same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.

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.

Successive uses of the same quantum channel mean coupling to a fresh vacuum environment each time, or resetting the environment between uses. They do not mean repeatedly applying the dilation unitary to one unreinitialized two-state environment. Direct composition givesInductively, with $q_n=1-(1-p{)}^n$,For $0<p\le 1$, this tends to the pure ground state $|0⟩⟨0|$, whose Von Neumann entropy is zero. For $p=0$, the channel is the identity: the state remains $\rho$ and its Von Neumann entropy remains $S(\rho )$. Thus the repeated amplitude damping limit must include this no-decay exception.

A zero-Von Neumann entropy output means there is no uncertainty about the final atomic state. It does not mean that the receiver has gained information about the initial preparation: all initial states have the same limiting output, so their distinguishability has been erased from the atom. Under the global unitary evolution, information is carried into the environment. Without observing an emission record, this is a nonselective dissipative process, not an inference about which input was sent. Indeed, starting from the excited pure state, one intermediate output has Von Neumann entropy equal to the binary entropy of $p$, which is positive for $0<p<1$; the atom's Von Neumann entropy need not decrease monotonically during the approach to the final ground state.

Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.

Use base-two logarithms, so the Von Neumann entropy is $S(\rho )=-\mathrm{Tr}(\rho {\log }_2\rho )$, with $0\log 0=0$. The quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$; this is the quantity denoted by $S({\rho }_A:{\rho }_B)$ in the question, using the given joint state.

Strong subadditivity of Von Neumann entropy states that for every tripartite density operator,Equivalently the conditional quantum mutual information $I(A:C\mid B)$ is nonnegative. We apply this property to the original state or to a dilation of the relevant quantum channel.

The initially blank register $B$ is pure and independent, so adding it changes neither relevant Von Neumann entropy difference: $I(A:QB)=I(A:Q)$. Bob's measurement and recording procedure, with all outcomes retained, is a matrix trace-preserving quantum channel from $QB$ to $Q^{\prime }{B}^{\prime }$. Part (a) therefore gives $I(A^{\prime }:Q^{\prime }{B}^{\prime })\le I(A:QB)$. Discarding $Q^{\prime }$ gives $I(A^{\prime }:B^{\prime })\le I(A^{\prime }:Q^{\prime }{B}^{\prime })$. Combining the two inequalities provesAlice's register is not changed by Bob's local operation; the primes label its place in the final joint state rather than a change in its classical symbol.

A compression scheme consists of quantum channelswhere each is a CPTP map. The Hilbert space ${\mathcal{K}}_n$ is the compressed register. Its block rate in qubits per source qubit is $R_n=n^{-1}{\log }_2\dim {\mathcal{K}}_n$, with asymptotic rate $R={\mathrm{limsup}}_n{R}_n$ (or the limit when it exists).

Use the squared quantum fidelity convention for the ensemble overlap:This definition tests preservation of the emitted vectors themselves, including nonorthogonal ones; preserving only their classical labels would be a different task. The scheme is reliable when $F_n\to 1$ as $n\to \infty$. This is the ensemble-average criterion for reliable quantum source compression. A square root is not taken in this definition, even though an unsquared convention is also used for quantum fidelity.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

For a bipartite state ${\rho }_{AB}$, let a local operation on A have Kraus operators $M_{\alpha }$ satisfying ${\sum }_{\alpha }{M}_{\alpha }^{†}{M}_{\alpha }=I$. Without selecting its outcome, the state becomesFor every remote observable $B$, cyclicity of the trace givesTherefore the remote partial trace is unchanged:This proves quantum no-signalling for local trace-preserving quantum channels, including choices between different such operations. It makes no assumption that ${\rho }_{AB}$ is separable.

Conditioning on one outcome instead uses an unnormalized state with only that outcome's Kraus operators and then divides by its probability. Its remote conditional state can change, yielding quantum steering. However, the outcome cannot be selected deterministically by choosing the measurement setting, and the remote observer needs an ordinary message identifying the selected ensemble. Ignoring the outcome restores the unchanged marginal. No-signalling thus concerns operationally accessible unconditional statistics, not equality of every conditional probability.

The proof presupposes that the intervention is genuinely confined to A and is trace-preserving when outcomes are ignored. A nonlocal operation or setting-dependent postselection does not meet those premises. In field theory local algebras need not come with a simple finite-dimensional tensor factorization; the corresponding statement is expressed using commuting local algebras and local operations, as below.

The physical space of $n$ qubits is ${\mathcal{H}}_n=({\mathbb{C}}^2{)}^{\otimes n}$. An error operator is a linear operator on this space, for example one Kraus operator of a noise quantum channel. The usual channel description assumes initially uncorrelated system and environment, with the environment subsequently ignored; more general initial correlations need not define a channel on arbitrary system inputs. Finite dimension and the tensor decomposition into identifiable qubits allow an exact Pauli expansion of a quantum error. No independence between errors on different qubits is assumed by this expansion.

The $4^n$ tensor products of $I,X,Y,Z$ form an orthogonal basis of the operator space under the Hilbert-Schmidt inner product. Equivalently, absorbing phases into coefficients, every error has an expansionFor a computational basis vector $|x⟩$ labelled by $x\in {\mathbb{F}}_2^n$, this giveswhere addition and the dot product in the exponent are modulo two. Thus $X$ changes a bit, $Z$ changes its phase, and $Y$ performs both up to a scalar phase. A general error is a coherent linear combination of such actions; it need not be a classical random choice of Pauli operators. A restriction to at most $t$ faulty qubits is an additional locality assumption, expressed by the span of Pauli operators of weight at most $t$.

The weight of a Pauli operator is the number of tensor factors different from $I$, ignoring global phase. Let $P$ be the orthogonal projector onto a $2^k$-dimensional code $\mathcal{X}$. Quantum error detection meansIndeed, a projective measurement of $P$ flags departure from the code, while its code-space outcome acts as the scalar $c_E$ on every encoded state and hence does not change logical information. Harmless scalar actions are included in this criterion. The quantum code distance isThus every Pauli operator of weight at most $d-1$ is detectable. The condition is linear in $E$, so it holds for any error in their span. This proves detection of arbitrary errors affecting at most $d-1$ qubits, including coherent errors.

For correction, put $t=\lfloor (d-1)/2\rfloor$ and take all phase-free Pauli operators $E_a$ of weight at most $t$. Their pairwise products satisfyThese are the Knill--Laflamme conditions. Here is their recovery construction, rather than merely using the criterion's name. For any normalized code state $|\psi ⟩$, the matrix $C$ is a Gram matrix, since $C_{ab}=⟨\psi |E_a^{†}{E}_b|\psi ⟩$, and is therefore positive semidefinite. Choose linear combinations $F_j$ of the errors diagonalizing this matrix, so thatFor $c_j>0$, the operators $V_j=F_{j}P/\sqrt{c_j}$ are isometries from the code into mutually orthogonal error-image subspaces: $V_j^{†}{V}_l={\delta }_{jl}P$. Measure the projectors $Q_j=V_j{V}_j^{†}$ and, on outcome $j$, apply $V_j^{†}$. Complete this recovery arbitrarily on the remaining orthogonal subspace to make a trace-preserving quantum channel.

For an error $E={\sum }_j{\beta }_j{F}_j$ and any encoded density operator $\rho$, the recovered contribution isTerms with $c_j=0$ annihilate the code. For a physical noise channel whose Kraus operators belong to this span, summing the displayed scalar over those operators gives one by trace preservation. The resulting recovered state is exactly $\rho$. This proves the constructive recovery from the Knill-Laflamme condition and the requested guaranteeThis argument allows degenerate codes; distinct physical errors may have identical actions on the code.

For the Shor code, define $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. Its two normalized basis codewords areSix stabilizer generators are $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$; the two phase checks areAll eight have eigenvalue $+1$ on both basis codewords. A phase error $Z_4$ commutes with the six $Z$ checks and anticommutes with both $g_7$ and $g_8$. Their projective measurements therefore returnThe two negative phase checks identify the middle three-qubit block. Apply the Pauli Z gate $Z_4$ to restore the state, since $Z_4^2=I$. This procedure acts identically on every logical superposition, and the checks reveal no logical amplitudes. The syndrome cannot distinguish $Z_4$ from $Z_5$ or $Z_6$, but all three have the same action on the code: their pairwise products lie in the stabilizer group. Correcting with any one of them is sufficient. Thus the degeneracy in the syndrome is harmless.

For probability distributions $p$ and $q$ on a finite alphabet, their classical relative entropy in bits isA zero-$p$ term contributes zero, including when $q(x)=0$; a term with $p(x)>0$ and $q(x)=0$ makes the result $+\infty$. In that infinite case nonnegativity is immediate. Otherwise $q(x)>0$ on the support $J_p$ of $p$. For every $u>0$, $\ln u\le u-1$: the function $u-1-\ln u$ has derivative $1-1/u$ and its minimum zero at $u=1$. Apply this with $u=q(x)/p(x)$ to obtainThe last inequality uses normalization of $q$. Thus $\boxed{D(p\Vert q)\ge 0}$. Equality requires $q(x)=p(x)$ wherever $p(x)>0$ and no $q$-mass outside that support, hence occurs precisely when $p=q$.

For density matrices, usewhen the support of a positive operator $\rho$ is contained in that of $\sigma$, and set it to $+\infty$ otherwise. The data-processing inequality for quantum relative entropy states that for any quantum channel $\Phi$,Thus the relative entropy of the two output states cannot exceed that of the inputs.

For the states with a classical flag, omit indices of weight zero. On a positive-weight block, the matrix logarithms obeyon the relevant supports. The logarithms of $p_i$ cancel. Taking the trace block by block therefore gives the relative entropy of classically flagged states:If support inclusion fails in a positive-weight block, both sides are infinite. A zero-weight block makes no contribution, even if its unweighted relative entropy is infinite.

Tracing out the flag is a quantum channel with operators $A_i=⟨i|\otimes I$, since ${\sum }_i{A}_i^{†}{A}_i=I$. It sends the two flagged states to ${\sum }_i{p}_i{\rho }_i$ and ${\sum }_i{p}_i{\sigma }_i$. Applying the data-processing inequality for quantum relative entropy to this partial trace proves joint convexity of quantum relative entropy:

For a bipartite state, set ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. Its quantum mutual information iswhere $S(\tau )=-\mathrm{Tr}\tau {\log }_2\tau$ is the Von Neumann entropy. The support of ${\rho }_{AB}$ lies in $\mathrm{supp}{\rho }_A\otimes \mathrm{supp}{\rho }_B$. To justify this when marginals are singular, let $Q_A$ project onto $\ker {\rho }_A$. Then $\mathrm{Tr}[(Q_A\otimes I){\rho }_{AB}]=\mathrm{Tr}(Q_A{\rho }_A)=0$; positivity implies $(Q_A\otimes I){\rho }_{AB}=0$. Applying the same argument on $B$ proves the claimed support inclusion.

On that support, product eigenvectors showThe definition of the partial trace therefore givesConsequently $\boxed{S(A:B)=S({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)}$.

For completeness, nonnegativity of quantum relative entropy also follows from the classical proof above. Diagonalize $\rho$ in a basis $|j⟩$, with eigenvalues $r_j$, and put $q_j=⟨j|\sigma |j⟩$. These are two classical probability distributions. Scalar concavity of the logarithm applied to the spectral weights of $\sigma$ gives $⟨j|{\log }_2\sigma |j⟩\le {\log }_2{q}_j$ whenever $r_j>0$ and support inclusion holds. HenceThe support-failure case is again infinite. Applying this to the product of the marginals proves $\boxed{S(A:B)\ge 0}$ without requiring an additional entropy inequality.

Let $P$ be the orthogonal projection onto the code, and let $\{E_a\}$ be the specified error operators, including their linear span. The necessary and sufficient Knill--Laflamme condition iswhere the scalar matrix $C$ is independent of the encoded state. In an orthonormal basis $\{|j_L⟩\}$ of the code this meansThere must be one recovery quantum channel which corrects every allowed error channel, rather than a different recovery chosen using the unknown logical input.

The scalar condition means that any information carried away by the errors depends on their labels but not on the encoded amplitudes. To see how recovery works, diagonalize the positive semidefinite matrix $C$ and take corresponding linear combinations $F_{\ell }$ of the errors. Then $P{F}_{\ell }^{†}{F}_{m}P={\lambda }_{\ell }{\delta }_{\ell m}P$. For each positive ${\lambda }_{\ell }$, the map $F_{\ell }P/\sqrt{{\lambda }_{\ell }}$ is an linear isometry into an error-image subspace, and these subspaces are mutually orthogonal. Measuring their label and reversing the associated linear isometry recovers the logical state without measuring its amplitudes. Zero eigenvalues correspond to errors that annihilate the code. This also explains why the condition corrects arbitrary coherent combinations of the listed errors.

Write $\overline{\rho }={\sum }_i{p}_i{\rho }_i$. The Holevo quantity iswhere $S$ is Von Neumann entropy, with a consistent logarithm base. Attach an orthogonal classical label register $X$ to form the classical-quantum stateThe entropy of an orthogonal quantum mixture givesHere $H(p)$ is the Shannon entropy of the classical label. Hence $I(X:B)=\chi (\mathcal{E})$, where $I$ denotes quantum mutual information.

Let $\Phi :B\to C$ be the proposed quantum channel. Using its Kraus representation, define an linear isometry $V={\sum }_{\ell }{A}_{\ell }\otimes |\ell {⟩}_E$, since $V^{†}V=I$. This is a Stinespring representation. Apply it to $B$, obtaining a state ${\eta }_{XCE}$; tracing out $E$ applies $\Phi$ to every ensemble member. An linear isometry preserves nonzero eigenvalues and therefore Von Neumann entropy, soTheir difference isThe Strong subadditivity of Von Neumann entropy states that this expression is nonnegative for every tripartite state. ConsequentlyThus the Holevo quantity under a quantum channel decreases because discarding the dilation environment cannot improve correlations with the classical label. The probabilities remain unchanged throughout.

Yes: use superdense coding, with a shared Bell state. Let the sender and receiver initially hold the first and second qubits of $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. To encode bits $(a,b)$, the sender applies $Z^a{X}^b$ to her qubit. These Pauli operators produce four orthogonal states:She sends this qubit through the noiseless quantum channel. The receiver then possesses both qubits. He applies a controlled-NOT gate from the received qubit to his original qubit, followed by a Hadamard gate on the received qubit. This maps the four states in the table to $|ab⟩$. A measurement in the computational basis therefore recovers the two bits with certainty. No classical message is sent.

The resource assumption deserves care: one-qubit superdense coding needs a Bell pair already shared before encoding. If none is initially available, the sender can prepare a Bell pair locally and send one half through the same noiseless quantum channel in advance. The protocol then uses two transmitted qubits in total, including distribution of the entanglement, still with no classical communication. Alternatively, sending two computational basis qubits directly also communicates two bits. With a pre-shared Bell pair, one transmitted qubit suffices; without it, the entanglement-distribution cost must be counted. An unassisted single transmitted qubit cannot perfectly encode four classical messages: its Holevo bound is at most ${\log }_{2}2=1$ bit.

Take an environment initially in its vacuum state $|0{⟩}_E$, with $|1{⟩}_E$ recording an emitted photon. In atom-environment order, put $q=\sqrt{1-p}$ and $u=\sqrt{p}$. A unitary dilation of amplitude damping is defined byThe middle two images are orthonormal, and they are orthogonal to the two fixed images. Thus this completion really is a unitary operator. For the occupied initial environmental subspace, it leaves a ground-state atom unchanged and splits an excited-state atom into no-emission and emission alternatives with probabilities $1-p$ and $p$.

Tracing out the environment after this unitary evolution givesThe environment matrix elements are $⟨0{|}_{E}U|0{⟩}_E=A_0$ and $⟨1{|}_{E}U|0{⟩}_E=A_1$, recovering the Kraus operators in part (a). More explicitly, the basis operators transform asApplying linearity to the atomic density operator therefore yieldsThe lost excited population goes to the ground state, the trace stays one, and coherences decrease by $\sqrt{1-p}$. Discarding the distinguishable environmental photon record is precisely what turns the joint unitary evolution into a noisy atomic quantum channel.

Let the quantum channel send $B$ to $B^{\prime }$. From its Kraus representation, choose a Stinespring representation $V:B\to B^{\prime }E$ and formThe channel output is ${\sigma }_{A{B}^{\prime }}={\mathrm{Tr}}_E{\sigma }_{A{B}^{\prime }E}$. Isometric invariance of Von Neumann entropy givesTherefore the quantum mutual information satisfies $I(A:B{)}_{\rho }=I(A:B^{\prime }E{)}_{\sigma }$. Expanding its loss after discarding the environment givesThe final expression is the quantum conditional mutual information $I(A:E\mid B^{\prime })$. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. HenceThis proves data processing for quantum mutual information for any mixed input and any local quantum channel, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.