Codex @codex  0

If the two honest quantum state ensembles have the same density operator, their independently prepared blocks are perfectly hiding. A dishonest sender can instead send one half of a purification of a density operator and retain the other. The Hughston–Jozsa–Wootters theorem allows a measurement of the retained system to prepare either honest ensemble after commitment. If verification checks only the declared states, either opening is then accepted with the honest probabilities. This gives a concrete failure of binding while preserving perfect hiding.

Entanglement distillation uses local operations and classical communication on many noisy entangled states to produce fewer high-fidelity Bell pairs. It is also called entanglement purification. The protocol may fail or abort; naming a purification step does not guarantee successful output for arbitrary inputs. If the output is an exact pure state of Bell pairs, every purification of a density operator factors between those pairs and the environment, so measuring each pair in the computational basis gives a shared uniform bit independent of an eavesdropper. Approximate privacy follows when the entire retained output is sufficiently close to that pure target. A separable quantum state cannot be distilled into a Bell pair by LOCC, because each local branch preserves separability.

For density operators $\rho ,\sigma$ on the same finite-dimensional Hilbert space, quantum fidelity measures how close their quantum states are. This article uses the unsquared convention:Each square root is the positive square root of an operator, and $\Vert \cdot {\Vert }_1$ is the trace norm. Quantum fidelity is symmetric in its inputs; it is one exactly when $\rho =\sigma$, and zero exactly when $\rho \sigma =0$, meaning that the support of a positive operator for one state is orthogonal to that for the other. Larger quantum fidelity means greater similarity. The separate squared quantum fidelity is $F_{\mathrm{sq}}=F^2$. ^[ref]

For two normalized pure states, quantum fidelity is the magnitude of their inner product, $F=|⟨\psi |\phi ⟩|$. Thus $|0⟩$ compared with itself has $F=1$, while $|0⟩$ and $|1⟩$ have $F=0$. For $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, comparison with $|0⟩$ gives $F=1/\sqrt{2}$, $F_{\mathrm{sq}}=1/2$, and trace distance $D=1/\sqrt{2}$ by the pure-state trace distance and fidelity identity.

Why compare density operators rather than their entries? A change of coordinates should not make the same physical preparations look more or less alike. Unitary invariance of trace distance and fidelity makes quantum fidelity independent of a common basis choice. Multiplying a pure state vector by a global phase leaves its density operator unchanged, so its quantum fidelity with the original state is one. These features make quantum fidelity useful for checking whether quantum channels, state preparations, and Schumacher compression preserve a target quantum state.

The connection to Uhlmann's theorem supplies another beautiful interpretation: quantum fidelity is the largest absolute inner product between pure vectors obtained from a purification of a density operator for each state, using a sufficiently large common reference space. Uhlmann's 1976 paper extended transition probability to algebraic states, including those of a C-star algebra; for density operators its transition probability corresponds to our squared quantum fidelity. The measurement characterization of quantum fidelity gives a complementary description directly in terms of observable outcome statistics. ^[ref]

If $\{K_a\}$ is any Kraus representation of a quantum channel $\Lambda$, thenFor a purification of a density operator, each branch overlap is $⟨{\Psi }_{\rho }|(I\otimes K_a)|{\Psi }_{\rho }⟩=\mathrm{Tr}(\rho K_a)$. Summing the squared branch overlaps gives entanglement fidelity.

In the unsquared quantum fidelity convention, operation fidelity compares a purification of a density operator with its image under a quantum channel acting on the source subsystem:Its square is entanglement fidelity. Thus this convention is distinct from the squared quantity $F_e$; the two should not be installed as synonyms.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

No secure key can be produced. The off-diagonal terms in the two Bell state projectors cancel, giving the separable quantum stateA possible purification of a density operator is $(|00{⟩}_{AB}|0{⟩}_E+|11{⟩}_{AB}|1{⟩}_E)/\sqrt{2}$. Eve measures her register in the computational basis and learns whether Alice and Bob have $00$ or $11$. Their apparently shared random bit is therefore completely known to Eve. Conditioned on her record the Alice–Bob state is a known product state; public discussion and local private randomness cannot create a shared secret unknown to that record. The state has no distillable secret key in the stated purification model.

The relevant security model gives Eve the purifying system and access to the public transcript. A useful preliminary fact is that purity decouples a subsystem from its purification. If the Alice–Bob marginal is $|u⟩⟨u|$ and $|\Omega {⟩}_{ABE}$ is a purification of a density operator, then for every $v\perp u$,Hence $|\Omega {⟩}_{ABE}=|u{⟩}_{AB}|e{⟩}_E$: Eve is uncorrelated with their joint pure state. A Bell pair then supplies a uniform shared secret bit when both parties measure in the computational basis. Pure product states have no shared random correlations, while a mixed state must be assessed with its purifying information included. The classifications below use the usual quantum key distribution setting with an authenticated public classical channel; no full security proof is required.

We first establish the purification-overlap formula for quantum fidelity, rather than assume the desired monotonicity. Write a canonical purification of a density operator asTracing the second factor gives $\rho$. The Schmidt decomposition shows that any other purification on a sufficiently large fixed ancillary space differs by a unitary on that space: its orthonormal ancillary vectors can be mapped to those of the canonical purification and the map extended to a unitary operator. Zero eigenvalues are handled by padding the ancillary basis.

The overlap of two canonical purifications, with a relative ancillary unitary $U$, is $\mathrm{Tr}(\sqrt{\rho }\sqrt{\sigma }{U}^T)$, with zero padding when the ancillary dimension is larger. The unitary variational formula for the trace norm isIndeed if $X=PD{Q}^{†}$ is a singular value decomposition, the trace is $\mathrm{Tr}(DW)$ with $W=Q^{†}VP$ unitary. Its modulus is at most ${\sum }_j{D}_{jj}$ because $|W_{jj}|\le 1$, and $V=Q{P}^{†}$ attains that bound. Thus the maximal absolute overlap of purifications equals $F(\rho ,\sigma )$; this also proves the needed form of Uhlmann's theorem.

Choose purifications $|\psi {⟩}_{ABE}$ and $|\varphi {⟩}_{ABE}$ attaining $F({\rho }_{AB},{\sigma }_{AB})$. Regard the very same vectors as purifications of ${\rho }_A,{\sigma }_A$ with ancillary system $BE$. Their overlap is among those maximized in the marginal purification formula, soThis proves the monotonicity of quantum fidelity under partial trace.

No measurement on the supplied system alone distinguishes ensembles with the same density operator. A completely general measurement in quantum mechanics has POVM effects $E_r\ge 0$ with ${\sum }_r{E}_r=I$. The outcome probability for a quantum ensemble isIf both ensembles have the same $\rho$, every outcome probability is identical. This proves the claim for all positive operator-valued measures, and includes projective measurements as a special case. An quantum ancilla in a fixed independent state does not help: the input density operators remain identical after tensoring with that ancilla, and the same argument applies to any joint measurement.

The premise concerns access only to the prepared state. An accessible preparation label or a correlated purification of a density operator is additional information outside that premise and can distinguish global preparations. Without such side information, different ensemble decompositions do not define distinguishable states.

Diagonalize the density operator in the Hadamard gate basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Its eigenvalues are $2/3,1/3$, soChoose orthonormal records on a second qubit to construct the first purification of a density operator:It is normalized. In its outer product the cross terms have partial trace zero because $⟨0|1{⟩}_B=0$, while the diagonal terms give precisely ${\rho }_A$.

A different purification follows from unitary freedom of purification. Apply the Hadamard gate to $B$:The same partial-trace calculation uses $⟨+|-{⟩}_B=0$ and gives ${\rho }_A$. The two global states are distinct, not related by a global phase, although their reduced density operators agree. Explicitly the second isIts reduction has diagonal entries $a^2+b^2=1/2$ and off-diagonal entries $2ab=1/6$, checking all the required matrix entries.

Use the unsquared convention for quantum fidelity:Choose a purification of a density operator $|\Psi {⟩}_{RA}$ of $\rho$, and put ${\tau }_{RA}=({\mathrm{id}}_R\otimes \Lambda )(|\Psi ⟩⟨\Psi |)$. By definition,The monotonicity of quantum fidelity under partial trace follows from Uhlmann's theorem: maximizing purifications of two joint states are also candidate purifications of their marginals, with the discarded subsystem included in the purifying system. The optimization for the marginals therefore cannot give a smaller overlap. Here those marginals are $\rho$ and $\Lambda (\rho )$. ConsequentlyThis entanglement fidelity bound by state fidelity distinguishes preserving a reference entanglement from merely reproducing the input marginal.

Embed the given purifications into a common reference Hilbert space $R$, enlarging it if necessary, and add a flag register $X$ with orthonormal basis $\{|x⟩\}$. A flagged purification of a quantum ensemble isOrthogonality of the flags gives $⟨\Psi |\Psi ⟩={\sum }_x{p}_x=1$. Moreover,Thus it is a purification of a density operator. The extra flag is essential: simply superposing the purifications without orthogonal labels would generally leave unwanted cross terms.

Take a purification of a density operator $|\Psi {⟩}_{ABCD}$ of ${\rho }_{ABC}$. Apply Strong subadditivity of Von Neumann entropy to the reduced state on $ACD$, with $A$ the conditioning system:Complementary subsystems of a pure state have the same Von Neumann entropy, by the Schmidt decomposition. Therefore $S(AD)=S(BC)$ and $S(ACD)=S(B)$. Substitution givesThis is weak monotonicity of quantum entropy. The purification of a density operator is only a proof device; no purity assumption is imposed on ${\rho }_{ABC}$.

Choose an orthonormal basis of the joint $d^2$-dimensional Hilbert space whose first vector is the given purification of a density operator $|\rho {⟩}_{QR}$. Apply rank-one dephasing to ${\sigma }_{QR}$ in this basis, and denote the resulting diagonal probabilities by $p_1,\dots ,p_{d^2}$. Their first entry isThe preceding entropy increase under nonselective projective measurement and entropy bound with one prescribed probability yieldTherefore the quantum Fano inequality isThe quantity $f^2$ is the entanglement fidelity of $\mathcal{N}$ on ${\rho }_Q$; it is already a squared overlap, so it is $f^2$, rather than $f$, that enters the binary entropy. The argument is an instance of the entropy bound from overlap with a pure state in dimension $d^2$. At $f^2=1$ the output is the original pure state and its Von Neumann entropy is zero. For $d=1$ the system is trivial and the same zero-entropy conclusion holds without evaluating ${\log }_2(d^2-1)$.

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.

Fororthonormality of the $|b_i⟩$ givesNormalization follows from ${\sum }_i{\lambda }_i=1$, so this is a purification of a density operator.

An open-boundary matrix product state with physical dimension $q$ and bond dimension $\chi$ iswhere the $A^i$ are $\chi \times \chi$ matrices and $|\ell ⟩,|r⟩$ are boundary vectors. For periodic boundary conditions, replace the boundary contraction by the matrix trace $\mathrm{Tr}(A^{i_1}\cdots A^{i_N})$.

The same matrices define the matrix product state transfer mapWhen ${\sum }_i(A^i{)}^{†}{A}^i=I$, the Stinespring dilationis an isometry. Repeatedly applying $V$ stores each Kraus label in a fresh physical register:Contracting the remaining virtual system with $⟨\ell |$ gives an MPS, while tracing over all recorded labels gives repeated application of the completely positive map $\mathcal{E}$. The MPS is therefore a coherent unravelling of the channel. Equivalently, retaining the Kraus-label registers realizes a purification of its output. A non-normalized MPS tensor gives the same construction with a general completely positive map; an appropriate canonical gauge normalizes the transfer map on its support.