Quantum information theory

Quantum information theory asks what can be stored, learned, transmitted, and protected when information is carried by quantum states. Its basic objects are density operators, measurement in quantum measurements, and quantum channels; its resources include qubits, entanglement, and classical communication. Specifying the task matters: preserving an unknown quantum state, recovering a classical bit, and estimating a probability amplitude are different demands on the same physical system.

A particularly beautiful feature is that knowing the whole need not mean knowing its parts. A Bell state is a completely specified pure state, yet either qubit alone has a maximally uncertain reduced density matrix. Conversely, two systems can share perfectly matching measurement records without entanglement. The worked Bell correlations versus a classical mixture comparison separates these possibilities using only two choices of Pauli measurement.

This viewpoint makes limitations useful design rules. Schumacher compression asks how much quantum memory a source needs; Holevo's theorem limits the classical information extractable from an encoding; quantum error correction protects states against noise. Quantum teleportation uses shared entanglement and classical communication to transfer an unknown state. Its resource accounting respects both quantum no-signalling and the no-cloning theorem.

Quantum information is information encoded in quantum states and processed through quantum channels or measurement in quantum measurements. A qubit supplies a two-dimensional state space; quantum phase and entanglement affect which operations and outcome statistics are possible. The Born rule connects the state description with the probabilities of classical measurement records.

The continuously varying probability amplitudes of a qubit do not supply a readable list of arbitrarily many classical bits. A measurement produces a classical record with state-dependent probabilities. Quantum state tomography estimates the state from many identically prepared systems. For a classical label encoded into one transmitted qubit, with no additional shared resource, Holevo's theorem bounds the mutual information available from a measurement by one bit. This bound concerns information about the label, rather than the number of possible outcome labels a measuring device can print. ^[ref]

Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

An encoding stores $n$ classical bits in $m$ qubits so that a receiver who later chooses a bit index can recover that bit with a specified success probability. The sender cannot tailor the encoding to the unknown requested index. This task permits approximate recovery of any one bit rather than simultaneous recovery of all bits, and must specify whether additional communication, shared randomness, or entanglement is allowed.

Encode signs $s_j=(-1{)}^{b_j}$ by the unit Bloch vector $(s_1,s_2,s_3)/\sqrt{3}$. Perform the requested Pauli measurement and interpret a positive outcome as zero. Every input and index has success $p_{\ast }$. To prove optimality, write each binary-measurement difference as $D_j=t_{j}I+v_j\cdot \sigma$, where positivity gives $|v_j|\le 1-|t_j|$. Averaging over all eight inputs cancels the $t_j$ terms. For any encoding vectors of norm at most one, the average success is at most $1/2+(1/48){\sum }_s|{\sum }_j{s}_j{v}_j|$. The Cauchy-Schwarz inequality bounds the sum by $8\sqrt{{\sum }_j|v_j{|}^2}\le 8\sqrt{3}$, proving the optimum even for mixed encodings and general binary measurements. The constant-success construction also maximizes worst-case success.

Classical communication transmits a distinguishable classical record, such as measurement bits, between parties. In quantum teleportation it tells the receiver which local correction to apply. Quantum no-signalling prevents shared entanglement alone from transmitting the record. A classical message cannot propagate across a spacelike separation in a relativistic protocol.

Remote state preparation uses a shared entangled state, local quantum measurements and classical communication to prepare a state known to the sender at the receiver. The sender can tailor the measurement to the desired state, so restricted state families can require less communication than quantum teleportation. A real great-circle family of qubit states admits one-bit remote preparation of real qubit states.

Prepare independent candidate blocks by one-bit heralding of a remote state preparation block, but send only the first successful block's index. Success eventually occurs with probability one; a prefix code for a rare geometric success uses at most $n+e/(e-1)$ expected bits per block. The expected singlet state consumption is $n{2}^n$. Alternatively, limiting the number of trials to $n{2}^n$ gives failure probability at most $e^{-n}$ and a fixed message length $n+{\log }_{2}n+O(1)$. Both versions approach one classical bit per remotely prepared qubit in their respective expected-length or high-success senses.

// Target: quantum-theory.bigb

For a target pure state $|\psi ⟩$ of a qubit, choose its orthogonal complement so thatMeasuring the sender's half of this singlet state in that basis leaves the receiver in $|\psi ⟩$ when the sender obtains $|{\psi }^{\perp }⟩$. The outcome has probability one half. One classical success flag heralds the exact preparation without conveying the target's classical description.

// Target: quantum-theory.bigb

Apply heralded remote preparation of an arbitrary qubit independently to $n$ shared singlet states. The sender reports one bit indicating whether every local outcome was successful. A positive flag certifies that the receiver's block is the desired product of $n$ pure states; its probability is $2^{-n}$. A negative flag need not identify which states failed, since the whole block can be discarded.

// Target: algebra.bigb

Alice holds the first two qubits and Bob the third of a three-vertex graph-state wire. Alice measures at $\alpha$ and $(-1{)}^r\beta$, with outcomes $r,s$, and sends these two bits. Bob applies $Z^r{X}^s$, the inverse of the known Pauli frame $X^s{Z}^r$, obtaining $U(\beta )U(\alpha )|+⟩$ up to global phase. Bob need not know the angles. Every branch succeeds, with branch probability $1/4$, and no postselection is needed. Alice's measurements and Bob's corrections constitute local operations and classical communication on the previously shared entangled resource.

For $|{\alpha }_{\theta }⟩=\cos \theta |0⟩+\sin \theta |1⟩$, Alice measures her half of a Bell state in the real basis $\{|{\alpha }_{\theta }⟩,|{\beta }_{\theta }⟩\}$, with $|{\beta }_{\theta }⟩=-\sin \theta |0⟩+\cos \theta |1⟩$. Bob receives one outcome bit and applies $I$ or $iY$. Since $iY|{\beta }_{\theta }⟩=|{\alpha }_{\theta }⟩$ for every $\theta$, the preparation is exact and the correction is independent of the unknown parameter.

Schumacher compression is quantum source coding into the typical subspace of ${\rho }^{\otimes n}$. Asymptotically reliable compression is possible at every qubit rate above the Von Neumann entropy $S(\rho )$.

A pure-state source emits independent states from a fixed quantum state ensemble. Its average block density operator is the tensor power of the one-use average. Unassisted reliable quantum source compression preserves the emitted quantum states, rather than only their classical labels. Schumacher compression gives the optimum asymptotic qubit rate $S(\rho )$ for such a source, with every larger rate achievable. Nonorthogonal equally likely pure states need not have entropy one: their classical labels have entropy one but their average density operator can have lower Von Neumann entropy.

For a density operator $\pi ={\sum }_i{q}_i|{\varphi }_i⟩⟨{\varphi }_i|$, the quantum typical subspace is spanned by product eigenvectors of ${\pi }^{\otimes n}$ whose eigenvalues $q_{i_1}\cdots q_{i_n}$ satisfyIt applies the classical weakly typical sequence definition to the spectrum. If $P_{\epsilon }^{(n)}$ is its orthogonal projection, thenThe first statement follows from the typical-set cardinality bounds, and the second from the weak law of large numbers. This subspace holds nearly all the source probability while using exponentially fewer dimensions than the whole space when $S(\pi )<{\log }_{2}d$.

For each fixed $\delta >0$, the quantum typical subspace projector selects eigenvalues of ${\pi }^{\otimes n}$ between $2^{-n(S(\pi )+\delta )}$ and $2^{-n(S(\pi )-\delta )}$. Its probability tends to one by the weak law of large numbers for the spectral information variable. If its probability is at least $1-ϵ$, its dimension is between $(1-ϵ)2^{n(S(\pi )-\delta )}$ and $2^{n(S(\pi )+\delta )}$. These statements give the size and concentration estimates behind Schumacher compression.

A quantum source-compression scheme has encoding and decoding quantum channels with compressed space ${\mathcal{K}}_n$ and rate ${\mathrm{limsup}}_n{n}^{-1}{\log }_2\dim {\mathcal{K}}_n$. For a pure-state source ensemble, average reliability means that the mean squared quantum fidelity after decoding tends to one. A stronger coherent criterion is $F_e({\pi }^{\otimes n},{\mathcal{D}}_n\circ {\mathcal{E}}_n)\to 1$, where entanglement fidelity also tests a purification reference. Encoding the quantum typical subspace achieves both criteria above $S(\pi )$. Schumacher's original quantum coding paper establishes quantum source coding.

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.

For a quantum state ensemble of pure vectors and an orthogonal projection $P$, measure $\{P,I-P\}$, preserve the successful subspace, and decode failure to a fixed normalized vector $\phi$. The resulting quantum channel is $\Lambda (X)=PXP+\mathrm{Tr}[(I-P)X]|\phi ⟩⟨\phi |$. For ${\alpha }_k=\Vert P{\psi }_k\Vert$, its squared quantum fidelity on the $k$th source vector is ${\alpha }_k^4+(1-{\alpha }_k^2)|⟨{\psi }_k|\phi ⟩{|}^2$. The inequality $a^2\ge 2a-1$ on $[0,1]$ proves the displayed average bound. It depends only on the source density operator's weight in $P$ and does not require the emitted vectors to be orthogonal.

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.

A classical-quantum state has block-diagonal form ${\rho }_{XB}={\sum }_x{p}_x|x⟩⟨x|\otimes {\rho }_x$. Its entropy is $H(p)+{\sum }_x{p}_{x}S({\rho }_x)$.

The classical-quantum state is block diagonal. If ${\lambda }_{xj}$ are the eigenvalues of ${\rho }_x$, its eigenvalues are $p_x{\lambda }_{xj}$. Inserting these into the definition of Von Neumann entropy gives the displayed sum of classical Shannon entropy and mean conditional quantum entropy. Its quantum mutual information is therefore the Holevo quantity $S({\sum }_x{p}_x{\rho }_x)-{\sum }_x{p}_{x}S({\rho }_x)$.

For an ensemble $\{p_x,{\rho }_x\}$ with average $\bar{\rho }$, the Holevo quantity is $\chi =S(\bar{\rho })-{\sum }_x{p}_{x}S({\rho }_x)=I(X:B)$ of the associated classical-quantum state.

Attach an orthogonal classical label register to an ensemble. Its quantum mutual information with the quantum system equals the ensemble Holevo quantity. A local quantum channel changes the ensemble states but preserves their labels and probabilities. Data processing for quantum mutual information then proves that the output Holevo quantity cannot exceed the input value.

The classical mutual information obtained by any measurement of a quantum ensemble is at most its Holevo quantity. This follows from the data-processing inequality for quantum relative entropy applied to the measurement channel.

For $k$ equiprobable quantum code states with average decoding error $ϵ<1$, data-processing inequality for quantum relative entropy applied to the binary test for quantum decoding success givesfor $k\ge 2$. Here $h_2$ is binary entropy. The case $k=1$ is trivial. Taking the supremum over allowed input distributions yields a code-size bound in terms of the largest Holevo quantity; at $ϵ=1$ the undivided inequality is vacuous.

A measure-and-prepare channel measures a POVM $\{M_x\}$ and outputs a prescribed state ${\sigma }_x$: $T(\rho )={\sum }_x\mathrm{Tr}(M_x\rho ){\sigma }_x$. Such channels are exactly the entanglement-breaking channels.

A measurement channel writes a POVM's outcome into an orthogonal classical register:For any input orthonormal basis $|j⟩$, the Kraus operators $K_{a,j}=|a⟩⟨j|\sqrt{E_a}$ realize this map and satisfy ${\sum }_{a,j}{K}_{a,j}^{†}{K}_{a,j}=I$. It is therefore a trace-preserving completely positive map.

For a bipartite input density operator ${\rho }_{AC}^x$ and a POVM $\{E_y\}$ on $A$, let $q(y|x)=\mathrm{Tr}[(E_y\otimes I){\rho }^x]$. For $q(y|x)>0$, the conditioned input on $C$ isIts positivity follows from the positive sandwich before the partial trace. A quantum channel on $C$ acts on these conditional density operators even if the original input is an entangled state.

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]

In the unsquared convention, $F({\sum }_x{p}_x{\rho }_x,{\sum }_x{p}_x{\sigma }_x)\ge {\sum }_x{p}_{x}F({\rho }_x,{\sigma }_x)$. Use Uhlmann's theorem to choose component purifications with nonnegative maximizing overlaps, and then construct matching flagged purifications of a quantum ensemble. Their overlap is the weighted sum on the right; the fidelity of the mixtures maximizes over all such overlaps.

Discarding a subsystem cannot decrease unsquared quantum fidelity: $F({\rho }_A,{\sigma }_A)\ge F({\rho }_{AB},{\sigma }_{AB})$. Maximizing purifications of the joint states are also candidates for purifications of the marginals, with the discarded subsystem included in the reference. Uhlmann's theorem then proves the inequality. Information loss makes the two states at least as similar.

If $M$ is a positive contraction and $p=⟨\psi |M^2|\psi ⟩>0$, the normalized successful post-measurement state is $M|\psi ⟩/\sqrt{p}$. Since $M^2\le M$,A likely outcome therefore preserves a pure input well in quantum fidelity when its measurement operator is positive. Positivity excludes a hidden unitary rotation that could otherwise change the state even for an outcome of probability one.

For normalized pure states $\rho =|\psi ⟩⟨\psi |$ and $\sigma =|\phi ⟩⟨\phi |$, the trace distance and quantum fidelity satisfyThe last two expressions distinguish the unsquared quantum fidelity from the squared quantum fidelity. Identical pure states give $D=0$, and orthogonal pure states give $D=1$. The equality need not hold for mixed states; the Fuchs-van de Graaf inequalities give bounds in that case.

Brief proof: for two distinct pure states, their two-dimensional span contains the nonzero part of $\rho -\sigma$. This restriction has trace zero and determinant $-(1-F^2)$, hence eigenvalues $\pm \sqrt{1-F^2}$. Its trace norm is twice the positive eigenvalue, giving trace distance $D=\sqrt{1-F^2}$. For identical pure states, $\rho -\sigma =0$ directly.

Compare the qubit pure statewith $|0⟩$, using the computational basis. Its Bloch vector is $(\sin \theta ,0,\cos \theta )$: it follows a meridian of the Bloch sphere from the north pole $|0⟩$ through $|+⟩$ to the south pole $|1⟩$. The inner product with $|0⟩$ is $\cos (\theta /2)$, nonnegative on this interval. Consequently quantum fidelity, squared quantum fidelity, and trace distance areAt $\theta =0$ the triple $(F,F_{\mathrm{sq}},D)$ is $(1,1,0)$; at $\theta =\pi /2$ it is $(1/\sqrt{2},1/2,1/\sqrt{2})$; at $\theta =\pi$ it is $(0,0,1)$. The angle between the Bloch vectors is $\theta$, while the quantum fidelity uses its half-angle. The pure-state trace distance and fidelity identity becomes the familiar relation ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$.

For equal prior probabilities, the Holevo–Helstrom theorem gives optimal quantum state discrimination success $P_{\mathrm{succ}}=(1+D)/2$. Along this meridian it rises from $1/2$ to one, with $(1+1/\sqrt{2})/2\approx 0.8536$ at $|+⟩$. This connects the geometric trace distance to the best possible decision from a single preparation; it is a different measurement task from the target-acceptance probability defining squared quantum fidelity.

// Target: quantum-information-theory.bigb

For any unitary $U$, the unique positive square root of an operator obeys $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$. Conjugating the positive operators appearing in trace distance and quantum fidelity therefore conjugates their square roots. Cyclicity of the trace removes the unitary factors, proving invariance of both quantities under simultaneous unitary transformations of the states.

// Target: quantum-information-theory.bigb

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

A POVM $\{E_i\}$ gives outcome probability distributions $p_i=\mathrm{Tr}(\rho E_i)$ and $q_i=\mathrm{Tr}(\sigma E_i)$ by the Born rule. Their classical overlap, the Bhattacharyya coefficient, is $B(p,q)={\sum }_i\sqrt{p_i{q}_i}$. Fuchs and Caves showed that minimizing this overlap over all POVMs gives the unsquared quantum fidelity:The same POVM is applied to either possible input quantum state. Each measurement in quantum mechanics has $B(p,q)\ge F$; some measurement attains equality, including when the density operators have zero eigenvalues. Thus quantum fidelity quantifies the smallest classical overlap that observation can reveal. For fidelity of commuting mixed states, measurement in a common eigenbasis attains the minimum. This operational interpretation compares statistics without assigning a probability that two mixed states are equal. ^{[ref] [ref]}

For all finite-dimensional density operators, including pure states and mixed states, the unsquared quantum fidelity and trace distance obeyIn the squared quantum fidelity convention this is $1-\sqrt{F_{\mathrm{sq}}}\le D\le \sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high quantum fidelity into a small operationally meaningful trace distance. If $F\ge 1-\epsilon$ with $0\le \epsilon \le 1$, then $D\le \sqrt{2\epsilon -{\epsilon }^2}$; if $D\le \delta$, then $F\ge 1-\delta$. The upper bound is attained by the pure-state trace distance and fidelity identity. For the fidelity of commuting mixed states example, the bounds read $0.0652\lesssim 0.3\lesssim 0.3551$. ^[ref]

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.

For a channel output $Q$ and complementary environment output $E$, coherent information is $I_c(\Lambda ,\rho )=S(Q)-S(E)=-H(R|Q)$ for a purification reference $R$.

For a purified input with reference $R$ and channel output $B$, $I(R:B)=S(R)+S(B)-S(RB)=S(\rho )+I_c(\Lambda ,\rho )$. The reference entropy is fixed through subsequent channels. Thus data-processing inequality for coherent information immediately implies data processing for quantum mutual information between that reference and successive outputs.

In a purified Stinespring dilation, coherent information satisfies $I_c=S(RE)-S(E)=S(\rho )-I(R:E)$. Nonnegativity of quantum mutual information implies $I_c\le S(\rho )$. Equality means that the reference and environment are uncorrelated, so the environment learns nothing about the purified input reference.

Purify the input on $RA$ and dilate a channel by $A\to BE$. The joint state $RBE$ is pure, so complementary entropies obey $S(B)=S(RE)$ and $S(RB)=S(E)$. Therefore coherent information is $I_c=S(B)-S(RB)=S(R\mid E)=-S(R\mid B)$. The positive environment conditional entropy and negative receiving-system conditional entropy are two equivalent forms.

A quantum channel on the receiving system cannot increase coherent information: $I(A⟩B{)}_{\rho }\ge I(A⟩B^{\prime }{)}_{\sigma }$. Dilate the channel to a linear isometry of Hilbert spaces $B\to B^{\prime }E$. The difference is $I(A:E|B^{\prime })$ of the dilated state, nonnegative by Strong subadditivity of Von Neumann entropy.

A quantum channel is anti-degradable when its receiver output can be simulated from the complementary environment output. Data processing then forces every coherent information to be nonpositive, and no-cloning prevents perfect quantum transmission.

Quantum relative entropy obeys $D({\sum }_i{p}_i{\rho }_i\Vert {\sum }_i{p}_i{\sigma }_i)\le {\sum }_i{p}_{i}D({\rho }_i\Vert {\sigma }_i)$. It follows by applying data processing to flagged block-diagonal states and discarding the flag.

A quantum ancilla is an auxiliary quantum system introduced to assist a state preparation, measurement, or channel use. Applying a channel to one part of an entangled system can reveal behavior invisible to inputs without an ancilla.

An ancilla qubit is a two-dimensional quantum ancilla. It is the quantum version of an ancilla bit; an auxiliary quantum register can contain several such qubits.

A positive linear map sends every positive semidefinite matrix to a positive semidefinite matrix. Positivity need not persist after tensoring with the identity map on an ancillary system; requiring that stronger property gives a completely positive map.

A linear map between ordered vector spaces is positive if it carries the positive cone of the domain into the positive cone of the codomain. For function spaces this means $f\ge 0⟹Tf\ge 0$. A Feller semigroup consists of such positive maps on continuous functions.

A decomposable positive map between complex matrix algebras has the form $\Phi ={\Phi }_1+{\Phi }_2\circ T$, where ${\Phi }_1,{\Phi }_2$ are completely positive maps and $T$ is the matrix transpose. Such a map cannot detect a positive partial transpose entangled state: applying it to one subsystem gives a sum of two positive operators.

Every positive linear map between $M_2(\mathbb{C})$ and $M_2(\mathbb{C})$, or between $M_2(\mathbb{C})$ and $M_3(\mathbb{C})$ in either direction, is a decomposable positive map. The low-dimensional result is the ingredient that makes the positive partial transpose criterion sufficient in two-qubit and qubit-qutrit systems. Woronowicz's low-dimensional positive-map paper establishes the $M_2$ to $M_3$ case; taking adjoints gives the reverse direction.

The $k$-reduction map is ${\Lambda }_k(X)=k\mathrm{Tr}(X)I-X$. It remains positive when tensored with the identity on pure inputs of Schmidt rank at most $k$, making it a witness for Schmidt number greater than $k$.

A linear map $\Phi$ between matrix algebras is completely positive when $\Phi \otimes {\mathrm{id}}_r$ is positive for every ancillary dimension $r$. In finite dimensions this is equivalent to positivity of its Choi matrix and to the existence of a Kraus representation.

A completely positive map with ${\sum }_{\alpha }{K}_{\alpha }^{†}{K}_{\alpha }\le I$ is a trace-nonincreasing quantum operation, representing an outcome branch with its probability retained in the trace. A deterministic quantum channel has equality and is trace preserving. Dividing a branch output by its trace gives a conditional state but generally produces a nonlinear update; that normalization is not part of the linear quantum operation.

In finite dimensions, a completely positive map has the form $\mathcal{N}(X)={\mathrm{Tr}}_E(VX{V}^{†})$ for $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q\prime }\otimes {\mathcal{H}}_E$. From a Kraus representation take $V={\sum }_j{K}_j\otimes |j⟩$. The observable-picture adjoint is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$. A general completely positive map need not have isometric $V$; it is a linear isometry of Hilbert spaces precisely when the map is trace preserving.

Every finite-dimensional completely positive map has $T(\rho )={\sum }_j{K}_j\rho K_j^{†}$. It is trace preserving exactly when ${\sum }_j{K}_j^{†}{K}_j=I$.

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 $1+\gamma$.

The operators $K_j$ in a Kraus representation are Kraus operators. For a measurement outcome $i$, its associated Kraus operators determine both the outcome probability and the conditional post-measurement state.

For normalized maximally entangled $|\varphi ⟩=d^{-1/2}{\sum }_j|jj⟩$, the Choi matrix of $T$ is $C_T=(T\otimes \mathrm{id})(|\varphi ⟩⟨\varphi |)$. Choi's theorem says $T$ is completely positive exactly when $C_T\ge 0$.

Use the unnormalized Choi matrix $J(\Phi )=(\Phi \otimes \mathrm{id})(|\tilde{\Psi }⟩⟨\tilde{\Psi }|)$, which has trace $d$ for a channel. A positive such matrix has a spectral decomposition with vectors $v_k=\sqrt{{\lambda }_k}{u}_k$. In paired bases define the Kraus operator $A_k$ by $(A_k{)}_{ij}=(v_k{)}_{ij}$, so $v_k=(A_k\otimes I){\sum }_j|j⟩|j⟩$. Partial contraction with a conjugate index vector gives $A_k|\varphi ⟩$, hence $\Phi (|\varphi ⟩⟨\varphi |)={\sum }_k{A}_k|\varphi ⟩⟨\varphi |A_k^{†}$. Rank-one projectors span the operator space, proving the Kraus representation on all inputs. Trace preservation is equivalent to ${\sum }_k{A}_k^{†}{A}_k=I$. This construction needs at most $\mathrm{rank}J$ nonzero Kraus operators.

A channel's Choi state is its normalized Choi matrix,It is a density operator and satisfies ${\mathrm{Tr}}_{\mathrm{out}}{C}_{\Lambda }=I_{\mathrm{in}}/d$. Some authors use an unnormalized Choi matrix of trace $d$, so reconstruction factors depend on the convention.

With normalized Choi state $C_{\Lambda }$ and the output factor first,Expand $C_{\Lambda }=d^{-1}{\sum }_{i,j}\Lambda (|i⟩⟨j|)\otimes |i⟩⟨j|$. The partial trace multiplies each coefficient by $\mathrm{Tr}(|i⟩⟨j|X^T)=X_{ij}$, reconstructing $\Lambda (X)$ by linearity.