Positive linear map

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.

A quantum channel is a linear completely positive trace-preserving map between operator algebras.

A memoryless quantum channel has independent identical noise on successive uses, described by the tensor-power map ${\Phi }^{\otimes n}$. This factorization does not restrict the encoder to product inputs: one can feed an entangled state across uses. A block code with equally likely messages $m$ chooses states ${\rho }_m$ and a decoder POVM $(D_m)$. Its average error is $1-|\mathcal{M}{|}^{-1}{\sum }_m\mathrm{Tr}[D_m{\Phi }^{\otimes n}({\rho }_m)]$, and its rate is $n^{-1}{\log }_2|\mathcal{M}|$. The classical capacity of a quantum channel allows such entangled block inputs; the one-use Holevo capacity gives the product-input capacity and requires regularization in general.

A quantum erasure channel transmits an input state unchanged with probability $1-p$ and replaces it by an orthogonal, recognizable erasure flag with probability $p$. Its output space has one more dimension than the input. This is a quantum channel for $0\le p\le 1$ and preserves all input coherences on the nonerased branch.

For input dimension $D$, an input of entropy $S(\rho )$ gives output entropy $h_2(p)+(1-p)S(\rho )$ and reference-output entropy $h_2(p)+pS(\rho )$. These follow from entropy of an orthogonal quantum mixture and purity of the input purification. Thus the quantum mutual information is $2(1-p)S(\rho )$, maximized at $\rho =I_D/D$. The entanglement-assisted classical capacity is consequently $2(1-p){\log }_{2}D$ bits per use; for a binary input it is $2(1-p)$.

This is the supremum of reliable classical bits per use of a memoryless quantum channel when sender and receiver share unlimited prior quantum entanglement. If $|{\psi }_{\rho }{⟩}_{RA}$ purifies the input, let ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The single-letter capacity formula is ${\max }_{\rho }[S(\rho )+S(\mathcal{N}(\rho ))-S({\omega }_{RB})]$, the maximum quantum mutual information between reference and output. The preshared resource is independent of the message.

A zero-temperature energy-relaxation channel transfers the excited population of a qubit to its ground state. Its Kraus operators are $A_0=\mathrm{diag}(1,\sqrt{1-p})$ and $A_1=\sqrt{p}|0⟩⟨1|$ for $0\le p\le 1$. They satisfy ${\sum }_j{A}_j^{†}{A}_j=I$. The excited population is multiplied by $1-p$ and coherences by $\sqrt{1-p}$. For nonzero $p$, it is not a unital quantum channel.

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 $\sqrt{1-p}$.

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 $1-p$.

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.

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.

The Holevo capacity of a quantum channel is the maximum one-use output Holevo quantity,It is the unassisted classical communication limit for product state encodings with collective output measurements. Arbitrary entangled inputs require the regularized expression in the Holevo-Schumacher-Westmoreland theorem.

Take independent ensembles for two quantum channels and their Cartesian product ensemble. The Von Neumann entropy of a tensor product is the sum of the two entropies, both for its average output and for each labelled output. Thus its Holevo quantity is the sum of the two individual quantities. Taking ensembles arbitrarily close to each supremum proves the displayed inequality without requiring maximizing ensembles to exist. It does not assert additivity for arbitrary channels.

A channel's Holevo capacity is additive with another channel when their joint capacity equals the sum of the individual capacities. If this holds with every other channel, the channel's tensor powers have capacity $n{\chi }^{\ast }(\Lambda )$ and regularization does not increase its unassisted classical capacity of a quantum channel. Additivity is a special property, not a general rule for quantum channels.

Every entanglement-breaking channel $\mathcal{E}$ has additive Holevo capacity with every finite-dimensional quantum channel $\mathcal{N}$. A measure-and-prepare channel factorization exposes a classical measurement outcome. The quantum mutual information balance identity and data processing for quantum mutual information then bound the joint output Holevo quantity by the sum of the individual capacities, using the conditional input ensemble after a local measurement for the other channel. Independent product ensembles attain the reverse inequality.

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

The unassisted classical capacity is the supremum of classical bit rates achievable with asymptotically vanishing message error. It allows entanglement among different channel inputs, but no preshared sender-receiver entanglement. The Holevo-Schumacher-Westmoreland theorem gives its regularized entropy expression.

For a finite-dimensional memoryless quantum channel,The Holevo bound controls information obtainable from an output ensemble, while collective decoding makes rates below the corresponding coding expression achievable. The Holevo capacity needs regularization because input states can be entangled across channel uses. Watrous's channel-coding chapter states and develops the classical-capacity theorem.

A random unitary channel has $T(\rho )={\sum }_j{p}_j{U}_j\rho U_j^{†}$. Its Choi matrix is a convex combination of maximally entangled pure states.

On a $d$-dimensional Hilbert space, let $X|j⟩=|j+1\mathrm{mod}d⟩$ and $Z|j⟩=e^{2\pi ij/d}|j⟩$. The $d^2$ unitaries $W_{k,m}=X^k{Z}^m$ form an orthogonal operator basis under the Hilbert-Schmidt inner product.

Averaging conjugation by all Heisenberg-Weyl operators completely depolarizes a system:Summing over phases $Z^m$ kills off-diagonal matrix elements, and averaging shifts $X^k$ makes all diagonal elements equal. Acting on subsystem $B$ of a bipartite operator gives $({\mathrm{Tr}}_{B}A)\otimes I_B/d$. This identity turns joint convexity of quantum relative entropy into monotonicity under partial trace.

A one-qubit Pauli channel applies $I,{\sigma }_x,{\sigma }_y,{\sigma }_z$ with probabilities $p_0,p_x,p_y,p_z$. It maps a Bloch vector diagonally by independently contracting or reversing its three components.

A bit-flip channel applies the Pauli X gate with probability $p$ and does nothing otherwise. If the classical error outcome is not recorded, its output is the convex mixture ${\mathcal{E}}_p(\rho )=(1-p)\rho +pX\rho X$, not a coherent superposition of the two alternatives.

// Target: quantum-information-theory.bigb

For pure input $|\psi ⟩$, squared quantum fidelity after the bit-flip channel is $(1-p)+p|⟨\psi |X|\psi ⟩{|}^2$. The last factor is the square of the input's Bloch $x$ component. Rotational symmetry of the uniform pure-qubit average gives $⟨r_x^2⟩=1/3$, because $r_x^2+r_y^2+r_z^2=1$. Hence the average squared fidelity is $1-2p/3$. Averaging an unsquared fidelity or over a different ensemble gives a different quantity.

// Target: quantum-mechanics.bigb

A qubit depolarizing channel shrinks every Bloch vector by the same factor: ${\Lambda }_p(\rho )=p\rho +(1-p)I/2$. In this retention-parameter convention it is completely positive for $-1/3\le p\le 1$; the usual mixture of unchanged and fully randomized states uses $0\le p\le 1$.

For the quantum depolarizing channel applying the identity with probability $1-p$ and each nonidentity Pauli operator with probability $p/3$, the entanglement-assisted classical capacity is the displayed number of bits per use. Here $H_4$ is Shannon entropy of the four listed probabilities. Its unassisted product-state capacity is $1-h_2(2p/3)$. Both vanish at $p=3/4$. If $q$ instead denotes replacement by $I/2$, then $p=3q/4$ and complete depolarization is at $q=1$. The original entanglement-assisted capacity paper gives the depolarizing-channel formula.

With Bloch vector retention $\eta \to 0$, the unassisted product-state capacity of a quantum depolarizing channel is ${\eta }^2/(2\ln 2)+O({\eta }^4)$. Its entanglement-assisted capacity is $3{\eta }^2/(2\ln 2)+O({\eta }^3)$, obtained by expanding Shannon entropy around the uniform four-outcome distribution. Consequently their limiting ratio is three, independent of the linear noise-parameter convention.

The displayed qubit quantum channel is completely positive for $0\le q\le 4/3$. Indeed its Pauli channel probabilities are $1-3q/4$ for the identity and $q/4$ for each nonidentity Pauli operator. In the range $q>1$ the coefficient $1-q$ is negative, so the displayed expression is not a convex mixture of the unchanged and maximally mixed states, even though the Pauli representation is a valid random unitary channel. Its Bloch vector is contracted and reversed by the factor $1-q$. The Holevo capacity of a qubit depolarizing channel remains $1-h_2(q/2)$ throughout this interval.

If the identity is applied with probability $p$ and the three nonidentity Pauli operators each with probability $(1-p)/3$, the Bloch vector retention factor is $(4p-1)/3$. Thus complete depolarization occurs at $p=1/4$, not at zero. The Holevo capacity of a qubit depolarizing channel becomes $1-h_2((2p+1)/3)$ in this identity-probability parametrization.

For the retention parameter $p$, the Holevo capacity isEvery output has Von Neumann entropy at least the value on a pure input, $h_2((1+p)/2)$. Two equiprobable orthogonal pure inputs attain this minimum entropy individually and have maximally mixed average output, attaining the bound. Additivity of Holevo capacity makes the unassisted classical capacity of a quantum channel equal to this value. King's depolarizing-channel capacity paper proves additivity with an arbitrary other channel.

For $\Phi (\rho )=p\rho +(1-p)I/2$, every output qubit has Von Neumann entropy at least $h((1-p)/2)$. A product input therefore has output entropy at least $nh((1-p)/2)$. The average output has entropy at most $n$, so its Holevo quantity is bounded by the displayed expression. This allows correlated classical message labels and joint output measurements, but requires product quantum inputs. The Holevo bound and Fano's inequality turn it into the corresponding asymptotic rate bound.

The phase-flip channel is ${\Lambda }_p(\rho )=(1-p)\rho +p{\sigma }_z\rho {\sigma }_z$. It multiplies the $x$ and $y$ components of the Bloch vector by $1-2p$ and leaves the $z$ component fixed.

Independent successive applications of the phase-flip channel $D_{ϵ}(\rho )=(1-ϵ)\rho +ϵZ\rho Z$ preserve diagonal entries and multiply off-diagonal entries by $(1-2ϵ{)}^n$. Hence $D_{ϵ}^n=D_{{ϵ}_n}$ with the displayed parameter, for nonnegative integers $n$. If $0<ϵ<1/2$, the limit is the completely dephasing channel, equivalently a nonselective projective measurement in the computational basis. A time-interval law can be iterated this way only under the memoryless composition assumption.

A dephasing channel suppresses off-diagonal entries of a density matrix in a preferred basis while preserving quantum state populations. Complete dephasing is identical to a projective measurement in that basis with its outcome discarded.

For the Bell correlations versus a classical mixture example, completely dephasing Alice's qubit in the computational basis maps ${\rho }_{\mathrm{Bell}}$ to ${\rho }_{\mathrm{mix}}$. It preserves their perfect $Z$ agreement and each reduced density matrix, yet destroys the $X$ correlation. This is why testing only populations can miss damage to a quantum memory: the stored quantum coherence may have disappeared while every preferred-basis measurement still looks correct. The joint Von Neumann entropy rises from zero to one bit, although each local entropy stays fixed.

An environment interaction preserving the computational populations multiplies a qubit's upper off-diagonal density entry by $\overline{c}$. A real phase-flip channel representation exists for every input precisely when $c$ is real, with probability $(1-c)/2$. For $c=r{e}^{i\phi }$, the channel is a phase-flip channel of probability $(1-r)/2$ followed by the diagonal unitary operator $\mathrm{diag}(1,e^{i\phi })$. The coherent phase cannot be discarded just because the populations are unchanged.

A quantum channel is unital when $T(I)=I$. Every random unitary channel is unital, while the converse fails in dimension at least three.

Choose input eigenvectors $u_i$ and output eigenvectors $v_j$ for a unital quantum channel. Positivity makes the displayed entries nonnegative. Trace preservation makes every column sum one; unitality makes every row sum one. Thus $A$ is a doubly stochastic matrix and the output spectrum satisfies $s=Ar$. The mixing matrix need not be a matrix of squared overlaps of a single unitary, even though its row and column constraints are the same.

A unital quantum channel mixes its input spectrum with a doubly stochastic matrix, so its output spectrum satisfies the majorization relation with its input spectrum. The reverse inequality is generally false: a pure state can be sent to a maximally mixed state. The direction expresses the fact that unital noise can make eigenvalues more uniform. Equality of the two spectra is possible, for example under a unitary channel.

The Werner–Holevo channel is $T(\rho )=(\mathrm{Tr}(\rho )I-{\rho }^T)/(d-1)$. Its normalized Choi matrix is the maximally mixed state on the antisymmetric subspace.

A quantum channel is strictly contractive in trace distance when it reduces the distance between every pair of distinct density operators by a uniform factor smaller than one.

A finite-dimensional quantum channel is primitive when some power maps every nonzero positive operator to a positive-definite operator. Equivalently, eigenvalue one is simple and no other eigenvalue lies on the unit circle.

A Stinespring dilation represents a quantum channel by an isometry $V={\sum }_i{K}_i\otimes |i⟩$ followed by tracing out the environment.

For a purification $|{\Psi }_{\rho }{⟩}_{RA}$, the entanglement fidelity of a channel $\Lambda$ isIt is independent of the chosen purification.

In the unsquared quantum fidelity convention, the fidelity between a purification and its channel output is the square root of entanglement fidelity. Tracing out the purifying reference leaves the input and output marginals. Monotonicity of quantum fidelity under partial trace therefore gives the bound. Preserving an input marginal is a weaker requirement than preserving its correlations with a reference.

For a quantum channel on a $d\ge 2$ dimensional system, purify the input to $|\Psi {⟩}_{QR}$ and set $\sigma =(\mathcal{N}\otimes \mathrm{id})(|\Psi ⟩⟨\Psi |)$. Its entanglement fidelity $F_e=⟨\Psi |\sigma |\Psi ⟩$ satisfies $S(\sigma )\le h(F_e)+(1-F_e){\log }_2(d^2-1)$. This is the entropy bound from overlap with a pure state in dimension $d^2$. For a trivial one-dimensional system the output entropy is zero.

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.

A Markovian continuous-time quantum channel obeys

A Lindblad dissipator is the non-Hamiltonian generator associated with a Lindblad operator $V$. It preserves the trace because cyclicity gives $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, cancelling the two anticommutator terms. It preserves Hermiticity and satisfies $\mathcal{D}[V](I)=[V,V^{†}]$; it need not preserve the identity or the purity of a density operator.

The Lindblad operators $L_{\alpha }$ describe dissipative channels in a Markovian open-system evolution. Their normalization fixes the corresponding rates, and their representation is not unique.

The Lindbladian $\mathcal{L}$ is the linear superoperator on density operators defined by the right-hand side of a Lindblad equation, so that $\dot{\rho }=\mathcal{L}(\rho )$.

For any initial density operator, average its time-independent Lindblad equation trajectory as ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$. These averages stay in the compact convex set of density operators. Along a convergent subsequence, $\mathcal{L}({\bar{\rho }}_T)=(\rho (T)-\rho (0))/T\to 0$, so the limit is stationary. Thus a finite-dimensional physical Affine Bloch equation has at least one physical equilibrium, even though an arbitrary affine real equation can have none.

The Lindbladian gap is the smallest positive decay rate $-\mathrm{Re}\lambda$ among nonzero eigenvalues $\lambda$ of a relaxing Lindbladian. It controls the slowest asymptotic exponential approach to the stationary state.

A finite-dimensional Lindbladian converges to one stationary state for every initial state exactly when zero is a simple eigenvalue and every other eigenvalue has strictly negative real part. A trivial commutant of the Hamiltonian and all jump operators and their adjoints is a standard irreducibility criterion leading to uniqueness under the usual finite-dimensional hypotheses.

A quantum-trajectory unravelling represents a Lindblad equation as an ensemble of stochastic pure-state histories. In one jump unravelling, a state evolves between jumps under the non-Hermitian effective Hamiltonian $H_{\mathrm{eff}}=H-\frac{i}{2}{\sum }_{\alpha }{L}_{\alpha }^{†}{L}_{\alpha }$, while a jump of type $\alpha$ sends it to a normalized multiple of $L_{\alpha }|\psi ⟩$.

A pure stationary state is dark when every jump operator acts on it as a scalar, after an allowed shift of the jump operators, and the corresponding effective Hamiltonian preserves its ray. Every trajectory that reaches such a state remains on that same ray.