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A bipartite density operator ${\rho }_{AB}$ has positive partial transpose when ${\rho }_{AB}^{T_B}\ge 0$ in a product basis. Positivity is independent of the choice of local bases, although the matrix representing the partial transpose changes.

If the state is a separable quantum state, write ${\rho }_{AB}={\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$. Taking its partial transpose givesIndeed, the matrix transpose preserves the nonnegative eigenvalues of each local density operator; a tensor product of positive semidefinite operators is positive, and so is their nonnegative sum. This proves the necessary direction of the positive partial transpose criterion.

Use the normalized Choi state $C_{\Lambda }=(\Lambda \otimes \mathrm{id})(|{\Phi }_d⟩⟨{\Phi }_d|)$, with $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$. If $\Lambda$ is an entanglement-breaking channel, its action on this particular bipartite input makes $C_{\Lambda }$ a separable quantum state.

Conversely, suppose $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, with local density operators ${\sigma }_a,{\tau }_a$. The Choi reconstruction formula givesBecause $\Lambda$ is trace preserving, ${\mathrm{Tr}}_A{C}_{\Lambda }=I/d$, so ${\sum }_a{E}_a=I$. Thus the $E_a$ form a POVM and the channel is a measure-and-prepare channel.

For any bipartite input ${\rho }_{AR}$, define the positive, possibly unnormalized reference operatorsIts output is ${\sum }_a{\sigma }_a\otimes R_a$. Since ${\sum }_a\mathrm{Tr}{R}_a=1$, normalizing each nonzero $R_a$ expresses this as a convex combination of product states. Hence every output is separable, proving the separable Choi-state criterion for entanglement breaking.

Apply the measure-and-prepare channel locally to an arbitrary bipartite density operator ${\rho }_{AR}$. The resulting state iswhereTo see that this is the correct output, the partial trace is cyclic for operators acting only on the traced subsystem, so the displayed expression equals ${\mathrm{Tr}}_A[(E_a\otimes I){\rho }_{AR}]$. Put $q_a=\mathrm{Tr}{R}_a$. The POVM completeness relation implies ${\sum }_a{q}_a=1$, and terms with $q_a=0$ vanish. Thereforeis a separable quantum state. Thus the channel is an entanglement-breaking channel, for every reference-system dimension.

Memorylessness means that the ensemble's average density operator is ${\rho }^{(n)}={\pi }^{\otimes n}$. Its spectral decomposition isThe product eigenvalues giveZero eigenvalues contribute zero to the Von Neumann entropy and are excluded from logarithmic typicality tests.

The quantum typical subspace is the span of eigenvectors whose eigenvalues satisfyIts orthogonal projection $P_{\epsilon }^{(n)}$ selects precisely the classical typical set for the eigenvalue distribution $q$. The typical-set cardinality bounds and the weak law of large numbers therefore givefor any fixed $\delta >0$ and all sufficiently large $n$.

Let $P_{+}$ be the maximizing projection in the variational characterization of trace distance. Use the binary POVM $\{P_{+},I-P_{+}\}$ and its measurement channel. The difference between its two output probability vectors iswhere $D=D(\rho ,\sigma )$ and the second equality uses $\mathrm{Tr}(\rho -\sigma )=0$. Since trace distance between diagonal density operators is half the L1 norm of their probability-vector difference,Thus a binary quantum measurement can preserve the distinguishability of this particular pair exactly. This is the trace-distance-preserving binary measurement, which may depend on the two states.

Choose the trace-distance-preserving binary measurement from the previous part. Let $p,q$ be its output probability distributions, so $\Vert p-q{\Vert }_1=2D(\rho ,\sigma )$. The data-processing inequality for quantum relative entropy givesFor diagonal density operators, the quantum expression is the classical Kullback-Leibler divergence. Applying the supplied Pinsker's inequality givesConsequently the quantum Pinsker inequality isThe $\ln 2$ factor converts natural-logarithm relative entropy to bits. If the relative entropy is infinite because its support condition fails, the inequality holds automatically.

Write a qubit density operator in Bloch vector form, $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. The quantum depolarizing channel maps $\mathbf{r}$ to $p\mathbf{r}$, so its output eigenvalues are $(1\pm p|\mathbf{r}|)/2$. The output Von Neumann entropy is minimized on pure inputs, where it equals $h_2((1+p)/2)$.

For any input ensemble, its output Holevo quantity is at most the maximum qubit entropy minus this minimum output entropy:Equality is attained by two equiprobable orthogonal pure inputs: their average output is $I/2$, while each output has the minimum entropy. ThusThis uses the question's retention parameter $p$; in the usual convex-mixture range, $0\le p\le 1$. The expression also holds throughout the full completely positive qubit range $-1/3\le p\le 1$, since $h_2((1+p)/2)=h_2((1+|p|)/2)$.

The supplied additivity of Holevo capacity gives ${\chi }^{\ast }({\Lambda }_{\mathrm{dep}}^{\otimes n})=n{\chi }^{\ast }({\Lambda }_{\mathrm{dep}})$. The Holevo-Schumacher-Westmoreland theorem expresses the unassisted classical capacity of a quantum channel as the regularized Holevo capacity, soTherefore entangled inputs across channel uses cannot increase this classical capacity.

By Donald's identity, the objective equalsThe nonnegativity of quantum relative entropy makes the last term at least zero, with equality exactly when $\rho =\overline{\omega }$. Hence the unique minimizing density operator is the ensemble average andFor the classical-quantum state ${\sigma }_{XB}$, its marginal on $X$ has entropy $H(p)$, its marginal on $B$ is $\overline{\omega }$, and its joint entropy is $H(p)+{\sum }_j{p}_{j}S({\omega }_j)$. Substituting in the definition of quantum mutual information givesThis is the relative-entropy barycenter of a quantum ensemble: the average state minimizes mean distinguishability from the ensemble. Its minimal value is exactly the Holevo quantity.

Every density operator can be written asTherefore $⟨\varphi |\rho |\varphi ⟩={\sum }_j{p}_j|⟨{\psi }_j|\varphi ⟩{|}^2\ge 0$, so $\rho$ is a positive semidefinite operator, and $\mathrm{Tr}\rho ={\sum }_j{p}_j=1$. A pure quantum state has a rank-one density operator $|\psi ⟩⟨\psi |$, equivalently ${\rho }^2=\rho$ and $\mathrm{Tr}({\rho }^2)=1$. A mixed quantum state is not pure and has $\mathrm{Tr}({\rho }^2)<1$ in finite dimension.

The Von Neumann equation and the Pauli matrix commutator identity giveso the Bloch vector obeysWriting $\omega =|\mathbf{\omega }|$ and $\hat{\mathbf{\omega }}=\mathbf{\omega }/\omega$, the assumption $\mathbf{a}\cdot \mathbf{\omega }=0$ givesand henceUnitary evolution rotates the Bloch vector and preserves its length. Since $|\mathbf{a}|<1$, the state remains mixed, whereas $|{\uparrow }_x⟩$ has the unit Bloch vector $(1,0,0)$. The system is therefore never definitely in $|{\uparrow }_x⟩$.

Let $q=({\mathbf{q}}_1,\dots ,{\mathbf{q}}_N)$ denote the particle positions. In the Ghirardi--Rimini--Weber model, the wavefunction obeys the ordinary Schrodinger equationbetween random collapses. Each particle $i$ has an independent Poisson process of collapse times with rate $\lambda$. At such a time the state jumps according towhere the random centre has probability density $\Vert L_i(\mathbf{X})\psi {\Vert }^2$. The normalization is arranged so that $\int d^{3}X{L}_i(\mathbf{X}{)}^2=I$.

The original GRW scales are approximatelyAn isolated microscopic particle is therefore exceedingly unlikely to collapse during a laboratory experiment. A macroscopic pointer containing about $N\sim 10^{23}$ relevant particles has total collapse rate $N\lambda \sim 10^7{\mathrm{s}}^{-1}$ and collapse time about $10^{-7}\mathrm{s}$. After a measurement interaction correlates different microscopic outcomes with pointer positions separated by much more than $r_C$, one constituent's localization suppresses all incompatible pointer branches. This GRW amplification mechanism produces one definite macroscopic outcome with probabilities given by the Born rule, while leaving ordinary microscopic unitary time evolution almost unchanged.

For one spatial coordinate, average over the random centre of one collapse. The resulting density operator has position-space kernelIts diagonal is unchanged, so $⟨x⟩$ and $⟨x^2⟩$ are unchanged. The first derivative of the Gaussian factor vanishes at $x=x^{\prime }$, so $⟨p⟩$ is also unchanged. Its second derivative does not vanish, and the position representation of the momentum operator givesIn three dimensions the increase in total ${\mathbf{p}}^2$ is $3{\hslash }^2/(2r_C^2)$, corresponding to kinetic-energy increase $3{\hslash }^2/(4m{r}_C^2)$ per collapse. These are ensemble statements: conditioning on one specified collapse centre can shift the position moments. Repetition at rate $\lambda$ predicts GRW spontaneous heating, so precision searches for anomalous bulk heating, spontaneous radiation, momentum diffusion, and loss of matter-wave interference test the model.

Let a projective measurement have mutually orthogonal projections $P_i$ satisfying ${\sum }_i{P}_i=I$. On a density operator $\rho$, the Born rule and the Lüders rule giveconditional on outcome $i$. On the first part of a bipartite quantum system, replace $P_i$ by $P_i\otimes I_2$.

The subsystems are isolated when there is no interaction term coupling them. Their Hamiltonian operator has the formso their subsequent unitary time evolution factorizes as $U_1\otimes U_2$.

The reduced density matrix of subsystem 1 iswhere the partial trace is characterized byfor every local observable $M_1$. Thus ${\rho }_1$ contains exactly the statistics accessible by measurements on subsystem 1.

Suppose a projective measurement $\{Q_j\}$ is performed on subsystem 2 and its outcome is not communicated. The resulting nonselective state isFor every $M_1$, the cyclic property of the trace and ${\sum }_j{Q}_j^2=I_2$ giveHence ${\rho }_1^{\prime }={\rho }_1$. No local projective measurement on subsystem 1 can reveal whether the remote unreported measurement occurred. This is quantum no-signalling, which prevents a choice made at a spacelike separation from transmitting information faster than light and makes the measurement formalism compatible with relativistic causality.

The proposed nondisturbing device would violate no information without disturbance. In an ordinary quantum instrument, let $M_{i\alpha }$ be the Kraus operators associated with classical output $i$. If every pure state $|\psi ⟩$ remains unchanged even conditional on the displayed outcome, every nonzero $M_{i\alpha }|\psi ⟩$ must be parallel to $|\psi ⟩$. A linear operator for which every vector is an eigenvector is a scalar multiple of the identity, so $M_{i\alpha }=c_{i\alpha }I$. Its output probabilityis independent of the state. It cannot equal $⟨\psi |P_i|\psi ⟩$ for arbitrary projectors. Equivalently, repeated nondisturbing samples would permit quantum state tomography of one specimen and then quantum cloning, contradicting ordinary quantum theory.

Such devices would not necessarily enable superluminal signalling. One consistent operational extension could make every sequence of outputs depend only on the local reduced density matrix and local settings. Since an unreported remote measurement leaves that matrix unchanged, all local device statistics would remain unchanged too. Other extensions could add nonlocal outcome-dependent rules and permit signalling, but that behavior is additional to the device specification rather than forced by it.

Write the spin-one-half Heisenberg antiferromagnet ason a bipartite lattice of coordination number $z$. A one-spin density operator is $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, where $\mathbf{r}$ is its Bloch vector and $|\mathbf{r}|\le 1$. In a two-sublattice product state,The minimum is $-1/4$, attained by pure antiparallel Bloch vectors, so the minimizing product state is a Néel state. In the $z\to \infty$ limit this mean-field approximation becomes exact and the ground-state energy per bond is therefore

The phrase “energy density” requires a coupling convention. For the unscaled Hamiltonian above, every site belongs to $z/2$ bonds andwhich diverges as $z\to \infty$. With the standard Kac normalizationthe finite energy density isIf the convention divides by the spatial dimension $d=z/2$ instead, the answer is $-J/4$ per site. A Hamiltonian written with ${\mathbf{\sigma }}_i\cdot {\mathbf{\sigma }}_j$ rather than ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ multiplies all these energies by four.

A density operator satisfiesAn $N\times N$ Hermitian matrix has $N^2$ real parameters, and the trace condition removes one, leavingreal free parameters.

If $A_H(t)=U(t{)}^{†}{A}_{S}U(t)$, equality of expectation values requiresTherefore

If the eigenvalues of a density operator are $p_j\ge 0$ with ${\sum }_j{p}_j=1$, thenwith equality exactly when one eigenvalue is one, namely for a pure state. In dimension $d$, Cauchy--Schwarz gives $\gamma \ge 1/d$, with equality for the maximally mixed state. On an infinite-dimensional space the universal lower bound is zero as an infimum.

In quantum channel discrimination, prepare a density operator ${\rho }_{HR}$ on the channel input $H$ and an optional quantum ancilla $R$. Under hypothesis $j\in \{1,2\}$ the output isUse a two-outcome quantum measurement $\{Q,I-Q\}$ and decide for $T_1$ on outcome $Q$. The conditional Type I error and Type II error areWith prior probabilities $p$ and $1-p$, symmetric Bayesian discrimination minimizes the average error $p\alpha +(1-p)\beta$ over the input and measurement.

Each rank-one density operator isTaking their tensor product and averaging over the $4^d$ independent choices of the fourth roots of unity givesThis is the claimed expansion in matrix elements.

A bipartite density operator is a separable quantum state when it has a convex combination decompositioninto product states. A state for which no such decomposition exists is an entangled state.

For a bipartite density operator ${\rho }_{SA}$, the reduced density matrix of $S$ isIt is characterized byfor every observable $M_S$ on $S$. Equivalently, in any orthonormal basis $\{|a_k⟩\}$ of $A$, the partial trace isand the result is independent of that basis.

For input pairs other than $11$, either local output is the equiprobable ensemble $\{|+⟩,|-⟩\}$; for input $11$, it is $\{|0⟩,|1⟩\}$. Both ensembles have the same density operator,Every local quantum measurement therefore has the same statistics for every remote input. The joint outputs have unusual correlations, but observing them requires the parties to compare results through an ordinary causal channel. Hence this device also does not necessarily allow superluminal signalling.

The exact spectral condition is that $0$ be a simple eigenvalue of $\mathcal{L}$ and that every other eigenvalue satisfy $\mathrm{Re}\lambda <0$. Then $e^{t\mathcal{L}}\rho$ converges for every initial density operator to the unique stationary state ${\rho }_{\ast }$ satisfying $\mathcal{L}({\rho }_{\ast })=0$. This is the unique stationary state of a Lindbladian condition; the smallest nonzero value of $-\mathrm{Re}\lambda$ is the Lindbladian gap.

A standard operator criterion is irreducibility: the only subspaces invariant under $H$, all $L_{\alpha }$, and all $L_{\alpha }^{†}$ are the zero and full spaces, equivalently their common commutant consists only of scalar multiples of the identity. With no nonzero imaginary-axis eigenvalues, this makes the semigroup relaxing. The spectral statement is the safest general answer because uniqueness of a fixed point alone does not exclude persistent oscillatory modes.