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For teleportation with an ambiguous Bell-state phase label, write $\rho =|\psi ⟩⟨\psi |$. The Bell-basis measurement outcomes ${\varphi }^{+}$ and ${\varphi }^{-}$ each occur with probability $1/4$. In either case a correct guess gives Bob $\rho$, while the wrong guess differs by the Pauli Z gate and gives $Z\rho Z$. The other two Bell states are identified correctly. Since Alice guesses each ambiguous label with probability $1/2$, the overall wrong-correction probability is $(1/2)(1/2)=1/4$. Averaging over the actual quantum measurement outcome and her guess gives the phase-flip channelConditional only on an ambiguous analyzer result, the output is $(\rho +Z\rho Z)/2$; that conditional channel has weight $1/2$ in the displayed unconditional result. There is no rejection of those rounds.

Use the teleportation fidelity for a pure state target, namely the squared quantum fidelityThis compares the input with the resulting mixed state; averaging state vectors before forming a density matrix would give a different, incorrect answer. Direct substitution givesThus the maximum is $1$, attained by $|0⟩$ and $|1⟩$ up to a global phase, because these are eigenstates of the Pauli Z gate. The minimum is $3/4$, attained by every equal-weight quantum superpositionon the equator of the Bloch sphere. The unsquared quantum fidelity is $\sqrt{F_{\psi }}$ if that alternative convention is used. No average over different input quantum states is needed for the requested state-dependent answer.

Use the source's north-pole assignment $|1⟩\leftrightarrow (0,0,1)$. Fix the ordered basis as $(|1⟩,|0⟩)$ and use the standard Pauli matrices in that order, soA pure state has $|\mathbf{r}|=1$: examples are $|1⟩$ at the north pole and $(|1⟩+|0⟩)/\sqrt{2}$ at $(1,0,0)$. An incoherent mixed statehas $\mathbf{r}=(0,0,2p-1)$ inside the Bloch sphere. At $p=1/2$ it is the origin, distinct from an equatorial pure state despite equal populations. The distinction is the off-diagonal quantum coherence of the density matrix.

A bipartite pure state is entangled when it cannot be written $|\alpha {⟩}_a\otimes |\beta {⟩}_b$. For mixed states, the corresponding definition excludes every convex mixture of product states, not just a single product state.

Apply a Hadamard gate to ion $a$, followed by the CNOT gate of part (d):The lower network in the figure shows this Bell pair preparation. Either ion alone has reduced density matrix $I/2$, whereas the joint quantum state is pure; hence the joint quantum state cannot be a product state. The entangling operation can be expanded into $H_b,G,Z_a,H_b$ exactly as above.

An ideal zero-photon number result projects onto the first term in part (b). ConsequentlyThis is a product state, not an entangled success. The remaining ideal outcome hasand leaves $|gg⟩$. In particular, simply discarding the field outcome does not reproduce the pure heralded Bell state.

The practical requirements are strong coherent coupling of both atoms to the same optical normal mode with stable relative coupling phases, preparation of $|ee;0⟩$, and interaction timing on the scale $g^{-1}$ before spontaneous emission or cavity loss destroys quantum coherence. The field must then be measured or extracted after halting the coherent exchange, with enough photon number resolution to distinguish one from two. Otherwise an inefficient detector can mistake a two-photon event for the desired herald. Even ideally the one-photon success probability is at most $1/3$, so trials must be repeated.

“No photons detected” implies $|ee⟩$ only for the ideal zero-number projection specified above. With loss or finite detector efficiency, no click also includes missed one- and two-photon events and generally leaves a conditional mixed state. Detecting a leaking photon during the interaction is also a different measurement protocol from the end-time projective measurement analyzed here.

A density operator represents a pure state exactly when it has rank one, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$. The spectra just found giveTheir wavefunctions are defined up to a global phase. The other two are mixed states, with purity $(3/4{)}^2+(1/4{)}^2=5/8$, and have no single-wavefunction representation as rank-one projectors.

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.

c1 is true for density-state controllability. Every qubit density operator can be written $\rho ={\lambda }_{2}I+({\lambda }_1-{\lambda }_2)|\psi ⟩⟨\psi |$ using its eigenvalues. A control taking one pure orthogonal projector to any other therefore also connects every pair of qubit density operators with the same spectrum. In Lie-algebra language, $\mathfrak{sp}(1)=\mathfrak{su}(2)$; there is no larger-dimensional symplectic exception. This does not supply arbitrary global phase: $SU(2)$ alone is already a counterexample to an implication involving strict operator controllability.

c2 is false. For every even $N=2\ell \ge 4$, the defining $Sp(\ell )$ action is transitive on unit vectors, so a system generating this algebra is pure-state controllable. It still preserves additional mixed-state invariants and lacks density operator controllability, as the preceding explicit four-dimensional example demonstrates. pure states have fewer orbit requirements than general mixed states.

A reachable set consists of states obtainable from one specified initial state using admissible inputs and allowed durations. Controllability means that every target in the declared state space is reachable from every initial state there; the declared space and time convention are essential.

For a closed quantum system, let $U_f$ solve the controlled Schrödinger equation with $U_f(0)=I$. The propagator and state reachable sets areA pure-state version acts on normalized vectors up to global phase. Pure-state controllability makes every pure target accessible; unitary operator controllability permits every desired coherent process, with global phase treated according to the definition.

For mixed states, density operator controllability means transitivity on each unitary orbit of a density operator, not access to all density operators regardless of spectrum. The Dynamical Lie algebra generated by $-i{H}_0,-i{H}_1,\dots$ describes the accessible connected unitary transformations under the usual unconstrained-time control assumptions. Conserved quantities or restricted controls can shrink the reachable set. Thus successful preparation of one state is weaker than implementation of an arbitrary gate on all inputs, and coherent controllability does not imply purification.

Density operator controllability cannot defeat unitary invariants: it grants access to the unitary orbit of a density operator, whose members have the same eigenvalues, rather than to every density operator. In the Hilbert-Schmidt inner product orthonormal normalization of this question,Quantum Hamiltonian evolution preserves this norm, so the target is not reachable from the stated initial state. Equivalently, their values of purity of a density operator are $1/2+1/4=3/4$ and $1/2+1/2=1$: the initial state is a mixed state and the target a pure state. Dissipation, an appropriate measurement protocol or an auxiliary system with additional resources would be needed to change the spectrum; freely controlling the closed-system Quantum Hamiltonian alone does not suffice.

Take the vectors in the quantum state ensemble to be normalized, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The corresponding density matrix isEach rank-one projector is a self-adjoint operator, so ${\rho }^{†}=\rho$. For every $|v⟩$ in the Hilbert space,proving that $\rho$ is positive semidefinite. The trace is $\mathrm{Tr}\rho ={\sum }_i{p}_i⟨{\psi }_i|{\psi }_i⟩=1$.

Conversely, diagonalize a self-adjoint operator $\rho$ in an orthonormal eigenbasis, writing $\rho ={\sum }_j{\lambda }_j|e_j⟩⟨e_j|$. Its positive semidefiniteness implies ${\lambda }_j=⟨e_j|\rho |e_j⟩\ge 0$, and its unit trace gives ${\sum }_j{\lambda }_j=1$. Thus $\{{\lambda }_j,|e_j⟩\}$ is a quantum state ensemble with precisely this density matrix. Terms with zero eigenvalue may be omitted. This argument includes rank-one pure states; an ensemble representation need not describe a genuinely mixed state.

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.

Apply $V^{-1}$ to the data and phase quantum registers, leaving the flag quantum ancilla untouched. This inverse needs no extra oracle assumption: the promised dyadic eigenvalues give $U^{2^n}=I$, hence $U^{-1}=U^{2^n-1}$. The inverse of each controlled unitary gate used in $V$ can therefore be built from repeated uses of the supplied controlled-$U$, and the known Hadamard gates and quantum Fourier transform gates can be reversed. Uncomputation is necessary to erase the eigenvalue label coherently. The resulting quantum state isA quantum measurement in the computational basis of the flag followed by postselection on one yieldsThis requires $A|b⟩\ne 0$. For a normalized input and nonnegative eigenvalues, the success probability of positive quantum spectral filtering satisfies $P_{\mathrm{success}}\ge {\lambda }_{\min }^2$. In the general inequality, equality holds precisely when the input is supported on the minimum-eigenvalue eigenspace. Omitting uncomputation and discarding the phase quantum register would instead leave a mixed state with diagonal weights proportional to $|{\beta }_j{|}^2{\lambda }_j^2$, rather than the desired coherent pure state.

The printed universal nonzero-success request needs a nonkernel-input hypothesis. An $n$-qubit Hermitian operator has $2^n$ eigenvalues, counted with multiplicity. All are distinct, and the printed dyadic grid contains exactly $2^n$ possible values. They therefore occupy the entire grid, including zero: this is a multiplicity-free complete dyadic spectrum, and ${\lambda }_{\min }=0$. Taking $n=1$, $A=\mathrm{diag}(0,1/2)$, and $|b⟩=|0⟩$ satisfies every printed spectral promise but gives $A|b⟩=0$. No normalized output vector exists, so no algorithm can deliver it with nonzero probability.

For every input outside the kernel, the procedure above has $P_{\mathrm{success}}>0={\lambda }_{\min }^2$, giving the requested strict bound on the meaningful domain. The corrected result is therefore exact quantum spectral filtering conditional on $A|b⟩\ne 0$, with the boxed probability. This does not require knowing the input probability amplitudes or making extra copies of an unknown quantum state. Arbitrarily small nonzero support outside the kernel gives arbitrarily small success probability, and a failed flag measurement disturbs the input; fresh independent trials cannot be assumed when only one unknown physical input is supplied.

A density operator describes both a quantum state prepared in one pure state and a statistical preparation whose members occupy different pure states. If normalized vectors $|{\psi }_i⟩$ occur with probabilities $p_i$, defineThe resulting operator is Hermitian, positive semidefinite and has trace one. Positivity follows because $⟨v|\rho |v⟩={\sum }_i{p}_i|⟨{\psi }_i|v⟩{|}^2\ge 0$. Conversely, the spectral theorem gives such a representation of any finite-dimensional operator with these properties, using its eigenvectors and eigenvalues. The matrix representing $\rho$ depends on the chosen basis; the operator and its predictions do not.

One operator gives every measurement probability. For a projective measurement with orthogonal projections ${\Pi }_a$ and outcomes $a$,For example, the first formula follows by averaging $⟨{\psi }_i|{\Pi }_a|{\psi }_i⟩$ over the preparation. The trace expression is basis independent. A general measurement replaces ${\Pi }_a$ by positive measurement effects that sum to the identity, without changing the probability formula. Thus $\rho$ specifies the statistics accessible to measurements on that system.

The distinction between pure and mixed states is especially transparent in the eigenvalues. A pure state has $\rho =|\psi ⟩⟨\psi |$, ${\rho }^2=\rho$, and $\mathrm{Tr}{\rho }^2=1$. A mixed state has at least two positive eigenvalues and $\mathrm{Tr}{\rho }^2<1$. For a qubit,pure states lie on the Bloch sphere; mixed states lie inside the Bloch ball. The maximally mixed qubit $\rho =I/2$ can be prepared by choosing equally between the two $z$ spin eigenstates, or equally between the two $x$ spin eigenstates. These ensembles are different preparation procedures but give the same density operator. No measurement on the resulting qubit alone can identify which ensemble was used.

quantum coherence must be distinguished from merely having two populated levels. In the $z$ basis, the state $|+x⟩$ hasBoth give equal probabilities for the two $z$ outcomes. A measurement along $x$ gives the positive result with probability one for the first state and one half for the second. The off-diagonal entries encode the relative phase information responsible for this difference. Whether a matrix entry is off diagonal is basis dependent; purity is not.

For an isolated system, each component evolves with the same unitary time-evolution operator. Therefore $\rho (t)=U(t)\rho (0)U^{†}(t)$ andUnitary evolution preserves the eigenvalues and purity of $\rho$. A density operator commuting with a time-independent Hamiltonian operator is stationary. A thermal equilibrium example is the canonical ensemble ${\rho }_{\beta }=e^{-\beta H}/Z$, where $Z=\mathrm{Tr}(e^{-\beta H})$ and $\beta =1/(k_{\mathrm{B}}T)$. expectation values then reproduce quantum statistical mechanics without assigning the system a definite energy eigenstate on every run.

A mixed state can also arise without classical randomness in the joint preparation. For a bipartite quantum system, the reduced density operator ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ gives every measurement probability accessible to subsystem $A$. For the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, the joint state is pure, whereas ${\rho }_A=I/2$. Its local mixedness reflects entanglement with the inaccessible system; the partial trace discards correlations that matter for joint measurements. More generally, interaction with an environment can suppress off-diagonal entries in a selected basis, producing decoherence while the combined system still evolves unitarily.

Finally, state evolution conditional on a measurement outcome differs from isolated evolution. An ideal projective measurement yielding $a$ updates the state by the Lüders rule $\rho \mapsto {\Pi }_a\rho {\Pi }_a/P(a)$. Discarding the outcome gives ${\sum }_a{\Pi }_a\rho {\Pi }_a$, which generally removes coherence between different eigenspaces. The Von Neumann entropy $-\mathrm{Tr}(\rho \ln \rho )$ quantifies mixedness: it vanishes for pure states and reaches $\ln d$ for $I/d$. Multiplying by $k_{\mathrm{B}}$ gives thermodynamic entropy units. These uses make the density operator a common language for preparation uncertainty, measurement, thermal states and entangled subsystems, while retaining their distinct physical origins.

Include Alice's detector and environment before taking a partial trace. Ideal measurement produces the global stateOrthogonality of the two record states removes their cross terms from the reduced state of the particles:This nonselective quantum measurement has changed the initially entangled singlet into a separable, classically anticorrelated reduced state. Its entropy is one bit, although the full particles-plus-records state is still pure under the assumed unitary dynamics. The reduced density matrix is an improper mixture; tracing the environment does not by itself assert a literal collapse of that global state.

The Born rule gives Alice outcomes zero and one with equal probability. Conditional on a recorded result, the selective quantum measurement assigns $|01⟩$ or $|10⟩$ to the particles. The unitary measurement coupling and orthogonal records explain decoherence in this basis; they do not supply an additional rule selecting a unique outcome. That is part of the measurement problem. In a collapse postulate the individual outcome is stochastic; other interpretations account for outcome experience differently while retaining these operational probabilities.

When Bob makes the same kind of measurement, the global state becomesTracing both laboratories gives the same reduced matrix, ${\rho }_{AB}^{(2)}={\rho }_{AB}^{(1)}$. Bob has added a correlated record, not restored the lost reduced coherence. If Alice's outcome is known, Bob's opposite outcome is certain in this ideal common basis. Without that record, his marginal is $I/2$, so his outcome remains equiprobable. Alice cannot choose her random result or use this correlation for faster-than-light signalling, consistently with quantum no-signalling.

These common-basis measurements alone do not distinguish quantum predictions from a local source sending opposite preassigned bits. To test that explanation, prepare fresh singlets and let the observers independently choose among incompatible measurement axes before recording each result. For singlet spin measurements along unit vectors $\mathbf{a},\mathbf{b}$, quantum mechanics predicts $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. The CHSH inequality bounds $|S|\le 2$ for Bell-local models with measurement-independent settings. Choosinggives $S=E(\mathbf{a},\mathbf{b})+E(\mathbf{a},{\mathbf{b}}^{\prime })+E({\mathbf{a}}^{\prime },\mathbf{b})-E({\mathbf{a}}^{\prime },{\mathbf{b}}^{\prime })=-2\sqrt{2}$. A suitable Bell test constrains that class of explanations. Independently chosen settings, control of detection bias and separation of the measurement events matter when interpreting an actual experiment.

If Alice first performs the fixed-axis irreversible measurement described above, the resulting mixed state has $E(\mathbf{a},\mathbf{b})=-a_z{b}_z$ and cannot supply that singlet violation. The incompatible settings must belong to the Bell measurement itself. Moreover, Bell violations do not choose among all interpretations of quantum mechanics: interpretations sharing the Born predictions remain observationally equivalent for this experiment. Objective-collapse modifications could instead require additional interference tests with specified departures from unitary dynamics.

A normalized pure state has $\rho =|\psi ⟩⟨\psi |$, and thereforeConversely, diagonalize a permissible density operator. Its eigenvalues obey $0\le {\lambda }_j\le 1$ and ${\sum }_j{\lambda }_j=1$, soEquality requires every eigenvalue to be zero or one; exactly one is one. Thus the idempotent matrix condition and purity one both characterize a pure state. A mixed state has at least two positive eigenvalues, ${\rho }^2\ne \rho$ and $\mathrm{Tr}{\rho }^2<1$. In dimension $d$ its purity is at least $1/d$.

A quantum state ensemble of normalized states $|{\psi }_i⟩$, with probabilities $p_i\ge 0$ and ${\sum }_i{p}_i=1$, has density operatorIt is a positive semidefinite Hermitian operator with trace one, and the Born rule assigns a measurement outcome with projector $\Pi$ the probability $\mathrm{Tr}(\rho \Pi )$.

A pure state has $\rho =|\psi ⟩⟨\psi |$, hence ${\rho }^2=\rho$ and purity of a density operator $\mathrm{Tr}({\rho }^2)=1$. Its only nonzero eigenvalue is one. A mixed state has at least two nonzero eigenvalues and $\mathrm{Tr}({\rho }^2)<1$. The spectral theorem proves the converse criteria, since the nonnegative eigenvalues sum to one. Different quantum state ensembles may give the same density operator; the mere presence of several terms in a preparation list does not make the state mixed if all positive-weight terms represent the same quantum state ray.

Taking the outer product for A and averaging the two orthogonal projections for B givesThe off-diagonal entries in ${\rho }_A$ retain quantum coherence between the $z$-basis states. They are absent from B's classical quantum state ensemble. In particular, $\mathrm{Tr}{\rho }_A^2=1$ and $\mathrm{Tr}{\rho }_B^2=1/2$: A is a pure state, B a mixed state.

Taking the partial trace over the left atom in the singlet state gives the right-hand reduced density matrixThus every measurement on the right stream alone has the same statistics as B or C. The joint pure state is compatible with locally mixed states; the correlations reside in the full density operator.

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.

The purity of a density operator is $\gamma =\mathrm{Tr}({\rho }^2)$. It equals one exactly for a pure state; in finite dimension $d$ it lies between $1/d$ and $1$. The minimum occurs exactly at the maximally mixed state $I/d$.

A worked qubit family interpolates between the mixed state $I/2$ and the pure state ${\rho }_{+}=|+⟩⟨+|$, where $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$:The eigenvectors are $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, with eigenvalues ${\lambda }_{\pm }=(1\pm p)/2$. Squaring and summing these eigenvalues givesHere $P_X(+)$ is the Born rule probability of the positive outcome in an $X$ projective measurement. The Bloch vector is $(p,0,0)$, since ${\rho }_p=(I+p{\sigma }_x)/2$ with ${\sigma }_x$ one of the Pauli matrices. At $p=0$ the purity is $1/2$; at $p=1/2$ it is $5/8$, with $P_X(+)=3/4$; at $p=1$ it is one. Within this family, $p=1$ is a pure state and every $p<1$ is a mixed state. Extending the formula to $p>1$ produces a negative eigenvalue $(1-p)/2$, so it ceases to be a density operator despite remaining Hermitian with trace one.

The Bloch vector length measures polarization: for any qubit density operator, $\gamma =(1+|\mathbf{r}{|}^2)/2$. This connects the interior of the Bloch ball with mixed states and its boundary, the Bloch sphere, with pure states. ^[ref]

For a short proof of the general bound, use the spectral theorem: the nonnegative eigenvalues ${\lambda }_j$ sum to one, so $\gamma ={\sum }_j{\lambda }_j^2$. ThenThe lower bound is attained only when all eigenvalues equal $1/d$; the upper bound is attained only when exactly one eigenvalue is nonzero, and hence equals one. Purity is therefore a simple quantitative test of mixedness, invariant under a unitary operator. It does not specify the entire density operator: even distinct pure states all have purity one. ^[ref]