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For a uniformly distributed pure qubit on the Bloch sphere, rotational symmetry gives $⟨r_z^2⟩=1/3$. ThusEquivalently, $p=|\alpha {|}^2$ is uniform on $[0,1]$ under this measure and ${\int }_0^1[p^2+(1-p{)}^2]dp=2/3$. The three positive coordinate-axis states mentioned in the source give $F_Z=1$, $F_X=F_Y=1/2$, whose average is the same because this fidelity depends only on the quadratic component $r_z^2$. This is a specified uniform-state average, not a statement about every possible prior ensemble.

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 resonant Rabi pi pulse givesIt swaps $|0⟩$ and $|1⟩$ with a common global phase, implementing the Pauli gate $X$, or NOT. Its Bloch sphere action is a half-turn about the $x$-axis.

Keep the Pauli matrix and north-pole conventions of part (a), and neglect Rabi detuning during each short pulse. Write $P=R_x(\pi /2)$ and $\varphi =\delta T$. During the first pulse the Bloch vector is $(0,-\sin \theta ,\cos \theta )$ for $0\le \theta \le \pi /2$. During free evolution,because the Hamiltonian operator generates $R_z(-\delta t)$. During the final pulse it isThese formulas specify every arc in the two Bloch sphere sketches. For $T=\pi /(2\delta )$ the path is north $\to -y\to -x$, and the last pulse leaves $-x$ fixed. For $T=\pi /\delta$ the path is north $\to -y\to +y\to$ north.

To test the complete quantum gate, multiply its unitary operators:It is proportional to the Hadamard gate when $\cos (\varphi /2)=-\sin (\varphi /2)$, hence the shortest positive waiting time isMore generally $\delta T=-\pi /2(\mathrm{mod}2\pi )$. With the alternative conventional ket definition $H_{01}$ specified in part (b)(i), the required matrix is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$, so $\boxed{T_{H_{01}}=\pi /(2\delta )}$ and $U_R=-i{H}_{01}$. The distinction concerns which basis order defines the named Hadamard gate; the evolution and the return population do not depend on that naming choice. Finally $r_z=-\cos \varphi$, so the actual return population isThe printed plus sign is inconsistent with the two identical pulses. At $T=0$, $U_R=P^2=-i{\sigma }_x$, so $|1⟩$ becomes $|0⟩$ and ${\rho }_{11}(0)=0$; the printed expression gives one. This counterexample is independent of Bloch sphere labeling. A plus sign would describe the complementary population ${\rho }_{00}$, or the return population with an opposite-sign final pulse. That changed sequence also changes the gate question and must not be silently substituted.

For $v=(a,b{)}^T$, multiplying each Pauli matrix between $v^{†}$ and $v$ givesAll three components are real. Moreover $|x|=(|a{|}^2+|b{|}^2)/2=\Vert v{\Vert }^2/2$. Thus normalized two-component spinors map to the radius-$1/2$ version of the Bloch sphere; the usual unit-radius convention multiplies these coordinates by two.

The four Pauli matrices form an orthogonal basis of $2\times 2$ matrices for the Hilbert-Schmidt inner product, since $\mathrm{Tr}({\sigma }_{\mu }{\sigma }_{\nu })=2{\delta }_{\mu \nu }$. A Hermitian density matrix therefore has real coefficients in this basis, and its trace-one condition fixes the identity coefficient. ThusThe multiplication rule for the Pauli matrices ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ implies $(n\cdot \sigma {)}^2=\Vert n{\Vert }^{2}I$. Its trace is zero, so for $n\ne 0$ its eigenvalues are $\pm \Vert n\Vert$; for $n=0$ it vanishes. HencePositivity is equivalent to $\Vert n\Vert \le 1$. Conversely every real vector in this unit ball defines a positive trace-one matrix by the boxed formula. Therefore $\Phi (\rho )=n$ is a one-to-one correspondence with the Bloch ball, with the displayed inverse. Each coordinate is linear in $\rho$, proving the affine property under any convex mixture.

A density matrix is pure precisely when its eigenvalues are $1,0$, equivalently when ${\rho }^2=\rho$. The eigenvalue formula makes this equivalent to $\Vert n\Vert =1$. Alternatively $\mathrm{Tr}{\rho }^2=(1+\Vert n{\Vert }^2)/2$. Thus the pure states correspond exactly to the Bloch sphere.

For $n\ne 0$, put $\hat{n}=n/\Vert n\Vert$. The spectral projectors for the larger and smaller eigenvalues areThey are orthogonal rank-one projections, have sum $I$, and give $\rho ={\lambda }_{+}{P}_{+}+{\lambda }_{-}{P}_{-}$. For normalized eigenvectors in this nondegenerate case they equal $|{\varphi }_1⟩⟨{\varphi }_1|$ and $|{\varphi }_2⟩⟨{\varphi }_2|$. Their Bloch vectors are therefore $\boxed{n_1=n/\Vert n\Vert ,n_2=-n/\Vert n\Vert }$, the antipodal eigenprojectors of a qubit density matrix.

At $n=0$, the state is $I/2$ and both eigenvalues coincide. There is no distinguished direction and the printed directional expression is undefined. Any orthonormal eigenbasis gives an antipodal pair of pure-state Bloch vectors, since its two projectors sum to $I$, but that pair can point in any direction. The nonzero-$n$ qualification is essential. The PDF denominators are $\Vert n\Vert$, not the converted TeX denominators involving $n_1,n_2$.

Fix the noise convention explicitly: $p$ is the total probability of a nonidentity Pauli error. The memoryless qubit quantum depolarizing channel iswith independent identical use on successive qubits, namely ${\mathcal{D}}_p^{\otimes n}$ on an $n$-qubit block. Its normalized Kraus operators are $\sqrt{1-p}I,\sqrt{p/3}X,\sqrt{p/3}Y,\sqrt{p/3}Z$.

Write the Bloch sphere representation of a density matrix as $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. Its eigenvalues are $(1\pm |\mathbf{r}|)/2$, proving exactly this positivity condition. Conjugation by one Pauli operator preserves that coordinate of the Bloch vector and reverses the other two. Summing the three conjugations givessoThe Bloch ball contracts isotropically, collapses to the maximally mixed state at $p=3/4$, and contracts with reversal for $p>3/4$. That last interval still represents a physical Pauli mixture. If instead $q$ is the probability of replacing the state by $I/2$, then $\mathcal{D}(\rho )=(1-q)\rho +qI/2$, with $p=3q/4$ and $\eta =1-q$; the usual replacement-probability range $0\le q\le 1$ stops at complete depolarization.

Let $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$, with endpoint limits. In bits per use, the unassisted product-state classical capacity and the entanglement-assisted capacity of a qubit depolarizing channel arewhere $H_4$ is Shannon entropy of the listed four probabilities. Equivalently $C_E=2+(1-p){\log }_2(1-p)+p{\log }_2(p/3)$. The first quantity is a classical capacity with product quantum inputs and potentially collective output measurements, not the quantum capacity. These are the requested capacity formulae; a proof of their coding theorems is not needed here. The maximally entangled input gives a Bell-diagonal joint state with those four probabilities, explaining the entropy appearing in $C_E$. The original entanglement-assisted capacity calculation gives the depolarizing-channel expression.

At $p=0$ the capacities are one and two. Both vanish at $p=3/4$. On the full Pauli-error interval they increase again beyond that point, reaching $1-h_2(2/3)$ and $2-{\log }_{2}3$ at $p=1$; on the replacement-probability plot both decrease to zero at $q=1$. The two panels make this convention distinction explicit.

For the high-noise capacity ratio for qubit depolarization, put $\eta =1-4p/3\to 0$. Taylor expansion of binary entropy around one half givesThe four probabilities for $C_E$ are $1/4+3\eta /4$ and three copies of $1/4-\eta /4$. For a perturbation ${\delta }_i$ of the uniform four-outcome distribution, with ${\sum }_i{\delta }_i=0$, entropy expansion givesHere ${\sum }_i{\delta }_i^2=3{\eta }^2/4$, hence $C_E=3{\eta }^2/(2\ln 2)+O({\eta }^3)$. ThereforeThe reciprocal ratio is $1/3$, and the same enhancement factor occurs as $q\to 1$ in the replacement convention.

There is a normalization error in the original PDF: the first term contains $I$ rather than $\rho$. For a normalized qubit density operator, that printed expression has trace $2(1-p)+p=2-p$, so it is not a quantum channel except at $p=1$. The intended depolarizing channel is the Pauli channel that leaves the input unchanged with probability $1-p$ and applies each nonidentity Pauli matrix with probability $p/3$, for $0\le p\le 1$. Its Kraus operators areand ${\sum }_{k=0}^3{K}_k^{†}{K}_k=I$, proving that it is trace-preserving and a completely positive map.

Write the input in its Bloch vector representation,Conjugation by a Pauli matrix keeps its own Bloch vector component and reverses the other two. More explicitly, ${\sigma }_k{\sigma }_j{\sigma }_k={\sigma }_j$ for $j=k$ and $-{\sigma }_j$ for $j\ne k$. HenceThe corrected quantum channel therefore givesThus its action isThe Bloch sphere is sent to a sphere of radius $|1-4p/3|$ about the origin of the Bloch ball. At $p=0$ the action is the identity. For $0<p<3/4$ the Bloch vector shrinks without changing direction. At $p=3/4$ every input becomes the maximally mixed density operator $I/2$. For $3/4<p\le 1$, the Bloch vector reverses direction and shrinks, reaching factor $-1/3$ at $p=1$. The reversal is compatible with complete positivity because this remains a probabilistic mixture of Pauli matrices. The parameter here is the total nonidentity-error probability, not the retention parameter used in some definitions of the quantum depolarizing channel.

For completeness, retain the PDF literally and call its output $F_p(\rho )$. The same calculation givesThis has no normalized-output Bloch ball interpretation as printed. If it is manually normalized, thenThat different transformation sends $p=0$ to $I/2$, rather than to the input, and agrees with the intended answer only at $p=1$. The literal and corrected conventions therefore cannot be silently identified.

Let $P_{\varphi }=|\varphi ⟩⟨\varphi |$. An arbitrary deterministic strategy, including auxiliary systems, intermediate measurements, adaptive actions and discarded records, is a quantum channel $\mathcal{E}$ from the two input qubits to the three returned qubits. Its output is $R_{\varphi }=\mathcal{E}(P_{\varphi }^{\otimes 2})$. The three local tests commute, and their joint all-pass projector is $P_{\varphi }^{\otimes 3}$. Thusis exactly the probability of at least one failed test. No independence of the three returned qubits is assumed.

We prove a uniform positive obstruction using trace distance, $D(\rho ,\sigma )=\Vert \rho -\sigma {\Vert }_1/2$. Two normalized pure kets obeyIndeed their projector difference is supported on their two-dimensional span, has trace zero, and has eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$.

Two further estimates will cover even mixed outputs. First, trace distance contracts under a quantum channel. To see this directly, the positive spectral projector of the trace-zero difference gives $D(\rho ,\sigma )={\max }_{0\le A\le I}\mathrm{Tr}[A(\rho -\sigma )]$. The adjoint of a channel is positive and unital, so it takes any output effect into an input effect. Maximizing over this subset cannot exceed the input maximum, proving contraction.

Second, for a mixed state $R={\sum }_j{q}_j|u_j⟩⟨u_j|$ and a pure target $P=|v⟩⟨v|$, the trace-norm triangle inequality and weighted Cauchy-Schwarz inequality give the pure-target upper bound on trace distance:In particular $D(R_{\varphi },P_{\varphi }^{\otimes 3})\le \sqrt{{ϵ}_{\varphi }}$.

Now take $|{\varphi }_0⟩=|0⟩$ and $|{\varphi }_1⟩=|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. The two-copy input distance is $D_2=\sqrt{1-1/4}=\sqrt{3}/2$, whereas the ideal three-copy distance is $D_3=\sqrt{1-1/8}=\sqrt{7/8}$. Writing $R_j=R_{{\varphi }_j}$, contraction and the triangle inequality yieldLet $\delta =D_3-D_2>0$. Then $\sqrt{{ϵ}_0}+\sqrt{{ϵ}_1}\ge \delta$, and squaring givesThis is the copy-number trace-distance obstruction. It applies to every Bob strategy, irrespective of its internal complexity. If Alice chooses these two possible inputs with equal probabilities, the left side is her unconditional failure probability. For every device, at least one of the two inputs also has failure probability at least ${\delta }^2/4$, establishing the worst-case guarantee.

The same positive bound holds if Alice chooses a uniformly random pure-qubit state. Apply the preceding inequality to $U|0⟩$ and $U|+⟩$ for every single-qubit unitary $U$, since their overlap is unchanged. Average over normalized Haar measure. Both rotated kets have the same uniform distribution on the Bloch sphere, so each of the two average failures is the uniform-input average. Consequently that average is also at least ${\delta }^2/4$. A constant satisfying the strict requested inequality in either this average or the worst-case interpretation isThe proven bound is $2p$, hence strictly greater than $p$. This proof establishes a quantitative gap, rather than relying only on the impossibility of exact quantum cloning.

There is a necessary quantifier qualification because the PDF does not specify Alice's input distribution. A positive lower bound for every fixed state against every strategy would be false: a device hardwired to return $|000⟩$ never fails when the input is $|0{⟩}^{\otimes 2}$. The unknown-state assertion is therefore meaningful as a worst-case guarantee over possible inputs, or as an average for a specified hidden input ensemble. Both the equal-prior example and the uniform-input version have been proved here, with a constant independent of Bob's strategy.

For a pure qubit $a|0⟩+b|1⟩$, the Bloch vector is $(2\mathrm{Re}(a^{\ast }b),2\mathrm{Im}(a^{\ast }b),|a{|}^2-|b{|}^2)$. Substitution givesBoth lie on the unit Bloch sphere, at the same azimuth and opposite heights. They are reflections across its equatorial plane, not antipodal vectors. Choose $\theta =\pi /3$ and $\phi =\pi /4$, so their coordinates are $(\sqrt{6}/4,\sqrt{6}/4,\pm 1/2)$. The following original sketch shows those vectors and their common meridian:

Use the optimal three-to-one qubit random access code. For Alice's bits $(b_1,b_2,b_3)$, put $s_j=(-1{)}^{b_j}$ and send the pure state with Bloch vectorThese are the eight vertices of a cube inscribed in the Bloch sphere. If an explicit ket is wanted, choose $\cos {\theta }_b=s_3/\sqrt{3}$ and $e^{i{\phi }_b}=(s_1+i{s}_2)/\sqrt{2}$, and prepare $\cos ({\theta }_b/2)|0⟩+e^{i{\phi }_b}\sin ({\theta }_b/2)|1⟩$.

Bob wanting bit $j$ performs a Pauli measurement of ${\sigma }_x,{\sigma }_y,{\sigma }_z$ respectively, reporting zero for eigenvalue $+1$ and one for $-1$. Its success for every input and every requested bit isOnly one chosen bit is extracted; the protocol does not recover all three simultaneously and uses no shared entanglement.

To prove that no one-qubit protocol gives a greater guaranteed probability, allow Alice arbitrary mixed encoding vectors $|r_b|\le 1$ and Bob any binary positive operator-valued measure. Write his decision difference as $D_j=E_{j,0}-E_{j,1}=t_{j}I+v_j\cdot \sigma$. Its eigenvalues lie in $[-1,1]$, implying $|v_j|\le 1-|t_j|\le 1$. On input $b$, success for bit $j$ is $[1+s_j(t_j+v_j\cdot r_b)]/2$. Average over all eight strings and three choices. The terms in $t_j$ cancel, leavingThe Cauchy-Schwarz inequality and cancellation of cross terms over all sign choices giveHence $p_{\mathrm{avg}}\le p_{\ast }$. Any worst-case guarantee is no greater than this average; the cube protocol attains it uniformly, proving optimality both for guaranteed success and for the uniform-input average. Biased prior information about the bits would define a different optimization problem.

Use $t_0=0$ as in the displayed carrier-phase convention. The drift is diagonal, and conjugation givesMultiplying by $A(t)\cos (\omega t+\varphi )$ and expanding the cosine into exponentials yieldsThe rotating-wave approximation averages away the components oscillating at $2\omega$, retainingThe envelope is slowly varying and the effective coupling is weak compared with the carrier frequency. The pulse phase selects the transverse rotation axis on the Bloch sphere; its pulse area determines the angle. Counter-rotating corrections are neglected under this approximation, not identically zero.

From the derived formula,so the action on the Bloch vector isFor $0<q<1$, every direction is contracted by the same positive factor. The pure-state Bloch sphere becomes a concentric sphere of radius $1-q$, and the Bloch ball contracts towards the maximally mixed density operator $I/2$. The lack of a preferred direction explains the name depolarizing channel.

At $q=1$ all inputs become $I/2$. For the allowed Pauli-error range $1<q<4/3$, the factor is negative, so there is also an antipodal reversal; the radius is $q-1\le 1/3$. Thus it would be incorrect to describe the whole printed $p\in (0,1)$ range as positive radial contraction.

Normalize the vector defining the first effect and write $E_1=\alpha |\psi ⟩⟨\psi |$. Since the two effects sum to $I$, the eigenvalues of $E_2=I-E_1$ are $1-\alpha$ along $\psi$ and $1$ on its orthogonal complement. For $E_2$ to have rank one, $\alpha =1$. Hence this binary rank-one qubit POVM is a projective measurement:The two Bloch vectors are antipodal on the Bloch sphere; the corresponding state vectors are orthogonal. The plot below chooses one representative unit vector. Rotating both directions gives every other case. The sphere represents the effect projectors' Bloch directions, rather than plotting their matrix entries.

In the computational basis, writeThe Kraus representation of the amplitude damping channel givesReading off its Bloch vector gives the Bloch-vector map of amplitude damping:Here $0\le p\le 1$. The shift in $s_z$ shows that this quantum channel is not unital unless $p=0$; its fixed ground state lies at the north pole, rather than the center of the Bloch sphere.

A pure spin-one-half state on the equator of the Bloch sphere has $⟨Z_k⟩=0$, so $|{\alpha }_k{|}^2=|{\beta }_k{|}^2=1/2$. Its azimuthal phase does not enter the conditional environment overlap, because the interaction is diagonal in $Z_k$. Substitution in the decoherence factor givesThe Zurek spin-bath model now has a real coherence factor: positive and negative values correspond to opposite relative phases, while suppression of coherence depends on $|z|$.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

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.

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.

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]