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The Pauli matrices are Hermitian matrices with zero trace, so a real Bloch vector makes $\varrho$ Hermitian and gives $\mathrm{tr}\varrho =1$. Their Pauli multiplication identity implies $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Consequently the two eigenvalues of $\mathbf{s}\cdot \mathbf{\sigma }$ are $\pm |\mathbf{s}|$, including the zero-vector case, andA density matrix must also be positive semidefinite. The displayed eigenvalues are nonnegative exactly when $\boxed{|\mathbf{s}|\le 1}$. Thus real coordinates alone are insufficient: the allowed vectors fill the Bloch ball. Its boundary consists of pure states, with eigenvalues one and zero; its center is the maximally mixed state $I/2$.

The Pauli multiplication identity gives $(\mathbf{s}\cdot \mathbf{\sigma }{)}^2=|\mathbf{s}{|}^{2}I$. Its eigenvalues are therefore $\pm |\mathbf{s}|$, and those of the density matrix areHermiticity and trace one are automatic; positive semidefiniteness holds exactly for $|\mathbf{s}|\le 1$. Thus physical qubit states form the Bloch ball, with pure states on the Bloch sphere and mixed states inside.

For the mixture of the two standard-basis states,These form the entire $z$-axis diameter, shown in the left panel of the diagram in part b. Generate the state by selecting $|0⟩$ with classical probability $p_0$ and $|1⟩$ with probability $p_1$, then discarding the classical selection record. A coherent superposition is generally a different state: the forgotten classical record removes off-diagonal coherence.

The six Pauli eigenstates have Bloch vectors $\pm {\mathbf{e}}_x$, $\pm {\mathbf{e}}_y$, $\pm {\mathbf{e}}_z$:Classical convex combinations of density operators produce the corresponding convex combinations of their Bloch vectors. Their convex hull is the stabilizer octahedron:The six vertices are pure Pauli states; the faces and interior represent their mixtures. To verify the reverse inclusion, give the vertex in the sign of each coordinate weight $|s_j|$ and use the leftover weight at the origin, itself the equal mixture of either opposite pair.

The output of a state of Bloch radius $u$ has eigenvalues $(1\pm ru)/2$. These are nonnegative for every $0\le u\le 1$ exactly whenIf $|r|>1$, a pure input already gives a negative eigenvalue. Conversely, $|r|\le 1$ preserves the Bloch ball, hence positivity for states and, by positive scaling, for all positive operators. This is positivity of the map alone; physical implementation on part of an entangled system needs complete positivity.

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 $n_1,n_2$.

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.

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.

The orthogonal projection onto the $+1$ eigenspace of the Pauli matrix ${\sigma }_z$ is $P_{+}=(I+{\sigma }_z)/2$. By the Born rule and the Bloch vector representation,Since $s_z=1/5$,The given vector has length $19/30<1$, so it lies inside the Bloch ball and indeed defines a valid mixed qubit state. For the physical spin operator $S_z=(\hslash /2){\sigma }_z$, the corresponding outcome is $+\hslash /2$ with the same probability.

For one fixed POVM repeated on independently prepared copies, the Born rule gives $m$ outcome probabilities with one normalization constraint. An arbitrary mixed qubit density operator has three independent real parameters,The probability map is affine in $\mathbf{r}$. It can be injective on the full Bloch ball only if $m-1\ge 3$. If its rank is smaller, a nonzero undetected direction gives two nearby physical Bloch vectors with the same probabilities. Hence $m\ge 4$ for an informationally complete POVM.

To attain this bound, take the four directionsThey are the vertices of a regular tetrahedron. The tetrahedral qubit POVM has effects $E_i=(I+{\mathbf{n}}_i\cdot \mathbf{\sigma })/4$, each positive and rank one, and ${\sum }_i{E}_i=I$. The observed probabilities areSince ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I_3/3$, they recover the state byThis is the minimum for a single fixed measurement family. If the wording permits changing measurement settings between groups of copies, three binary projective measurements of the three Pauli observables also suffice, each with $m=2$. That procedure is informationally complete three-observable qubit tomography, and it does not contradict the four-outcome bound for one fixed POVM. Tomography estimates probabilities statistically from many copies; a finite sample does not determine an arbitrary continuous state exactly.

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.

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]

A quantum state describes a preparation through its predicted measurement in quantum mechanics probabilities. In the Hilbert space formulation, a pure state is a quantum state ray: a normalized vector representative $|\psi ⟩$ satisfies $⟨\psi |\psi ⟩=1$, and representatives differing by a global phase describe the same quantum state. ^[ref]

Both pure states and mixed states are represented by a density operator on the chosen complex Hilbert space:Thus $\rho$ is Hermitian, positive semidefinite and of unit trace; in infinite dimension it must also be a trace-class operator. For a normalized pure state, $\rho =|\psi ⟩⟨\psi |$ is a rank-one orthogonal projection. A mixed state has no single such vector representative. Its density matrix is the matrix of $\rho$ in a chosen orthonormal basis. ^[ref] For a discrete POVM with effects $E_a\ge 0$ and ${\sum }_a{E}_a=I$, the Born rule readsThese constraints and Born rule probabilities describe arbitrary finite-dimensional preparations. ^[ref]

Randomly choosing normalized pure states with probabilities $w_j$ and forgetting the label givesThis quantum state ensemble averages probabilities; a quantum superposition adds probability amplitudes. Several ensembles can give the same density operator. Quantum entanglement can make a reduced density matrix mixed while the joint quantum state is pure. ^[ref]

The preparation, the choice of basis, and the measurement are separate ingredients. A wavefunction $\psi (x)=⟨x|\psi ⟩$ is the position representation of a pure state; a general mixed state requires a density operator. For a closed quantum system, a unitary operator evolves it as $\rho \mapsto U\rho U^{†}$, preserving its eigenvalues and purity of a density operator. ^[ref] The Bloch sphere and Bloch ball give the qubit geometry; quantum state tomography connects repeated preparations to an estimated density matrix; projective measurement treats the conditional state after an outcome.

Quantum state tomography estimates a density operator from measurement frequencies on many independently and identically prepared systems. One fixed projective measurement reveals only its basis populations, so further measurement settings are generally needed. For a qubit, writeThe three Pauli matrices thus turn measured expectation values into the Bloch vector. This reconstruction uses the Bloch ball representation of arbitrary qubit density operators. ^[ref].

Divide the prepared ensemble into groups measured in the $X$, $Y$ and $Z$ bases. For example, positive-outcome frequencies $0.8,0.5,0.9$ give the estimate $\mathbf{r}=(0.6,0,0.8)$ andHere $|\mathbf{r}|=1$, so the reconstructed state is the pure state $(3|0⟩+|1⟩)/\sqrt{10}$. Its $Z$ frequencies alone also fit the diagonal mixed state with populations $0.9,0.1$; the $X$ data distinguish them. For the angular pure state in measurement in quantum mechanics, $r_x=\sin \theta \cos \varphi$, $r_y=\sin \theta \sin \varphi$ and $r_z=\cos \theta$, so $Y$ resolves the sign of the phase that $X$ alone cannot.

Finite samples introduce statistical uncertainty: with $N$ independent binary trials in one basis, the frequency has variance $p(1-p)/N$. At $p=1/2$ and $N=100$, its standard deviation is $0.05$. Fluctuations can place a naive Bloch vector estimate outside the Bloch ball; a physical reconstruction must enforce positivity of the density operator. Repeated measurements on one unprepared specimen sample its changing conditional state, rather than independent copies of its original state. Multiple bases are needed in this projective protocol; a single fixed informationally complete POVM, such as the tetrahedral qubit POVM, is an alternative. It still requires an ensemble of trials.

A qubit state specifies measurement probabilities, rather than a pre-existing answer to every possible measurement. For a pure state, normalized vectors differing by a global phase represent the same physical state:Equivalently, the pure state is a ray: the one-dimensional complex linear subspace spanned by any nonzero representative. Dividing a nonzero vector by its norm fixes its length, but leaves the global phase free. The density operator is unchanged because $(e^{i\gamma }|\psi ⟩)(e^{i\gamma }|\psi ⟩{)}^{†}=|\psi ⟩⟨\psi |$. In contrast, multiplying only one nonzero component by $e^{i\varphi }$ changes a relative quantum phase and can change measurement probabilities. ^[ref].

After choosing a global phase, every pure state of a qubit can be writtenAt $\theta =0$ or $\pi$, the parameter $\varphi$ does not distinguish physical states: only one component survives. Elsewhere it is the relative quantum phase. The corresponding unit Bloch vector is $\mathbf{r}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ on the existing Bloch sphere. Its angles describe a space of states; for a photon polarization qubit they are not the photon's position or direction of travel. ^[ref].

The Born rule in the computational basis gives $P(0)={\cos }^2(\theta /2)$ and $P(1)={\sin }^2(\theta /2)$. These probabilities discard $\varphi$. In the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, adding or subtracting the probability amplitudes instead givesThis last equality follows by substituting the angular representative. In the basis $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, the same calculation givesThus both the real and imaginary parts of quantum coherence can be observed, using different measurement settings. For example, $(|0⟩+|1⟩)/\sqrt{2}$ and $(|0⟩-|1⟩)/\sqrt{2}$ both give $P(0)=P(1)=1/2$, but the first gives $P_X(+)=1$ and the second $P_X(+)=0$. A global phase could never cause this difference.

For a general qubit state, write its density matrix in the computational basis asThese are exactly the Hermitian, unit-trace, positive semidefinite constraints. Positivity means every eigenvalue is nonnegative, not that every matrix entry is nonnegative. The general definition is given in density matrix. ^[ref].

For this two-by-two matrix, the explicit constraint follows from its determinant, $\det \rho =p(1-p)-|c{|}^2$: with trace one, a nonnegative determinant and real eigenvalues rule out a negative eigenvalue. Its eigenvalues areThe pure states have $|c{|}^2=p(1-p)$ and eigenvalues $1,0$; the mixed states have strict inequality and two positive eigenvalues. For a normalized representative, $p=|\alpha {|}^2$ and $c=\alpha {\beta }^{\ast }$. A statistical preparation $\rho ={\sum }_j{w}_j|{\psi }_j⟩⟨{\psi }_j|$, with $w_j\ge 0$ and ${\sum }_j{w}_j=1$, averages probabilities, rather than adding probability amplitudes. Such a preparation is mixed if the positive-weight vectors do not all represent the same ray. A reduced density matrix of an entangled pair can also be mixed even when the pair has a pure state. ^[ref].

For a concrete check, the pure state $(\sqrt{3}|0⟩+|1⟩)/2$ and the mixed state obtained by preparing $|0⟩$ with probability $3/4$ and $|1⟩$ with probability $1/4$ haveBoth have trace one and the same computational-basis measurement probabilities. Their eigenvalues are respectively $(1,0)$ and $(3/4,1/4)$; the respective values of their purity of a density operator $\mathrm{Tr}({\rho }^2)$ are $1$ and $5/8$. Their Hadamard basis plus probabilities are respectively $1/2+\sqrt{3}/4\approx 0.9330$ and $1/2$. By contrast, setting $p=3/4,c=1/2$ would give determinant $-1/16$ and eigenvalue $(2-\sqrt{5})/4<0$: normalized diagonal entries alone do not make a valid qubit state.

The same constraints become especially beautiful in the Bloch ball representation. With the Pauli matrices $\mathbf{\sigma }=(X,Y,Z)$ and the identity operator $I$,The eigenvalues are $(1\pm |\mathbf{r}|)/2$, so the Bloch sphere is precisely the pure state boundary, and $I/2$ is the maximally mixed centre. For each axis $j$, the Born rule gives $P_j(\pm )=(1\pm r_j)/2$. This unifies a geometric picture, matrix positivity, and experimental frequencies in one description. For the equatorial family in the phase diagram, $p=1/2,c=e^{-i\varphi }/2$ and $\mathbf{r}=(\cos \varphi ,\sin \varphi ,0)$; replacing it by $I/2$ erases the transverse components and every fringe. See quantum state tomography for how independent ensembles measured along the three axes estimate these components. ^[ref].

The convex hull of the six Pauli eigenstates in the Bloch ball. A mixture has a Bloch vector obeying the displayed inequality. Conversely, assign weights $|s_x|,|s_y|,|s_z|$ to the corresponding signed vertices and distribute the leftover weight over any opposite pair; this realizes every vector satisfying the inequality. Single-qubit Clifford gates permute its vertices, while a T gate can take a vertex to a pure state outside it.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.