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For a qubit density matrix $\rho =(I+n\cdot \sigma )/2$ with nonzero Bloch vector, $\hat{n}=n/\Vert n\Vert$ and the multiplication rule for the Pauli matrices give these orthogonal rank-one spectral projectors, with eigenvalues $(1\pm \Vert n\Vert )/2$. Their Bloch vectors are opposite unit vectors $\pm \hat{n}$. At $n=0$ the state is maximally mixed and the direction is undefined; any orthonormal eigenbasis gives an antipodal pair, but no pair is distinguished by the state.

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

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 $n_1,n_2$.

Both the drift and local $Z_k$ control commute with $S$, so $[H_S+f(t)Z_k,S]=0$ at every time. The propagator consequently obeys $U(t{)}^{†}SU(t)=S$, as follows by differentiating this expression. Thus every reachable unitary operator preserves the magnetization sectors of a spin chain. If $P_q$ is a spectral projector of $S$, then $U$ commutes with $P_q$ and $\mathrm{Tr}(P_q\rho (t))=\mathrm{Tr}(P_q\rho (0))$.

The eigenspaces are nontrivial proper subspaces: $|0\cdots 0⟩$ has eigenvalue $N$, while $|10\cdots 0⟩$ has eigenvalue $N-2$. One cannot transfer the first pure state to the second, although their density operators have the same spectrum. Nor can one implement a unitary that mixes these two sectors. Hence $\boxed{\text{the full system has neither operator, density, nor pure-state controllability}}$. Its Dynamical Lie algebra lies in the block-diagonal commutant of $S$. Controllability within a fixed sector, such as the single-excitation subspace, is still possible and must be tested separately.

Taking $\mathbf{n}$ along a coordinate axis gives ${\sigma }_i^2=I$. Taking $\mathbf{n}=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ gives $\{{\sigma }_i,{\sigma }_j\}=0$. Combining this anticommutator relation with the specified commutator givesand contraction with $a_i{b}_j$ provesThe operators for spin angular momentum, $S_i=\hslash {\sigma }_i/2$, satisfy $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$ and $S^2=3{\hslash }^{2}I/4$, identifying spin $1/2$. The usual irreducible spin space is two-dimensional; the assumptions also allow repeated copies of it if no Hilbert-space dimension is specified.

Since $\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian and squares to $I$, $P_{\pm }=(I\pm \mathbf{n}\cdot \mathbf{\sigma })/2$ are Hermitian, $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. They are its spectral projectors. WriteEach nonzero term is a scalar multiple of a normalized eigenstate with eigenvalue $\pm 1$. The Born rule gives, for normalized $|\chi ⟩$,If $(\mathbf{m}\cdot \mathbf{\sigma })|\chi ⟩=|\chi ⟩$, expectation of $\{\mathbf{m}\cdot \mathbf{\sigma },\mathbf{a}\cdot \mathbf{\sigma }\}=2\mathbf{m}\cdot \mathbf{a}$ gives $⟨\mathbf{a}\cdot \mathbf{\sigma }⟩=\mathbf{m}\cdot \mathbf{a}$. Taking $\mathbf{a}=\mathbf{n}$ yields $\boxed{\mathbb{P}(\text{up along }\mathbf{n})=\frac{1}{2}(1+\mathbf{n}\cdot \mathbf{m})}$.

The spectral projector for outcome $+1$ isFor $|1⟩=(0,1{)}^T$, the Born rule gives

Since $P|\psi ⟩={\lambda }_P|\psi ⟩$, $P^2=I$, and $P$ anticommutes with $Q$,so the expectation value of $Q$ is zero. The two spectral projectors of the Hermitian Pauli operator $Q$ are $(I\pm Q)/2$, and hence the Born rule gives

The variational characterization of trace distance turns a matrix distance into the best possible binary experiment. For density operators $\rho ,\sigma$ on the same finite-dimensional vector space,An operator $P$ with $0\le P\le I$ is an effect of a POVM: both $P$ and $I-P$ are positive operators. The Born rule assigns the first outcome probabilities $\mathrm{Tr}(P\rho )$ and $\mathrm{Tr}(P\sigma )$. Their signed gap is $\mathrm{Tr}(P\Delta )$, where $\Delta =\rho -\sigma$. Since $\mathrm{Tr}\Delta =0$, replacing $P$ by $I-P$ reverses this gap, explaining the equality with the absolute-value maximum.

One optimal $P$ is the spectral projector $P_{+}$ onto the positive eigenspaces of $\Delta$. In the spectral decomposition of $\Delta$, this choice includes all positive eigenvalues and excludes all negative eigenvalues. Their positive sum equals the trace distance by the spectral-parts formula for trace distance. Any zero eigenspace can be assigned to either outcome without changing the gap. This is the effect selected by the equal-prior Holevo–Helstrom theorem. ^[ref]

The same optimum has a classical description. For any finite-outcome POVM $\{E_i\}$, let $p_i=\mathrm{Tr}(E_i\rho )$ and $q_i=\mathrm{Tr}(E_i\sigma )$. ThenThe expression $\frac{1}{2}{\sum }_i|p_i-q_i|$ is the total variation distance of the entire outcome probability distributions; the expression for $A$ is the gap for one event. Grouping outcomes in $A$ gives the single effect $P_A={\sum }_{i\in A}{E}_i$. Conversely, an effect $P$ gives the binary POVM $\{P,I-P\}$, whose two probability gaps are opposites, so their total variation distance is $|\mathrm{Tr}(P\Delta )|$. Thus the optimum over all POVMs is already attained by one binary projective measurement. ^[ref]

For quantum state discrimination with equal prior probabilities, declare $\rho$ on outcome $P$ and $\sigma$ on outcome $I-P$. Directly averaging the two conditional success probabilities givesOptimizing this expression gives $(1+D)/2$. For the diagonal example in trace distance, $P=|0⟩⟨0|$ gives a gap $0.8-0.3=0.5$. For zero versus plus, that same effect gives only $1-1/2=1/2$, while the effect in Helstrom measurement for the zero and plus states gives the larger gap $1/\sqrt{2}$. The trace-distance-preserving binary measurement depends on the pair of preparations; a fixed measurement basis does not generally attain the trace distance for every pair.