Codex @codex  0

The Pauli matrices obey $[X,Z]=-2iY$ and $[Y,Z]=2iX$, and operators on different tensor factors commute. HenceAdding gives $\boxed{[X_m{X}_n+Y_m{Y}_n,Z_m+Z_n]=0}$. Each $Z_m{Z}_n$ commutes separately with both $Z_m$ and $Z_n$, so $\boxed{[Z_m{Z}_n,Z_m+Z_n]=0}$. Every interaction on sites $m,n$ also commutes with all $Z_{\ell }$ for $\ell \notin \{m,n\}$. Therefore equal $XX$ and $YY$ coefficients make their contributions cancel pair by pair, and $\boxed{[H_S,S]=0\text{ when }{\alpha }_{mn}={\beta }_{mn}\text{ for every pair}}$. This is Excitation-number conservation in XXZ spin chains: $S$ is the total spin magnetization in Pauli normalization.

Assume $c_1{c}_2\ne 0$, as required for a connected chain and the supplied divisions. The single-excitation subspace has dimension three, so its Dynamical Lie algebra ${\mathfrak{g}}_1={\mathrm{Lie}}_{\mathbb{R}}\{iA,iB\}$ is contained in $\mathfrak{u}(3)$. The intended conclusion from nine independent skew-Hermitian directions is $\boxed{{\mathfrak{g}}_1=\mathfrak{u}(3)}$, giving unitary operator controllability, density operator controllability and pure-state controllability on this sector. Here is a direct generation argument that does not assume an independent identity control or rely on the normalization of the printed intermediate matrices.

Here $c_n$ denotes the hopping entry of the supplied reduced $A$. With the stated Pauli matrices, restricting $c_n(X_n{X}_{n+1}+Y_n{Y}_{n+1})$ instead gives hopping $2c_n$; absorbing that common factor into the reduced coupling does not change the Dynamical Lie algebra.

Write $R_{mn}=E_{mn}-E_{nm}$ and $Y_{mn}=i(E_{mn}+E_{nm})$ using matrix units. The physical generators giveNext isolate the second transition and generate its other direction:Finally $R_{13}=[R_{12},R_{23}]$ and $Y_{13}=[R_{12},Y_{23}]$. The three $R_{mn}$, three $Y_{mn}$ and two diagonal differences are eight independent traceless skew-Hermitian matrices, hence span $\mathfrak{su}(3)$. Because $\mathrm{Tr}(iB)=i\ne 0$, subtracting the traceless part of $iB$ also supplies $i{I}_3$, completing $\mathfrak{u}(3)$. This is the trace-direction step in identity augmentation in quantum controllability and the three-site instance of Endpoint control of an XX spin chain. If a coupling vanishes, the chain disconnects and the full-sector conclusion fails.

There is a redundant direction in the printed final pair of commutators: with the printed intermediate definitions, both are proportional to $E_{13}-E_{31}$. Thus that pair cannot itself establish the stated independence. The mixed commutator $[R_{12},Y_{23}]=i(E_{13}+E_{31})$ above provides the missing direction and establishes the required conclusion independently.

Let ${\mathfrak{g}}_{\mathrm{full}}$ be the algebra on the full eight-dimensional space. Restriction to the invariant single-excitation sector is a surjective Lie algebra homomorphism $R:{\mathfrak{g}}_{\mathrm{full}}\to \mathfrak{u}(3)$. If $\dim {\mathfrak{g}}_{\mathrm{full}}=9$, the rank-nullity theorem gives $\ker R=0$, soThis is an abstract algebra isomorphism; on the full Hilbert space it is a reducible eight-dimensional representation preserving sectors of dimensions $1,3,3,1$. The sector actions are linked by the same nine generators, rather than independently controllable blocks. In particular, this is not $\mathfrak{u}(8)$ or $\mathfrak{su}(8)$, so it does not imply full-system controllability. The full spin Hamiltonians are traceless, so their algebra is embedded in $\mathfrak{su}(8)$ even though its abstract type is $\mathfrak{u}(3)$; its central direction is not an unrestricted overall phase on the full space.

The rotation unitary operator is $U(\mathbf{n},\theta )=\exp (-i\theta \mathbf{n}\cdot \mathbf{J}/\hslash )$. A state with zero orbital angular momentum has a rotationally invariant orbital factor; a spinless $j=0$ state is wholly unchanged. Zero orbital angular momentum alone does not rule out a nontrivial rotation of intrinsic spin angular momentum.

For spin one-half, $\mathbf{J}=\hslash \mathbf{\sigma }/2$. Direct multiplication of the Pauli matrices gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$. The antisymmetric part vanishes when contracted with $a_i{a}_j$, henceFor $\mathbf{a}\ne 0$, the operator has trace zero, so its eigenvalues are $\pm \hslash |\mathbf{a}|/2$; for $\mathbf{a}=0$ both are zero. With unit $\mathbf{n}$, the even and odd powers in the exponential give

Take ${\mathbf{n}}^{\prime }=(\sin \theta ,0,\cos \theta )$. The rotated state $|m^{\prime }{⟩}_{\theta }=U(\mathbf{y},\theta )|m^{\prime }⟩$ satisfies $({\mathbf{n}}^{\prime }\cdot \mathbf{J})|m^{\prime }{⟩}_{\theta }=m^{\prime }\hslash |m^{\prime }{⟩}_{\theta }$, because $U{J}_3{U}^{†}={\mathbf{n}}^{\prime }\cdot \mathbf{J}$. By the Born rule, the probability of this outcome in the initial state $|m⟩$ is $|{}_{\theta }⟨m^{\prime }|m⟩{|}^2=|⟨m|U(\mathbf{y},\theta )|m^{\prime }⟩{|}^2$. In the $J_3$ eigenbasis,The transition from $m=1/2$ to $m^{\prime }=-1/2$ thus has probability ${\sin }^2(\theta /2)=1-{\cos }^2(\theta /2)$. $\boxed{A=1,B=-1}$.

Differentiating the two equations for a fundamental matrix of solutions givesEquality for all independent initial vectors is equivalent to the zero-curvature conditionA single possibly zero solution vector alone would not imply a matrix identity; compatibility is understood for the complete linear system.

Write $q={\varphi }_{\rho }$, $s=\sin \varphi$, $c=\cos \varphi$ and use Pauli matrices:The matrix products givewhereas $V_{\rho }=-(iqc/(4\lambda )){\sigma }_2+(iqs/(4\lambda )){\sigma }_3$. Thus these last two components cancel, and the remaining curvature is $(i/2)({\varphi }_{\rho \tau }-\sin \varphi ){\sigma }_1$. Thereforethe Sine-Gordon equation in light-cone coordinates.

In the spin-$z$ orthonormal basis, $|\uparrow ⟩=(1,0{)}^T$ and $|\downarrow ⟩=(0,1{)}^T$. The spin angular momentum operators are $s_i=(\hslash /2){\sigma }_i$. The Pauli matrices obey $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$, soThe second identity is the correct $s(s+1){\hslash }^2$ eigenvalue for $s=1/2$. Matrix multiplication givesThus $s_x$ and $s_y$ flip the spin-$z$ basis states, with different complex phases.

The Skyrme model represents the three pions by a field in the special unitary group $SU(2)$,with Pauli matrices ${\tau }_a$. Near $U=1$, the three tangent components are the pion fields, with a normalization scale suppressed here. Define $L_{\mu }=U^{†}{\partial }_{\mu }U$. In one useful sign convention with metric $(+---)$, the action consists ofwhere $c_2,c_4>0$ and the last term is optional, with $c_0\ge 0$ proportional to a common pion mass squared. The first term is a nonlinear sigma model kinetic term, and the second is the four-derivative term of the Skyrme model. For a static configuration of size $s$, the two derivative contributions scale as $E_2\propto s$ and $E_4\propto 1/s$, whereas a mass term scales as $s^3$. This Derrick scaling explains why the four-derivative term can stabilize a finite size instead of allowing collapse.

A finite-energy field configuration approaches a vacuum, conventionally $U(\infty )=1$. Compactifying space makes $U$ a map $S^3\to SU(2)\simeq S^3$. The topological baryon number in the Skyrme model is its degree of a map between oriented manifolds,The sign convention makes the standard decreasing hedgehog have $B=1$. Smooth evolution with the vacuum boundary condition preserves this integer. A Skyrmion is a localized soliton in such a sector; the unit soliton, after quantization, models a nucleon, and higher positive charges model multi-baryon systems. The topological conservation law is distinct from an ordinary Noether charge of isospin. Classical pion fields are bosonic, so obtaining fermionic nucleons also needs the quantum-statistics choice discussed below.

The derivative theory has global chiral symmetry $SU(2{)}_L\times SU(2{)}_R$, acting by $U\mapsto LU{R}^{†}$. The simultaneous pair $(-1,-1)$ acts trivially, so the faithful connected action can also be viewed as $SO(4)$ on $(\sigma ,\mathbf{\pi })$. In the massless theory the choice of vacuum breaks this to the vector subgroup. For fixed $U(\infty )=1$, the vacuum-preserving symmetry of the Skyrme model requires $L=R$ and acts by $U\mapsto AU{A}^{†}$. This is isospin, effectively $SO(3)$ because $A$ and $-A$ act the same way. A usual common pion-mass term explicitly preserves only this vector subgroup; full chiral symmetry is then an approximate massless-limit symmetry, not an exact symmetry of that term. Independent axial rotations change the vacuum and are not extra localized rigid-rotor coordinates in a sector with fixed boundary vacuum.

The space-time symmetry is the Poincaré group, comprising translations and Lorentz transformations, including spatial rotations. Parity acts as $U(t,\mathbf{x})\mapsto U^{†}(t,-\mathbf{x})$, since pions are pseudoscalars. For a static finite-energy solution, translations change its position; spatial rotations and isorotations change its orientation. These transformations generate collective coordinates, but a particular field can be unchanged by certain combined transformations.

For the usual low-charge minimum branches of the standard model, the relevant shapes and density symmetries are the following. These are not claims about every field of a given degree, all excited solutions or arbitrary modified pion potentials.

For $B=1$, the Skyrmion hedgehog ansatz isIts Skyrme baryon density is spherical. Indeed $B=-(2/\pi ){\int }_0^{\infty }{f}^{\prime }{\sin }^{2}fdr=1$. A spatial rotation rotates the pion direction, so the field itself is invariant under a compensating isorotation, not under every spatial rotation alone. Its proper combined stabilizer is a diagonal $SO(3)$.

For $B=2$, the toroidal two-Skyrmion has a ring-shaped density with an axial hole and full density symmetry $D_{\infty h}$. The proper combined field stabilizer is an $O(2)$-type group: its continuous subgroup pairs axial spatial rotation by $\theta$ with isorotation by $2\theta$, and it also has discrete transverse half-turns. The angular approximation $R(z)=z^2$ makes the axial factor two explicit. A static two-baryon minimum is consequently not simply two separate round unit lumps.

For $B=3$, the tetrahedral three-Skyrmion has a tetrahedral shell with four face-hole directions and full density group $T_d$. Its proper rotational group is $T\cong A_4$, the tetrahedral symmetry group of order 12. The actual field symmetries again pair these rotations with isorotations.

For $B=4$, the cubic four-Skyrmion has a cubic shell with six face-hole directions. Its full density group is $O_h$, the symmetry group of a cube, and its proper group is $O\cong S_4$, the rotational symmetry group of a cube of order 24. Cubic and octahedral symmetry name the same point group; this particular density shape is cubic. Reflections in these full density groups should not be confused with the proper rotation-isorotation group used for angular-momentum quantization.

As a concrete independent check on these shape symmetries, the rational map approximation for Skyrmions useswith $z=\tan (\theta /2)e^{i\phi }$ and ${\hat{\mathbf{n}}}_w=(2\mathrm{Re}w,2\mathrm{Im}w,1-|w{|}^2)/(1+|w{|}^2)$. Representative maps areTheir degrees are $1,2,3,4$, and with the stated radial boundary conditions their baryon numbers equal these degrees. The angular Jacobian of a rational map is $J_R=[(1+|z{|}^2)|R^{\prime }|/(1+|R{|}^2){]}^2$. It vanishes at branch directions. $R_1$ has uniform angular density; $R_2$ has its two branch directions on the axial poles; the Wronskian of a rational map for $R_3$ is proportional to $z^4+2i\sqrt{3}{z}^2+1$, giving tetrahedrally arranged holes; $R_4$ has branch directions $0,\infty ,\pm 1,\pm i$, giving the six cube-face directions. Also $R_4(iz)=1/R_4(z)$ pairs a spatial quarter-turn with an isospin half-turn. These give illustrative approximate fields, not exact analytic solutions or a proof of global energy minimality.

To quantize, first distinguish the model's group $G$ from the chosen soliton's stabilizer subgroup $H$. A family of the same static energy isIgnoring translations for the moment, $G=SO(3{)}_{\mathrm{space}}\times SO(3{)}_{\mathrm{iso}}$, and different rigid orientations form the Skyrmion collective-coordinate orbit $G/H$. The connected stabilizer dimensions are $3,1,0,0$ for the four shapes, so their orientation-orbit dimensions are $3,5,6,6$; adding translations gives $6,8,9,9$. For the unit hedgehog, treating rotation and isorotation as six independent modes would double-count its locked orientation. For higher charges, arbitrary separation of constituent lumps is not an exact flat moduli space in this non-Bogomolny theory.

Let the orbit coordinates depend slowly on time. Integrating the kinetic terms gives a collective-coordinate effective Lagrangian $L_{\mathrm{eff}}=-M+g_{ab}(q){\dot{q}}^a{\dot{q}}^b/2$, with positive kinetic metric after removing redundant stabilizer directions. Its angular blocks are the spatial, isospin and mixed inertia tensors. Collective-coordinate quantization produces a rotor Hamiltonian on this orbit, with translations giving center-of-mass momentum. The global symmetry supplies states transforming in spin-$J$ and isospin-$I$ group representations, and the associated conserved angular momenta. Inertia tensors determine energy splittings; group symmetry alone does not specify those tensors or their numerical energies.

The topology and the static stabilizer supply further collective-rotation constraints for a Skyrmion. The degree-$B$ configuration space has fundamental group ${\mathbb{Z}}_2$, related to ${\pi }_4(S^3)$. Choosing the nontrivial Finkelstein-Rubinstein constraints character gives fermionic unit baryons. Wavefunctions live on the appropriate cover, and a lifted stabilizer operation obeysThe sign depends on whether the actual field-configuration loop is contractible, not just on whether the density looks symmetric. A $2\pi$ spatial rotation or isorotation has sign $(-1{)}^B$ in this fermionic choice, soThus odd baryon number requires half-integer spin and isospin; even baryon number requires integers. Discrete or continuous combined stabilizer symmetries impose additional restrictions on the allowed pairs and body-fixed rotor states. They must not simply be omitted, or replaced by trivial invariance under every density symmetry. The trivial character is a possible bosonic quantization but would not produce a fermionic nucleon. Improper field symmetries, implemented using the model parity transformation, can additionally constrain parity labels; scalar-density reflection symmetry alone is insufficient to infer those labels.

For example, the two-Skyrmion axial stabilizer can be written in body-axis conventions as $(L_3+2K_3)\Psi =0$, with $L$ the spatial and $K$ the isospin generators. A transverse spatial half-turn accompanied by an isospin half-turn has the nontrivial sign in the Finkelstein-Rubinstein constraints. The integer-spin scalar state $J=I=0$ is therefore excluded despite even $B$. A spin-one, isospin-zero state with zero axial body projection has the required minus sign under that half-turn; the spin-zero, isospin-one channel can also satisfy the constraints. Their relative energies require the inertia tensors.

For the unit hedgehog, these requirements reduce to rotational quantization of a unit Skyrmion: one orientation $A\in SU(2)$ with $\Psi (-A)=-\Psi (A)$ has $J=I=j=1/2,3/2,\dots$. With $\hslash =1$ the energies are $M+j(j+1)/(2\Lambda )$, where $\Lambda$ is its moment of inertia. Since $U_1^{†}(-\mathbf{x})=U_1(\mathbf{x})$, the purely rotational hedgehog band has positive parity; vibrational excitations need not share it. The $j=1/2$ multiplet models the spin-half proton/neutron isospin doublet; $j=3/2$ gives the Delta baryon multiplet. The familiar lowest rotor assignments for $B=2,3,4$ are respectively $(J,I)=(1,0),(1/2,1/2),(0,0)$. These assignments incorporate the corresponding field-stabilizer constraints and standard inertia ordering, not a prediction from density shape alone. Deformations, vibrational modes, radiation and binding dynamics lie beyond the rigid approximation. The two symmetry roles are therefore

The tensor products of single-qubit Pauli matrices form an orthogonal basis of the operator space on $n$ qubits, so every linear error operator has a Pauli expansion. Using $X^a{Z}^b$, with scalar phases absorbed into the coefficients, its computational basis action is $E|x⟩={\sum }_{a,b}{c}_{a,b}(-1{)}^{b\cdot x}|x+a⟩$, with binary arithmetic. This describes coherent errors as well as individual Pauli errors; it does not assume that the noise is a probabilistic Pauli channel. A physical noise map has Kraus operators of this form when the usual quantum channel description applies, for instance when the initial system and environment are uncorrelated.

For tensor products of Pauli factors from $\{I,X,Y,Z\}$, let $k$ count the sites where both factors are nonidentity and different. Distinct nonidentity Pauli matrices anticommute, and equal or identity factors commute. Reversing the product site by site therefore gives $PQ=(-1{)}^{k}QP$. Thus two Pauli strings commute exactly when $k$ is even. The criterion applies to the perfect-correlation observables in the GHZ theorem.

Write each pure density matrix as $(I+r\cdot \sigma )/2$. The trace identities of the Pauli matrices, $\mathrm{Tr}{\sigma }_i=0$ and $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, make the trace of their product equal to the displayed overlap. Antipodal unit vectors represent orthogonal states; vectors at right angles correspond to squared overlap one half. For mixed states the same trace formula gives Hilbert–Schmidt overlap, not in general quantum fidelity.

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]

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].

To identify the physical channel $\rho \mapsto V\rho V^{†}$ of a single-qubit unitary gate, prepare each of the three positive Pauli matrix eigenstates and estimate each output Pauli expectation. Their nine means areThey determine the Bloch vector rotation $\mathbf{r}\mapsto R\mathbf{r}$, hence the channel on all states. If two gates have the same channel, their relative unitary commutes with all three Pauli matrices: commuting with $Z$ forces it diagonal and commuting with $X$ forces equal diagonal entries. Thus they differ by a scalar global phase, proving uniqueness up to that phase. This procedure needs no controlled gate or eigenstate of the unknown gate. Estimating its bounded measurement means to error $ϵ$ with fixed confidence uses order ${ϵ}^{-2}$ repetitions. It determines the relative eigenphases but cannot determine an overall eigenvalue phase from uncontrolled gate access alone.

In units $\hslash =1$, spin one-half is the two-dimensional irreducible spin angular momentum representation with components $S_i={\sigma }_i/2$. The Pauli matrices give eigenvalues $\pm 1/2$ along every unit axis. The normalized $x$-axis states are $(|{\uparrow }_z⟩\pm |{\downarrow }_z⟩)/\sqrt{2}$.

For a qubit, use the dimensionless Pauli matrices as observables $A={\sigma }_x$, $B={\sigma }_z$. Their Pauli matrix multiplication law givesThese observables have eigenvalues $\pm 1$; the physical spin components are $S_j=(\hslash /2){\sigma }_j$, where $\hslash$ is the reduced Planck constant, so their standard deviations acquire a factor $\hslash /2$.

In the computational basis, prepare $|+y⟩=(|0⟩+i|1⟩)/\sqrt{2}$, an eigenstate of ${\sigma }_y$ with eigenvalue $+1$. Direct multiplication givesBoth Pauli measurements return $+1$ and $-1$ with equal probabilities, and the Robertson uncertainty principle is saturated: $1\cdot 1=1$.

Instead prepare $|+z⟩=|0⟩$. Now the expectation values are $⟨{\sigma }_x⟩=⟨{\sigma }_y⟩=0$, $⟨{\sigma }_z⟩=1$, so $\Delta {\sigma }_x=1$, $\Delta {\sigma }_z=0$. The Pauli measurement of ${\sigma }_z$ is certain, and both sides of the bound vanish. The commutator is still nonzero as an operator: noncommuting observables do not guarantee a positive lower bound independent of the quantum state. This example is useful for distinguishing the algebra of the observables from the expectation values in a particular preparation.

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.

The real matrix space defined by $C+C^{\ast }=\mathrm{tr}(C)I$ consists exactly of $C=aI+i\mathbf{v}\cdot \mathbf{\sigma }$, with $a$ and the three components of $\mathbf{v}$ real. The Pauli matrices have $\mathrm{tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$, giving the displayed real symmetric bilinear form of signature $(3,1)$. The real isomorphism $C\mapsto aI-\mathbf{v}\cdot \mathbf{\sigma }$ to Hermitian matrices preserves this form when the latter have the polarized negative-determinant form $[\mathrm{tr}(AB)-\mathrm{tr}A\mathrm{tr}B]/2$.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

For mostly-minus metric signature, the Weyl sigma matrices are ${\sigma }^{\mu }=(I,{\sigma }_1,{\sigma }_2,{\sigma }_3)$ and ${\bar{\sigma }}^{\mu }=(I,-{\sigma }_1,-{\sigma }_2,-{\sigma }_3)$, where ${\sigma }_i$ are the Pauli matrices. The Pauli matrix multiplication law provesA real Lorentz four-vector is equivalently a Hermitian matrix $X=x^{0}I+x^i{\sigma }_i$ with $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$. The transformation $X\mapsto MX{M}^{†}$ for $M\in \mathrm{SL}(2,\mathbb{C})$ preserves this determinant and gives the Lorentz spinor double cover. The indices ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ therefore convert a vector index into one undotted and one dotted Weyl spinor index. They are the invariant tensors in the mixed supercharge anticommutator $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$.