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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 independent resource pairs have density matrices $v_j|{\Phi }^{+}⟩⟨{\Phi }^{+}|+(1-v_j)I_4/4$. An ideal complete Bell-basis measurement, followed by outcome-dependent Pauli gates, produces a chosen Bell state with white-noise visibility $v_1{v}_2$: either white-noise input makes the remote pair maximally mixed. If the reported label is reliable with probability $q$ and otherwise uniformly random, averaging the erroneous corrections makes that branch maximally mixed too, so $v_{\mathrm{out}}=v_1{v}_{2}q$. A second independent detector of the same quality during quantum teleportation adds another factor $q$ to the teleported Bloch vector.

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

Two positive rank-one effects summing to $I$ in dimension two must be complementary orthogonal projections. Write the first as $\alpha |\psi ⟩⟨\psi |$ with normalized $\psi$. The second has eigenvalues $1-\alpha$ and $1$, so its rank-one condition forces $\alpha =1$. The two unit Bloch vectors are antipodal, and this POVM is a projective measurement.

The Bloch sphere is the unit sphere of Bloch vectors, corresponding to pure qubit states through $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$. A local unitary change of qubit basis induces a rotation of this sphere, with its common phase physically irrelevant. Mixed states occupy the interior of the unit ball.

Applying the amplitude damping channel to the Bloch vector form of a qubit density operator gives an affine map. Transverse components contract by $\sqrt{1-p}$; the longitudinal component contracts by $1-p$ and shifts toward the ground-state north pole. Reading the entries of the output density operator proves the map. The displacement distinguishes this channel from centered depolarization.

For $\mathbf{d}\ne 0$, put $\hat{\mathbf{d}}=\mathbf{d}/|\mathbf{d}|$. The lower-band orthogonal projector is $P_{-}=(I-\hat{\mathbf{d}}\cdot \mathbf{\sigma })/2$. Applying the Pauli matrix multiplication law to $i\mathrm{Tr}(P_{-}[{\partial }_{k_x}{P}_{-},{\partial }_{k_y}{P}_{-}])$ givesThus the Chern number is the signed number of times the Brillouin zone covers a two-sphere under $\hat{\mathbf{d}}$. The lower energy eigenstate's actual Bloch vector is $-\hat{\mathbf{d}}$. The Berry curvature has the displayed positive sign for this lower band and the stated Berry connection; switching to the upper band changes its sign.

A useful local calculation exposes what each valley contributes. Write $p=\hslash v|\mathbf{q}|$. The linear Dirac Hamiltonian givesEach unbounded continuum cone gives a half-integer in this calculation. It is not a separately realizable half-integer Chern number of an isolated lattice energy band. The full honeycomb lattice supplies both valleys and the large-momentum completion. Starting with a trivial large-$|M|$ state, each single-valley mass crossing changes the lattice integer by the corresponding unit. There are no other band touchings. This establishes the exact lattice resultfor $t_2>0$, excluding the boundaries. This sign belongs to our explicit bond vectors, orientation and occupied-band convention.

For an original worked example, choose $t_1=1$, $t_2=0.08$, $\varphi =\pi /2$ and $M=0$. Then $\mu \simeq 0.415692$, $(m_{+},m_{-})=(-\mu ,+\mu )$, both valley gaps are $2\mu \simeq 0.831384$, and $C=-1$. Choosing instead $t_2=0$, $M=\mu$ gives exactly the same valley-gap magnitudes but $C=0$. The first state's two valley contributions reinforce; the second state's cancel.

An independent eigenvector-overlap calculation on periodic $31\times 31$, $61\times 61$ and $121\times 121$ momentum grids gives $C=-1$ for the first example and $C=0$ for the second. Reversing $\varphi$ in the first example gives $C=+1$. Direct integration of the analytic lattice Berry curvature agrees. For the overlap calculation one can remove the embedding phase from $f$, replacing it by $1+e^{i\mathbf{k}\cdot {\mathbf{a}}_1}+e^{i\mathbf{k}\cdot {\mathbf{a}}_2}$; this makes the cell-basis Hamiltonian periodic and prevents a false boundary-seam contribution. The local Berry curvature can depend on this basis convention; the integrated Chern number agrees. All mass-zero parameters are excluded before evaluating any integer.

The displayed qubit quantum channel is completely positive for $0\le q\le 4/3$. Indeed its Pauli channel probabilities are $1-3q/4$ for the identity and $q/4$ for each nonidentity Pauli operator. In the range $q>1$ the coefficient $1-q$ is negative, so the displayed expression is not a convex mixture of the unchanged and maximally mixed states, even though the Pauli representation is a valid random unitary channel. Its Bloch vector is contracted and reversed by the factor $1-q$. The Holevo capacity of a qubit depolarizing channel remains $1-h_2(q/2)$ throughout this interval.

Choose an isolated pair of energy levels, with ground state $|g⟩=(0,1{)}^T$ and excited state $|e⟩=(1,0{)}^T$. Use the Pauli matrices in the ordered basis $(|e⟩,|g⟩)$, so ${\sigma }_z=\mathrm{diag}(1,-1)$. After subtracting the mean energy, prescribe the laboratory Hamiltonian operatorThe unperturbed energy gap is $\hslash {\omega }_0$, the drive has angular frequency $\omega$, and $\varphi$ fixes its oscillation phase at the chosen time origin. All of ${\omega }_0$, $\omega$, the Rabi angular frequency $\Omega$ and the Rabi detuning $\delta$ are in radians per second; ordinary frequency is angular frequency divided by $2\pi$. In particular, $\Omega$ here is the coefficient multiplying the cosine in $H_{\mathrm{lab}}/\hslash$, not twice that coefficient.

Write $|{\psi }_{\mathrm{lab}}⟩=R(t)|{\psi }_{\mathrm{rot}}⟩$ with $R(t)=\exp (-i\omega t{\sigma }_z/2)$. Transforming the Schrödinger equation gives $H_{\mathrm{rot}}=R^{†}{H}_{\mathrm{lab}}R-i\hslash R^{†}\dot{R}$. The rotating-wave approximation drops terms at angular frequency $2\omega$, leavingThis approximation requires $\Omega ,|\delta |\ll {\omega }_0,\omega$, and an interaction duration over which omitted corrections remain small. It also assumes that other energy levels can be neglected. Resonance alone does not justify either assumption. An exactly co-rotating drive has no discarded counter-rotating component; the linearly oscillating laboratory drive above does.

The Pauli matrix multiplication law gives $K^2={\Omega }_{\mathrm{eff}}^{2}I$. Separating even and odd terms of the matrix exponential therefore proveswith the continuous limit $U=I$ when $\Omega =\delta =0$. Starting from $|g⟩$, the upper probability amplitude is $-i{e}^{-i\varphi }(\Omega /{\Omega }_{\mathrm{eff}})\sin ({\Omega }_{\mathrm{eff}}t/2)$. The Born rule givesThe rotating transformation only changes the quantum phases of these basis states, so their laboratory populations are the same. Rabi detuning tilts the rotation axis of the Bloch vector: it increases the oscillation rate to ${\Omega }_{\mathrm{eff}}$ but caps the transferred population at ${\Omega }^2/({\Omega }^2+{\delta }^2)$. A fixed drive oscillation phase cancels from this single-pulse population when the initial state is $|g⟩$. It still fixes the relative quantum phase of a resulting quantum superposition and matters when another pulse supplies a phase reference.

For equiprobable qubit states with Bloch vectors $r,s$, their density-matrix difference is $(r-s)\cdot \sigma /2$, with eigenvalues $\pm |r-s|/2$. An effect $0\le E\le I$ declaring the first state yields success $1/2+\mathrm{Tr}[E({\rho }_r-{\rho }_s)]/2$. Selecting its positive-eigenvalue eigenspace maximizes this trace and proves the displayed result. Measure along $r-s$ and attach the corresponding state labels; if the vectors coincide, no measurement improves on guessing.

Choose a Hilbert-Schmidt inner product orthonormal Hermitian operator basis with final element $I/\sqrt{N}$. A density operator has real coefficients $r_k=\mathrm{Tr}({\sigma }_k\rho )$ and fixed final coefficient $1/\sqrt{N}$. The remaining coefficients form its generalized Bloch vector. Positivity restricts these real vectors to a compact convex body; for $N>2$ this body is not the entire Euclidean ball allowed by a purity bound.

For $P_{\alpha }=\mathrm{diag}(1,e^{i\alpha })$, conjugation by the Hadamard gate turns a relative quantum phase into an $x$-axis Bloch vector rotation. In the convention $\rho =(I+\mathbf{P}\cdot \mathbf{\sigma })/2$, the map is $(P_x,P_y,P_z)\mapsto (P_x,P_y\cos \alpha -P_z\sin \alpha ,P_y\sin \alpha +P_z\cos \alpha )$. An $x$-polarized input is unchanged, whereas a $z$-polarized input gives a sinusoidal population difference. A maximally mixed input remains maximally mixed under every unitary time evolution.

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.

Every Hermitian matrix of size two can be written uniquely as $M=dI+b_x{\sigma }_x+b_y{\sigma }_y+b_z{\sigma }_z$ with real coefficients. For $M=\left(\begin{array}{cc}u& w\\ w^{\ast }& z\end{array}\right)$, take $d=(u+z)/2$, $b_z=(u-z)/2$, $b_x=\mathrm{Re}w$ and $b_y=-\mathrm{Im}w$. Alternatively, $d=\mathrm{tr}M/2$ and $b_i=\mathrm{tr}(M{\sigma }_i)/2$ follow from the trace orthogonality of the Pauli matrices and the identity matrix.

The Pauli matrix multiplication law gives $(\mathbf{b}\cdot \mathbf{\sigma }{)}^2=|\mathbf{b}{|}^{2}I$, hence the two eigenvalues are $d\pm |\mathbf{b}|$. For $\mathbf{b}\ne 0$, the corresponding pure-state Bloch vectors are $\pm \mathbf{b}/|\mathbf{b}|$.

With Bloch vector retention $\eta \to 0$, the unassisted product-state capacity of a quantum depolarizing channel is ${\eta }^2/(2\ln 2)+O({\eta }^4)$. Its entanglement-assisted capacity is $3{\eta }^2/(2\ln 2)+O({\eta }^3)$, obtained by expanding Shannon entropy around the uniform four-outcome distribution. Consequently their limiting ratio is three, independent of the linear noise-parameter convention.

A POVM is informationally complete if its outcome probabilities determine every input density operator uniquely. Its effects must span the real space of Hermitian operators. For a qubit, normalization leaves at most $m-1$ independent probabilities from $m$ outcomes; an arbitrary Bloch vector has three real parameters, so one fixed informationally complete measurement requires at least four outcomes. Using several differently oriented binary measurements is a different experimental setting and can also perform quantum state tomography.

Different ensemble members accumulating different deterministic phases can shorten their averaged Bloch vector even when each member remains in a pure state. Averaging static frequency offsets gives an effective dephasing channel. It need not represent irreversible phase loss in each individual system; an appropriate refocusing pulse can reverse static offsets. Changing offsets during the sequence requires a temporal noise model.

A polarizing beam splitter converts a photon polarization qubit to two paths. Rotating one arm's photon polarization to match the other and recombining at a balanced beam splitter gives $P_1(\theta )=1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$. Measurements at phases zero and $\pi /2$ recover the transverse Bloch vector components; direct counts in the two separated arms recover the population difference. All three components, rather than a fringe alone, determine an arbitrary photon polarization density matrix.

// Target: quantum-theory.bigb

This one-parameter family maps every Bloch vector $\mathbf{s}$ to $r\mathbf{s}$. It is unital, trace preserving and Hermiticity preserving for every real $r$. It is a positive map exactly when $|r|\le 1$, because the output eigenvalues of a state are $(1\pm r|\mathbf{s}|)/2$. Its Choi matrix has eigenvalues $(1+3r)/2$ and $(1-r)/2$ with multiplicities one and three in the trace-two convention, so Choi's theorem on completely positive maps gives complete positivity exactly for $-1/3\le r\le 1$. In that range the Kraus representation is a mixture of conjugations by $I,X,Y,Z$ with probabilities $(1+3r)/4,(1-r)/4,(1-r)/4,(1-r)/4$. The physically allowed maps are qubit depolarizing channels, including a limited range of negative Bloch scaling.

For $H=-\gamma B_z{S}_z-\gamma B_x{S}_x\cos (\omega t)$, choose ${\omega }_z=\gamma B_z>0$ and nominal resonance $\omega ={\omega }_z$. Starting from up spin, the rotating-wave approximation gives the normalized Bloch vector in the title, with signed $\Omega =\gamma B_x/2$. It follows by transforming the spinor $(\cos (\Omega t/2),i\sin (\Omega t/2){)}^T$ from the interaction picture back to the Schrödinger picture. Thus weak transverse driving combines slow meridional Rabi oscillations with fast Larmor precession.

This result requires $|B_x|\ll |B_z|$ and cannot be extrapolated as a controlled strong-drive formula. The exact equation is $\dot{\mathbf{s}}=\gamma \mathbf{s}\times (B_x\cos \omega t,0,B_z)$. For $|\gamma B_z|t\ll 1$, neglecting the bias gives the controlled leading solutionThe fixed-axis driving generators commute at all times, making this solution exact when the bias is zero at any prescribed drive frequency. At nominal resonance in a strong-drive experiment, the initial rotation scale is $\Theta \simeq \gamma B_{x}t$, twice the incorrectly extrapolated rotating-wave rate. Bias effects must be retained on longer times.