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A qubit depolarizing channel shrinks every Bloch vector by the same factor: ${\Lambda }_p(\rho )=p\rho +(1-p)I/2$. In this retention-parameter convention it is completely positive for $-1/3\le p\le 1$; the usual mixture of unchanged and fully randomized states uses $0\le p\le 1$.

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.

Compare the qubit pure statewith $|0⟩$, using the computational basis. Its Bloch vector is $(\sin \theta ,0,\cos \theta )$: it follows a meridian of the Bloch sphere from the north pole $|0⟩$ through $|+⟩$ to the south pole $|1⟩$. The inner product with $|0⟩$ is $\cos (\theta /2)$, nonnegative on this interval. Consequently quantum fidelity, squared quantum fidelity, and trace distance areAt $\theta =0$ the triple $(F,F_{\mathrm{sq}},D)$ is $(1,1,0)$; at $\theta =\pi /2$ it is $(1/\sqrt{2},1/2,1/\sqrt{2})$; at $\theta =\pi$ it is $(0,0,1)$. The angle between the Bloch vectors is $\theta$, while the quantum fidelity uses its half-angle. The pure-state trace distance and fidelity identity becomes the familiar relation ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$.

For equal prior probabilities, the Holevo–Helstrom theorem gives optimal quantum state discrimination success $P_{\mathrm{succ}}=(1+D)/2$. Along this meridian it rises from $1/2$ to one, with $(1+1/\sqrt{2})/2\approx 0.8536$ at $|+⟩$. This connects the geometric trace distance to the best possible decision from a single preparation; it is a different measurement task from the target-acceptance probability defining squared quantum fidelity.

// Target: quantum-information-theory.bigb

The ensemble Bloch vector of a two-level quantum state. With $\rho =(I+\mathbf{P}\cdot \mathbf{\sigma })/2$, full polarization has $|\mathbf{P}|=1$ and is pure, while zero polarization has $\rho =I/2$ and is maximally mixed. This internal-state ensemble polarization differs from the polarization of a light wave used to drive it.

Choose the Pauli matrices with ${\sigma }_z=\mathrm{diag}(1,-1)$, and let $\omega >0$ be an angular frequency. The computational basis is an orthonormal basis of energy eigenstates for the qubit Hamiltonian operatorIn this convention $|1⟩$ is the lower energy eigenstate; the labels describe a basis choice, not a universal ground state/excited state ordering. Start in the Hadamard basis pure state $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. ThenThe first factor in the second expression is a global phase. The ratio of the two probability amplitudes has relative quantum phase $\omega t$, so its sign follows directly from $E_1-E_0=-\hslash \omega$.

Multiplying this density matrix by each Pauli matrix and taking its trace gives the Bloch vector:The Bloch vector travels counterclockwise around the equator of the Bloch sphere, from $+x$ toward $+y$ when viewed from $+z$. A computational-basis measurement always has $P_Z(0)=P_Z(1)=1/2$, while the Hadamard basis and ${\sigma }_y$ eigenbasis measurements giveAt $\omega t=\pi /2$, the quantum state is $|+i⟩=(|0⟩+i|1⟩)/\sqrt{2}$ up to a global phase, giving $P_Y(+)=1$. At $\omega t=\pi$, it is $|-⟩$ up to a global phase, giving $P_X(+)=0$. Both retain equal energy measurement probabilities.

At $t=2\pi /\omega$, the time-evolution operator is $U(t)=-I$, so $|\psi (t)⟩=-|+⟩$. The physical quantum state ray, its density operator, and every measurement's probabilities have returned. The representative vector returns exactly only at $t=4\pi /\omega$, when $U(t)=I$. The minus sign is a global phase for this isolated evolution; comparing an evolved branch with an unevolved branch through a controlled unitary gate can make it a relative quantum phase, as explained in global phase of an uncontrolled quantum gate is unobservable.

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 Rabi detuning is the unperturbed transition angular frequency minus the drive angular frequency in this convention. Its sign controls the longitudinal rotation of the Bloch vector; the single-pulse population formula is even in $\delta$. Reversing the definition of Rabi detuning requires reversing its sign in the rotating-frame Hamiltonian operator as well.

For a rotating-frame two-state Hamiltonian operator $H=\hslash \mathbf{W}\cdot \mathbf{\sigma }/2$, the Bloch vector obeys $\dot{\mathbf{R}}=\mathbf{W}\times \mathbf{R}$. Thus $\mathbf{W}$ specifies a rotation axis and angular speed. With detuning defined as drive frequency minus transition frequency and the lower state at positive $R_z$, a phase-zero transverse drive can give $\mathbf{W}=(\Omega ,0,\delta )$. Changing basis or detuning convention requires changing the corresponding signs consistently.

Ramsey interferometry uses two coherent pulses with a free interval between them: the first creates a quantum superposition, the interval accumulates relative quantum phase, and the second converts that quantum phase into measurement probabilities. Its attraction is that an ordinary population readout can reveal the quantum coherence hidden from an energy measurement during the free interval.

Here is an ideal qubit model with an explicit pulse convention. Let $R_y(\theta )=\exp (-i\theta {\sigma }_y/2)$, prepare $|0⟩$, apply $R_y(\pi /2)$, evolve freely with $U_{\varphi }=\exp (-i\varphi {\sigma }_z/2)$, and apply the inverse pulse $R_y(-\pi /2)$ before a computational-basis measurement. The first pulse gives $|+⟩$. Direct matrix multiplication givesThus the Born rule gives $P(0)={\cos }^2(\varphi /2)$ and $P(1)={\sin }^2(\varphi /2)$. Before the second pulse, the same computational-basis measurement would give $1/2$ for both outcomes for every $\varphi$. At $\varphi =2\pi /3$, the second pulse changes these probabilities to $1/4$ and $3/4$. If the two pulses have the same sign instead, the output labels interchange: the pulse convention is part of the prediction.

In a pulse experiment the phase is measured relative to the oscillator supplying the coherent pulses. In a rotating frame, write the free Hamiltonian operator as $H_{\mathrm{rot}}=\hslash \delta {\sigma }_z/2$ with signed angular frequency difference $\delta$ in the chosen labeling convention. A free interval $T$ then accumulates $\varphi =\delta T$; changing the second pulse's phase changes the analysis direction. This comparison makes Ramsey interferometry useful for measuring frequency differences and diagnosing loss of quantum coherence.

For example, suppose the first pulse has prepared a density matrix with equal populations and transverse Bloch vector length $0\le V\le 1$:The same inverse analysis pulse yields $P(0)=(1+V\cos \varphi )/2$. The pure state case $V=1$ gives the full ideal fringe; the mixed state $I/2$ gives the flat result $1/2$. This is a controlled mathematical comparison, not a fit to an experiment. Experimental contrast alone does not identify quantum decoherence: imperfect preparation, pulse errors and readout errors can also reduce it.

Under repeated identical amplitude damping channels, all excited population eventually relaxes to the ground state for any fixed $0<p\le 1$. The Bloch vector approaches $(0,0,1)$. For $p=1$ the limit is reached in one action, while $p=0$ is the identity channel and preserves every input. Keeping the zero-damping endpoint separate prevents an incorrect universal relaxation claim.

A one-qubit rotation gate is $R_{\alpha }(\theta )=e^{-i\theta {\sigma }_{\alpha }/2}$, where ${\sigma }_{\alpha }$ is a Pauli matrix. It rotates the Bloch vector by angle $\theta$ about the corresponding axis.

The phase gate with angle $\pi /2$. It rotates the Bloch vector by $\pi /2$ around $z$ and is a Clifford gate.

For four uniformly distributed Bell states, measure one qubit along Bloch vector $\mathbf{n}$ and the other along $z$. The sign within each ${\Phi }^{\pm }$ or ${\Psi }^{\pm }$ pair is invisible and the displayed mutual information is at most one bit. A local Hadamard gate before a computational detector changes $n_z$ from one to zero and destroys the remaining information. An entangling Bell decoder falls outside this product-measurement restriction and recovers two bits.

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.

For a spin one-half system, an ideal projective measurement along a unit vector $\mathbf{n}$ measures ${\sigma }_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{\sigma }$. Its eigenvalues are $\pm 1$, corresponding to physical spin angular momentum components $\pm \hslash /2$, and its orthogonal projections are ${\Pi }_{\pm }=(I\pm {\sigma }_{\mathbf{n}})/2$. For density matrix $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, the Born rule and the Pauli matrix multiplication law give the displayed probabilities. Thus a spin measurement along one direction tests only that component of the Bloch vector; measurements along three independent directions determine the entire density matrix.

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.

Use the rotating frame and positive gyromagnetic ratio convention of free induction decay. Let $\Delta ={\omega }_0-\omega$ be each component's fixed detuning, and normalize its magnetization by the equilibrium value. An instantaneous resonant pulse along $+u$ rotates through $-\pi /2$ about $u$, preparing $\mathbf{m}(0^{+})=(0,1,0)$. A second pulse along $+v$ rotates through $-\pi$ about $v$. These ideal pulses require a drive much stronger than the relevant detunings and negligible relaxation during either pulse. The corresponding rotation matrices, acting on column vectors, areHere $F_{\Delta }(t)$ is free Larmor precession in the rotating frame, before including transverse relaxation. Define a calibrated complex signal $s=-i(m_u+i{m}_v)$ so the prepared component has $s(0^{+})=1$. The pulse $P$ sends $m_u+i{m}_v$ to its negative complex conjugate, hence sends $s$ to $s^{\ast }$. Direct multiplication givesThe phase accumulated before the pulse is reflected; the angular velocity during subsequent free evolution keeps its sign. At $2\tau$ every fixed detuning has zero residual phase. Equivalently, $P{F}_{\Delta }(t)P=F_{\Delta }(-t)$, so $F_{\Delta }(\tau )P{F}_{\Delta }(\tau )=P$. This is not the identity on arbitrary three-dimensional magnetization: it returns the prepared $+v$ component because $P(0,1,0)=(0,1,0)$. The two transverse pulse axes specified here also fix the receiver sign of the echo.

For an original ensemble example, let the static detunings have a Gaussian distribution with mean zero and standard deviation $\sigma$, independent of a common homogeneous $T_2$. Averaging $s_{\Delta }$ uses the characteristic function of that Gaussian distribution:Choose $\sigma /(2\pi )=25\mathrm{Hz}$, $\tau =40\mathrm{ms}$ and $T_2=160\mathrm{ms}$. Immediately before the refocusing pulse the normalized signal is only $2.08\times 10^{-9}$, yet $S(80\mathrm{ms})=e^{-1/2}=0.60653$. A one-pulse free induction decay at 80 ms would instead be about $3.11\times 10^{-35}$. These chosen numbers describe a calculation, not a measured trace. Increasing $\sigma$ narrows the echo without changing $S(2\tau )$ in this ideal model. Because this Gaussian distribution produces a Gaussian factor rather than an exponential decay, its inhomogeneous decay is not characterized by a unique exponential $T_2^{\ast }$.

The exact phase-refocusing time is $2\tau$. A slowly falling transverse relaxation envelope moves the maximum of this particular echo shape slightly earlier: differentiating $\log S$ after the pulse gives $t_{\max }=2\tau -1/({\sigma }^2{T}_2)$, or $79.747\mathrm{ms}$ here. Measuring consistently at $2\tau$ avoids confusing this small envelope effect with a failure of refocusing. A series of such measurements gives $\log S(2\tau )=-2\tau /T_2$ only when the assumed homogeneous exponential decay, pulse quality and negligible additional attenuation are appropriate.

The ideal projective measurement of a spin-one-half component along a unit axis $\mathbf{n}$. The two outcomes have Born rule probabilities $⟨\psi |P_{\pm }|\psi ⟩$. For an input with Bloch vector $\mathbf{r}$, they are $(1\pm \mathbf{n}\cdot \mathbf{r})/2$. This model distinguishes a mathematical spin measurement from engineering a literal free-electron beam splitter.

Choose the four unit Bloch vectors at the vertices of a regular tetrahedron, so ${\sum }_i{\mathbf{n}}_i=0$ and ${\sum }_i{\mathbf{n}}_i{\mathbf{n}}_i^T=4I/3$. The displayed four positive rank-one effects sum to $I$. On input $(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, their probabilities are $p_i=(1+\mathbf{r}\cdot {\mathbf{n}}_i)/4$, giving the reconstruction $\mathbf{r}=3{\sum }_i{p}_i{\mathbf{n}}_i$. This realizes the smallest fixed qubit informationally complete POVM.

The phase gate with angle $\pi /4$. It rotates the Bloch vector by $\pi /4$ around $z$, and $T^2=S$. It is not a Clifford gate: it takes the positive $X$ eigenstate to a state with vector $(1/\sqrt{2},1/\sqrt{2},0)$, outside the stabilizer octahedron.