Bloch Vector

Purpose

Every one-qubit density operator has the Pauli expansion

$$\rho = \frac12 \left( I + \mathbf r\cdot\boldsymbol\sigma \right),$$

where

$$\boldsymbol\sigma = (X,Y,Z), \qquad \mathbf r = (r_x,r_y,r_z)\in\mathbb R^3.$$

The components are measurable Pauli expectations:

$$r_i = \operatorname{Tr}(\rho\sigma_i) = \langle\sigma_i\rangle.$$

Physicality is exactly

$$\lVert\mathbf r\rVert\leq1.$$

The full unit ball is the Bloch ball; only its pure-state boundary is the Bloch sphere. The canonical operational treatment is Bloch Sphere for Quantum Information, while the density-operator derivation is Bloch Sphere for Density Operators.

At a glance

Task	Formula
State from vector	$\rho=(I+\mathbf r\cdot\boldsymbol\sigma)/2$
Vector from state	$r_i=\operatorname{Tr}(\rho\sigma_i)$
Physicality	$\lVert\mathbf r\rVert\leq1$
Eigenvalues	$\lambda_\pm=(1\pm\lVert\mathbf r\rVert)/2$
Purity	$\operatorname{Tr}(\rho^2)=(1+\lVert\mathbf r\rVert^2)/2$
Determinant	$\det\rho=(1-\lVert\mathbf r\rVert^2)/4$
Pauli-axis measurement	$p_\pm=(1\pm\hat{\mathbf n}\cdot\mathbf r)/2$
Qubit overlap	$\operatorname{Tr}(\rho\sigma)=(1+\mathbf r\cdot\mathbf s)/2$
Trace distance	$D(\rho,\sigma)=\lVert\mathbf r-\mathbf s\rVert/2$
Unitary action	$\mathbf r\mapsto R\mathbf r$
Channel action	$\mathbf r\mapsto T\mathbf r+\mathbf t$

Here $\sigma$ in the overlap row denotes a second density operator, not a Pauli matrix; its Bloch vector is $\mathbf s$.

Matrix and vector forms

With the standard computational-basis convention

$$Z\lvert0\rangle = +\lvert0\rangle, \qquad Z\lvert1\rangle = -\lvert1\rangle,$$

the matrix is

$$\rho = \frac12 \begin{pmatrix} 1+r_z & r_x-ir_y\\ r_x+ir_y & 1-r_z \end{pmatrix}.$$

Conversely, for

$$\rho = \begin{pmatrix} a & c\\ c^* & 1-a \end{pmatrix},$$

the Bloch coordinates are

$$r_x = 2\operatorname{Re}c, \qquad r_y = - 2\operatorname{Im}c, \qquad r_z = 2a-1.$$

The minus sign in $r_y$ follows from

$$Y = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}.$$

A different computational-basis ordering or $Y$ convention changes coordinate signs, so matrix entries should not be converted by memory alone.

Physicality, spectrum, and rank

Using

$$(\mathbf r\cdot\boldsymbol\sigma)^2 = \lVert\mathbf r\rVert^2I,$$

the eigenvalues are

$$\lambda_\pm = \frac{ 1\pm r }{2}, \qquad r \equiv \lVert\mathbf r\rVert.$$

Positivity requires $\lambda_-\geq0$, giving $r\leq1$. The determinant is

$$\det\rho = \lambda_+\lambda_- = \frac{1-r^2}{4}.$$

The important cases are:

Radius	Spectrum	State type
$r=0$	$(1/2,1/2)$	Maximally mixed
$0<r<1$	Two positive eigenvalues	Rank-two mixed
$r=1$	$(1,0)$	Pure
$r>1$	One negative eigenvalue	Not a density operator

Hermiticity and trace one alone do not guarantee physicality. A candidate tomographic matrix outside the ball needs a physical estimator or uncertainty analysis, not an interpretation as a valid state.

Purity, determinant, and entropy

The purity is

$$\gamma(\rho) \equiv \operatorname{Tr}(\rho^2) = \frac{1+r^2}{2}.$$

Therefore

$$\frac12 \leq \gamma(\rho) \leq 1.$$

For a qubit,

$$\gamma(\rho) = 1-2\det\rho.$$

The radius is recovered from purity by

$$r = \sqrt{ 2\gamma(\rho)-1 }.$$

With logarithm base two, the von Neumann entropy is

$$S(\rho) = h_2 \left( \frac{1+r}{2} \right),$$

where

$$h_2(p) = - p\log_2p - (1-p)\log_2(1-p).$$

Entropy decreases from one bit at the center to zero on the surface. Radius, purity, determinant, and entropy determine the same one-qubit spectrum, but they do not identify the preparation ensemble or noise mechanism.

Pure-state parameterization

Up to global phase, every pure qubit can be written

$$\lvert\psi\rangle = \cos\left(\frac{\theta}{2}\right) \lvert0\rangle + e^{i\phi} \sin\left(\frac{\theta}{2}\right) \lvert1\rangle.$$

Its unit Bloch vector is

$$\mathbf r = \left( \sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta \right).$$

The half angles arise because normalized spinors modulo global phase form $\mathbb{CP}^1$, which maps to the two-sphere. The states at antipodal points are orthogonal, not physically identical.

The six cardinal states are:

State	Bloch vector
$\lvert0\rangle$	$(0,0,1)$
$\lvert1\rangle$	$(0,0,-1)$
$\lvert+\rangle=(\lvert0\rangle+\lvert1\rangle)/\sqrt2$	$(1,0,0)$
$\lvert-\rangle=(\lvert0\rangle-\lvert1\rangle)/\sqrt2$	$(-1,0,0)$
$\lvert{+i}\rangle=(\lvert0\rangle+i\lvert1\rangle)/\sqrt2$	$(0,1,0)$
$\lvert{-i}\rangle=(\lvert0\rangle-i\lvert1\rangle)/\sqrt2$	$(0,-1,0)$

Convex mixtures

If

$$\rho = \sum_k p_k\rho_k, \qquad p_k\geq0, \qquad \sum_kp_k=1,$$

then the Bloch vectors average affinely:

$$\mathbf r = \sum_k p_k\mathbf r_k.$$

This proves convexity of the Bloch ball. It does not make the decomposition unique: the center, for example, is an equal mixture of either pair of antipodal pure states. An interior point can also be the reduced state of an entangled system.

The Bloch vector specifies the density operator, not a hidden classical direction and not one preferred ensemble decomposition.

Measurements along an axis

For a unit vector $\hat{\mathbf n}$, the Pauli observable

$$\sigma_{\mathbf n} \equiv \hat{\mathbf n}\cdot\boldsymbol\sigma$$

has effects

$$\Pi_\pm = \frac12 \left( I \pm \hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

The Born probabilities are

$$p_\pm = \operatorname{Tr}(\rho\Pi_\pm) = \frac12 \left( 1 \pm \hat{\mathbf n}\cdot\mathbf r \right).$$

The mean and variance are

$$\langle\sigma_{\mathbf n}\rangle = \hat{\mathbf n}\cdot\mathbf r,$$ $$\operatorname{Var}(\sigma_{\mathbf n}) = 1- \left( \hat{\mathbf n}\cdot\mathbf r \right)^2.$$

For an ideal rank-one projective measurement, the conditional postmeasurement Bloch vector is $+\hat{\mathbf n}$ or $-\hat{\mathbf n}$ according to the outcome. A general POVM effect or noisy readout need not correspond to a unit axis or that state-update rule.

Arbitrary qubit observables

Every Hermitian qubit operator can be written

$$A = a_0I + \mathbf a\cdot\boldsymbol\sigma, \qquad a_0\in\mathbb R, \qquad \mathbf a\in\mathbb R^3.$$

Its expectation is

$$\langle A\rangle_\rho = a_0 + \mathbf a\cdot\mathbf r.$$

Its eigenvalues are

$$a_\pm = a_0 \pm \lVert\mathbf a\rVert.$$

Using the Pauli product identity gives

$$\operatorname{Var}_\rho(A) = \lVert\mathbf a\rVert^2 - (\mathbf a\cdot\mathbf r)^2.$$

This formula includes projective Pauli measurements when $a_0=0$ and $\lVert\mathbf a\rVert=1$.

State comparison geometry

Let

$$\rho = \frac12(I+\mathbf r\cdot\boldsymbol\sigma), \qquad \sigma = \frac12(I+\mathbf s\cdot\boldsymbol\sigma).$$

Their Hilbert–Schmidt overlap is

$$\operatorname{Tr}(\rho\sigma) = \frac12 \left( 1+\mathbf r\cdot\mathbf s \right).$$

The trace distance has the qubit shortcut

$$D(\rho,\sigma) = \frac12 \lVert\rho-\sigma\rVert_1 = \frac12 \lVert\mathbf r-\mathbf s\rVert.$$

Using this site’s squared Uhlmann-fidelity convention,

$$\begin{aligned} F(\rho,\sigma) = \frac12 \Big[ 1 &+ \mathbf r\cdot\mathbf s \\ &+ \sqrt{ (1-r^2)(1-s^2) } \Big]. \end{aligned}$$

If at least one state is pure, the square-root term vanishes and

$$F(\rho,\sigma) = \operatorname{Tr}(\rho\sigma).$$

For pure states whose Bloch directions subtend an angle $\Gamma$,

$$F = \frac{ 1+\cos\Gamma }{2} = \cos^2 \left( \frac{\Gamma}{2} \right).$$

The determinant and Bloch shortcuts are special to dimension two. The canonical state-comparison conventions are on Fidelity and Trace Distance.

Unitary rotations

A Pauli-axis rotation is

$$U_{\hat{\mathbf n}}(\vartheta) = \exp \left( - \frac{i\vartheta}{2} \hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

Under

$$\rho' = U\rho U^\dagger,$$

the Bloch vector transforms as

$$\mathbf r' = R_{\hat{\mathbf n}}(\vartheta) \mathbf r,$$

where $R_{\hat{\mathbf n}}(\vartheta)\in SO(3)$ is the active right-hand-rule rotation for this sign convention.

Rodrigues’ formula is

$$\begin{aligned} R_{\hat{\mathbf n}}(\vartheta)\mathbf r ={}& \mathbf r\cos\vartheta \\ &+ (\hat{\mathbf n}\times\mathbf r) \sin\vartheta \\ &+ \hat{\mathbf n} (\hat{\mathbf n}\cdot\mathbf r) (1-\cos\vartheta). \end{aligned}$$

Unitaries preserve radius, purity, entropy, and distances between Bloch vectors. The matrices $U$ and $-U$ produce the same Bloch rotation, expressing the two-to-one map from $SU(2)$ to $SO(3)$.

Hamiltonian dynamics

For

$$H = h_0I + \frac{\hbar}{2} \boldsymbol\Omega\cdot\boldsymbol\sigma,$$

the von Neumann equation

$$\dot\rho = - \frac{i}{\hbar} [H,\rho]$$

becomes

$$\dot{\mathbf r} = \boldsymbol\Omega \times \mathbf r.$$

The identity term $h_0I$ changes only state-vector global phase and does not move the density operator. If $\boldsymbol\Omega$ is constant, the vector precesses rigidly about that axis with angular speed $\lVert\boldsymbol\Omega\rVert$.

Changing the exponential sign, using a passive frame rotation, or swapping the $Y$ convention reverses apparent rotation directions. State the convention before comparing control pulses.

Channels as affine maps

Every trace-preserving linear map on one-qubit operators acts as

$$\mathbf r' = T\mathbf r + \mathbf t,$$

with a real $3\times3$ matrix $T$ and real vector $\mathbf t$. A channel is unital exactly when

$$\mathbf t=\mathbf0.$$

Complete positivity imposes constraints beyond the visual condition that an affine image fits inside the ball. A drawn ellipsoid is not by itself a channel certificate.

Common conventions include:

Dephasing

$$(r_x,r_y,r_z) \longmapsto (\lambda r_x,\lambda r_y,r_z).$$

Transverse coherence contracts while computational-basis populations remain fixed.

Depolarization

For

$$\mathcal D_p(\rho) = (1-p)\rho + p\frac I2,$$

one has

$$\mathbf r' = (1-p)\mathbf r.$$

Other sources parameterize depolarization differently, so the channel definition should accompany $p$.

Amplitude damping

With $\lvert0\rangle$ at $z=+1$ and decay probability $\gamma$,

$$\begin{aligned} r_x' &= \sqrt{1-\gamma}\,r_x, \\ r_y' &= \sqrt{1-\gamma}\,r_y, \\ r_z' &= (1-\gamma)r_z+\gamma. \end{aligned}$$

This channel translates the ball toward the ground-state pole and is nonunital for $\gamma>0$.

One-qubit tomography

Ideal calibrated measurements of $X$, $Y$, and $Z$ estimate

$$\mathbf r = \left( \langle X\rangle, \langle Y\rangle, \langle Z\rangle \right).$$

If $N_i^\pm$ are outcome counts for Pauli setting $i$, a linear estimate is

$$\widehat r_i = \frac{ N_i^+-N_i^- }{ N_i^++N_i^- }.$$

Finite data can produce

$$\lVert\widehat{\mathbf r}\rVert>1.$$

That does not indicate a physical state beyond the Bloch ball. It indicates statistical fluctuation, model mismatch, or calibration error. A physical reconstruction may use constrained maximum likelihood or Bayesian estimation, and should report uncertainty and state-preparation-and-measurement assumptions.

Three Pauli settings are informationally complete for one qubit. They are not enough to reconstruct an arbitrary many-qubit state, whose density operator has $4^n-1$ independent real parameters.

Worked matrix diagnostic

Consider

$$\rho = \begin{pmatrix} 0.7 & 0.2-0.1i\\ 0.2+0.1i & 0.3 \end{pmatrix}.$$

Its Bloch vector is

$$\mathbf r = (0.4,0.2,0.4), \qquad r=0.6.$$

Hence

$$\lambda_\pm = \frac{1\pm0.6}{2} = (0.8,0.2),$$

and

$$\operatorname{Tr}(\rho^2) = \frac{1+0.6^2}{2} = 0.68.$$

The radius check verifies positivity without a separate matrix diagonalization.

Scope and limitations

The Bloch ball is a complete state representation only for one qubit. Reduced one-qubit Bloch vectors do not encode multipartite correlations or entanglement. For example, every qubit of a Bell state has $\mathbf r=\mathbf0$, although the joint state is pure and maximally entangled.

Generalized generator expansions exist in dimension $d>2$, but their positivity regions are not ordinary Euclidean balls. A vector with an allowed norm can still correspond to a nonpositive higher-dimensional matrix.

Bloch axes are coordinate conventions. They coincide with physical spatial directions only when the experimental encoding and measurement calibration establish that relation. A trajectory in the ball describes state evolution but does not uniquely identify a Hamiltonian, channel, postselection rule, or microscopic noise source.

Common mistakes

• Calling every mixed-state point a point on the Bloch sphere rather than in the Bloch ball.
• Treating $\mathbf r$ as a two-component state vector or a hidden variable.
• Missing the minus sign in $r_y=-2\operatorname{Im}\rho_{01}$ for the stated $Y$ convention.
• Checking Hermiticity and trace but not $\lVert\mathbf r\rVert\leq1$.
• Using the pure-state surface for a mixed state.
• Confusing a state direction with a measurement axis.
• Omitting the factor of $1/2$ in $e^{-i\vartheta\hat{\mathbf n}\cdot\boldsymbol\sigma/2}$ and doubling the intended Bloch rotation.
• Treating every ball contraction as a completely positive channel.
• Calling amplitude damping unital or purely dephasing.
• Using local Bloch vectors as a complete multi-qubit description.
• Comparing fidelity formulas without checking squared versus root convention.
• Interpreting an unphysical linear-tomography estimate as a valid density operator.

Exercises

1. Reconstruct and test a state

For

$$\rho = \begin{pmatrix} 0.6 & 0.3i\\ -0.3i & 0.4 \end{pmatrix},$$

find the Bloch vector, eigenvalues, and purity. Is the state physical?

Solution

Here $a=0.6$ and $c=0.3i$, so

$$\mathbf r = \left( 0, -0.6, 0.2 \right).$$

Its radius is

$$r = \sqrt{0.36+0.04} = \sqrt{0.4}.$$

Therefore

$$\lambda_\pm = \frac{ 1\pm\sqrt{0.4} }{2},$$

which are both nonnegative. The purity is

$$\operatorname{Tr}(\rho^2) = \frac{1+0.4}{2} = 0.7.$$

The state is physical because $r<1$.

2. Tilted-axis measurement

A state has $\mathbf r=(0,0,r)$ and is measured along

$$\hat{\mathbf n} = \frac{ \hat{\mathbf x}+\hat{\mathbf z} }{\sqrt2}.$$

Find the two outcome probabilities.

Solution

The projection is

$$\hat{\mathbf n}\cdot\mathbf r = \frac r{\sqrt2}.$$

Therefore

$$p_\pm = \frac12 \left( 1 \pm \frac r{\sqrt2} \right).$$

For $r=0$, both outcomes have probability one half; for $r=1$, the state is not an eigenstate of the tilted measurement, so neither probability is one.

3. Positive z rotation

Apply

$$U = e^{-i(\pi/4)Z}$$

to the $+x$ state. Find the final Bloch vector.

Solution

The unitary is

$$U = \exp \left( - \frac{i}{2} \frac{\pi}{2} Z \right),$$

so it rotates the Bloch vector by $+\pi/2$ about $z$. The active right-hand-rule rotation sends

$$(1,0,0) \longmapsto (0,1,0).$$

The final state is $\lvert{+i}\rangle$ up to global phase.

4. Amplitude damping of the center

Apply amplitude damping with probability $\gamma$ to $\rho=I/2$. Find the output vector and purity, and explain what this says about unitality.

Solution

The input vector is $\mathbf0$. The affine map gives

$$\mathbf r' = (0,0,\gamma).$$

Its purity is

$$\operatorname{Tr} \left[ (\rho')^2 \right] = \frac{ 1+\gamma^2 }{2}.$$

For $\gamma>0$, the maximally mixed state does not remain at the center:

$$\mathcal A_\gamma(I/2) \neq I/2.$$

Therefore amplitude damping is nonunital.

Canonical explanations

• Bloch Sphere for Quantum Information
• Bloch Sphere for Density Operators
• Bloch Sphere Geometry
• Bloch Sphere for Spin
• Pauli Matrices
• Single-Qubit Gates
• Quantum Channels and Noise
• State Tomography

Related lookup pages

• Pauli Matrices Table
• Spin-Half Matrices
• Fidelity
• Trace Distance
• Purity
• Von Neumann Entropy

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• J. Preskill, Lecture Notes for Physics 229: Quantum Information and Computation, California Institute of Technology.
• D. F. V. James, P. G. Kwiat, W. J. Munro, and A. G. White, “Measurement of qubits,” Physical Review A 64, 052312 (2001).