Bloch Sphere for Quantum Information

The Bloch sphere turns one-qubit calculations into geometry: measurement probabilities become projections, unitary gates become rotations, and noise deforms the set of accessible states. The full ball, including its interior, represents mixed as well as pure states. Its three coordinates describe one qubit exactly; they do not describe correlations in a larger register.

Helpful background. Density operators supplies positivity and the Born rule, while Pauli matrices supplies the operator basis used below.

Bloch coordinates and one-qubit states

The Bloch ball is an exact three-real-parameter representation of every one-qubit density operator:

$$\begin{gathered} \rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right), \\ \mathbf r\in\mathbb R^3, \qquad \lVert\mathbf r\rVert\leq1. \end{gathered}$$

The vector $\mathbf r$ is the Bloch vector. Pure states lie on the unit sphere $\lVert\mathbf r\rVert=1$, mixed states lie inside it, and the maximally mixed state is the center. Its components are experimentally meaningful Pauli expectation values:

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

In quantum information, the picture is more than a state visualization. It is an operational dictionary:

• a state corresponds to its vector $\mathbf r$;
• a projective measurement corresponds to an axis $\hat{\mathbf n}$;
• a one-qubit unitary corresponds to a proper rotation of $\mathbf r$;
• a one-qubit channel corresponds to an affine map $\mathbf r\mapsto T\mathbf r+\mathbf t$.

For the underlying geometry, see Bloch coordinates; for their relation to physical spin directions and spinors, see the spin Bloch sphere.

From a Density Matrix to Three Measurable Numbers

Pauli expansion

The identity and Pauli matrices form an orthogonal basis for $2\times2$ operators under the Hilbert–Schmidt inner product:

$$\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}, \qquad \operatorname{Tr}(\sigma_i)=0.$$

Consequently, every trace-one Hermitian matrix has a unique expansion

$$\rho = \frac12 \left( I+r_xX+r_yY+r_zZ \right).$$

The coefficients are recovered by traces:

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

In the computational basis,

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

Thus $r_z=\rho_{00}-\rho_{11}$ is the computational-basis population imbalance. The transverse coordinates satisfy $r_x=2\operatorname{Re}\rho_{01}$ and $r_y=-2\operatorname{Im}\rho_{01}$; the minus sign follows from the usual Pauli $Y$ matrix. A basis label such as $Z$ is a convention fixed by the chosen encoding and control frame. It need not be a literal spatial direction.

Positivity makes a ball

The eigenvalues of $\rho$ are

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

The condition $\rho\geq0$ is therefore exactly

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

The purity is

$$\operatorname{Tr}(\rho^2) = \frac{1+\lVert\mathbf r\rVert^2}{2}.$$

It ranges from $1/2$ at the center to $1$ on the surface. The radius measures purity for a one-qubit state; it does not by itself identify a temperature, a noise source, or a preparation procedure.

Pure states on the surface

After choosing the computational basis, a pure state can be parameterized as

$$|\psi(\theta,\phi)\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi} \sin\frac{\theta}{2}|1\rangle.$$

Its Bloch vector is

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

The polar angle controls the population imbalance, and the azimuthal angle records relative phase. Global phase is absent because the density operator

$$|\psi\rangle\langle\psi|$$

is unchanged by $|\psi\rangle\mapsto e^{i\chi}|\psi\rangle$. The derivation from normalized rays is given in Bloch Sphere for Spin and Projective Hilbert Space.

Six cardinal states

The eigenstates of $X$, $Y$, and $Z$ define six useful reference points.

Direction	State	Pauli eigenvalue
$+z$	$\lvert0\rangle$	$Z=+1$
$-z$	$\lvert1\rangle$	$Z=-1$
$+x$	$\lvert+\rangle=(\lvert0\rangle+\lvert1\rangle)/\sqrt2$	$X=+1$
$-x$	$\lvert-\rangle=(\lvert0\rangle-\lvert1\rangle)/\sqrt2$	$X=-1$
$+y$	$\lvert+i\rangle=(\lvert0\rangle+i\lvert1\rangle)/\sqrt2$	$Y=+1$
$-y$	$\lvert-i\rangle=(\lvert0\rangle-i\lvert1\rangle)/\sqrt2$	$Y=-1$

These states are common calibration targets. The names $x$, $y$, and $z$ refer to Pauli coordinates in the logical frame. A hardware implementation must establish which pulses and readout settings realize those coordinates.

Mixed States and Convexity

If preparations $\rho_a$ occur with classical probabilities $p_a$, the averaged state is

$$\rho=\sum_a p_a\rho_a.$$

Bloch vectors average in the same way:

$$\mathbf r=\sum_a p_a\mathbf r_a.$$

The Bloch ball is therefore convex. Mixing surface points produces an interior point unless all weight lies on the same pure state.

The decomposition is not unique. For example,

$$\frac I2 = \frac12 \left( |0\rangle\langle0| + |1\rangle\langle1| \right)$$

and

$$\frac I2 = \frac12 \left( |+\rangle\langle+| + |-\rangle\langle-| \right).$$

Both ensembles give $\mathbf r=\mathbf0$. A point in the ball specifies the density operator, not a unique story about which pure state “really occurred.” It can also be the reduced state of a qubit entangled with another system, with no ignorance ensemble preferred by the formalism.

Two $x$–$z$ sections of the Bloch ball. In (a), a projective measurement depends on the projection of $\mathbf r$ onto the unit direction $\hat{\mathbf n}$. In (b), amplitude damping with $\gamma=0.6$ maps the ball to the shaded ellipsoid: its center moves to $z=0.6$, its transverse radii become $\sqrt{0.4}$, and its vertical radius becomes $0.4$. The ground-state pole remains fixed. All lengths use the same scale.

Measurement Axes

Projective measurement along an axis

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

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

has eigenvalues $\pm1$. Its projectors are

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

The Born rule gives

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

The expectation value is the signed projection

$$\langle\sigma_{\hat{\mathbf n}}\rangle = p_+-p_- = \hat{\mathbf n}\cdot\mathbf r.$$

This is the operational meaning of the angle between the state and measurement vectors. Parallel pure-state and measurement axes give a certain $+1$ outcome; antiparallel axes give a certain $-1$ outcome; perpendicular axes give equal probabilities.

State axes and measurement axes are different objects

The same three-dimensional coordinates describe a state vector and a measurement direction, but they play different roles. The state belongs to the preparation model. The axis belongs to the measurement effect. If both are unknown, observed probabilities alone do not determine which one is miscalibrated.

A real detector may be biased or unsharp. A binary effect can be expanded as

$$E_+ = \alpha I+\mathbf m\cdot\boldsymbol\sigma,$$

where $0\le E_+\le I$ requires $0\le\alpha\le1$ and $\lVert\mathbf m\rVert\le\min(\alpha,1-\alpha)$. The other outcome has effect $I-E_+$. The resulting probability

$$p(+|\rho) = \alpha+\mathbf m\cdot\mathbf r$$

need not correspond to an ideal projective axis. Measurement tomography explains how to reconstruct unknown detector effects.

One-Qubit State Tomography

If ideal $X$, $Y$, and $Z$ measurements are calibrated, repeated outcomes estimate

$$\begin{aligned} r_x&=p_X(+)-p_X(-), \\ r_y&=p_Y(+)-p_Y(-), \\ r_z&=p_Z(+)-p_Z(-). \end{aligned}$$

For counts $N_{i,+}$ and $N_{i,-}$ in basis $i$,

$$\widehat r_i = \frac{N_{i,+}-N_{i,-}} {N_{i,+}+N_{i,-}}.$$

These estimates fluctuate. Independent linear inversion can produce

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

which corresponds to no positive density operator. Finite sampling and errors in the measurement model can produce such an estimate. Constrained least squares, maximum likelihood, or Bayesian estimation can enforce physicality, but the estimator then has its own bias and uncertainty properties.

For $N_i$ independent shots of a calibrated Pauli measurement, $\widehat r_i$ has variance $(1-r_i^2)/N_i$. This sampling uncertainty is separate from systematic error in the measurement axis or readout. If the preparations and measurements are both unknown, a reconstructed vector can absorb errors from either; positivity alone does not validate the calibration. State tomography develops physical estimators and their uncertainty.

Three Pauli settings are informationally complete for one-qubit state tomography, but tomography scales poorly for many qubits. An arbitrary $n$-qubit density matrix has $4^n-1$ real parameters. The single-qubit sphere does not remove that scaling.

Rotations and Gates

Unitary gates rotate the ball

Ignoring an overall phase, every one-qubit unitary can be written

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

Conjugation by $U$ maps

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

to a rotation in Bloch-coordinate space:

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

The rotation preserves

$$\lVert\mathbf r'\rVert = \lVert\mathbf r\rVert,$$

so a unitary cannot purify or mix an isolated qubit. It moves pure states along the surface and mixed states on a sphere of fixed radius.

The factor of $1/2$ in the exponential matters. A Bloch-vector rotation by angle $\vartheta$ is generated by $\vartheta\hat{\mathbf n}\cdot\boldsymbol\sigma/2$. The underlying spinor changes sign under a $2\pi$ rotation, while its density operator and Bloch vector return after $2\pi$. Spin rotations derives the SU(2)-to-SO(3) relation.

Familiar gates as rotations

Up to global phase:

Gate family	Bloch action
$X$	Rotation by $\pi$ about $x$
$Y$	Rotation by $\pi$ about $y$
$Z$	Rotation by $\pi$ about $z$
$R_x(\vartheta)$	Rotation by $\vartheta$ about $x$
$R_y(\vartheta)$	Rotation by $\vartheta$ about $y$
$R_z(\vartheta)$	Rotation by $\vartheta$ about $z$
$H$	Rotation by $\pi$ about $(\hat{\mathbf x}+\hat{\mathbf z})/\sqrt2$, up to phase

Single-qubit gates connects these rotations to matrices, decompositions, and physical controls.

Schrödinger and Heisenberg pictures

One may rotate the state while holding the measurement axis fixed:

$$\mathbf r\mapsto R\mathbf r,$$

or hold the state fixed and rotate the observable oppositely:

$$\hat{\mathbf n}\mapsto R^{-1}\hat{\mathbf n}.$$

The probability is unchanged because

$$\hat{\mathbf n}\cdot R\mathbf r = \left( R^{-1}\hat{\mathbf n} \right)\cdot\mathbf r.$$

This distinction is practical. A laboratory may implement an $X$ measurement by applying a basis-change pulse and then using a fixed $Z$ readout. Whether the diagram shows a rotated state or rotated measurement axis depends on the chosen description.

Composition order

Rotations generally do not commute:

$$R_x(\alpha)R_y(\beta) \neq R_y(\beta)R_x(\alpha).$$

On a state column, the rightmost gate acts first. Bloch diagrams can hide this convention when arrows are drawn without an explicit time order. For pulse sequences, state the active rotation convention, axis orientation, and multiplication order.

Noisy Channels as Affine Maps

Every trace-preserving, Hermiticity-preserving linear map on one-qubit operators acts affinely on the Bloch vector:

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

where $T$ is a real $3\times3$ matrix and $\mathbf t\in\mathbb R^3$. A physical quantum channel must be completely positive as well as trace preserving; this imposes constraints stronger than merely mapping the Bloch ball into itself.

A channel is unital when

$$\mathcal E(I)=I.$$

In Bloch coordinates this means

$$\mathbf t=\mathbf0.$$

Unital channels leave the center fixed, though they may contract or anisotropically distort the ball. Nonunital channels translate the center.

Dephasing

Define the dephasing family

$$\begin{aligned} \mathcal D_\lambda(\rho) &= \frac{1+\lambda}{2}\rho + \frac{1-\lambda}{2}Z\rho Z, \\ -1\leq\lambda\leq1. \end{aligned}$$

Its Bloch action is

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

At $\lambda=0$, transverse coherence vanishes and the sphere collapses to the $z$-axis segment. Populations are unchanged. Negative $\lambda$ includes an additional phase inversion; in many dynamical dephasing models $\lambda$ decays from $1$ toward $0$. Dephasing channel relates this map to Kraus operators and coherence times.

Depolarization

For the convention

$$\mathcal P_q(\rho) = (1-q)\rho+q\frac I2, \qquad 0\leq q\leq1,$$

the Bloch vector transforms isotropically:

$$\mathbf r\longmapsto(1-q)\mathbf r.$$

The ball contracts toward the center. Other sources use different depolarizing parameters, so the channel definition should accompany any quoted error probability.

Amplitude damping

With $|0\rangle$ chosen as the ground state and damping 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 contracts and translates the ball toward $+z$. It is not unital because

$$\frac I2 \longmapsto \frac12 \left( I+\gamma Z \right).$$

Amplitude-Damping Channel derives the channel and explains its physical interpretation.

Positivity on one qubit is insufficient

A simple counterexample is transposition: $\rho\mapsto\rho^{\mathsf T}$ reflects $r_y\mapsto-r_y$ and preserves the Bloch ball, but it is not completely positive. Applied to half of an entangled pair it can produce a negative eigenvalue, as the final exercise shows. Ruskai, Szarek, and Werner characterize the allowed qubit-channel contractions and translations. The general tests are developed in quantum channels.

Geometric Diagnostics

Several one-qubit quantities become elementary functions of Bloch vectors.

Trace distance

For qubit states $\rho$ and $\sigma$ with vectors $\mathbf r$ and $\mathbf s$,

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

Thus optimal one-shot distinguishability is related to Euclidean separation in the ball. Antipodal pure states have distance $1$ and are orthogonal. Trace distance states the general definition and discrimination theorem.

Fidelity

Using the squared Uhlmann-fidelity convention,

$$F(\rho,\sigma) = \frac{ 1+\mathbf r\cdot\mathbf s }{2} + \frac{ \sqrt{1-\lVert\mathbf r\rVert^2} \sqrt{1-\lVert\mathbf s\rVert^2} }{2}.$$

If $\rho=|\psi\rangle\langle\psi|$ is pure, this reduces to

$$F(\rho,\sigma) = \langle\psi|\sigma|\psi\rangle = \frac{1+\mathbf r\cdot\mathbf s}{2}.$$

Some sources call $\sqrt F$ the fidelity, so conventions must be stated. Fidelity fixes the site’s formula convention.

Entropy and radius

Because the eigenvalues are $(1\pm r)/2$ with $r=\lVert\mathbf r\rVert$, the von Neumann entropy is

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

where $h_2$ is binary entropy. Entropy decreases monotonically from one bit at the center to zero on the surface. This simple radial dependence is special to one qubit.

Worked Operational Examples

Ramsey phase as azimuthal motion

Prepare $|0\rangle$, apply a $\pi/2$ pulse about $y$, and obtain the $+x$ state. A relative phase accumulation

$$U_\phi = \exp \left( -\frac{i\phi Z}{2} \right)$$

rotates the vector to

$$\mathbf r_\phi = (\cos\phi,\sin\phi,0).$$

An $X$ measurement gives

$$p_X(+|\phi) = \frac{1+\cos\phi}{2},$$

while a $Y$ measurement gives

$$p_Y(+|\phi) = \frac{1+\sin\phi}{2}.$$

The phase is invisible to a direct $Z$ measurement because $r_z=0$ throughout. This is why a final analysis pulse is needed to convert phase into a population difference.

Dephasing as loss of contrast

If the same equatorial state undergoes $\mathcal D_\lambda$,

$$\mathbf r_\phi \longmapsto \lambda (\cos\phi,\sin\phi,0).$$

The signed fringe amplitude is multiplied by $\lambda$. Contrast, defined as the nonnegative modulation depth, is multiplied by $|\lambda|$; a negative $\lambda$ additionally shifts the fringe by $\pi$. The radial contraction summarizes the loss of coherence for this state family, but it does not by itself identify whether the cause was stochastic detuning, entanglement with an environment, averaging over drift, or another mechanism.

A calibrated measurement

Suppose a prepared state has

$$\mathbf r=(0.6,0,0.8),$$

which is pure because $\lVert\mathbf r\rVert=1$. Measuring along

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

gives

$$p_+ = \frac12 \left[ 1+\frac{1.4}{\sqrt2} \right] \approx0.995.$$

The near-certain outcome reflects geometric alignment, not simultaneous pre-existing $X$ and $Z$ values.

Limitations of the Picture

It is exact only for one qubit

An $n$-qubit density matrix requires $4^n-1$ real parameters. One may draw a reduced Bloch vector for each qubit, but those vectors omit correlations and entanglement. For the Bell state

$$|\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2},$$

each qubit has $\mathbf r=\mathbf0$, yet the joint state is pure and maximally entangled. Two centered local balls do not describe the global state.

Generalized Bloch vectors are not ordinary balls

For a qudit, one can expand a state in $d^2-1$ traceless generators. Positivity does not fill a simple Euclidean ball when $d>2$. The elegant equivalence

$$\rho\geq0 \quad\Longleftrightarrow\quad \lVert\mathbf r\rVert\leq1$$

is special to $d=2$ under the standard normalization.

The coordinates are expectation values measured on separate, similarly prepared specimens, not three simultaneously readable properties of one specimen. Their axes depend on the computational basis and control frame. The same observed trajectory can also have several microscopic explanations; geometry by itself does not identify an environment or a preparation ensemble.

Exercises

1. Reconstruct a state from Pauli data

Ideal measurements give

$$\begin{aligned} p_X(+)&=0.65, \\ p_Y(+)&=0.30, \\ p_Z(+)&=0.75. \end{aligned}$$

Find the Bloch vector and density matrix. Is the estimate physical?

Solution

For a Pauli measurement,

$$r_i=2p_i(+)-1.$$

Therefore

$$\mathbf r=(0.30,-0.40,0.50).$$

Its length is

$$\begin{aligned} \lVert\mathbf r\rVert &= \sqrt{0.09+0.16+0.25} \\ &= \sqrt{0.50} <1, \end{aligned}$$

so it is physical. The density matrix is

$$\rho = \frac12 \begin{pmatrix} 1.50&0.30+0.40i \\ 0.30-0.40i&0.50 \end{pmatrix}.$$

Equivalently,

$$\rho = \begin{pmatrix} 0.75&0.15+0.20i \\ 0.15-0.20i&0.25 \end{pmatrix}.$$

2. Diagnose an unphysical linear estimate

Tomographic linear inversion gives

$$\widehat{\mathbf r}=(0.8,0.6,0.3).$$

Why is this not a valid qubit state, and what should an analyst do?

Solution

The squared length is

$$\lVert\widehat{\mathbf r}\rVert^2 = 0.8^2+0.6^2+0.3^2 = 1.09>1.$$

The smaller eigenvalue

$$\widehat\lambda_- = \frac{ 1-\sqrt{1.09} }{2}$$

is negative, so the reconstructed matrix is not positive semidefinite.

The analyst should not simply interpret the vector as “more than pure.” They should retain the raw counts, quantify statistical and calibration uncertainty, and use a documented physical estimator such as constrained least squares, maximum likelihood, or a Bayesian method. The choice of estimator and correction procedure must be reported.

3. Measure along a tilted axis

A qubit is in the $+x$ state, $\mathbf r=(1,0,0)$. It is measured along

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

Find $p_+$ and $p_-$.

Solution

The dot product is

$$\hat{\mathbf n}\cdot\mathbf r = \frac1{\sqrt2}.$$

Hence

$$\begin{aligned} p_+ &= \frac12 \left( 1+\frac1{\sqrt2} \right), \\ p_- &= \frac12 \left( 1-\frac1{\sqrt2} \right). \end{aligned}$$

Numerically, $p_+\approx0.854$ and $p_-\approx0.146$.

4. Follow a gate rotation

The initial Bloch vector is $\mathbf r=(1,0,0)$. Apply

$$R_z\left(\frac\pi2\right) = \exp\left(-\frac{i\pi Z}{4}\right).$$

Find the final state and the probabilities of $X$, $Y$, and $Z$ outcomes.

Solution

A positive $\pi/2$ rotation about $z$ sends $+x$ to $+y$:

$$\mathbf r'=(0,1,0).$$

The final state is

$$|+i\rangle = \frac{|0\rangle+i|1\rangle}{\sqrt2}$$

up to global phase. Therefore

$$\begin{aligned} p_X(+)&=1/2, \\ p_Y(+)&=1, \\ p_Z(+)&=1/2. \end{aligned}$$

5. Dephasing and purity

The $+x$ state passes through $\mathcal D_\lambda$ with $0\leq\lambda\leq1$. Find its output Bloch vector and purity. Evaluate the result at $\lambda=0$ and $\lambda=1$.

Solution

The input vector is $(1,0,0)$, so

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

Its purity is

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

At $\lambda=1$, the state remains pure with purity $1$. At $\lambda=0$, the output is the center of the equatorial chord,

$$\rho'=\frac I2,$$

with purity $1/2$. Complete $Z$ dephasing of the equal superposition produces an equal incoherent mixture of $|0\rangle$ and $|1\rangle$.

6. A ball-preserving map that is not a channel

Show that transposition sends $(r_x,r_y,r_z)$ to $(r_x,-r_y,r_z)$. Apply transposition to only the second qubit of $|\Phi^+\rangle\langle\Phi^+|$, where $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. Why does the result rule out transposition as a physical channel on an arbitrary input?

Solution

Transposing the Pauli expansion leaves $X,Z$ unchanged and sends $Y$ to $-Y$, so every one-qubit density matrix remains positive. But in the ordered basis $00,01,10,11$, partial transposition gives

$$(\operatorname{id}\otimes\mathsf T)(|\Phi^+\rangle\langle\Phi^+|) =\frac12 \begin{pmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 \end{pmatrix}.$$

The antisymmetric vector $(|01\rangle-|10\rangle)/\sqrt2$ has eigenvalue $-1/2$. A channel must remain positive when an untouched reference system is included. Transposition fails this requirement even though its one-qubit picture looks harmless.

For larger systems, continue to density operators for quantum information. BB84 uses the four $X$ and $Z$ cardinal states as a communication alphabet.

References

• Bengtsson, I., and K. Życzkowski. Geometry of Quantum States, 2nd edition. Cambridge University Press, 2017. doi:10.1017/9781139207010.
• Bloch, F. “Nuclear induction.” Physical Review 70, 460–474 (1946). doi:10.1103/PhysRev.70.460.
• Blume-Kohout, R. “Optimal, reliable estimation of quantum states.” New Journal of Physics 12, 043034 (2010). doi:10.1088/1367-2630/12/4/043034.
• Fano, U. “Description of states in quantum mechanics by density matrix and operator techniques.” Reviews of Modern Physics 29, 74–93 (1957). doi:10.1103/RevModPhys.29.74.
• Nielsen, M. A., and I. L. Chuang. Quantum Computation and Quantum Information, 10th anniversary edition. Cambridge University Press, 2010. doi:10.1017/CBO9780511976667.
• Ruskai, M. B., S. Szarek, and E. Werner. “An analysis of completely-positive trace-preserving maps on $M_2$.” Linear Algebra and its Applications 347, 159–187 (2002). doi:10.1016/S0024-3795(01)00547-X; author manuscript.
• Watrous, J. The Theory of Quantum Information. Cambridge University Press, 2018. Author’s manuscript and errata.