The Bloch Sphere

The state of a single qubit looks, at first, like a small pile of complex numbers: two amplitudes, four real degrees of freedom, tangled together by rules about normalization and phase. The Bloch sphere sweeps that clutter away. Once you notice that the overall phase of a state can never be measured and that the total probability is pinned to one, exactly two real numbers survive — and two numbers are precisely what it takes to name a point on the surface of a sphere. The abstract algebra of amplitudes becomes a globe you can point at. Named for Felix Bloch, whose magnetization vector for nuclear spins is the same object in other language, the picture turns every question about a qubit into geometry. A pure state $|\psi\rangle = \cos\tfrac{\theta}{2}|0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}|1\rangle$ is a bearing $\phi$ and a latitude $\theta$ : north pole $|0\rangle$ , south pole $|1\rangle$ , equator the balanced superpositions. Mixed states — states of partial ignorance — fill the interior through $\rho = \tfrac12(I + \vec r\cdot\vec\sigma)$ , the radius of the Bloch vector measuring purity, the dead centre being total ignorance. The payoff arrives with dynamics. Every single-qubit gate is a rigid rotation of the ball; every measurement is a projection onto an axis; every Born-rule probability is a cosine. This article builds each correspondence, verifies the numbers, and shows why the innocuous $\tfrac12$ in the half-angle is the same "spinor" factor that makes a spin return to itself only after two full turns.

Key takeaways

The Bloch sphere maps every single-qubit state to a point in a unit ball in $\mathbb{R}^3$ : pure states on the surface ( $|\vec r|=1$ ), mixed states strictly inside ( $|\vec r|<1$ ), the maximally mixed state $I/2$ dead centre.
A pure qubit $|\psi\rangle = \cos\tfrac{\theta}{2}|0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}|1\rangle$ is the unit vector $\vec r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ — a latitude $\theta$ and a longitude $\phi$ .
The density operator is $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ , where the Bloch vector $\vec r=(\langle X\rangle,\langle Y\rangle,\langle Z\rangle)$ collects the three Pauli expectation values.
The radius is a purity meter, $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\vec r|^2)$ ; global phase is invisible, so orthogonal states sit at antipodes and geometric angles are twice Hilbert-space angles.
Every single-qubit unitary is a rotation of the ball, $U=e^{-i\alpha\,\hat n\cdot\vec\sigma/2}$ ; every measurement is a projection, $P(+\hat m)=\tfrac12(1+\hat m\cdot\vec r)$ .
Key idea: Strip away global phase and normalization from a qubit and exactly two real parameters remain — encode them as the polar and azimuthal angles of a point on the unit sphere. Density operators then fill the ball via $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ , turning algebra into geometry.

Where this lives today

The Bloch sphere is not a teaching cartoon experimentalists outgrow — it is the working coordinate system of the qubit laboratory. Single-qubit tomography measures $\langle X\rangle,\langle Y\rangle,\langle Z\rangle$ on many identical copies, assembles them into $\vec r$ , and reconstructs $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ ; the radius is read directly as a fidelity budget. On a superconducting or trapped-ion processor, calibrating a gate means calibrating a rotation: a Rabi experiment drives the Bloch vector around an in-plane axis, its oscillation period fixing the pulse amplitude, while a Ramsey experiment lets it precess about $\hat z$ , the fringe frequency fixing the qubit detuning. Decoherence is diagnosed geometrically: the transverse shrink rate gives $T_2$ , the longitudinal relaxation gives $T_1$ , and a channel's noise is exactly the affine map by which it contracts and displaces the ball. The same sphere reappears wholesale in optics as the Poincaré sphere for the polarization of light — one geometry, two physical incarnations, because both describe a two-level system.

The physical picture

Think of a globe. Put the definite state $|0\rangle$ at the north pole and $|1\rangle$ at the south. Every other pure state is a definite latitude and a definite longitude: the polar angle $\theta$ says how much $|0\rangle$ versus $|1\rangle$ the state carries, and the azimuth $\phi$ sets the relative phase between the two amplitudes. The equator is the set of perfectly balanced superpositions — $|+\rangle$ and $|-\rangle$ on one axis, $|i\rangle$ and $|{-i}\rangle$ on the other. Nothing about the globe records the overall phase of the state, because nothing physical ever does. Because that phase is invisible, you are forced onto the half-angle: the amplitudes are $\cos\tfrac{\theta}{2}$ and $e^{i\phi}\sin\tfrac{\theta}{2}$ , so a $180^\circ$ trip on the sphere is a $90^\circ$ turn between the underlying rays. Orthogonal states end up antipodal, not perpendicular — $|0\rangle$ and $|1\rangle$ are pole to pole. The first figure fixes the geography and carries a number: a state at latitude $\theta=60^\circ$ predicts a $75\%$ outcome before any algebra.

Figure

A pure qubit as a point on the unit sphere. The polar angle θ sets the 0-versus-1 weight and the azimuth φ sets the relative phase. Here θ = 60°, φ = 90° gives Bloch vector r = (0, 0.87, 0.50) and outcome probability P(0) = cos²(θ⁄2) = 0.75

Purity is radius. A surface point is pure and maximally informative; slide inward and you mix in classical ignorance; the dead centre $I/2$ reads $50/50$ along every axis at once. Decoherence is literally the Bloch vector shrinking toward the centre, which is why $|\vec r|$ is a faithful purity meter through $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\vec r|^2)$ . The second figure is a cross-section of the ball with the numbers attached.

Figure

A cross-section of the Bloch ball. The radius r fixes purity through Tr(ρ²) = ½(1 + r²): the surface r = 1 is a pure state (Tr ρ² = 1), the interior point r = 0.79 is a mixture (Tr ρ² = 0.81), and the centre r = 0 is the maximally mixed state I⁄2 (Tr ρ² = 0.5)

Finally, gates are rigid rotations of this ball. The Pauli $X$ spins it $180^\circ$ about the $x$ -axis, swapping the poles; a phase gate spins it about $z$ ; the Hadamard is a $180^\circ$ turn about $\hat n=(1,0,1)/\sqrt2$ that trades $\hat x$ and $\hat z$ . Composing gates is composing rotations, and the whole toolkit of Euler angles comes for free.

The core equations

The parametrization is the starting point: after removing the unphysical global phase and imposing normalization, a pure state is fixed by two angles. The convention takes the $|0\rangle$ amplitude to be real and non-negative.

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

where: $\theta$ — polar angle, $\theta\in[0,\pi]$ ; $\theta=0$ gives $|0\rangle$ , $\theta=\pi$ gives $|1\rangle$ ; $\phi$ — azimuthal angle, $\phi\in[0,2\pi)$ , sets the relative phase; $e^{i\phi}$ — relative phase; the global phase has been fixed by making the $|0\rangle$ amplitude real. Reading it: latitude controls the $Z$ -bias, longitude controls the phase. The three real coordinates of the point on the sphere are nothing but the three Pauli expectation values — the numbers a tomographer measures directly. This is the bridge between operators and geometry.

r_x=\langle X\rangle,\quad r_y=\langle Y\rangle,\quad r_z=\langle Z\rangle \qquad (2)

where: $\langle A\rangle=\mathrm{Tr}(\rho A)$ — expectation value of observable $A$ in state $\rho$ ; $r_z=\langle Z\rangle=|a|^2-|b|^2$ — the population imbalance between $|0\rangle$ and $|1\rangle$ . Reading it: the Bloch vector is a table of measurement averages, one per axis. Assembling those three numbers reconstructs the entire density operator, because $\{I,X,Y,Z\}$ is a complete basis for $2\times2$ Hermitian matrices. This one line works for pure and mixed states alike.

\rho = \tfrac{1}{2}\left(I + \vec{r}\cdot\vec{\sigma}\right) = \tfrac{1}{2}\left(I + r_x X + r_y Y + r_z Z\right) \qquad (3)

where: $\rho$ — density operator, $2\times2$ , Hermitian, $\mathrm{Tr}\,\rho=1$ , $\rho\ge0$ ; $\vec r=(r_x,r_y,r_z)\in\mathbb{R}^3$ with $|\vec r|\le1$ — the Bloch vector; $\vec\sigma=(X,Y,Z)$ — the Pauli triple. Reading it: half the identity plus a "tilt" in the direction $\vec r$ ; a longer tilt means a more definite state. The single number that separates pure from mixed is the length of the Bloch vector, and it enters purity as a clean quadratic. This is the arithmetic behind "radius equals purity".

\mathrm{Tr}(\rho^2) = \tfrac{1}{2}\left(1 + |\vec{r}|^2\right) \qquad (4)

where: $\mathrm{Tr}(\rho^2)$ — purity, $1$ for pure states, $\tfrac12$ for the maximally mixed state; $|\vec r|^2=r_x^2+r_y^2+r_z^2$ — squared radius. Reading it: purity slides linearly in $|\vec r|^2$ from $\tfrac12$ at the centre to $1$ on the surface. Measurement collapses to a dot product. To measure the observable $\hat m\cdot\vec\sigma$ along any axis $\hat m$ , project the Bloch vector onto that axis. The Born rule becomes trigonometry.

P(+\hat{m}) = \tfrac{1}{2}\left(1 + \hat{m}\cdot\vec{r}\right) \qquad (5)

where: $\hat m$ — unit measurement axis; $+\hat m$ — the $+1$ eigenstate of $\hat m\cdot\vec\sigma$ ; $P(+\hat m)$ — probability of the $+1$ outcome. Reading it: the more the state points along the measurement axis, the more likely the $+1$ result; perpendicular gives an even split. Dynamics is rotation. Any single-qubit unitary is, up to an irrelevant global phase, generated by a Pauli combination, and its action on the ball is a rigid rotation by $\alpha$ about the axis $\hat n$ .

U=e^{-i\alpha\,\hat{n}\cdot\vec{\sigma}/2}=\cos\tfrac{\alpha}{2}\,I - i\sin\tfrac{\alpha}{2}\,\hat{n}\cdot\vec{\sigma} \qquad (6)

where: $\hat n$ — unit rotation axis in $\mathbb{R}^3$ ; $\alpha$ — rotation angle; conjugation $\rho\mapsto U\rho U^\dagger$ rotates $\vec r$ by $\alpha$ about $\hat n$ . Reading it: the half-angle $\alpha/2$ inside the exponential is the spinor double cover — the state needs $\alpha=4\pi$ to return, the ball only $2\pi$ .

Derivation AWhy a qubit needs exactly two real numbers, and why the half-angle

The goal is to show, from first principles, that a general normalized single-qubit state collapses from four real degrees of freedom down to two physically meaningful ones, and that the natural coordinates for those two are a polar angle $\theta$ appearing as a half-angle in the amplitudes together with an azimuth $\phi$ . We start from the most general complex superposition, spend the normalization constraint and the unobservability of global phase, and read off the surviving parametrization — making explicit at each step which physical fact removes which parameter.

Step 1. Write the most general state $|\psi\rangle = a|0\rangle + b|1\rangle$ with $a=|a|e^{i\alpha}$ and $b=|b|e^{i\beta}$ , so there are four real numbers on the table: two magnitudes and two phases. Why: a qubit lives in $\mathbb{C}^2$ , and an unconstrained complex 2-vector has exactly four real components — this is the raw material we must whittle down.

Step 2. Impose normalization $|a|^2+|b|^2=1$ ; one real constraint removes one degree of freedom, leaving three. Why: only unit vectors represent physical states under the Born rule, so the normalization sphere is not optional — it is what makes probabilities sum to one.

Step 3. Multiply by the global phase $e^{-i\alpha}$ to get $|\psi\rangle\sim |a|\,|0\rangle + |b|e^{i(\beta-\alpha)}|1\rangle$ ; since $e^{-i\alpha}$ is physically undetectable, only the relative phase $\phi=\beta-\alpha$ survives, removing one more parameter and leaving two. Why: no experiment can distinguish $|\psi\rangle$ from $e^{i\gamma}|\psi\rangle$ , because every probability $|\langle x|\psi\rangle|^2$ is phase-blind — this is the single fact that makes the sphere two-dimensional rather than three.

Step 4. Solve the remaining magnitude constraint with a single angle by setting $|a|=\cos\tfrac{\theta}{2}$ and $|b|=\sin\tfrac{\theta}{2}$ ; then $\theta$ ranges over $[0,\pi]$ as $|a|$ runs from $1$ to $0$ , and pairing it with $\phi\in[0,2\pi)$ gives spherical coordinates on $S^2$ . Why: the half-angle is forced, not chosen — it is what lets the full polar sweep $\theta:0\to\pi$ carry the amplitude all the way from $|0\rangle$ to $|1\rangle$ , and it is exactly why orthogonal states $|0\rangle,|1\rangle$ land at antipodes $180^\circ$ apart while being only $90^\circ$ apart as rays. *As a quick check: with $\theta=60^\circ,\phi=90^\circ$ the amplitudes are $\cos30^\circ=0.866$ and $i\sin30^\circ=0.5i$ , giving $\vec r=(0,\sqrt3/2,1/2)$ and $P(0)=\cos^2 30^\circ=0.75$ , consistent with Figure 1.*

Go deeper · Derivation B · 1 minThe density operator is the Bloch ball: positivity becomes a radius bound

Derivation BThe density operator is the Bloch ball: positivity becomes a radius bound

The goal is to establish the affine isomorphism between qubit density operators and points of the closed unit ball, and to prove that the single physical requirement $\rho\ge0$ is exactly equivalent to the geometric bound $|\vec r|\le1$ . We expand an arbitrary Hermitian matrix in the Pauli basis, fix two coefficients with the trace condition, diagonalize to get the eigenvalues in terms of the radius, and then let positivity speak — the whole taxonomy of pure, mixed, and forbidden states falls out of one inequality.

Step 1. Any $2\times2$ Hermitian matrix expands as $\rho=c_0 I+\vec c\cdot\vec\sigma$ with real $c_0,\vec c$ , because $\{I,X,Y,Z\}$ is a real basis for the four-dimensional space of Hermitian $2\times2$ matrices. Why: a density operator must be Hermitian, and Hermiticity is precisely the statement that it lives in this real span — no complex coefficients appear.

Step 2. The trace condition $\mathrm{Tr}\,\rho=1$ gives $2c_0=1$ , so $c_0=\tfrac12$ ; defining $\vec r=2\vec c$ yields $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ . Why: $\mathrm{Tr}\,\rho=1$ encodes "probabilities sum to one" and it pins the identity component completely, leaving only the three-vector $\vec r$ free.

Step 3. Diagonalize: $\hat r\cdot\vec\sigma$ has eigenvalues $\pm1$ , so $\rho$ has eigenvalues $\lambda_\pm=\tfrac12(1\pm|\vec r|)$ , equivalently $\det\rho=\tfrac14(1-|\vec r|^2)$ with $\mathrm{Tr}\,\rho=1$ . Why: the eigenvalues are the outcome probabilities in $\rho$ 's own eigenbasis, and expressing them through $|\vec r|$ is what converts an operator inequality into a scalar one.

Step 4. Demand positivity $\lambda_\pm\ge0$ ; the smaller eigenvalue $\lambda_-=\tfrac12(1-|\vec r|)\ge0$ holds exactly when $|\vec r|\le1$ , with equality giving $\lambda_-=0$ — a rank-one projector, i.e. a pure state on the surface. Why: $\rho\ge0$ is the last axiom of a valid state, and it draws the boundary of the ball; points with $|\vec r|>1$ have a negative eigenvalue and are not states at all. *As a consistency check: $\det\rho=\tfrac14(1-|\vec r|^2)\ge0$ together with $\mathrm{Tr}\,\rho=1$ reproduce $\lambda_\pm=\tfrac12(1\pm|\vec r|)$ ; at $|\vec r|=1$ one eigenvalue vanishes, confirming a pure projector.*

Go deeper · Derivation C · 1 minGates are rotations and measurements are projections

Derivation CGates are rotations and measurements are projections

The goal is to close the dictionary between quantum operations and Euclidean geometry: to show that the unitary $e^{-i\alpha\hat n\cdot\vec\sigma/2}$ rotates the Bloch vector by $\alpha$ about $\hat n$ , and that measuring $\hat m\cdot\vec\sigma$ gives the outcome probability $\tfrac12(1+\hat m\cdot\vec r)$ . We exponentiate the Pauli generator using its involution property, note the induced action on $\vec r$ through conjugation, and then trace the state against the eigenprojector to recover the Born rule as a dot product.

Step 1. Because $\hat n$ is a unit vector, $(\hat n\cdot\vec\sigma)^2=I$ ; feeding this into the exponential series, even powers give $I$ and odd powers give $\hat n\cdot\vec\sigma$ , so $e^{-i\alpha\hat n\cdot\vec\sigma/2}=\cos\tfrac{\alpha}{2}I-i\sin\tfrac{\alpha}{2}\,\hat n\cdot\vec\sigma$ . Why: the involution collapses an infinite series into a two-term Euler identity — the same trick that turns $e^{i\alpha}$ into $\cos+i\sin$ , now for matrices.

Step 2. Conjugate a Pauli operator: using $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$ , one finds $U(\hat m\cdot\vec\sigma)U^\dagger=(R\hat m)\cdot\vec\sigma$ where $R\in SO(3)$ is rotation by $\alpha$ about $\hat n$ ; hence $\rho\mapsto U\rho U^\dagger$ sends $\vec r\mapsto R\vec r$ . Why: this is the two-to-one homomorphism $SU(2)\to SO(3)$ made concrete — the qubit's half-angle $\alpha/2$ becomes the ball's full angle $\alpha$ , so a $2\pi$ state rotation is only a $\pi$ ball rotation.

Step 3. For measurement, the $+1$ eigenprojector of $\hat m\cdot\vec\sigma$ is $\Pi_+=\tfrac12(I+\hat m\cdot\vec\sigma)$ ; then $P(+\hat m)=\mathrm{Tr}(\rho\,\Pi_+)=\tfrac14\mathrm{Tr}[(I+\vec r\cdot\vec\sigma)(I+\hat m\cdot\vec\sigma)]=\tfrac12(1+\hat m\cdot\vec r)$ , since $\mathrm{Tr}[(\vec r\cdot\vec\sigma)(\hat m\cdot\vec\sigma)]=2\,\vec r\cdot\hat m$ and single Pauli traces vanish. Why: the Born rule collapses to the projection of $\vec r$ onto the measurement axis, so probabilities become cosines of the angle between state and apparatus — the geometric heart of the picture. *Three explicit checks underpin the derivation: $(\hat n\cdot\vec\sigma)^2=I$ (the involution used in Step 1); the cross-trace $\mathrm{Tr}[(\vec r\cdot\vec\sigma)(\hat m\cdot\vec\sigma)]=2\vec r\cdot\hat m$ (used in Step 3); and $R_x(\pi/2)$ acting on $|0\rangle$ sends $\hat z\mapsto-\hat y$ , i.e. $|0\rangle\mapsto(|0\rangle-i|1\rangle)/\sqrt2=|{-i}\rangle$ , matching the worked example.*

Limiting cases

The radius and angles encode the full state taxonomy in a single table.

Case	Result	Meaning
Pure state	$\vert\vec r\vert = 1,\ \ \rho=\vert\psi\rangle\langle\psi\vert$	On the surface; $\mathrm{Tr}(\rho^2)=1$ ; maximal single-qubit information.
Maximally mixed	$\vec r = 0,\ \ \rho=\tfrac12 I$	The centre; every axis reads $50/50$ ; $\mathrm{Tr}(\rho^2)=\tfrac12$ , the minimum.
General mixed	$0<\vert\vec r\vert<1$	Strictly interior; a convex mix of pure states; $\mathrm{Tr}(\rho^2)=\tfrac12(1+\vert\vec r\vert^2)\in(\tfrac12,1)$ .
Antipodal points	$\vec r\to-\vec r\ \Leftrightarrow\ \langle\psi\vert\chi\rangle=0$	Orthogonal pure states are diametrically opposite: $\vert0\rangle/\vert1\rangle$ , $\vert+\rangle/\vert-\rangle$ , $\vert i\rangle/\vert{-i}\rangle$ .
Pure dephasing	$r_x,r_y\to0$ , $r_z$ fixed	Equator collapses to the polar axis over $T_2$ ; off-diagonal coherences decay.
Energy relaxation	$\vec r\to$ ground point along $\hat z$	Amplitude damping over $T_1$ drives the state toward $\|0\rangle$ (or the thermal point).

Common trapOrthogonal means antipodal, not perpendicular

The half-angle $\theta/2$ trips up nearly everyone. The states $|0\rangle$ and $|1\rangle$ are orthogonal in Hilbert space — their inner product is zero — yet on the Bloch sphere they sit at the north and south poles, $180^\circ$ apart, not $90^\circ$ apart. Geometric angles on the sphere are twice the Hilbert-space angles between rays: two pure states whose Bloch vectors make an angle $\gamma$ have squared overlap $|\langle\psi|\chi\rangle|^2=\cos^2(\gamma/2)=\tfrac12(1+\cos\gamma)$ , so orthogonality ( $\text{overlap}=0$ ) means $\gamma=\pi$ , i.e. antipodal. If you ever find yourself placing $|+\rangle$ at $90^\circ$ from $|0\rangle$ and expecting them to be orthogonal, you have dropped the factor of two.

Worked examples

1 · A mixed state from a preparation recipe (fully worked)

Step 1. Convert each pure ingredient to its Bloch vector: $|0\rangle$ sits at $\hat z=(0,0,1)$ and $|+\rangle$ at $\hat x=(1,0,0)$ . The Bloch vector of a probabilistic mixture is the same convex combination of the ingredient vectors, so $\vec r=0.75(0,0,1)+0.25(1,0,0)=(0.25,\,0,\,0.75)$ . The length is $|\vec r|=\sqrt{0.25^2+0.75^2}=\sqrt{0.625}=\mathbf{0.791}$ — safely inside the ball, so this is a genuine mixture.

Step 2. Purity follows from the radius: $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\vec r|^2)=\tfrac12(1+0.625)=\mathbf{0.8125}$ . As a cross-check, form $\rho=\begin{pmatrix}0.875&0.125\\0.125&0.125\end{pmatrix}$ and compute $\mathrm{Tr}(\rho^2)=0.875^2+2(0.125^2)+0.125^2=0.8125$ — the two routes agree.

Step 3. The eigenvalues are $\lambda_\pm=\tfrac12(1\pm|\vec r|)=\tfrac12(1\pm0.791)=\{\mathbf{0.895},\,\mathbf{0.105}\}$ , so the state is effectively a $90/10$ mixture of two antipodal pure states lying along $\pm\hat r$ . This is the spectral decomposition read straight off the geometry.

Step 4. The $Z$ -measurement probability is a projection onto $\hat z$ : $P(0)=\tfrac12(1+\hat z\cdot\vec r)=\tfrac12(1+0.75)=\mathbf{0.875}$ . No amplitude algebra was needed — just a dot product. For contrast, an $X$ -measurement gives $P(+)=\tfrac12(1+0.25)=0.625$ and a $Y$ -measurement gives $P(+i)=\tfrac12(1+0)=0.5$ . Result: $\vec r=(0.25,0,0.75)$ , purity $0.8125$ , eigenvalues $\{0.895,0.105\}$ , $P(0)=0.875$ .

2 · A gate as a rotation of the ball

Apply the rotation $R_x(\pi/2)=e^{-i(\pi/2)X/2}$ to $|0\rangle$ . Where does the state land, and what does a $Y$ -measurement then read?

Step 1. Evaluate the gate with equation (6): $R_x(\pi/2)=\cos\tfrac{\pi}{4}I-i\sin\tfrac{\pi}{4}X=\tfrac{1}{\sqrt2}(I-iX)$ . Acting on $|0\rangle$ gives $\tfrac{1}{\sqrt2}(|0\rangle-i|1\rangle)=\mathbf{|{-i}\rangle}$ .

Step 2. Read it geometrically: $|0\rangle$ is the north pole $\hat z=(0,0,1)$ , and a $+90^\circ$ rotation about $\hat x$ sends $\hat z\mapsto-\hat y$ . So the output Bloch vector is $\mathbf{(0,-1,0)}$ — a point on the equator, exactly the $|{-i}\rangle$ state, confirming the amplitude calculation without matrices.

Step 3. A $Y$ -measurement projects onto $\hat y$ : $P(+i)=\tfrac12(1+\hat y\cdot\vec r)=\tfrac12(1-1)=\mathbf{0}$ , so the outcome is deterministically $-i$ . A $Z$ -measurement, by contrast, would read $P(0)=\tfrac12(1+0)=0.5$ — a perfect coin, because the state now lies on the equator, $90^\circ$ from the $Z$ -axis. Result: $R_x(\pi/2)|0\rangle=|{-i}\rangle$ , Bloch vector $(0,-1,0)$ , and the $Y$ -outcome is $-i$ with certainty.

Physics of the portrait

The Bloch sphere arrived, in a sense, forty years early — under a different name, for a different job, in the hands of a physicist mapping the precession of nuclear spins.

Origins · 1 minFelix Bloch and how this was discovered

Felix Bloch

1905–1983 · quantum theory of solids and nuclear magnetic resonance

Felix Bloch was born in Zurich in 1905 and became Werner Heisenberg's first doctoral student, producing in 1928 the theorem on electrons in periodic potentials — Bloch's theorem, Bloch waves — that founded the quantum theory of solids. Forced out of Germany in 1933, he settled at Stanford and turned to the magnetism of nuclei. In 1946 he wrote down the phenomenological equations of nuclear induction, describing how a macroscopic magnetization vector precesses, tips, and relaxes in a field — the Bloch equations — and detected nuclear magnetic resonance in bulk matter, simultaneously with Edward Purcell at Harvard; the two shared the 1952 Nobel Prize. That magnetization vector is precisely the Bloch vector of a single spin- $\tfrac12$ : its tip traces the surface of a sphere for a pure state and shrinks inward as relaxation mixes it. Half a century later, when quantum information needed a picture for a qubit, it borrowed Bloch's sphere wholesale — the poles relabelled $|0\rangle$ and $|1\rangle$ , the precession relabelled a phase gate, the relaxation relabelled decoherence. Bloch, later the first Director-General of CERN, had built the geometry of the qubit before the qubit had a name.

History

1927

— Pauli introduces the spin matrices for the electron

1928

— Bloch's theorem founds the quantum band theory of solids

1938

— Rabi detects nuclear magnetic resonance in molecular beams

1946

— Bloch formulates the equations of nuclear induction and the magnetization vector

1952

— Bloch and Purcell share the Nobel Prize for nuclear magnetic resonance

2000

— Nielsen and Chuang make the Bloch sphere the standard picture of a qubit

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, §1.2 and §4.2 (Bloch sphere; single-qubit gates as rotations) · Preskill, Quantum Information lecture notes, Ch. 2 (density operators and the Bloch ball) · Watrous, The Theory of Quantum Information, §2.1 · Wilde, Quantum Information Theory, Ch. 3 (qubit and Bloch representation) · Kitaev, Shen & Vyalyi, Classical and Quantum Computation, §8 · Mermin, Quantum Computer Science, Ch. 1. The key relations are: (i) the affine isomorphism $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$ with eigenvalues $\tfrac12(1\pm|\vec r|)$ and $\det\rho=\tfrac14(1-|\vec r|^2)$ ; (ii) purity $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\vec r|^2)$ , which follows from $(\vec r\cdot\vec\sigma)^2=|\vec r|^2 I$ ; (iii) the Born-rule projection $P(+\hat m)=\tfrac12(1+\hat m\cdot\vec r)$ via $\Pi_+=\tfrac12(I+\hat m\cdot\vec\sigma)$ ; (iv) the gate exponential $e^{-i\alpha\hat n\cdot\vec\sigma/2}=\cos\tfrac\alpha2 I-i\sin\tfrac\alpha2\,\hat n\cdot\vec\sigma$ . The worked numbers ( $\vec r=(0.25,0,0.75)$ , purity $0.8125$ , eigenvalues $\{0.895,0.105\}$ ; $R_x(\pi/2)|0\rangle=|{-i}\rangle$ ) were verified symbolically and numerically. The Pauli identities $\mathrm{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$ and $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$ are standard.

Builds on: The Qubit and Its State Space · Dirac Bra-Ket Notation · Global and Relative Phase · Pauli Operators · Density Operators. Leads to: Born Rule · Projective Measurements · Expectation Values and Variance · Observables and Hermitian Operators.

Ask the tutor: Want to watch a specific gate sequence steer the Bloch vector, or see how tomography rebuilds $\rho$ from measured $\langle X\rangle,\langle Y\rangle,\langle Z\rangle$ ? Ask and we will trace it on the sphere.

Why the half-angle θ/2 instead of θ?

Because the amplitudes must satisfy $|a|^2+|b|^2=1$ , the natural substitution $|a|=\cos\tfrac{\theta}{2}$ , $|b|=\sin\tfrac{\theta}{2}$ solves the constraint with one parameter while letting $\theta$ sweep the full polar range $[0,\pi]$ . The physical consequence is the whole personality of the picture: geometric angles on the sphere are twice the Hilbert-space angles between rays. Orthogonal states, which are $90^\circ$ apart as rays, appear $180^\circ$ apart — antipodal — on the sphere. Drop the half and $|0\rangle$ and $|1\rangle$ would land only a quarter-turn apart, and every measurement probability would come out wrong.

Why does global phase never appear on the sphere?

Because it is not physical. Multiplying a state by $e^{i\gamma}$ leaves every measurable quantity untouched: each outcome probability is $|\langle x|\psi\rangle|^2$ , and the phase cancels in the modulus. Two state vectors that differ only by an overall phase are the same physical state — they are the same ray in Hilbert space — so a faithful picture must not distinguish them. The Bloch sphere achieves this by parametrizing rays, not vectors: fixing the $|0\rangle$ amplitude to be real and non-negative spends the global-phase freedom, and only the relative phase $\phi$ survives as the azimuth. What is invisible in the lab is invisible on the sphere.

What is the difference between a surface point and an interior point?

Radius, and with it purity. A surface point has $|\vec r|=1$ and is a pure state $\rho=|\psi\rangle\langle\psi|$ with $\mathrm{Tr}(\rho^2)=1$ — maximal information, a definite answer along at least one axis. An interior point has $|\vec r|<1$ and is a genuine mixture, $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\vec r|^2)<1$ , carrying classical uncertainty; its eigenvalues $\tfrac12(1\pm|\vec r|)$ are the probabilities in the state's own basis and spread apart toward the surface. The very centre, $\vec r=0$ , is the maximally mixed state $I/2$ : nothing is known, every axis a fair coin.

Why is the maximally mixed state at the centre, and basis-independent?

Average any pair of antipodal pure states equally and you land at the origin: $\tfrac12|0\rangle\langle0|+\tfrac12|1\rangle\langle1|=\tfrac12 I$ , and identically $\tfrac12|+\rangle\langle+|+\tfrac12|-\rangle\langle-|=\tfrac12 I$ . The centre is the one point with no preferred direction, reached by an equal mixture along any axis whatsoever — that basis-independence is exactly what "maximally mixed" means. Physically it is total ignorance: measured along $X$ , $Y$ , or $Z$ it reads $50/50$ every time. It is also the fixed point toward which strong decoherence drives every state, which is why $\vec r$ shrinking to zero is the signature of a qubit losing all its quantum information.

Why is a gate a rotation, and what is the two-to-one business?

A single-qubit unitary can always be written $U=e^{-i\alpha\hat n\cdot\vec\sigma/2}$ up to global phase, and conjugating the Bloch operator by it, $\rho\mapsto U\rho U^\dagger$ , rotates the Bloch vector by angle $\alpha$ about the axis $\hat n$ . So the group of qubit unitaries $SU(2)$ maps onto the group of ball rotations $SO(3)$ — but two-to-one, because $U$ and $-U$ give the same rotation. The visible sign of this is the half-angle inside the exponential: rotating the state by $\alpha=2\pi$ multiplies it by $-1$ (a $360^\circ$ turn does not bring a qubit home), while the ball has already returned after $\alpha=2\pi$ maps to a $2\pi$ rotation of $\vec r$ . This "spinor double cover" is the same factor of two that makes a spin- $\tfrac12$ particle require two full turns to come back to itself.

Does the Bloch sphere work for two or more qubits?

No — and the reason matters. The sphere is exhausted by a single qubit: two real parameters pure, three mixed. Two qubits already need six real parameters for a pure state and fifteen for a mixed one, with no faithful three-dimensional ball. Worse, the headline two-qubit feature, entanglement, is exactly what a single Bloch sphere cannot show: an entangled pair has each qubit individually in a mixed reduced state inside its own ball, while the correlations that make the pair special live in the relationship between the two, on neither sphere. Two Bloch spheres side by side never display the entanglement — that needs the joint correlation matrix or full two-qubit tomography.

Is the Bloch sphere the same as the Poincaré sphere?

Mathematically yes, physically a cousin. The Poincaré sphere describes the polarization of light, a two-level system exactly like a qubit: its poles are right- and left-circular polarization, its equator the linear polarizations, its interior partial polarization — the optical analogue of mixedness. The Stokes parameters play the role of the Bloch components, and a wave plate acts as a rotation just as a gate does. The identity is no coincidence: any two-state quantum or wave system carries the same $SU(2)$ structure, so it wears the same sphere, and intuition carries freely between quantum information and polarization optics.