The Bloch Sphere

A qubit \alpha|0\rangle + \beta|1\rangle hides behind two complex amplitudes — four real numbers in total. That is impossible to draw directly. But two of those four are invisible: normalisation (|\alpha|^2 + |\beta|^2 = 1) removes one, and the overall global phase — multiplying the whole state by e^{i\gamma} — can never be measured, removing another. Strip those two away and only two real angles remain. Two angles is exactly what it takes to name a point on a globe — so every pure qubit state is a single point on the surface of a sphere, the Bloch sphere, and the whole of single-qubit quantum mechanics becomes geometry you can hold in your hand.

Latitude and longitude of a state

Writing the amplitudes with those two angles gives the standard form

|\psi\rangle = \cos\tfrac{\theta}{2}\,|0\rangle + e^{i\varphi}\sin\tfrac{\theta}{2}\,|1\rangle,

where the polar angle \theta \in [0, \pi] is the latitude (measured down from the top) and the azimuthal angle \varphi \in [0, 2\pi) is the longitude. The point on the unit sphere has the familiar spherical coordinates (\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta). The poles are the classical states and the equator is where superposition is strongest:

the north pole (\theta = 0) is |0\rangle;
the south pole (\theta = \pi) is |1\rangle;
the equator (\theta = \tfrac{\pi}{2}) holds the equal superpositions — |{+}\rangle, |{-}\rangle, |{+}i\rangle, |{-}i\rangle — as \varphi sweeps round.

Steer the state yourself

Drag the box to rotate the sphere, and move \theta and \varphi to fly the state vector around it. Watch how \theta = 0 parks you on |0\rangle at the top, while sliding \theta to 90^\circ drops you onto the equator and \varphi then just spins you around it.

every pure qubit state is one point on a unit sphere;
|\psi\rangle = \cos\tfrac{\theta}{2}|0\rangle + e^{i\varphi}\sin\tfrac{\theta}{2}|1\rangle: \theta latitude, \varphi longitude;
poles are |0\rangle (north) and |1\rangle (south); the equator holds the equal superpositions;
global phase is dropped (it is unobservable) — which is exactly why two angles suffice;
a single-qubit unitary gate acts as a rotation of the whole sphere.

Worked example: placing a state

Where does |{+}\rangle = \tfrac{1}{\sqrt2}\,|0\rangle + \tfrac{1}{\sqrt2}\,|1\rangle live? Match it to the standard form: we need \cos\tfrac{\theta}{2} = \tfrac{1}{\sqrt2}, so \tfrac{\theta}{2} = 45^\circ and \theta = 90^\circ — on the equator. The amplitudes are both positive and real, so e^{i\varphi} = 1 and \varphi = 0. Its coordinates are therefore

(\sin 90^\circ\cos 0^\circ,\ \sin 90^\circ \sin 0^\circ,\ \cos 90^\circ) = (1, 0, 0),

the positive x-axis — which is exactly where |{+}\rangle is conventionally drawn. Swapping the sign to |{-}\rangle makes \varphi = 180^\circ, sending it to the opposite side, (-1, 0, 0).

It looks like a typo — why \theta/2 and not \theta? Because we want opposite points on the sphere to be orthogonal states. The poles |0\rangle and |1\rangle are antipodal — a full 180^\circ apart on the globe — yet as quantum states they are perpendicular (their inner product is zero). The half-angle does the bookkeeping: a 180^\circ journey across the sphere corresponds to only a 90^\circ turn in the state, taking |0\rangle to an orthogonal |1\rangle. Physicists call this the "spin-½" behaviour — you have to turn a qubit through 720^\circ in state-space to bring it fully back to where it started.

The Bloch sphere is a picture of a single qubit — and only a pure one. It is tempting to imagine drawing two qubits as two dots on two spheres, but that quietly throws away the most important quantum resource of all: entanglement. An entangled pair has no individual state to plot; the two-qubit state simply does not factor into "this qubit here, that qubit there". And remember what the sphere leaves out: the unobservable global phase. It is a wonderful map of one qubit, not a universal picture of a quantum computer.