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Quantum Mechanics Core Recommended

The Bloch Sphere

Picturing every possible single-qubit state as a point on a sphere

By the end of this topic you'll be able to

Explain why every pure single-qubit state can be drawn as a point on a unit sphere
Locate |0⟩, |1⟩, |+⟩, |−⟩, and the phase states on the sphere
Distinguish what rotating the Bloch vector corresponds to physically

A qubit's state has two complex amplitudes — four real numbers — but normalization removes one degree of freedom, and the overall phase of the state is unobservable: |ψ⟩ and e^{iφ}|ψ⟩ give identical measurement statistics for every conceivable measurement. Strip away those two, and exactly two real parameters remain — enough to locate a single point on a sphere.

|0⟩|1⟩
Polar angle θ90°
Phase angle φ0°
P(0) = 50%P(1) = 50%

Binney and Skinner derive the general form directly from the requirement that a measurement along an arbitrary direction n, with polar coordinates (θ,φ), return a definite value: the state certain to give +½ along n works out to a specific combination of sin(θ/2) and cos(θ/2), each carrying its own phase factor. Written in the computing convention as |ψ⟩ = cos(θ/2)|0⟩ + e^{iφ}sin(θ/2)|1⟩, θ is the polar angle from the north pole and φ the azimuthal angle around the equator.

|0⟩ occupies the north pole (θ=0) and |1⟩ the south pole (θ=π). The entire equator (θ=π/2) is filled with equal-superposition states distinguished purely by phase: |+⟩ sits at φ=0, |−⟩ at φ=π, and every other phase gives another point around the same circle.

A calculation of exactly this kind is what gave Binney and Skinner their result for a spin prepared along z and then measured along a tilted direction n: the probability of the opposite outcome comes out to sin²(θ/2), where θ is the angle between z and n. It vanishes when the two directions coincide and rises to one half when they are perpendicular — precisely the geometric statement that a measurement's outcome probabilities depend only on the angle between preparation and measurement axes.

This geometry makes the earlier claim that 'phase is invisible to direct measurement' visually obvious: a measurement in the computational basis reads off only θ, the polar angle, never φ. Two states that share a value of θ but differ in φ are measured identically in that basis, even though they are different states in every other respect.

That difference stops being hidden the instant a phase-sensitive gate is applied before measurement — which is exactly the mechanism behind quantum interference, and the reason a second Hadamard can turn an invisible phase into a visible, measurable flip.

Gates become rotations of this arrow: X is a half-turn about the x-axis that exchanges the poles (a bit flip), Z is a half-turn about the z-axis that leaves the poles fixed but rotates the equator by π (a phase flip), and H rotates the north pole onto the equator.

Once a gate is pictured as a rotation of an arrow on a sphere rather than an abstract matrix multiplication, single-qubit circuits stop feeling like symbol manipulation and start feeling geometric and, with practice, intuitive.

One limitation is worth flagging early: this is a picture of a single qubit's pure state only. It does not extend to two or more entangled qubits, and it needs a genuine modification — the density matrix, a few topics ahead — before it can represent a mixed state, where the corresponding point moves inside the sphere rather than sitting on its surface.

Try It Yourself

Worked Example

A qubit sits at Bloch angles θ = 60°, φ = 90°. Write out its state in α|0⟩+β|1⟩ form, and find the probability of measuring 0.

  1. 1Use the Bloch-vector form: |ψ⟩ = cos(θ/2)|0⟩ + e^{iφ}sin(θ/2)|1⟩, with θ = 60° so θ/2 = 30°.
  2. 2cos(30°) = √3/2, and sin(30°) = 1/2, so the amplitudes are α = √3/2 and β = e^{iπ/2}·(1/2) = i/2 (using φ = 90° = π/2 radians, and e^{iπ/2} = i from the Euler's-formula derivation in Complex Numbers & Dirac Notation).
  3. 3So |ψ⟩ = (√3/2)|0⟩ + (i/2)|1⟩.
  4. 4P(0) = |α|² = |√3/2|² = 3/4. (Check: P(1) = |i/2|² = 1/4, and 3/4+1/4 = 1. ✓)
Answer

|ψ⟩ = (√3/2)|0⟩ + (i/2)|1⟩, and P(measure 0) = 3/4 — note the phase φ affected β but never entered P(0), exactly as the geometry predicts.

Reference

Bloch vector formθ = polar angle, φ = azimuthal (phase) angle
|0⟩North pole
|1⟩South pole
|+⟩ = (|0⟩+|1⟩)/√2Equator, reference phase
|−⟩ = (|0⟩−|1⟩)/√2Equator, opposite phase

Quick Check

Why can a single pure qubit state be fully described by just two real numbers (θ, φ)?

What does the Hadamard gate do to a Bloch vector starting at the north pole?

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