University Physics IV · Frontiers of Modern Physics · 15.1

Two-State Systems & the Bloch Sphere

Almost every quantum system you will ever actually control is a two-level system in disguise. This is the toolkit for handling one: normalise it, throw away the global phase, read the two angles that survive off the Bloch sphere, and reach for the density matrix the moment the state stops being pure.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A qubit is not a new kind of object. It is what any quantum system becomes once every level but two can be ignored: spin-½ in a field, a photon's polarisation, the lowest two rungs of a superconducting circuit. All of them live in a two-dimensional complex Hilbert space, so the state is a|0⟩ + b|1⟩ — four real numbers, two of which are not physical. Normalisation |a|² + |b|² = 1 removes one; the global phase removes another, because e^(iγ)|ψ⟩ returns identical probabilities in every basis and identical expectation values for every observable.

What survives is two real parameters, and the natural way to hold them is as angles: a = cos(θ/2), b = e^(iφ) sin(θ/2), a point on a sphere. The half-angle is not decoration. It is what forces orthogonal states to be antipodal, and it is why a 2π rotation returns −|ψ⟩. Unitary evolution is then a rigid rotation of that sphere, so gates cannot move a point off the surface — which is already a warning that the surface cannot hold ignorance.

Entangle the system with anything you do not track and no ket describes it; the density matrix ρ = ½(I + r⋅σ) does, with |r| < 1 putting the state inside the ball. The whole picture is rented from the truncation it began with: populate a third level and the sphere is a lie.

Simple definition: A qubit is a quantum system whose accessible states span a two-dimensional Hilbert space, so every pure state is a normalised superposition a|0⟩ + b|1⟩ carrying exactly two physically meaningful real parameters.
Example: |ψ⟩ = cos 30°|0⟩ + e^(i⋅40°) sin 30°|1⟩ sits at θ = 60°, φ = 40°: a measurement in the computational basis returns |1⟩ with probability sin² 30° = 0.250, and ⟨σ_z⟩ = cos 60° = 0.500.

Pure state in two angles|ψ⟩ = cos(θ/2)|0⟩ + e^(iφ) sin(θ/2)|1⟩

Four real numbers in a and b fall to two once the norm and the global phase are spent.

θ from 0 to π measured from +z, φ from 0 to 2π about z; both dimensionless

Statistics in the computational basisP(0) = cos²(θ/2), P(1) = sin²(θ/2), ⟨σ_z⟩ = cos θ

φ is absent here — the relative phase stays invisible until you measure off the z axis.

Probabilities dimensionless and summing to 1; ⟨σ_z⟩ lies between −1 and +1

Bloch vector from three averagesr = (⟨σₓ⟩, ⟨σ_y⟩, ⟨σ_z⟩) = (sin θ cos φ, sin θ sin φ, cos θ)

Three measured averages fix the state outright: that is single-qubit tomography in one line.

Dimensionless; |r| = 1 for a pure state, |r| ⟨1 for a mixed one, |r| = 0 at the centre

Density matrix and purityρ = ½(I + r⋅σ), Tr ρ² = ½(1 + |r|²)

Separates a superposition from a coin toss: both give P(0) = ½, only one has |r| = 1.

ρ is 2×2, Hermitian, Tr ρ = 1; purity runs from ½ when mixed to 1 when pure

Born rule along any axisP(±n) = ⟨ψ|½(I ± n⋅σ)|ψ⟩ = ½(1 ± r⋅n) = cos²(Θ/2)

One expression covers every basis, and halving Θ is where Malus's law comes from.

n a unit vector; Θ the angle between r and n on the sphere, in degrees or radians

A gate is a rotation of the ballU = exp(−iα n⋅σ/2) turns r by α about n

Unitary evolution cannot change |r|, so no gate on the qubit alone can make it mixed.

α in radians; U(2π) = −I, yet r comes back unmoved

Nothing is born a qubit

A spin-½ in a magnetic field genuinely has a two-dimensional state space, and so does the polarisation of a photon in one momentum mode. Those are exact. Everything else is a truncation. A transmon carries a whole ladder of levels and is usable only because the ladder is anharmonic: the 0→1 transition sits near 5 GHz and the 1→2 transition some 200–300 MHz below it, so a drive tuned to one is detuned from the other. That detuning sets a speed limit — a pulse short enough to carry 200 MHz of bandwidth will drive the second transition too, which is why single-qubit gates run in tens of nanoseconds rather than fractions of one. Two more conditions hide in the same place. The splitting must beat the temperature: at 5 GHz and 20 mK, hf/k_BT = 12.0, so the thermal population of |1⟩ is about e⁻¹² = 6 × 10⁻⁶. And nothing in the experiment may push the system out of the chosen pair. Once it leaks, the whole construction below is simply wrong, because there is no third pole to draw.

Four real numbers, two of them fiction

Write |ψ⟩ = a|0⟩ + b|1⟩ with a and b complex: four real numbers. Normalisation |a|² + |b|² = 1 spends one. The global phase spends another, and it is worth seeing why rather than being told. Every prediction quantum mechanics makes is either a probability |⟨m|ψ⟩|² or an expectation ⟨ψ|A|ψ⟩. Replace |ψ⟩ by e^(iγ)|ψ⟩ and the factor enters once directly and once conjugated, so it cancels exactly — in every basis, for every observable. A quantity no experiment can reach is not part of the state. What does not cancel is the relative phase between a and b, because there the two amplitudes interfere with each other. |+⟩ = (|0⟩ + |1⟩)/√2 and |−⟩ = (|0⟩ − |1⟩)/√2 differ by that sign alone; both give P(0) = ½ in the z basis, yet they are orthogonal and an x measurement tells them apart every time.

The half-angle, and what the sphere is not

Carry the two survivors as angles: |ψ⟩ = cos(θ/2)|0⟩ + e^(iφ) sin(θ/2)|1⟩. Then θ = 0 is |0⟩ at the north pole, θ = π is |1⟩ at the south, and θ = π/2 is the equator, where |+⟩, |−⟩ and the two circular states live. The half-angle carries the whole construction. It makes orthogonal states antipodal, 180° apart on a sphere on which nothing else is 180° apart, and in general |⟨χ|ψ⟩|² = cos²(Θ/2) for a Bloch angle Θ. The same factor of two says a physical rotation by 2π is only a half-turn for the ket: exp(−iπ n⋅σ) = −I, so a spin-½ comes back with its sign reversed while the sphere records nothing at all. That sign is not a bookkeeping ghost. Send one arm of a neutron interferometer through a 2π rotation and recombine it with an unrotated arm, and constructive interference becomes destructive — the 4π periodicity measured in 1975.

Predicting statistics in the basis you chose

A measurement of n⋅σ has eigenvalues ±1 with projectors ½(I ± n⋅σ), so the Born rule collapses to P(±n) = ½(1 ± r⋅n) — one line covering every basis at once. Two readings follow. Along z, r⋅z = cos θ, giving P(0) = cos²(θ/2) and P(1) = sin²(θ/2), with φ gone: the relative phase is invisible in the very basis it is defined against. Along any axis with a transverse component φ returns in full. Take the pure state θ = 60°, φ = 0. Measuring z gives P(0) = cos² 30° = 0.750. Measuring x gives P(+) = ½(1 + sin 60°) = ½(1 + 0.866) = 0.933. Turn φ to 180° and leave θ alone: the z result does not move, while the x result drops to 0.067. Written as P = cos²(Θ/2) with Θ the angle between r and n, this is Malus's law for photons — the bench angle between two polarisers is half the Bloch angle between the states.

Gates are rigid rotations, and that is a limit

Unitary evolution preserves norms and inner products, so it cannot change |r|: every gate is a rotation of the ball about some axis, U = exp(−iα n⋅σ/2) turning r by α about n. X is a half-turn about x and swaps the poles. Z is a half-turn about z, adding π to φ and leaving every z probability untouched. The Hadamard is a half-turn about (x + z)/√2, which exchanges the z and x axes and so converts a phase into a population — the reason it bookends nearly every algorithm. In hardware these are pulses: a resonant drive of Rabi frequency Ω sweeps θ at rate Ω, so with Ω/2π = 25 MHz a π pulse takes 1/(2 × 25 MHz) = 20 ns to carry |0⟩ to |1⟩, and half of it prepares |+⟩. Because the motion is rigid, a point on the surface stays on the surface for ever. Nothing acting on the qubit alone can make it mixed, which tells you before any experiment that decoherence has to involve something outside it.

When no ket will do

Prepare |+⟩ a thousand times, or flip a fair coin and prepare |0⟩ or |1⟩. Both give heads and tails in the z basis, and they are not the same state — no ket at all describes the second. The density matrix does: ρ = Σ pᵢ|ψᵢ⟩⟨ψᵢ|, and for one qubit every such object is ρ = ½(I + r⋅σ) with |r| ≤ 1. The superposition has r = (1, 0, 0) on the surface; the coin has r = 0 at the centre, ρ = ½ I. Measure x and they part company completely, 1 against ½. The diagonal of ρ holds the populations and the off-diagonal entries ρ₀₁ = ½(rₓ − i r_y) hold the coherences, which is why dephasing is written as those entries decaying: T₂ drives rₓ and r_y to zero while T₁ relaxes r_z toward its thermal value. Purity Tr ρ² = ½(1 + |r|²) runs from 1 on the surface to ½ at the centre, and measuring ⟨σₓ⟩, ⟨σ_y⟩ and ⟨σ_z⟩ fixes r, and so ρ, outright.

Change one variable at a time

Make the relationship visible.

Interactive model

State polar angle θ60 °

Analyser angle α from z90 °

Bloch length |r|1.00

Hold θ at 60° and sweep the analyser: the bars only reach 1 and 0 when n lines up with r, and they cross at 0.5 when n is perpendicular to it. Then pull |r| from 1.00 down to 0.00 — the arrow shortens, both bars slide to 0.5, and every axis becomes a coin toss.

P(+n)0.933

MEAN ⟨n⋅σ⟩0.866

P(0) ALONG z0.750

PURITY Tr ρ²1.000

Live interpretationP(+n): 0.933. MEAN ⟨n⋅σ⟩: 0.866. P(0) ALONG z: 0.750. PURITY Tr ρ²: 1.000

Catch the common trap

Explain before calculating.

Misconception check

The Bloch sphere is a map of the state space, not a picture of a small arrow of angular momentum, and reading it as ordinary space produces two errors at once. The first is angles. Orthogonal states are antipodal, so |0⟩ and |1⟩ sit 180° apart, and two states 90° apart — |0⟩ and |+⟩ — are not orthogonal at all, since |⟨+|0⟩|² = cos² 45° = ½. Every Hilbert-space angle is doubled on the sphere. The second is the arrow. For a spin-½ the vector drawn to the north pole is ⟨σ⟩, an expectation of length 1, not the spin itself, whose magnitude is √(s(s+1))ħ = 0.866ħ while its z component is only 0.5ħ — no state has the spin actually pointing along z. And the drawn arrow is blind to the minus sign a 2π rotation puts on the ket, which is real and interferometrically measurable.

One qubit is prepared as (|0⟩ + |1⟩)/√2 every run. A second is prepared by flipping a fair coin and setting it to |0⟩ or |1⟩. Which measurement tells the two preparations apart?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA qubit is prepared as |ψ⟩ = ½|0⟩ + (√3/2) e^(iπ/3)|1⟩. Check that it is normalised, read off its Bloch angles, then give P(1), ⟨σ_z⟩ and the full Bloch vector.

Norm: |a|² = (½)² = 0.250 and |b|² = (√3/2)² = 0.750, summing to 1.000, so the state is already normalised.
Angles: cos(θ/2) = 0.500 gives θ/2 = 60° and θ = 120°. The phase carried by b is φ = π/3 = 60°. Note the amplitude of |0⟩ was already real and positive, so no global phase had to be stripped first.
Computational-basis statistics: P(1) = sin²(θ/2) = sin² 60° = 0.750, and ⟨σ_z⟩ = P(0) − P(1) = 0.250 − 0.750 = −0.500, which is cos 120° as it must be.
Bloch vector: r = (sin 120° cos 60°, sin 120° sin 60°, cos 120°) = (0.866 × 0.500, 0.866 × 0.866, −0.500) = (0.433, 0.750, −0.500).
Check purity: |r|² = 0.1875 + 0.5625 + 0.2500 = 1.0000, so |r| = 1 and the point sits on the surface — a pure state, as a ket must give.

Answerθ = 120°, φ = 60°, P(1) = 0.750, ⟨σ_z⟩ = −0.500, r = (0.433, 0.750, −0.500) with |r| = 1.

MediumFor the family |ψ(φ)⟩ = (|0⟩ + e^(iφ)|1⟩)/√2, show that the z statistics do not depend on φ, then find P(+x) at φ = 0°, 60°, 90° and 180° and say what the last value means.

Angles: cos(θ/2) = 1/√2 gives θ/2 = 45° and θ = 90° for every φ, so the whole family lies on the equator.
z statistics: P(0) = cos² 45° = 0.500 and ⟨σ_z⟩ = cos 90° = 0, independently of φ. A z measurement is a fair coin whatever the phase is.
Bloch vector: r = (sin 90° cos φ, sin 90° sin φ, cos 90°) = (cos φ, sin φ, 0), so φ simply walks the point around the equator.
x statistics: P(+x) = ½(1 + rₓ) = ½(1 + cos φ). At φ = 0° this is 1.000; at 60° it is ½(1 + 0.500) = 0.750; at 90° it is 0.500; at 180° it is 0.000.
Read the endpoints: φ = 0° is |+⟩ and φ = 180° is |−⟩, which are orthogonal. Turning φ through 180° has carried the state to a perfectly distinguishable one without moving a single z probability.

AnswerP(0) = 0.500 for every φ; P(+x) = ½(1 + cos φ) = 1.000, 0.750, 0.500 and 0.000 at φ = 0°, 60°, 90° and 180°. Relative phase is measurable — just not in the z basis.

HardTomography on a single qubit returns ⟨σₓ⟩ = 0.48, ⟨σ_y⟩ = −0.36 and ⟨σ_z⟩ = 0.60. Decide whether the state is pure, write ρ, give its purity and eigenvalues, and say which entries dephasing attacks.

Bloch length: |r|² = 0.48² + 0.36² + 0.60² = 0.2304 + 0.1296 + 0.3600 = 0.7200, so |r| = 0.849. That is less than 1, so the state is mixed and no ket describes it.
Density matrix from ρ = ½(I + r⋅σ): the diagonal is ½(1 ± r_z) = 0.800 and 0.200, and ρ₀₁ = ½(rₓ − i r_y) = ½(0.48 + 0.36i) = 0.24 + 0.18i. So ρ = [[0.800, 0.24 + 0.18i], [0.24 − 0.18i, 0.200]], with Tr ρ = 1 and ρ Hermitian.
Purity: Tr ρ² = ½(1 + |r|²) = ½(1.7200) = 0.860, against 1 for a pure state and 0.500 for the completely mixed one.
Eigenvalues: (1 ± |r|)/2 = (1 ± 0.849)/2 = 0.924 and 0.076. Checking, 0.924² + 0.076² = 0.860, matching the purity, so the state is a 92.4 : 7.6 mixture of the two states along ±r.
What dephasing sees: P(0) = ½(1 + 0.60) = 0.800 comes from r_z alone, while the off-diagonal 0.24 ± 0.18i carries rₓ and r_y. Pure dephasing drives those two to zero over T₂, collapsing r onto the z axis and the purity to ½(1 + 0.36) = 0.680, without changing P(0) at all.

Answer|r| = 0.849 < 1, so mixed; ρ = [[0.800, 0.24 + 0.18i], [0.24 − 0.18i, 0.200]]; purity 0.860; eigenvalues 0.924 and 0.076; dephasing kills the off-diagonal entries and leaves P(0) = 0.800.