Chapter 04

The Bloch Sphere

Two real angles fix any qubit, so the whole space of states is the surface of a ball. This one picture turns abstract amplitudes into geometry you can point at.

↩ before you start · keep these handy
·From Ch. 2: a qubit is α|0⟩ + β|1⟩, and only the relative phase between α and β is physical.
·From Ch. 3: ⟨Z⟩ = P(0) − P(1) is the long-run average of a measurement.
·cos and sin read an angle off a circle; cos0°=1, cos90°=0. We only ever plug angles in — no trig identities needed.
🔑 symbol decoder · every new mark, in plain words
θ“theta” — the polar angle, measured down from the north pole |0⟩. Like latitude. It sets the odds. φ“phi” — the azimuth, how far around the vertical axis. Like longitude. It is the phase. polestop = |0⟩, bottom = |1⟩ — the only two points a classical bit is allowed. equatorthe ring halfway down, where P(0) = P(1) — all the even superpositions. ⟨X⟩ ⟨Y⟩ ⟨Z⟩the arrow’s three coordinates — each is the average result of measuring along that axis. rthe arrow’s length. A pure qubit has r = 1 (on the surface); noise shrinks it inside the ball.
§1

Every qubit is a point on a globe

The north pole is |0⟩, the south pole is |1⟩, and the equator holds the even superpositions, differing only by phase. Drag the arrow — every point on the surface is a legal state. A classical bit only ever gets the two poles.

💡 everyday picture

Any city on Earth is pinned by just two numbers: latitude (how far down from the north pole) and longitude (how far around). The Bloch sphere is exactly that map for a qubit: θ is the latitude and φ is the longitude. One warning the analogy makes obvious: this globe is a map of states, not a place in the room — its “north” is the state |0⟩, not a direction you could walk.

▸ dragbloch.point
θ = {{ thetaDeg }}°
polar · tilt from |0⟩
φ = {{ phiDeg }}°
azimuth · phase
P(0) {{ p0pct }}%
P(1) {{ p1pct }}%
drag horizontally → phase φ
drag vertically → mix θ
recapEvery qubit is one point on the ball’s surface; a classical bit is stuck at the two poles.
§2 · the mathematics

From amplitudes to angles

Drop the invisible global phase and any qubit can be written with just two angles — a polar angle θ and an azimuth φ:

|ψ⟩ = cos(θ/2) |0⟩ + e^{iφ} sin(θ/2) |1⟩

The factor of ½ is why opposite points on the sphere — like |0⟩ and |1⟩ — are orthogonal states even though they're 180° apart in space.

✎ worked example · a point at θ = 60°, φ = 0°
1.plug into |ψ⟩ = cos(θ/2)|0⟩ + e^{iφ}sin(θ/2)|1⟩ with θ/2 = 30°, φ = 0
2.cos30° = 0.866, sin30° = 0.5 → |ψ⟩ = 0.866|0⟩ + 0.5|1⟩
3.odds: P(0) = 0.866² = 0.75, P(1) = 0.5² = 0.25 — a point three-quarters of the way up toward |0⟩
recapTwo angles — tilt θ and phase φ — name any qubit; the ½ factor makes opposite poles orthogonal.
§3 · the mathematics

The three coordinates are expectations

The arrow's x, y, z components aren't decoration — they are the average measurement results along three axes (the Pauli observables X, Y, Z):

⟨X⟩
{{ bx }}
⟨Y⟩
{{ by }}
⟨Z⟩
{{ bz }}

Drag the arrow and watch all three update. A pure qubit always sits on the surface, where ⟨X⟩² + ⟨Y⟩² + ⟨Z⟩² = 1; noise would pull it inside the ball.

recapThe arrow's x, y, z are the average X, Y, Z measurements; pure states sit on the surface where the length is 1.
⚠ common misconceptions

“The Bloch sphere is a picture of real space.” It is not. It is an abstract map of states: the three axes are measurement averages (⟨X⟩, ⟨Y⟩, ⟨Z⟩), not left/up/forward in the room. The arrow does not say where the qubit is — it says what it is.

“Spinning the arrow around the vertical (changing φ) changes the odds.” No. P(0) and P(1) depend only on the height (the angle θ). Slide around a circle of latitude and the Z-odds never move — all that changes is the invisible phase.

“|0⟩ and |1⟩ point opposite ways, so they're only half-different.” They are perfectly different — fully orthogonal. The trick is the half-angle: states use θ/2, so 180° apart on the globe is only 90° apart in state-space, exactly the right angle for “completely distinguishable.”

✓ you can now
place any qubit on the sphere from its two angles, and read a surface point back as a state
explain why the poles are the classical 0 and 1 and the equator holds the even superpositions
read the arrow's height as ⟨Z⟩, and know that pure states live on the surface (r = 1)
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