Interactive Math & Science Lessons

Bloch Sphere Explorer

single-qubit states, basis measurements, and gate rotations on the Bloch sphere

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At δ = Ï€ the two halves cancel exactly 🖖

On the equator the state is an even split of |0⟩ and |1⟩, so a Z measurement is a coin toss whatever the phase does. The phase still decides everything — it just surfaces in the X and Y bases. The phasor panel makes it visible: it draws A = 1 + eiδ and sweeps δ by itself, as an animation rather than a control. The amplitude is |A| = 2cos(δ/2), so the probability follows cos²(δ/2): a full 100% at δ = 0, exactly 50% at δ = π/2, and a clean zero at δ = π, where two unit phasors point opposite ways and sum to nothing. Identical populations, opposite outcome — that cancellation is the resource every quantum algorithm spends.

One qubit fits on a globe 🖖

A single qubit's entire state is just one point on this sphere's surface. The north pole is |0⟩, the south pole is |1⟩, and everything between them is a superposition. Latitude (θ) sets how the odds split between 0 and 1 — the poles are certainty, the equator is a 50/50 coin — while longitude (φ) sets the phase. The practical takeaway: measuring in the Z basis simply asks which hemisphere your point leans toward.

It takes 720° to come home 🖖

Notice the half-angle: the state uses cos(θ/2), not cos(θ). That factor of two is the fingerprint of a deep link — 3D rotations form the group SO(3), but a qubit's real state space is SU(2), which wraps around it twice. The consequence is startling: rotate a qubit a full 360° and its state vector flips to -|ψ⟩ instead of returning home; only a second turn, a full 720°, truly restores it. Every electron shares this spinor behavior, dramatized by Dirac's belt trick.

Problem solved in full

  1. A qubit at θ = 90° reading 50/50 like a coin 5 steps

    A qubit at θ = 90° reads 50/50. That looks like a coin. Work out why it is not — and where the certainty went.

    1. A pure qubit state is two angles on a sphere. θ sets how the probability splits between |0⟩ and |1⟩; φ sets a phase that the Z measurement cannot see at all.

    2. At θ = 90° the two cosines are equal, so both outcomes are exactly half. Measured in the Z basis this really is indistinguishable from a fair coin.

    3. But the state has more in it than those two probabilities. The Bloch vector points along +x, and its length is 1.

    4. Length 1 means the state is pure — it sits on the surface of the sphere. A genuinely random coin would be a mixed state at the centre, with length 0, and both give 50/50 in Z.

    5. The difference shows up in another basis. This state is |+⟩, and measuring it along x returns + with certainty, every time.

    Answer

    The tool prints 50.00% for both outcomes with a Bloch vector of (1, 0, 0). The lesson is that 50/50 is a statement about the basis you chose, not about the state. A coin is random in every basis; this qubit is random in one and completely determined in another, and the surface of the sphere is where that distinction lives. Switch the basis selector from Z to X and watch the same state read 100/0 without anything about it having changed.

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