Superposition & the Bloch Sphere

A qubit in superposition is written $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, where the amplitudes $\alpha$ and $\beta$ are numbers that encode how much of each basis state is present. The one rule they must obey is normalization:

$|\alpha|^2 + |\beta|^2 = 1$

That constraint is what makes the probabilities add up to 1 — measuring yields $0$ with probability $|\alpha|^2$ and $1$ with probability $|\beta|^2$.

Drag the sphere below to rotate it in 3D, and use the θ and φ sliders to move the qubit — from $|0\rangle$ at the north pole, through the even-superposition equator, to $|1\rangle$ at the south pole. The amplitudes and probabilities update live, and their squares always sum to exactly 1.

The Bloch sphere

Because the amplitudes are constrained, a single qubit only has two real degrees of freedom, and they map perfectly onto a point on the surface of a sphere — the Bloch sphere. Using a polar angle $\theta$ and an azimuthal phase $\varphi$:

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

• The north pole ($\theta = 0$) is $|0\rangle$; the south pole ($\theta = \pi$) is $|1\rangle$.
• The equator ($\theta = \pi/2$) holds the even superpositions, where $|\alpha|^2 = |\beta|^2 = \tfrac12$.
• Moving around the equator changes only the phase $\varphi$, which does not affect single measurement probabilities but matters for interference.

The visualizer above is the full sphere: $\theta$ sets how much $|0\rangle$ versus $|1\rangle$, while $\varphi$ rotates the state around the equator — that is the phase, invisible to a single measurement but central to interference.

Why the half-angle?

Notice the $\theta/2$. A qubit and the same qubit rotated by a full $2\pi$ are physically indistinguishable, so a $180°$ trip from $|0\rangle$ to $|1\rangle$ on the sphere corresponds to only a $90°$ change in the underlying state. The half-angle bookkeeping keeps the geometry honest: opposite points on the sphere ($|0\rangle$ and $|1\rangle$) are perfectly distinguishable, i.e. orthogonal.

Gates are rotations

This geometric view pays off immediately: every single-qubit gate is just a rotation of the Bloch sphere. The Hadamard gate sends $|0\rangle$ (north pole) to a point on the equator — exactly the even superposition with $\alpha = \beta = \tfrac{1}{\sqrt2} \approx 0.707$. Other gates spin the vector about different axes. Thinking in rotations makes it obvious why applying a gate twice can return you to the start.

Takeaways

• A qubit state $\alpha|0\rangle + \beta|1\rangle$ must satisfy $|\alpha|^2 + |\beta|^2 = 1$ so probabilities sum to 1.
• Every single-qubit state is a point on the Bloch sphere: poles are $|0\rangle$/$|1\rangle$, the equator is even superposition.
• The phase $\varphi$ is invisible to a single measurement but drives interference; single-qubit gates are rotations of the sphere.

References

• Nielsen & Chuang, Quantum Computation and Quantum Information — §1.2 (qubits and the Bloch sphere)
• Qiskit Textbook — Representing qubit states