The Bloch Sphere Complete Guide: Visualizing Qubits, Relative Phase, and Entanglement Collapse

1. The Mathematical Formulation

Any pure qubit state in two-dimensional complex Hilbert space $\mathbb{C}^2$ is written as a linear superposition of computational basis states $|0\rangle$ and $|1\rangle$:

Because a global phase factor $e^{i\gamma}$ has no observable physical consequences (it cancels out when computing measurement probabilities $|\langle \phi | \psi \rangle|^2$), we can factor out the global phase and parameterize the remaining degrees of freedom with two real angles:

where the angles span the spherical domain:

• $0 \le \theta \le \pi$ — Polar latitude angle (determines relative measurement probabilities $P(|0\rangle) = \cos^2(\theta/2)$ and $P(|1\rangle) = \sin^2(\theta/2)$).
• $0 \le \phi < 2\pi$ — Azimuthal longitudinal angle (controls quantum relative phase between $|0\rangle$ and $|1\rangle$).

2. The 6 Cardinal Basis States on the Sphere

The Cartesian coordinates $(x, y, z)$ of the Bloch vector are calculated directly from spherical coordinates:

3. Pure States, Mixed States, and Why Entanglement Collapses the Sphere

One of the most profound insights in quantum information science is that only pure quantum states live on the surface of the Bloch sphere ($|\vec{r}| = 1$). Mixed states—which describe statistical ensembles or entangled subsystems—live strictly inside the sphere ball ($|\vec{r}| < 1$).

The general density matrix $\rho$ of a single qubit is parameterized by its Bloch vector $\vec{r} = (r_x, r_y, r_z)$:

What Happens During Quantum Entanglement?

Consider the maximally entangled Bell pair $|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$. If we trace out qubit B to inspect the state of qubit A alone ($\rho_A = \text{Tr}_B(|\Phi^+\rangle\langle\Phi^+|)$), we obtain:

The trace purity $\text{Tr}(\rho^2)$ for this maximally mixed state is:

The length of the Bloch vector is given by $|\vec{r}| = \sqrt{2\text{Tr}(\rho^2) - 1}$. When $\text{Tr}(\rho^2) = 0.5$, $|\vec{r}| = 0$!