Bloch Sphere Revisited (Full Geometry)

Overview

Earlier, we introduced the Bloch sphere as a way to represent any single-qubit pure state (up to global phase) with two angles $(\theta,\phi)$. This page deepens that picture.

You will learn:

• how to interpret $(\theta,\phi)$ as latitude/longitude of a point,
• how measurement probabilities are read geometrically (as angles between directions),
• and what the Bloch sphere does and does not represent.

No gates, no matrices—just geometry and measurement.

Intuition

A pure qubit state has two meaningful continuous degrees of freedom (after normalization and ignoring global phase). A point on a sphere also needs two parameters. That match is not a coincidence: the Bloch sphere is a map from states to geometry.

Here are three geometric ideas to keep in mind:

1. States are directions. Up to global phase, a pure state corresponds to a direction on the sphere.

2. Measurements choose an axis. Measuring a qubit in some basis means choosing an axis on the sphere; the two outcomes correspond to the two opposite endpoints of that axis.

3. Probabilities depend on angles. The closer the state-direction is to an outcome-direction, the higher the probability of that outcome.

This is the geometric version of “overlap squared.”

Formal Description

Bloch-sphere parameterization

A general single-qubit pure state can be written as

$$|\psi\rangle = \cos\!\left(\tfrac{\theta}{2}\right)|0\rangle + e^{i\phi}\,\sin\!\left(\tfrac{\theta}{2}\right)|1\rangle,$$

where:

• $\theta \in [0,\pi]$ controls how much weight is on $|0\rangle$ vs $|1\rangle$.
• $\phi \in [0,2\pi)$ is the relative phase between the basis components.

The corresponding point on the unit sphere (the Bloch vector) is

$$\mathbf{r}=(x,y,z)=(\sin\theta\cos\phi,\;\sin\theta\sin\phi,\;\cos\theta).$$

This is just spherical coordinates.

Measurement as choosing a direction

Any two-outcome projective measurement for a qubit can be viewed as choosing a unit vector (a direction) $\mathbf{n}$ on the sphere. The two outcomes correspond to the two opposite points $\pm\mathbf{n}$.

Geometrically:

• “Outcome +” corresponds to the state aligned with $\mathbf{n}$.
• “Outcome −” corresponds to the state aligned with $-\mathbf{n}$.

Probabilities from angles

Let $\gamma$ be the angle between the state direction $\mathbf{r}$ and the measurement axis direction $\mathbf{n}$.

Then the probability of the “+” outcome is

$$P(+) = \cos^2\!\left(\tfrac{\gamma}{2}\right),$$

and the probability of the “−” outcome is

$$P(-) = \sin^2\!\left(\tfrac{\gamma}{2}\right).$$

In words: angle 0 means certainty (probability 1), angle $\pi$ means certainty of the opposite outcome, and angle $\pi/2$ means a fair 50/50.

You can connect this to the computational basis by taking $\mathbf{n}$ to be the $z$-axis:

• $\gamma=\theta$ (the state’s polar angle)
• so $P(0)=\cos^2(\theta/2)$ and $P(1)=\sin^2(\theta/2)$

which matches the standard Bloch-sphere form.

What the Bloch sphere does NOT represent

The Bloch sphere is a map for single-qubit pure states.

• It does not represent two-qubit states (entanglement requires a higher-dimensional description).
• It does not represent mixed states as points on the surface; mixed states correspond to points inside the sphere (the “Bloch ball”), which we introduced indirectly via density matrices.
• It is not a literal physical sphere in space; it is a geometric representation of the state description.

Worked Example

Consider the state with

$$\theta=\tfrac{\pi}{3},\quad \phi=0.$$

Then

$$|\psi\rangle = \cos\!\left(\tfrac{\pi}{6}\right)|0\rangle + \sin\!\left(\tfrac{\pi}{6}\right)|1\rangle = \tfrac{\sqrt{3}}{2}|0\rangle + \tfrac{1}{2}|1\rangle.$$

Measurement in the computational basis

Here the measurement axis is the $z$-axis, and the angle between the state direction and the north pole is $\gamma=\theta=\pi/3$.

So

$$P(0)=\cos^2\!\left(\tfrac{\theta}{2}\right)=\cos^2\!\left(\tfrac{\pi}{6}\right)=\left(\tfrac{\sqrt{3}}{2}\right)^2=\tfrac{3}{4}.$$

And

$$P(1)=\tfrac{1}{4}.$$

Measurement along the opposite direction

If you instead choose the measurement axis to be the south pole (the $-z$ direction), then the relevant angle is $\gamma' = \pi-\theta = 2\pi/3$.

So the probability of the “aligned” outcome with $-z$ is

$$\cos^2\!\left(\tfrac{\gamma'}{2}\right)=\cos^2\!\left(\tfrac{\pi}{3}\right)=\tfrac{1}{4},$$

which matches the earlier probability of outcome 1.

This illustrates the rule: changing the measurement axis changes which outcome direction you call “+,” but the geometry stays consistent.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Now that we can talk about measurement geometrically, we’ll introduce a more general language for measurement results: observables and expectation values. This will help you reason about measurement outcomes as numerical quantities, not just “0 or 1.”