Rotations on the Bloch Sphere

Overview

Once you can picture a single-qubit state as a point on the Bloch sphere, the next question is: what does it mean to change that state?

This page introduces the idea that many single-qubit state changes can be understood as rotations of the Bloch-sphere point around an axis. We will keep it conceptual: no matrices and no circuits. The goal is to build intuition for “state evolution” that will later become the language of gates.

Intuition

On the Bloch sphere, a pure state is a point on the surface. A “rotation” means you move that point along the surface while keeping it on the sphere.

Why should rotations appear at all?

• A normalized state stays normalized under valid evolution.
• Global phase is physically irrelevant.
• What remains is a 2-parameter geometry (the sphere surface), so the most natural continuous motions are rotations.

You can think of three special directions (axes) in 3D space:

• The $z$-axis points through $|0\rangle$ (north) and $|1\rangle$ (south).
• The $x$-axis and $y$-axis lie in the equatorial plane.

Rotating around different axes changes different aspects of the state:

• Rotations around the $z$-axis change the relative phase between $|0\rangle$ and $|1\rangle$.
• Rotations around axes in the equator change the weights (probabilities) of $|0\rangle$ vs $|1\rangle$.

The key intuition: the sphere turns “complex amplitudes” into “geometry.”

Formal Description

Write a general qubit state in Bloch form:

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

• $\theta \in [0,\pi]$ controls latitude (how much probability weight is on $|0\rangle$ vs $|1\rangle$).
• $\phi \in [0,2\pi)$ controls longitude (the relative phase).

A conceptual “rotation” is a transformation that updates $(\theta,\phi)$ in a way consistent with a rigid rotation of the Bloch vector

$$(x,y,z)=(\sin\theta\cos\phi,\;\sin\theta\sin\phi,\;\cos\theta).$$

Z-axis rotations and phase

A particularly important case is rotating around the $z$-axis by an angle $\delta$.

Geometrically: the point moves around the equator direction, changing longitude $\phi$ while keeping latitude $\theta$ the same.

In coordinates, that means:

$$\theta' = \theta,\quad \phi' = \phi + \delta.$$

So the state becomes

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

Notice what changed: the magnitude of each amplitude stayed the same, but the relative phase changed.

Rotations that change probabilities

If you change $\theta$, you change the probabilities of measuring 0 vs 1 in the computational basis:

$$P(0)=\cos^2\!\left(\tfrac{\theta}{2}\right),\quad P(1)=\sin^2\!\left(\tfrac{\theta}{2}\right).$$

So any transformation that changes $\theta$ changes those probabilities. On the Bloch sphere, that corresponds to moving the point north/south.

Worked Example

Start with the state on the +$x$ direction (equator at longitude 0):

$$|\psi\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle).$$

This corresponds to $\theta=\pi/2$ and $\phi=0$.

Now apply a conceptual rotation around the $z$-axis by $\delta=\pi/2$. The rule above says $\phi' = \phi+\delta = \pi/2$ and $\theta$ stays $\pi/2$.

So the rotated state is

$$|\psi'\rangle = \tfrac{1}{\sqrt{2}}\left(|0\rangle + e^{i\pi/2}|1\rangle\right)=\tfrac{1}{\sqrt{2}}(|0\rangle + i|1\rangle).$$

What changed and what did not?

• Probabilities in the computational basis stayed $1/2$ and $1/2$ (because $\theta$ did not change).
• The relative phase changed (because $\phi$ changed).

Geometrically, you moved along the equator from +$x$ to +$y$.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Rotations describe how states can change. Now we will turn to what we can observe: measurement. Next is the measurement postulate—the rule that maps a state to probabilistic outcomes and updates the state after an outcome occurs.