Bloch Sphere Representation

Overview

So far, a qubit has been a state vector

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,$$

with complex amplitudes $\alpha$ and $\beta$. This page adds a geometric viewpoint: every single-qubit pure state corresponds to a point on a sphere, called the Bloch sphere (after we ignore global phase).

The goal is not to memorize formulas. The goal is to understand what information a qubit state contains and how that information can be pictured.

Intuition

There are four real degrees of freedom in the pair $(\alpha,\beta)$ because each complex number has a real part and an imaginary part.

But not all of those degrees of freedom describe physically different states:

• Normalization removes one degree of freedom because $|\alpha|^2 + |\beta|^2 = 1$.
• Global phase removes one more degree of freedom because $|\psi\rangle$ and $e^{i\gamma}|\psi\rangle$ describe the same physical state.

That leaves two meaningful continuous parameters—exactly what you need to specify a point on a sphere surface (like latitude and longitude).

The Bloch sphere is that sphere: it is a way to encode those two parameters as angles.

The computational basis states sit at the poles:

• North pole: $|0\rangle$
• South pole: $|1\rangle$

States with “equal weight” on $|0\rangle$ and $|1\rangle$ lie on the equator; their longitude depends on relative phase.

Formal Description

Start with a normalized qubit state

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,\quad |\alpha|^2 + |\beta|^2 = 1.$$

Because overall (global) phase is unobservable, we can choose a convenient representative of the same physical state by factoring out a global phase so that the amplitude of $|0\rangle$ is real and nonnegative.

With that choice, every single-qubit pure state can be written in the standard Bloch-sphere form:

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

Here is what each symbol means:

• $\theta$ is a real angle in $[0,\pi]$. It controls the relative weights of $|0\rangle$ and $|1\rangle$.
• $\phi$ is a real angle in $[0,2\pi)$. It controls the relative phase between the $|0\rangle$ and $|1\rangle$ components.
• $e^{i\phi}$ is a complex number of magnitude 1 (a pure phase factor).

You can read off the measurement probabilities in the computational basis immediately:

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

Notice that $\phi$ does not affect these probabilities for computational-basis measurement. That does not mean $\phi$ is meaningless; it means $\phi$ becomes visible when you compare or combine states (interference) or measure in a different basis.

To connect angles to a point on the unit sphere, define the Bloch vector

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

This is exactly the usual spherical-coordinate parameterization of the unit sphere.

Worked Example

Let

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

This has equal magnitudes on $|0\rangle$ and $|1\rangle$, so it lies on the equator. In Bloch form we match

$$\cos\!\left(\tfrac{\theta}{2}\right)=\tfrac{1}{\sqrt{2}},\quad \sin\!\left(\tfrac{\theta}{2}\right)=\tfrac{1}{\sqrt{2}},$$

which implies $\theta=\pi/2$.

The relative phase is $i = e^{i\pi/2}$, so $\phi=\pi/2$.

The corresponding point is

$$(x,y,z)=(\sin(\pi/2)\cos(\pi/2),\;\sin(\pi/2)\sin(\pi/2),\;\cos(\pi/2))=(0,1,0).$$

So this state is on the equator pointing along the +$y$ direction.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

The Bloch sphere turns “state vectors” into points. Next we will interpret changes of state geometrically as rotations of that point around axes of the sphere—still without introducing matrices or circuits.