Born Rule (Worked Examples)

Overview

The measurement postulate says measurement outcomes are probabilistic. The Born rule is the specific rule that converts a quantum state into those probabilities.

This page states the Born rule in a clean mathematical form and then works several examples step-by-step, with explicit connections to the amplitudes $\alpha$ and $\beta$.

Intuition

A state vector is not directly observable. What you can observe are measurement outcomes.

Born’s rule is the bridge: it tells you how “how much of the state points along an outcome direction” becomes a probability.

In everyday geometry, you can project a vector onto an axis and take its length. In quantum mechanics, you project the state onto a basis state, and the squared magnitude of that overlap becomes the probability.

Formal Description

Choose a measurement basis $\{|b_0\rangle,|b_1\rangle\}$ for a qubit. (The computational basis is the special case $|b_0\rangle=|0\rangle$, $|b_1\rangle=|1\rangle$.)

Let the state be $|\psi\rangle$.

Born rule (qubit version). The probability of obtaining outcome $b_k$ is

$$P(b_k) = |\langle b_k|\psi\rangle|^2.$$

Here:

• $\langle b_k|\psi\rangle$ is an inner product (an “overlap”) between the basis state and the current state.
• $|\cdot|$ denotes complex magnitude.
• Squaring the magnitude produces a real number between 0 and 1.

Computational basis specialization. If

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

then

$$\langle 0|\psi\rangle = \alpha,\quad \langle 1|\psi\rangle = \beta,$$

so

$$P(0)=|\alpha|^2,\quad P(1)=|\beta|^2.$$

This is the simplest and most-used case.

Worked Example

We will do three examples.

Example 1: Real amplitudes

Let

$$|\psi\rangle = \tfrac{3}{5}|0\rangle + \tfrac{4}{5}|1\rangle.$$

Born rule in the computational basis gives:

$$P(0)=\left|\tfrac{3}{5}\right|^2=\tfrac{9}{25},\quad P(1)=\left|\tfrac{4}{5}\right|^2=\tfrac{16}{25}.$$

Example 2: Complex amplitude

Let

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

Compute:

$$P(0)=\left|\tfrac{1+i}{2}\right|^2=\tfrac{1}{4}|1+i|^2=\tfrac{1}{4}(1^2+1^2)=\tfrac{1}{2}.$$

And:

$$P(1)=\left|\tfrac{1}{\sqrt{2}}\right|^2=\tfrac{1}{2}.$$

The phase in $1+i$ did not change the computational-basis probabilities, but it can matter for other measurements.

Example 3: Same probabilities, different state

Compare

$$|\psi_+\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle),\quad |\psi_-\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle-|1\rangle).$$

Born rule in the computational basis gives $P(0)=P(1)=1/2$ for both.

This reinforces an important lesson: probabilities in one basis do not uniquely determine the state. Relative phase can be invisible to one measurement but visible to another.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

Born’s rule works in any measurement basis, not only $\{|0\rangle,|1\rangle\}$. Next we will explain what a measurement basis is and how choosing a different basis changes what your measurement “means.”