The Born Rule

From amplitudes to probabilities

Quantum mechanics does not predict which single outcome a measurement will yield — it predicts the probabilities of all possible outcomes. The rule that connects the mathematical description of a quantum state to those probabilities is called the Born rule, named after Max Born who proposed it in 1926.

A quantum state is described by a wavefunction $\psi$ . For a continuous variable such as position, $\psi(x)$ is a complex-valued function of $x$ . For a system with a discrete set of outcomes — like the two-outcome spin measurement of a spin- $\tfrac{1}{2}$ particle — the state is a superposition

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

where $\alpha$ and $\beta$ are complex numbers called probability amplitudes.

The Born rule states:

The probability of obtaining outcome $|0\rangle$ upon measurement is $P(0) = |\alpha|^2$ , and the probability of obtaining outcome $|1\rangle$ is $P(1) = |\beta|^2$ .

More generally, if the state is written as a superposition $|\psi\rangle = \sum_n c_n |n\rangle$ , then

$P(\text{outcome } n) = |c_n|^2.$

The absolute-value-squared $|c_n|^2$ takes the complex amplitude $c_n = a + bi$ and returns the real non-negative number $a^2 + b^2$ , which is a valid probability.

Normalization is essential

For the probabilities to sum to one — as any complete set of probabilities must — the state must satisfy the normalization condition:

$\sum_n |c_n|^2 = 1 \quad \text{(discrete)}, \qquad \int_{-\infty}^{\infty} |\psi(x)|^2\, dx = 1 \quad \text{(continuous)}.$

This is not a separate postulate but a direct consequence of demanding that the total probability of getting some outcome equals $1$ .

As a quick check, consider $|\psi\rangle = \tfrac{1}{\sqrt{5}}|0\rangle + \tfrac{2}{\sqrt{5}}|1\rangle$ . The two squared amplitudes are

$\left|\frac{1}{\sqrt{5}}\right|^2 + \left|\frac{2}{\sqrt{5}}\right|^2 = \frac{1}{5} + \frac{4}{5} = 1. \checkmark$

Applying the rule

To find the probability of a specific outcome, isolate its amplitude and square the modulus:

$P(0) = \left|\frac{1}{\sqrt{5}}\right|^2 = \frac{1}{5} = 0.2, \qquad P(1) = \left|\frac{2}{\sqrt{5}}\right|^2 = \frac{4}{5} = 0.8.$

The procedure works identically when amplitudes are complex. For example, if $\alpha = \tfrac{3}{5}i$ , then $|\alpha|^2 = \left(\tfrac{3}{5}\right)^2 = \tfrac{9}{25}$ , since the modulus of $ri$ is $r$ for real $r$ .

The continuous case

For a particle moving in one dimension the state is described by a wavefunction $\psi(x)$ , and the Born rule becomes an integral: the probability of finding the particle in the interval $[a, b]$ is

$P(a \le x \le b) = \int_a^b |\psi(x)|^2\, dx.$

The function $|\psi(x)|^2$ is called the probability density: it is non-negative everywhere and integrates to $1$ over all space. The connection to the discrete case is direct — replace the sum over outcomes with an integral over positions.

Try it

This is a numerical exercise — your code should return a number. Given the state

$|\psi\rangle = \frac{1}{\sqrt{5}}|0\rangle + \frac{2}{\sqrt{5}}|1\rangle,$

apply the Born rule to compute the probability of obtaining outcome $|0\rangle$ .

Run your code to see the quantum state.