Expectation Value

When you measure a quantum system many times, you do not always get the same answer — you sample from a probability distribution. The single number that best summarises that distribution is its expectation value (also called the mean or average), written $\langle X \rangle$ .

Definition for a discrete distribution

Let $X$ be a random variable that can take values $x_1, x_2, \ldots, x_n$ with probabilities $p_1, p_2, \ldots, p_n$ (where $p_i \ge 0$ and $\sum_i p_i = 1$ ). The expectation value is

\langle X \rangle = \sum_{i=1}^{n} x_i \, p_i.

Every term is an outcome weighted by how likely it is; summing all weighted outcomes gives the long-run average you would observe over many trials.

A worked example

Consider a fair coin that pays $+1$ for heads and $-1$ for tails, each with probability $1/2$ :

\langle X \rangle = (+1)\cdot\tfrac{1}{2} + (-1)\cdot\tfrac{1}{2} = 0.

The expectation value is $0$ even though $0$ is not a possible outcome — the mean describes the centre of the distribution, not a guaranteed result.

Why it matters in quantum mechanics

In quantum mechanics, a measurement of an observable $\hat{A}$ on a state $|\psi\rangle$ produces one of the eigenvalues $a_k$ , each with probability $p_k = |\langle a_k|\psi\rangle|^2$ . The expectation value of $\hat{A}$ is therefore

\langle \hat{A} \rangle = \sum_k a_k \, p_k = \langle\psi|\hat{A}|\psi\rangle,

where the second equality is the standard inner-product form derived by substituting the spectral decomposition of $\hat{A}$ . This quantity predicts the average of many identical measurements; it is one of the most-used numbers in quantum physics.

Try it

This is a numerical exercise — return a number. A fair six-sided die has outcomes $1, 2, 3, 4, 5, 6$ , each with probability $1/6$ . What is the expectation value of a single roll?

Deriving it directly:

\langle X \rangle = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5.

Run your code to see the quantum state.