Expectation of Position

The Born rule tells us that $|\psi(x)|^2 \, dx$ is the probability of finding a particle in the interval $[x, x+dx]$ . Once we have that probability density, we can ask a natural question: on average, where is the particle?

The expectation value of position

The average value of any quantity whose probability density is $P(x)$ is computed by weighting each possible outcome by its probability. For position this gives

\langle x \rangle = \int_{-\infty}^{\infty} x \, |\psi(x)|^2 \, dx.

This integral is called the expectation value of $x$ . It is not a prediction of what a single measurement will return — individual measurements are random — but rather the mean of many repeated measurements on identically prepared systems.

A worked example

Consider the normalized wave function

\psi(x) = \sqrt{3}\, x, \quad 0 \le x \le 1, \qquad \psi(x) = 0 \text{ elsewhere.}

Check normalization. We need $\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1$ . Since $\psi$ is zero outside $[0,1]$ :

\int_0^1 |\psi(x)|^2 \, dx = \int_0^1 3x^2 \, dx = 3 \cdot \frac{x^3}{3} \bigg|_0^1 = 1. \checkmark

Compute $\langle x \rangle$ . Substituting $|\psi(x)|^2 = 3x^2$ :

\langle x \rangle = \int_0^1 x \cdot 3x^2 \, dx = 3 \int_0^1 x^3 \, dx = 3 \cdot \frac{x^4}{4} \bigg|_0^1 = \frac{3}{4}.

The expected position is $\langle x \rangle = 0.75$ . Notice that this is not the midpoint of the interval $[0,1]$ : the probability density $3x^2$ grows toward $x = 1$ , shifting the mean to the right of $0.5$ .

What the expectation value is (and is not)

The expectation value $\langle x \rangle$ depends only on the state $\psi$ , not on any particular measurement. Before a measurement, the particle does not have a definite position; the wave function describes a genuine spread. The spread is captured by the standard deviation

\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2},

which will become central when we study the Heisenberg uncertainty principle.

Try it

This is a numerical exercise — your code should return a number, not a circuit. Use the result derived above for the wave function $\psi(x) = \sqrt{3}\,x$ on $[0,1]$ .

Run your code to see the quantum state.