Probability in an Interval

The Born rule says that a particle with (normalised) wavefunction $\psi(x)$ is found in the interval $[a, b]$ with probability

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

The integrand $|\psi(x)|^2$ is the probability density — it is always real and non-negative, and it integrates to 1 over all space by the normalisation condition.

A worked example: the infinite square well

Consider a particle confined to a one-dimensional box of width $L$ . Inside the box the walls are infinitely high, so the wavefunction must vanish at both walls. The ground state (lowest energy, $n = 1$ ) satisfying this boundary condition and the normalisation condition is

$\psi_1(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{\pi x}{L}\right), \qquad 0 \le x \le L,$

and $\psi_1(x) = 0$ everywhere outside the box. You can confirm normalisation directly:

$\int_0^L |\psi_1|^2\, dx = \frac{2}{L}\int_0^L \sin^2\!\!\left(\frac{\pi x}{L}\right)dx.$

Using $\sin^2\theta = \frac{1}{2}(1 - \cos 2\theta)$ the integral becomes $\frac{2}{L}\cdot\frac{L}{2} = 1$ . Good.

Finding the probability in an interval

Take $L = 1\,\text{m}$ for simplicity. The probability of finding the particle in the left third of the box, $0 \le x \le \tfrac{1}{3}$ , is

$P = \int_0^{1/3} 2\sin^2(\pi x)\, dx.$

Apply the identity $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ :

$P = \int_0^{1/3} \bigl(1 - \cos(2\pi x)\bigr)\, dx = \left[x - \frac{\sin(2\pi x)}{2\pi}\right]_0^{1/3}.$

Evaluating the boundary terms and using $\sin(2\pi/3) = \sqrt{3}/2$ :

$P = \frac{1}{3} - \frac{\sin(2\pi/3)}{2\pi} = \frac{1}{3} - \frac{\sqrt{3}/2}{2\pi} = \frac{1}{3} - \frac{\sqrt{3}}{4\pi} \approx 0.1955.$

Why this matters

Every measurable quantity in quantum mechanics ultimately reduces to an integral of this form — or a sum of such integrals for discrete spectra. Mastering the mechanics of evaluating $\int |\psi|^2\,dx$ over a chosen region is therefore a foundational skill for all subsequent quantum calculations.

Try it

This is a numerical exercise — return a number. The ground state of an infinite square well of width $L = 1\,\text{m}$ has wavefunction $\psi(x) = \sqrt{2}\sin(\pi x)$ for $0 \le x \le 1$ . Compute the probability of finding the particle in the left third of the well ( $0 \le x \le \tfrac{1}{3}$ ).

Run your code to see the quantum state.