Probability Distributions in the Well

The wave function $\psi_n(x)$ of a particle in a box encodes everything about the particle's quantum state, but it is not itself a measurable quantity. What the Born rule tells us is that

$P(x \in [a, b]) = \int_a^b |\psi_n(x)|^2\, dx,$

the probability of finding the particle anywhere in the interval $[a, b]$ is the integral of $|\psi_n(x)|^2$ , the probability density. Computing this density — and understanding how its shape changes with the quantum number $n$ — is the subject of this lesson.

The normalized eigenfunctions

In the previous lessons we established that the normalized energy eigenfunctions of the infinite square well of length $L$ are

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

with $\psi_n(x) = 0$ outside the well. The factor $\sqrt{2/L}$ ensures that $\int_0^L |\psi_n(x)|^2\,dx = 1$ , which is required by probability conservation.

Squaring the wave function

The probability density for the $n$ -th eigenstate is obtained by squaring:

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

This is always non-negative, as every probability density must be. Notice that $|\psi_n|^2$ is not the same as $\psi_n$ itself — $\psi_n$ oscillates between positive and negative values, but its square is always $\geq 0$ .

Using the half-angle identity $\sin^2\theta = \frac{1}{2}(1 - \cos 2\theta)$ , we can write

$|\psi_n(x)|^2 = \frac{1}{L}\!\left[1 - \cos\!\left(\frac{2n\pi x}{L}\right)\right].$

This form makes it easy to integrate: the cosine term averages to zero over any whole number of half-periods, leaving a uniform density $1/L$ for the average.

What the density looks like

For the ground state ( $n = 1$ ) the density has a single maximum at the centre of the well, $x = L/2$ , where $\sin^2(\pi \cdot \tfrac{L}{2} / L) = \sin^2(\pi/2) = 1$ , giving $|\psi_1(L/2)|^2 = 2/L$ . The density tapers to zero at both walls. The particle is most likely to be found near the middle.

For the first excited state ( $n = 2$ ), the density is $|\psi_2(x)|^2 = (2/L)\sin^2(2\pi x/L)$ . This has two equal peaks, one at $x = L/4$ and one at $x = 3L/4$ , and a node — a zero — at the centre $x = L/2$ . Finding the particle exactly at the midpoint has probability zero. The number of interior nodes of $|\psi_n|^2$ is $n - 1$ .

Computing a probability from the density

To find the probability of locating the particle in some sub-interval $[a, b]$ , integrate the density over that interval. For the ground state in a well of length $L$ :

$P_1(a, b) = \int_a^b \frac{2}{L}\sin^2\!\left(\frac{\pi x}{L}\right)dx.$

Substituting $u = \pi x / L$ and using $\sin^2 u = \frac{1}{2}(1 - \cos 2u)$ :

= \frac{1}{L}\!\left[x - \frac{L}{2\pi}\sin\frac{2\pi x}{L}\right]_a^b.$$ As a concrete check, take $a = 0$, $b = L/2$ (left half of the box). Then $$P_1\!\left(0,\,\tfrac{L}{2}\right) = \frac{1}{L}\!\left[\frac{L}{2} - \frac{L}{2\pi}\sin(\pi)\right] = \frac{1}{L}\cdot\frac{L}{2} = \frac{1}{2}.$$ The result is $1/2$, exactly as expected from the symmetry of $\sin^2$ about the midpoint of its half-period. The ground-state density is symmetric about $x = L/2$, so each half must carry exactly half the total probability. ## The correspondence-principle limit For large $n$ the density $|\psi_n|^2 = (2/L)\sin^2(n\pi x/L)$ oscillates $n$ times inside the box. Its spatial average over any region large compared with the oscillation scale $L/n$ approaches $1/L$ — uniform across the box. This is exactly the classical prediction: a particle bouncing back and forth at constant speed spends equal time at every position. Quantum mechanics reduces to classical mechanics in the limit of large quantum numbers, in agreement with the **correspondence principle** first articulated by Bohr. ## Try it This is a numerical exercise — `return` a number. For a box of length $L = 1$, the ground-state probability density is $|\psi_1(x)|^2 = 2\sin^2(\pi x)$. What is the probability of finding the particle in the left half of the box ($0 \leq x \leq 0.5$)? <RunnableExample />