A matter wave is described by a wave function $\psi(x)$. But what is waving? Max Born's answer (1926) is that $\psi$ tells you where the particle is likely to be found. You can't predict where one electron will land, but you can predict the pattern that thousands of them make, exactly. The tool is the definite integral.
Key idea
$|\psi(x)|^2$ is a probability density: the probability of finding the particle between $a$ and $b$ is $$P(a \le x \le b) = \int_a^b |\psi(x)|^2\,dx.$$ The particle must be somewhere, so a wave function is normalized: $\int_{-\infty}^{\infty}|\psi|^2\,dx = 1$. $\psi$ itself can be negative or even complex; only $|\psi|^2$ is a probability.
Why it works
For light, the intensity of a wave is proportional to the square of its amplitude, and in the photon picture the intensity at a spot counts how many photons arrive there. Born carried the same idea over to matter: the chance of finding a particle in a small interval $dx$ is $|\psi|^2dx$, in the same way that the number of photons arriving is set by the wave's amplitude squared. This is why the double-slit pattern for electrons builds up dot by dot into the $|\psi|^2$ pattern. Since $|\psi|^2dx$ is a pure number, $|\psi|^2$ has units of 1/length.
Worked examples
Worked example: A uniform wave function
A particle's wave function is $\psi = A$ for $0 < x < L$ and zero elsewhere. Find $A^2$ and the probability of finding the particle in the first quarter of the region.
- Normalize: $\int_0^L A^2\,dx = A^2L = 1$, so $A^2 = \frac{1}{L}$.
- $P = \int_0^{L/4}\frac{1}{L}\,dx = \frac{1}{4}$.
Answer: $A^2 = 1/L$; $P = 0.25$
Worked example: A particle in a box
The ground state of a particle in a box is $\psi = \sqrt{2/L}\,\sin(\pi x/L)$ for $0 < x < L$. Find the probability that it is in the left quarter, $0 < x < L/4$.
- $P = \frac{2}{L}\int_0^{L/4}\sin^2\frac{\pi x}{L}\,dx$. Use $\sin^2u = \frac{1 - \cos 2u}{2}$.
- $P = \frac{1}{L}\left[x - \frac{L}{2\pi}\sin\frac{2\pi x}{L}\right]_0^{L/4} = \frac{1}{4} - \frac{1}{2\pi}\sin\frac{\pi}{2}$.
- $P = 0.250 - 0.159 = 0.091$.
Answer: About 0.091, less than a quarter: the particle prefers the middle
Worked example: Normalize a wave function
Find $A^2$ so that $\psi = A\,x(L - x)$ on $0 < x < L$ is normalized.
- $\int_0^L A^2x^2(L - x)^2\,dx = A^2\int_0^L (L^2x^2 - 2Lx^3 + x^4)\,dx$.
- $= A^2\left(\frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5}\right) = A^2\frac{L^5}{30}$.
- Set it equal to 1: $A^2 = \frac{30}{L^5}$.
Answer: $A^2 = 30/L^5$
Common mistake
Treating $\psi$ itself as the probability. $\psi$ can be negative (as in the $n = 2$ box state) or complex, and a probability can't be. Square its magnitude first. A check: a probability density must never be negative, and its integral over all space must be 1.
Common mistake
Reading $|\psi(x)|^2$ as the probability of being exactly at $x$. It is a density: you need an interval and an integral, $|\psi|^2dx$. The chance of being at one exact point is zero.