The Wavefunction and the Born Rule

Fire electrons one at a time through a pair of narrow slits and let each one land on a screen. The first arrives at a single, sharp spot — a particle, unmistakably. The second lands somewhere else. So does the third. There is no way to predict where any individual electron will strike. But wait, and let a hundred thousand of them pile up, and a pattern emerges out of the randomness: bright bands and dark bands, an interference pattern, the signature of a wave. Each electron went through as a wave of possibility and arrived as a particle at a random place — yet the statistics of that randomness are exact, repeatable, and calculable.

What is the thing that waves, then, if a single electron is a point? It is not the electron's matter smeared out in space. It is a field of probability. Quantum mechanics answers the question "where is the particle?" not with a location but with a function — the wavefunction \psi(x) — and a rule for turning that function into the odds of finding the particle here rather than there. That rule is the Born rule, and it is the bridge between the smooth, deterministic mathematics of quantum theory and the sharp, random clicks of a real detector. This page is about that single idea: how \psi encodes probability, and how you get the probabilities back out.

Amplitude, not probability

The state of a particle moving in one dimension is described by its wavefunction \psi(x), a complex number attached to every point x. Complex, note — \psi can be negative, it can even be imaginary, and those signs and phases are not decoration: they are what let wavefunctions interfere, adding and cancelling to make the bright and dark bands. Because it can be negative or complex, \psi(x) itself cannot be a probability — there is no such thing as a -0.3 chance of anything. Physicists call \psi(x) a probability amplitude.

Max Born's insight, in 1926, was that to get a genuine probability you square the modulus of the amplitude. The quantity

|\psi(x)|^2 = \psi^*(x)\,\psi(x)

is real, is never negative (you multiplied a complex number by its own conjugate), and it is the probability density for the particle's position. That last word — density — is the whole subtlety, and the next section is devoted to getting it right.

Why "density"? Because you must integrate

Here is the trap that catches everyone once. The value |\psi(x)|^2 at a single point is not the probability of finding the particle at that point. The probability of the particle sitting at one exact mathematical point — one real number out of a continuum — is zero. What |\psi(x)|^2 gives you is a density: multiply it by a small width dx and you get the small probability of finding the particle in the little slice between x and x + dx:

dP = |\psi(x)|^2\,dx.

It works exactly like mass. The mass density of a rod (kilograms per metre) is not a mass; you only get a mass once you multiply by a length, or integrate along a stretch of rod. In the same way |\psi|^2 is "probability per metre", and only becomes a probability when summed over a region:

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

This immediately fixes the units of \psi. In one dimension |\psi|^2\,dx must be a pure (dimensionless) probability, so |\psi|^2 has units of 1/\text{length}, and \psi itself carries the strange-looking units of 1/\sqrt{\text{length}}. If someone hands you a wavefunction with no units of length in sight, they have quietly set a length scale to 1.

The figure below shows the difference in the flesh. The upper curve is a wave-packet amplitude \psi(x) — notice how it swings positive and negative, wiggling above and below the axis. The lower curve is the probability density |\psi(x)|^2 obtained by squaring it: always positive, humped where the particle is most likely to be found, and zero at the nodes where the amplitude crosses through zero. Drag the width slider to squeeze or spread the packet and watch both curves respond together.

Squeeze the packet narrow and the density spikes high and thin — a well-localised particle. Let it spread wide and the density flattens into a low, broad plateau — a particle whose position is barely pinned down at all. The total shaded area under |\psi|^2, however, never changes: it is always exactly 1. That constancy is the last piece of the rule.

Normalisation: the particle is somewhere

If |\psi|^2 is the probability density for position, then adding up the probability over all of space must give the certainty that the particle is somewhere. Certainty is a probability of 1. So every physically legitimate wavefunction must obey the normalisation condition:

\int_{-\infty}^{\infty} |\psi(x)|^2\,dx = 1.

This does real work. The Schrödinger equation is linear, so if \psi solves it then so does C\psi for any constant C — the equation fixes the shape of the wavefunction but leaves its overall size undetermined. Normalisation is what pins down that constant: you choose C so that the total area under |\psi|^2 comes to exactly one. Let's do it.

Example 1 — normalising a "box" state. A particle is known to be somewhere in a box of length L, with a flat, equal chance of being anywhere inside it and no chance outside. Its (real) wavefunction is a constant on the box and zero elsewhere:

\psi(x) = \begin{cases} C, & 0 \le x \le L, \\ 0, & \text{otherwise.} \end{cases}

Impose normalisation. The density is |\psi|^2 = C^2 inside the box, so

\int_{-\infty}^{\infty} |\psi|^2\,dx = \int_0^L C^2\,dx = C^2 L \overset{!}{=} 1 \quad\Longrightarrow\quad C = \frac{1}{\sqrt{L}}.

The normalised amplitude is \psi = 1/\sqrt{L} — and there are those units, 1/\sqrt{\text{length}}, appearing automatically. The density is |\psi|^2 = 1/L, a flat "probability per unit length" of one-over-the-box. Notice it did not matter what shape the box was as long as its area came out to one.

Example 2 — a probability from a sub-interval. Staying with that box state, what is the chance of finding the particle in the left quarter, between 0 and L/4? Integrate the density over just that slice:

P\!\left(0 \le x \le \tfrac{L}{4}\right) = \int_0^{L/4} \frac{1}{L}\,dx = \frac{1}{L}\cdot\frac{L}{4} = \frac{1}{4}.

A one-in-four chance — exactly what "equally likely to be anywhere in the box" ought to give for a quarter of the box. For a uniform state the probability of any sub-interval of width w is simply w/L, its fraction of the whole.

Example 3 — a sloping (triangular) amplitude. Now a state that is not uniform: an amplitude that grows linearly across the box,

\psi(x) = C\,x \quad (0 \le x \le L), \qquad \psi = 0 \text{ outside.}

Normalise it. The density is |\psi|^2 = C^2 x^2, so

\int_0^L C^2 x^2\,dx = C^2\,\frac{L^3}{3} \overset{!}{=} 1 \quad\Longrightarrow\quad C = \sqrt{\frac{3}{L^3}}.

Because the amplitude — and hence the density — grows with x, the particle is far more likely to be found near the right-hand wall than the left. Let's check the odds of it being in the left half, 0 \le x \le L/2:

P\!\left(0 \le x \le \tfrac{L}{2}\right) = \int_0^{L/2} \frac{3}{L^3}\,x^2\,dx = \frac{3}{L^3}\cdot\frac{(L/2)^3}{3} = \frac{(L/2)^3}{L^3} = \frac{1}{8}.

Only a one-in-eight chance of being in the left half — so a seven-in-eight chance of being in the right half. The lopsided amplitude has made the particle strongly prefer the region where |\psi|^2 is large, exactly as the Born rule promises.

Not the electron's substance, and not any physical stuff you could catch in a net — what waves is the amplitude \psi, a field of possibility. In the two-slit experiment the amplitude passes through both slits, and the two contributions add: where they arrive in step they reinforce and |\psi|^2 is large (a bright band), where they arrive out of step they cancel and |\psi|^2 is nearly zero (a dark band). Crucially, the cancellation needs the amplitude to take opposite signs — this is why \psi must be allowed to go negative and complex, and why you can never build interference out of probabilities alone, which are stubbornly positive. The single electron then "chooses" a landing spot at random, weighted by |\psi|^2. Wave on the way, particle on arrival; the wave was never the electron, but the bookkeeping of its chances.

Two traps in one. The first: reading \psi(x) straight off as "the chance of the particle being at x." It is not — it is an amplitude, it can be negative or complex, and you must square its modulus before probability even enters the conversation. Amplitude first, probability only after the squaring.

The second, subtler trap: thinking that once you have squared it, |\psi(x)|^2 is now "the probability of being at x." Still no. |\psi(x)|^2 is a probability density — probability per unit length — with units of 1/\text{length}, not a bare number between 0 and 1. A density can even be bigger than one (squeeze the packet in the figure and watch the peak climb past 1) without anything being wrong, because it is not yet a probability. You only get an honest probability — a dimensionless number that obeys 0 \le P \le 1 — after you integrate the density over a stretch of x. Square, then integrate. Skip either step and the numbers stop meaning anything.