The Born interpretation: P(x, t) = |ψ(x, t)|²
Probability lives in the area under |ψ|², not its height.
Until now ψ has been a wave with no stated meaning. Here it gets one: the rule that turns a complex function into numbers a detector can check. Every later calculation in this course ends the same way, so learn it cleanly now: take the squared modulus, integrate over the region you care about, compare with counts from repeated runs.
Build the model
No instrument ever returns ψ. |ψ|² predicts how often each detector fires.
1The big idea
The wavefunction is complex, and no instrument returns its value; a position measurement returns a position, one click per run. Born's rule links the two: |ψ(x, t)|² = ψ*ψ is a probability per unit length, so the chance that a measurement at time t finds the particle between a and b is the area ∫ₐᵇ |ψ|² dx. The rule is read off experiment, not derived: detectors register whole particles at single points while the accumulated pattern has the shape of a wave's intensity, and the squared modulus is the natural candidate because it is real, never negative, and blind to an overall phase. Three consequences follow at once: ψ carries units of m⁻¹⁄² in one dimension, the total area must be 1, and multiplying ψ by e(iα) changes nothing observable while the relative phase between two pieces of ψ can change everything. What the rule predicts is a frequency over many identical preparations, pinned down ever more tightly as N grows; it never says where the next particle lands. It stops being the whole story at energies where particles are created and destroyed, where no one-particle density is conserved, and it is silent on how ψ changes when a detection happens.
2In one sentence
The Born rule says |ψ(x, t)|² dx is the probability that a position measurement at time t finds the particle between x and x + dx, so the probability for any interval is the integral of |ψ|² across it.
Example. In the state ψ = √(2/L) sin(πx/L) with L = 1.0 nm, |ψ|² at the centre is 2/L = 2.0 nm⁻¹, so the chance of finding the particle within 0.010 nm of the centre is about 2.0 nm⁻¹ × 0.020 nm = 0.040: roughly 400 detections in 10 000 identical runs.
3A detector records clicks, never ψ
Send electrons through a double slit one at a time and every run ends the same way: one pixel fires, at one place, registering one whole electron. No run returns a value of ψ(x, t), and none returns a fraction of an electron. Only after thousands of runs does a pattern appear, with the shape a wave would make and the density of clicks following |ψ|² as optical intensity follows |E|². That rules out the reading Schrödinger first favoured, in which e|ψ|² is the electron's charge smeared through space: a packet spread over a micrometre would then deliver a fraction of a charge to every pixel it covers, and no detector has ever recorded one. Born's 1926 rule keeps the wave and changes what it describes: ψ is an amplitude, and |ψ(x, t)|² is the probability per unit length of a detection at x at time t. His paper first tied the probability to the amplitude itself; a footnote added in proof corrected it to the square.
4Height is a density; area is a probability
|ψ(x, t)|² is a probability per unit length. The probability of finding the particle between a and b is the area ∫ₐᵇ |ψ|² dx, and when the window Δx is narrow compared with the distance over which |ψ|² changes, the area reduces to height × width, P ≈ |ψ(x₀, t)|² Δx. Take a Gaussian packet |ψ(x)|² = e(−x²/(2σ²))/(√(2π) σ) with σ = 0.10 nm. Its peak density is 3.99 nm⁻¹, or 3.99 × 10⁹ m⁻¹: not a probability, and no cause for alarm. For |x| ≤ 0.010 nm, height × width gives 3.99 nm⁻¹ × 0.020 nm = 0.0798, against 0.0797 exactly. Widen the window to |x| ≤ σ and height × width claims 0.798 while the true area is 0.683, because the height is no longer flat across the window. The probability of any single exact point is zero.
5Why ψ carries the inverse square root of a length
Because |ψ|² dx is a pure number and dx is a length, |ψ|² has units of m⁻¹ and ψ has units of m⁻¹⁄². In three dimensions the probability in a volume element is |ψ(r, t)|² d³r, so ψ carries m⁻³⁄². That is a free check on every wavefunction: ψ = √(2/L) sin(πx/L) passes, since √(2/L) is an inverse root length and the sine is a pure number, while (2/L) sin(πx/L) fails on sight. The numbers themselves depend on the unit. For L = 0.50 nm the prefactor is 2.0 nm⁻¹⁄², or 6.3 × 10⁴ m⁻¹⁄², and the peak density 2/L is 4.0 nm⁻¹, or 4.0 × 10⁹ m⁻¹. Switching from nm to m multiplies ψ by 3.2 × 10⁴ and |ψ|² by 10⁹, and leaves every probability exactly where it was. The same accounting fixes the total: the particle is certainly somewhere, so the area under |ψ|² over all x must be 1.
6An overall phase is invisible; a relative phase is not
Multiply ψ by e(iα), with α a real constant, and |e(iα)ψ|² = e(−iα)e(iα)ψ*ψ = |ψ|² at every x, so no position statistic changes, and the same holds for every measurement in this course. ψ, −ψ and iψ are the same state, and a plot of Re ψ alone shows a convention rather than the particle: multiplying by i turns Re ψ into −Im ψ. Phase matters when it differs between the pieces of a sum. For ψ = ψ₁ + e(iφ)ψ₂, |ψ|² = |ψ₁|² + |ψ₂|² + 2 Re(ψ₁* ψ₂ e(iφ)). Where ψ₁ and ψ₂ are both real and equal to 1.0 nm⁻¹⁄², the density is 4.0 nm⁻¹ at φ = 0, 2.0 nm⁻¹ at φ = π/2 and zero at φ = π, whereas adding the two densities would give 2.0 nm⁻¹ every time. That cross term makes interference fringes, and it is why the rule is applied to the total ψ, never to its parts one at a time.
7A prediction of frequencies, never of one run
The Born rule speaks about an ensemble: many systems prepared identically, each measured once. For any interval it predicts the fraction of runs that land there, and it is silent about which runs will. Each run is a yes-or-no trial with probability P, so N runs give NP detections on average with a binomial scatter √(NP(1 − P)). With P = 0.683 and N = 1000, expect 683 ± 15; with N = 10⁶, 683 000 ± 465, a relative scatter of 0.07%. That shrinking scatter is how the rule is tested, and three-slit experiments that look for interference beyond the pairwise cross terms of |ψ₁ + ψ₂ + ψ₃|² have found none within their precision. What the rule does not supply is a position the particle already had before the click: in standard quantum mechanics the spread belongs to the state, not to our ignorance. What a detection does to ψ is a question for Unit 4.
8Four checks, the classical limit, and where the rule stops
Check every Born-rule answer four ways. Normalisation: probabilities for intervals that cover all of space add to 1. Dimensions: |ψ|² times a length is a pure number. Limiting cases: a narrow window gives |ψ|²Δx, and a window covering everything gives 1. Classical correspondence: a particle bouncing between walls a distance L apart at constant speed spends equal times in equal lengths, so its density is 1/L. The state ψₙ = √(2/L) sin(nπx/L) gives P(0 ≤ x ≤ L/4) = 1/4 − sin(nπ/2)/(2nπ): 0.091 for n = 1, 0.264 for n = 11 and 0.242 for n = 21, closing on the classical 0.250 as n grows. The rule has limits of its own. It describes one non-relativistic particle; where particles are created and destroyed, a one-particle density is not conserved. Two particles need a joint density |ψ(x₁, x₂, t)|². And it says where detections happen, not how ψ changes when they do.
Key equations
What each relation says, and the units it is written in.
Born rule for position
dP = |ψ(x, t)|² dx
For a slice Δx narrow compared with the distance over which |ψ|² changes, this becomes height × width: P ≈ |ψ(x₀, t)|² Δx.
|ψ|² in m⁻¹, dx in m, dP a pure number; t is the instant at which the position is measured
Probability for an interval
P(a ≤ x ≤ b, t) = ∫ₐᵇ |ψ(x, t)|² dx
Probability lives in the area, never in the height: a density of 4.0 nm⁻¹ is legal, a probability of 4.0 is not.
a and b in m; ψ normalised so the area over all x is 1, so P is dimensionless and lies between 0 and 1
Squared modulus of a complex ψ
|ψ|² = ψ*ψ = (Re ψ)² + (Im ψ)²
Squaring ψ itself gives a complex number; only ψ*ψ is real. For ψ = (1.2 − 0.5i) nm⁻¹⁄², |ψ|² = 1.44 + 0.25 = 1.69 nm⁻¹.
ψ* is the complex conjugate; Re ψ and Im ψ both carry m⁻¹⁄², so |ψ|² is real, ≥ 0 and in m⁻¹
Units of the wavefunction
[ψ] = L⁻¹⁄² on a line, L⁻³⁄² in space
A one-line test of any proposed ψ: √(2/L) sin(πx/L) passes, while (2/L) sin(πx/L) fails before a single integral is done.
ψ in m⁻¹⁄² or m⁻³⁄², because |ψ|² dx and |ψ|² d³r must be pure numbers
Global phase
|e(iα) ψ|² = |ψ|²
ψ, −ψ and iψ are one physical state. Only a phase that differs between the pieces of a superposition can change a probability.
α a real constant in rad, the same at every x
Counts over N identical runs
n = NP ± √(NP(1 − P))
This is what the rule actually predicts. The relative scatter falls as 1/√N, and no single run is predicted at all.
n detections in the interval out of N runs; P from the Born integral; the ± is one binomial standard deviation
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Turn α through 360°: Re ψ and Im ψ trade places while |ψ|², the box and every readout stay put. Narrow σ below 0.40 nm and the peak rises past the dashed 1 nm⁻¹ line, which is legal. Widen w: the box passes 1, the area never does.
Live results
- PEAK DENSITY |ψ(0)|²
- 0.798 nm⁻¹
- AREA P(|x| < w)
- 0.683
- BOX 2w|ψ(0)|²
- 0.798
- EXPECTED IN 1000 RUNS
- 683
Catch the common trap
The mistake this lesson is built to prevent.
The commonest slip is to read the height of |ψ|² as a probability, as in “the particle is at x₀ with probability |ψ(x₀)|²”. It fails three ways. The units are wrong: |ψ|² is measured in m⁻¹ and a probability has none, so the number even changes when you switch from nm to m. It can exceed 1 with nothing amiss: an electron spread evenly across a 0.20 nm region has density 5.0 nm⁻¹ everywhere inside it. And the probability of a single exact point is zero, because a point has no width. Only an area is a probability: ∫|ψ|² dx over an interval, or |ψ(x₀)|² Δx when the interval is narrow enough that the height is flat across it.
Test it: what changes?
- Peak density |ψ(0)|² rises as 1/σ
- Peak may exceed 1 nm⁻¹, which is legal
- Probability inside ±w rises
- Total area stays 1
- Re ψ and Im ψ curves change
- |ψ|² unchanged at every x
- Window probability and expected counts unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is prepared the same way 20 000 times. Near x₀ its wavefunction has magnitude |ψ(x₀)| = 1.5 × 10⁴ m⁻¹⁄², nearly constant over a 50 pm stretch. A detector accepts only the band from x₀ − 10 pm to x₀ + 10 pm. How many detections should the band record, and by how much should that total vary from one batch of 20 000 runs to the next?
View solution →
- Probabilities come from |ψ|², never from ψ itself: |ψ(x₀)|² = (1.5 × 10⁴ m⁻¹⁄²)² = 2.25 × 10⁸ m⁻¹. Units check: a density per metre, as a one-dimensional |ψ|² must be.
- The band is Δx = 20 pm = 2.0 × 10⁻¹¹ m wide and |ψ|² is flat across it, so height × width is accurate: P = 2.25 × 10⁸ m⁻¹ × 2.0 × 10⁻¹¹ m = 4.5 × 10⁻³, a pure number between 0 and 1.
- Expected detections in N = 20 000 runs: NP = 20 000 × 4.5 × 10⁻³ = 90.
- Each run either lands in the band or does not, so the count is binomial: √(NP(1 − P)) = √(90 × 0.9955) = 9.5. Limiting check: halve the band and P halves with it, to 2.25 × 10⁻³.
Answer. About 90 detections per batch, scattering by roughly ±9 (one standard deviation) from batch to batch; P = 4.5 × 10⁻³ per run.
MediumExample 2
At t = 0 an electron confined to 0 ≤ x ≤ L, with L = 0.60 nm, is prepared in ψ(x) = √(2/L) sin(πx/L) [cos(kx) + i sin(kx)] with k = 5.0 nm⁻¹, and ψ = 0 outside. Find the probability that a position measurement at t = 0 finds it in the left third of the region, and compare with a classical particle bouncing between the walls at constant speed.
View solution →
- Dimensions first: √(2/L) = √(2/0.60 nm) = 1.83 nm⁻¹⁄², an inverse root length, and the bracket is a pure number, so ψ has the units a one-dimensional wavefunction needs.
- Re ψ = √(2/L) sin(πx/L) cos(kx) and Im ψ = √(2/L) sin(πx/L) sin(kx), so |ψ|² = (Re ψ)² + (Im ψ)² = (2/L) sin²(πx/L). The phase kx, and with it the value of k, drops out of every position probability at this instant.
- Normalisation check: ∫₀ᴸ (2/L) sin²(πx/L) dx = (2/L)(L/2) = 1, so areas under |ψ|² are probabilities as they stand.
- Left third: P = (2/L) ∫₀(L/3) sin²(πx/L) dx = (2/L)[x/2 − (L/(4π)) sin(2πx/L)] from 0 to L/3 = 1/3 − sin(2π/3)/(2π) = 0.3333 − 0.1378 = 0.196.
- Classical check: at constant speed the particle spends equal times in equal lengths, giving 1/3. The quantum value is smaller because |ψ|² vanishes at the wall and peaks at the centre; for sin(nπx/L) with large n the answer closes on 1/3. L has cancelled, as it must in a ratio of lengths.
Answer. At t = 0, P = 0.196 for the left third, independent of k and of L; a classical particle at constant speed would give 0.333.
HardExample 3
At t = 0 a particle confined to 0 ≤ x ≤ L, with L = 0.80 nm, is prepared in ψ = [ψ₁(x) + e(iφ) ψ₂(x)]/√2, where ψ₁ = √(2/L) sin(πx/L) and ψ₂ = √(2/L) sin(2πx/L). Find the probability Pleft that a position measurement at t = 0 detects it in the left half as a function of the relative phase φ, show that an overall phase changes nothing, find the φ between 0 and π that gives Pleft = 0.70, and predict the left-half count in 400 identical runs at that φ.
View solution →
- ψ₁ and ψ₂ are real, so ψ*ψ = ½[ψ₁ + e(−iφ)ψ₂][ψ₁ + e(iφ)ψ₂] = ½[ψ₁² + ψ₂² + 2ψ₁ψ₂ cos φ]. Multiplying all of ψ by e(iα) multiplies ψ* by e(−iα), and the two cancel: only the relative phase φ survives.
- Normalisation check: over the whole region ∫ψ₁² dx = ∫ψ₂² dx = 1 and ∫₀ᴸ ψ₁ψ₂ dx = 0, so the total probability is ½(1 + 1 + 0) = 1 for every φ.
- Left half: by symmetry ∫₀(L/2) ψ₁² dx = ∫₀(L/2) ψ₂² dx = ½. The cross term, with u = πx/L, is (2/π)∫₀(π/2) sin u sin 2u du = (4/π)∫₀(π/2) sin²u cos u du = 4/(3π) = 0.4244.
- So Pleft = ½[½ + ½ + 2(0.4244) cos φ] = 0.500 + 0.424 cos φ: 0.924 at φ = 0, 0.500 at φ = π/2 and 0.076 at φ = π. L has cancelled, and every value lies between 0 and 1, as a probability must. This is Pleft at t = 0 only: ψ₁ and ψ₂ belong to different energies, so φ drifts once time runs and Pleft sloshes with it.
- Set 0.500 + 0.4244 cos φ = 0.70: cos φ = 0.4712, so φ = 1.08 rad = 61.9°.
- At that φ, N = 400 runs give NP = 280 left-half detections, with binomial scatter √(400 × 0.70 × 0.30) = 9.2.
Answer. At t = 0, Pleft = 0.500 + 0.424 cos φ, unchanged by any overall phase; Pleft = 0.70 at φ = 1.08 rad (61.9°), where 400 runs give about 280 ± 9 left-half detections.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Square the modulus, integrate over the interval, compare with counts.
P(a ≤ x ≤ b) = ∫ₐᵇ |ψ|² dx for normalised ψ, with |ψ|² = ψ*ψ a density in m⁻¹. Use |ψ(x₀)|² Δx only when |ψ|² is flat across the window. An overall phase changes nothing; relative phases between pieces do. Counts over N runs scatter by √(NP(1 − P)).