
Optional unit · institution-dependent · Interactive lesson
Wavefunctions, Probability & Uncertainty
A matter wave does not say where the electron is. Squaring it says where the electron is likely to be found, which is a weaker claim and a stranger one. Here you learn to normalise an amplitude, read a probability off an area, add amplitudes rather than probabilities, and price what localising a particle costs in momentum.
UPII-15.06 · Lesson 6 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
The wave carries probability, and only its square is measurable.
The key idea
A wavefunction is a complex amplitude Ψ(x, t) whose modulus squared |Ψ(x, t)|² is the probability per unit length of finding the particle at x, so every probability is an area under |Ψ|² and never a value of Ψ itself.
Example. For an electron in the ground state of a 1.00 nm box, Ψ₁ = √(2/L) sin(πx/L), the chance of finding it in the middle tenth of the box is ∫|Ψ₁|² dx = 0.198 — almost double the 0.100 an evenly spread particle would give.
The complete physical model and assumptions
De Broglie supplied a wavelength but never said what was waving. Born's answer is the model: Ψ(x, t) is a complex probability amplitude, not itself observable, and the observable built from it is |Ψ|², a probability per unit length. Three things follow, and they are this topic. Because |Ψ|² is a density, a number comes out only after integrating over an interval, so a single point carries zero probability; and because the particle is somewhere, ∫|Ψ|² dx = 1 over all x, which fixes the one free constant in any trial Ψ and forces Ψ towards zero far away — the condition that later quantises bound states. Because amplitudes superpose while probabilities do not, two routes to one point add as Ψ₁ + Ψ₂, and the cross term makes fringes no sum of probabilities can imitate. And because a packet narrow in x can only be assembled from many wavelengths, each carrying p = h/λ, the widths of the position and momentum distributions obey Δx Δpₓ ≥ ℏ/2 — a property of the state, fixed before any instrument is switched on. The model's edges are worth naming: it predicts the distribution of outcomes and never the outcome of one run, the rule for what a measurement leaves behind is bolted on rather than derived, and a single spinless non-relativistic Ψ is only the first sketch of a theory that later needs spin, many particles and fields.
1A wave of probability, not a wave of stuff
Ψ is complex, and no instrument reads a complex number, so Ψ is not the thing you measure. What you measure is |Ψ|² = Ψ*Ψ, which is real, never negative, and — this is the part that trips people — a density rather than a probability. In one dimension Ψ carries units of m(−1/2) precisely so that |Ψ|² comes out in m⁻¹, a probability per metre. The optical analogy helps as far as it goes: a light wave's field E is not what a screen records, the irradiance goes as E², and for single photons that irradiance is the arrival probability. One more feature comes free. Multiplying the whole of Ψ by e(iα) leaves |Ψ|² untouched, so an overall phase is unobservable. A phase difference between two parts of Ψ is a different matter entirely, and the rest of this lesson turns on it.
2Normalising turns a shape into a distribution
A candidate wavefunction usually arrives with its shape known and its height free. The particle has to be found somewhere, so ∫|Ψ|² dx = 1 taken over all x, and that single condition fixes the height. Take the ground state of a box of width L, Ψ₁ = A sin(πx/L) inside and zero outside. Then ∫₀ᴸ A² sin²(πx/L) dx = A²L/2, using the average value ½ of sin² over a half-cycle, so A²L/2 = 1 and A = √(2/L). For L = 1.00 nm that is A = √(2 ÷ 1.00 × 10⁻⁹ m) = 4.47 × 10⁴ m(−1/2) — an odd-looking unit that is exactly right, since A² times a length must be dimensionless. Normalisation also disqualifies candidates: a pure plane wave e(ikx) has |Ψ|² = 1 everywhere and an infinite integral, so it is a useful idealisation of a beam, not a state of one particle.
3A density is not a probability
At the centre of that 1.00 nm box the density is |Ψ₁|² = 2/L = 2.0 × 10⁹ m⁻¹. A number vastly larger than 1 is no contradiction, because it is a probability per metre and only becomes a probability once multiplied by a length. Over a window Δx narrow enough that the density hardly changes, P ≈ |Ψ(x)|² Δx: a 0.020 nm window at the centre holds 2.0 × 10⁹ m⁻¹ × 2.0 × 10⁻¹¹ m = 0.040, a 4.0% chance. For a wider window you must integrate. The middle tenth of the box, from 0.45L to 0.55L, gives ∫|Ψ₁|² dx = 0.198, not the 0.100 of an evenly spread particle, because the ground-state density is humped in the middle. Two habits follow: check that a probability you have quoted lies between 0 and 1, and never compare two densities without saying over what width.
4Add the amplitudes, then square
Send electrons one at a time at a double slit. If Ψ₁ and Ψ₂ are the amplitudes at a screen point P for the two openings, the amplitude there is Ψ₁ + Ψ₂ and the arrival rate goes as |Ψ₁ + Ψ₂|² = |Ψ₁|² + |Ψ₂|² + 2|Ψ₁||Ψ₂| cos φ, where φ is the phase difference set by the path difference. Classical bookkeeping — this electron went through slit 1 or slit 2, so add the two probabilities — keeps only the first two terms and predicts a featureless blur. The cross term is the fringes. With equal amplitudes it swings the rate from 4|Ψ₁|² at a maximum to exactly zero at a minimum, while the classical sum sits flat at 2|Ψ₁|²; averaged across the screen the two agree, so interference redistributes electrons without creating them. Arrange any measurement that reveals which slit was used and φ stops being defined, the cross term averages away, and the fringes go with it.
5What a single measurement actually gives
One electron makes one mark. It does not arrive smeared over the screen in the shape of |Ψ|², and nothing in the theory says which mark it will make. The distribution appears only when the same preparation is repeated thousands of times and the marks are histogrammed, which is why an experiment quoting a probability must say what the preparation was. The result also changes the state. Detect the electron inside a 0.020 nm window and an immediate second measurement finds it in that window again with near certainty: the state that follows the detection is one localised there, not the spread-out Ψ you started with. So Δx is not a confession of ignorance about a particle that secretly had a position. It is the width of the distribution the preparation produces, and preparing a narrower one is a physical act with a price.
6Narrow in position means wide in momentum
That price is the uncertainty relation, and its origin is wave analysis rather than clumsiness. A single wavelength extends forever; to build a bump of width Δx you must add waves over a range of wavenumbers Δk, and the mathematics of Fourier sums gives Δx Δk ≥ ½. Each component carries p = ℏk, so Δx Δpₓ ≥ ℏ/2. A Gaussian packet is the one shape that reaches equality; every other shape does worse. Put numbers to it: hold an electron inside Δx = 0.10 nm, about an atomic radius, and Δpₓ ≥ 1.055 × 10⁻³⁴ ÷ (2 × 1.0 × 10⁻¹⁰) = 5.3 × 10⁻²⁵ kg m s⁻¹, a speed spread of 5.8 × 10⁵ m s⁻¹ and a kinetic energy scale near 0.95 eV. Atoms are stable at the size they are because squeezing the electron further costs more kinetic energy than the extra Coulomb attraction returns.
Key equations
Read the units, assumptions and limits before substituting.
Born rule
|Ψ(x, t)|² = Ψ*Ψ = probability per unit length
Converts an unobservable complex amplitude into the thing a detector counts.
Ψ in m(−1/2) in one dimension, so |Ψ|² is in m⁻¹Probability in an interval
P(a ≤ x ≤ b) = ∫ₐᵇ |Ψ(x, t)|² dx
Probability is an area, which is why one exact point carries none.
a and b in m; P is a pure number between 0 and 1Normalisation
∫ |Ψ|² dx = 1, over all x
One equation promotes a shape into a probability distribution.
fixes the constant A in a trial Ψ = A f(x); needs Ψ → 0 as |x| → ∞Superposition of amplitudes
|Ψ₁ + Ψ₂|² = |Ψ₁|² + |Ψ₂|² + 2|Ψ₁||Ψ₂| cos φ
The cos φ term is the interference that adding probabilities cannot make.
φ is the phase difference of the two amplitudes at that point, in radSpread of the position distribution
Δx = √(⟨x²⟩ − ⟨x⟩²)
The width the state predicts, not the error bar of a ruler.
⟨f⟩ = ∫ f(x) |Ψ|² dx, the average over the distribution; Δx in mPosition-momentum uncertainty
Δx Δpₓ ≥ ℏ/2
A floor the Gaussian packet touches and no preparation beats.
ℏ = h/2π = 1.055 × 10⁻³⁴ J s; Δx in m, Δpₓ in kg m s⁻¹Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag Δx from 0.30 nm down to 0.10 nm: the position hump grows three times narrower and three times taller while the momentum hump does the exact reverse, so the product never moves. Then slide x₀ — the packet shifts, but neither width changes and the momentum panel does not stir.
Live results
- POSITION SPREAD Δx
- 0.20 nm
- MOMENTUM SPREAD Δp
- 2.64 10⁻²⁵ kg m s⁻¹
- PRODUCT Δx Δp
- 0.527 10⁻³⁴ J s
- PEAK DENSITY |Ψ|²
- 1.99 nm⁻¹
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Position spread Δx: 0.20 nm; Packet centre x₀: 0.00 nm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The uncertainty principle is not a complaint about clumsy instruments. The familiar story — the photon you need in order to see the electron kicks it, spoiling the momentum you were measuring — leaves the impression that the electron really had a definite x and a definite p, and the apparatus merely hid them. But Δx and Δpₓ are computed from Ψ before any measurement happens: they are the widths of the two distributions the state itself predicts, tied together because a packet narrow in x can only be built by adding many wavelengths, each with its own p = h/λ. A state of one definite momentum is a single wavelength filling all space, Δx unbounded. Build a perfect detector and nothing improves — prepare the same state again and the spreads still satisfy Δx Δpₓ ≥ ℏ/2.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A crude model of an electron trapped against a barrier gives Ψ(x) = A x for 0 ≤ x ≤ 2.0 nm, with Ψ = 0 outside. Find A, then find the probability that the electron is in the outer tenth of the region, 1.8 nm ≤ x ≤ 2.0 nm.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Normalisation: ∫₀ᴸ |Ψ|² dx = 1 with |Ψ|² = A²x², so ∫₀ᴸ A²x² dx = A²L³/3 = 1 and A = √(3/L³).
- With L = 2.0 nm = 2.0 × 10⁻⁹ m, L³ = 8.0 × 10⁻²⁷ m³ and 3/L³ = 3.75 × 10²⁶ m⁻³, so A = 1.9 × 10¹³ m(−3/2). The unit is Ψ per length, as it must be if A x is to come out in m(−1/2).
- The normalised density is |Ψ|² = 3x²/L³, so P(0.9L ≤ x ≤ L) = ∫ 3x²/L³ dx = [x³/L³] from 0.9L to L = 1 − 0.9³.
- 1 − 0.729 = 0.271. An evenly spread particle would give 0.100 in the same window, so this state leans hard on the far wall.
Answer. A = 1.9 × 10¹³ m(−3/2), and P(1.8 nm ≤ x ≤ 2.0 nm) = 0.271, or 27%.
MediumExample 2
A pulsed source prepares electrons whose position spread along the beam is Δx = 0.50 nm. (a) Find the smallest momentum spread consistent with that preparation and the speed spread it implies. (b) Find the energy scale (Δpₓ)²/2mₑ in eV and compare it with the 25 meV of thermal energy at 300 K. (c) Repeat the speed estimate for a 1.0 g bead localised to 1.0 μm.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- The bound is Δx Δpₓ ≥ ℏ/2 with ℏ = 1.055 × 10⁻³⁴ J s, so the smallest allowed spread is Δpₓ = ℏ/(2Δx).
- Δpₓ = 1.055 × 10⁻³⁴ ÷ (2 × 0.50 × 10⁻⁹) = 1.055 × 10⁻²⁵ kg m s⁻¹, which is 1.1 × 10⁻²⁵ kg m s⁻¹ to two significant figures.
- Δv = Δpₓ/mₑ = 1.055 × 10⁻²⁵ ÷ 9.11 × 10⁻³¹ = 1.2 × 10⁵ m s⁻¹: confining an electron to half a nanometre leaves its speed uncertain by over a hundred kilometres per second.
- Energy scale: (Δpₓ)²/(2mₑ) = (1.055 × 10⁻²⁵)² ÷ (2 × 9.11 × 10⁻³¹) = 6.11 × 10⁻²¹ J, and dividing by 1.602 × 10⁻¹⁹ J eV⁻¹ gives 0.038 eV.
- 38 meV sits above the 25 meV of thermal energy at 300 K, so confinement on this scale is not a detail buried in thermal noise — it sets how the device behaves.
- The bead: Δpₓ = 1.055 × 10⁻³⁴ ÷ (2 × 1.0 × 10⁻⁶) = 5.3 × 10⁻²⁹ kg m s⁻¹, so Δv = 5.3 × 10⁻²⁹ ÷ 1.0 × 10⁻³ = 5.3 × 10⁻²⁶ m s⁻¹, which no experiment could ever resolve.
Answer. Δpₓ ≥ 1.1 × 10⁻²⁵ kg m s⁻¹, Δv ≈ 1.2 × 10⁵ m s⁻¹ and (Δpₓ)²/2mₑ ≈ 0.038 eV; for the bead Δv ≈ 5.3 × 10⁻²⁶ m s⁻¹.
HardExample 3
In an electron interferometer the second path is attenuated so that its amplitude at a screen point P is 0.60 times the first path's. With path 1 alone open, P records 1.20 × 10³ electrons per second. Find the rate from path 2 alone, the maximum and minimum rates as the phase difference is swept, the ratio between them, and the visibility V = (Rₘₐₓ − Rₘᵢₙ)/(Rₘₐₓ + Rₘᵢₙ).
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Rates follow |Ψ|², so path 2 alone gives (0.60)² × 1.20 × 10³ = 0.36 × 1.20 × 10³ = 432 s⁻¹.
- Write |Ψ₁| = a and |Ψ₂| = 0.60a. With both open, |Ψ₁ + Ψ₂|² = a²(1 + 0.36 + 2 × 0.60 cos φ) = a²(1.36 + 1.20 cos φ), where a² corresponds to 1.20 × 10³ s⁻¹.
- Maximum at cos φ = +1: Rₘₐₓ = 2.56 × 1.20 × 10³ = 3.07 × 10³ s⁻¹.
- Minimum at cos φ = −1: Rₘᵢₙ = 0.16 × 1.20 × 10³ = 1.92 × 10² s⁻¹. The dark fringe is not empty, because unequal amplitudes cannot cancel completely.
- Ratio Rₘₐₓ/Rₘᵢₙ = 2.56/0.16 = 16, and V = 1.20/1.36 = 0.88, so cutting one path to 60% of its amplitude still leaves a strong pattern.
- Bookkeeping check: adding probabilities would give a flat 1.20 × 10³ + 432 = 1.63 × 10³ s⁻¹. The bright fringe runs 88% above that and the dark fringe 88% below, and since cos φ averages to zero across the screen the flat rate is what survives averaging.
Answer. 432 s⁻¹ from path 2 alone; Rₘₐₓ = 3.07 × 10³ s⁻¹ and Rₘᵢₙ = 1.92 × 10² s⁻¹, a ratio of 16 : 1, with visibility V = 0.88.
Quick check
Select an answer and explain it before checking.
Review & continue
A wavefunction is a complex amplitude Ψ(x, t) whose modulus squared |Ψ(x, t)|² is the probability per unit length of finding the particle at x, so every probability is an area under |Ψ|² and never a value of Ψ itself.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.