Lesson 3 of 7
Wavefunctions and probability
I can represent a particle state by a wavefunction, interpret amplitude squared as probability density, including an atomic electron cloud, and normalise square and sinusoidal profiles.Syllabus
Syllabus 9478, 19(f). Represent a particle state by a wavefunction, interpret amplitude squared as probability density, including an atomic electron cloud, and normalise square and sinusoidal profiles.
Make a guess. You are not marked.
A particle's wavefunction has amplitude 2A near point X and amplitude A near point Y, over equal small intervals. How does the chance of detecting the particle near X compare with near Y?
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Four times as likely, because probability goes with amplitude squared.
The probability of detection in a small interval is proportional to the square of the wavefunction's amplitude there. (2A)2 = 4A2, which is four times A2.
A wavefunction represents a particle's state. Its amplitude is not a displacement of matter. Squared amplitude gives a probability density; area under that density gives an interval probability.
For a real wavefunction ψ, density is ψ2. More generally it is the modulus squared, |ψ|2. A negative real amplitude is allowed, but its squared density is nonnegative. Reversing the sign of the whole real wavefunction leaves its density unchanged; relative signs matter when combining contributions.
In one dimension, the probability of a position interval is the area under |ψ|2 over that interval. Normalisation sets the total area over all possible positions to 1. A density value alone is not a probability: it has an inverse-length unit and can be numerically greater than 1 without contradiction.
| Coordinate x | Amplitude ψ | Density |ψ|2 |
|---|---|---|
| m | m-1/2 | m-1 |
| nm | nm-1/2 | nm-1 |
Density multiplied by a width in the matching unit is dimensionless. An atomic electron cloud instead represents a three-dimensional spatial probability distribution, not a classical orbit. Its full volume density has unit m-3, and probability comes from a volume, rather than a one-dimensional area under a curve.
Square-profile normalisation
Determine the height before finding an interval area
Let ψ = C for 0 < x < 4.00 nm, with zero amplitude outside. For this constant profile:
C = 1/√(4.00 nm) = 0.500 nm-1/2
|ψ|2 = 0.250 nm-1
Between 1.00 and 2.50 nm the width is 1.50 nm, giving:
Square the amplitude first, then find the density area. Squaring the area under ψ gives a different result with the wrong units. Choosing C = -0.500 nm-1/2 would give the same probabilities.
In metre units the same amplitude is 1.58114 × 104 m-1/2, and density is 2.50 × 108 m-1. The interval width is then 1.50 × 10-9 m, so its probability remains 0.375.
This abrupt square profile is an ideal normalisation exercise. The values at its isolated endpoints do not change its area. It is not an infinite-well stationary state.
Check your understanding
A normalised real wavefunction is 0.500 nm^(-1/2) for 0 < x < 4.00 nm and zero outside. What is the probability of finding the particle between 1.00 and 2.50 nm?
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0.375.
Square the amplitude first: density is 0.250 nm-1. Its area over the 1.50 nm interval is 0.250 x 1.50 = 0.375, a dimensionless probability.
Sinusoidal-profile normalisation
Use the mean square over the whole interval
Now let ψ = C sin(πx/L) within 0 < x < L, where L = 4.00 nm, and zero outside. The mean of sin2 over this half-wave is 1/2:
C = √(2/L) = 0.707107 nm-1/2
|ψ|2 = 0.500 sin2(πx/4.00) nm-1
The last expression uses x in nm, making πx/4.00 a dimensionless angle in radians. Symmetry gives probability 0.500 in each half of the interval. The central half, 1.00 to 3.00 nm, contains more probability than the combined two outer quarters because its density is higher on average.
The different coefficient from the square profile is essential. Using 1/√L here would leave total probability 1/2, not 1.
Check your understanding
For 0 < x < L, psi = C sin(pi x/L), with L = 4.00 nm and zero amplitude outside. The mean of sin2 over this interval is 1/2. Which positive C normalises the state?
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0.707 nm^(-1/2).
C2 L/2 = 1 gives C = sqrt(2/L). With L entered in nm, C = sqrt(2/4.00) = 0.707 nm^(-1/2), and density times a width in nm is dimensionless.
Square profile: signed amplitude
This discontinuous square profile is an ideal normalization exercise, not an infinite-well stationary state. Values at single endpoints do not change an interval area.
Square profile: density and interval probability
Shaded width: 1.50 nm. Density: 0.250 per nm. Their product is probability 0.375, with no unit. The total width of 4.00 nm gives total probability 1.
Sinusoidal profile: signed amplitude
The coefficient is sqrt(2/L), with L = 4.00 nm. Amplitude is not a material displacement, and area under this amplitude graph is not probability.
Sinusoidal profile: density and interval probability
Only area under density gives an interval probability. The central 1-3 nm area is 0.818309886...; the total density area is 1. The shaded area is not the square of the area under amplitude.
Paper 4 practice: spreadsheet
Generate the sine model and refine its area estimate
Download the probability starter workbook. Its Coarse, Fine and Finer sheets contain coordinates for this generated 4.00 nm model, not experimental observations. The calculation cells are blank for you to complete.
All three sheets use A for x / nm, B for ψ / nm-1/2, C for density / nm-1 and D for interval probability. G2 contains L = 4; I2 contains the coordinate step. H2 is the blank normalisation coefficient. Enter these first formulas on each sheet:
| Cell | Formula |
|---|---|
| H2 | =SQRT(2/$G$2) |
| B2 | =$H$2*SIN(PI()*A2/$G$2) |
| C2 | =B2^2 |
| D2 | =(A3-A2)*(C2+C3)/2 |
Keep the fixed references when copying. B and C have a value at every coordinate. D is the trapezium area between two consecutive coordinates, so its last formula belongs one row earlier:
| Sheet and step | Coordinates; B/C fill | D interval fill |
|---|---|---|
| Coarse: 0.5 nm | A2:A10; B2:C10 | D2:D9 |
| Fine: 0.25 nm | A2:A18; B2:C18 | D2:D17 |
| Finer: 0.125 nm | A2:A34; B2:C34 | D2:D33 |
Plot B against numeric A and C against numeric A in separate XY graphs, with both vertical quantities and units labelled. The sine argument is in radians. For area, density in nm-1 must multiply widths in nm; mixing an unchanged nm-1 density with metre widths would change the answer incorrectly.
Use a blank cell for each of the following sums. Preserve the stored values while formatting display precision.
| Sheet | Probability sums |
|---|---|
| Coarse | Full: =SUM(D2:D9)1-3 nm: =SUM(D4:D7) |
| Fine | Full: =SUM(D2:D17)1-3 nm: =SUM(D6:D13) |
| Finer | Full: =SUM(D2:D33)1-3 nm: =SUM(D10:D25) |
Compare the three area estimates
The central-interval estimates are 0.8017766953, 0.8142087183 and 0.8172865746, respectively. Compare them with the supplied exact model area 1/2 + 1/π = 0.8183098862. The shortfalls are about 0.0165332, 0.00410117 and 0.00102331.
The full-range sums happen to equal 1, apart from numerical round-off, on all three unrounded regular grids. That coincidence does not make each partial area exact. The exact smooth density and the straight-sided trapezia are different representations.
Refining the step reduces a numerical approximation error here. It is distinct from experimental uncertainty and from the quantum position-momentum width relation. Every interval probability must stay between 0 and 1; a dimensional density has no such numerical upper limit.