Normalising ψ and reading probabilities off it
The equation fixes ψ's shape; normalisation fixes its size.
The Schrödinger equation hands you ψ with its size left blank. Here you fill it in: make the total probability 1, do the Gaussian or polynomial integral that demands, then turn ψ into numbers by integrating |ψ|² across an interval. Along the way you meet the functions that refuse, and what physicists do with them.
Build the model
The equation fixes the shape of ψ; the total probability fixes its size.
1The big idea
The Schrödinger equation is linear and homogeneous, so it can fix the shape of ψ but never its size: if ψ is a solution, so is Aψ for any complex constant A. The Born rule supplies the missing condition. Because |ψ|² is a probability per unit length and the particle is certainly somewhere, ∫|ψ|² dx over all x must equal 1, and that one number fixes |A|; the phase of A cancels from every |ψ|² and is chosen to make A real and positive. Most normalisations then come down to two families of integrals: polynomials confined to a region, handled by expanding or by the beta formula, and Gaussians, handled by ∫e(−x²/σ²) dx = σ√π and its moments. Once normalised, ψ answers any question of the form 'how likely is x between a and b?' by an integral of |ψ|² over that interval, and each answer can be checked: the whole line gives 1, a symmetric state gives ½ on each side, and no answer exceeds 1. For a real potential the Schrödinger equation keeps the integral constant, so normalising at t = 0 is enough. The procedure fails when the integral diverges. A ψ whose tail falls no faster than |x|⁻¹⁄², and above all a plane wave e(ikx), has no normalising constant. Plane waves are too useful to drop, so they are given the delta normalisation of the free-particle unit or confined to a box of length L with A = 1/√L; neither makes a plane wave a state, only the limit of wave packets that are.
2In one sentence
Normalising ψ means choosing the constant that multiplies it so that ∫|ψ|² dx over all x equals 1; after that, the integral of |ψ|² over any interval is the probability of finding the particle there.
Example. For ψ = A e(−x²/(2σ²)) with σ = 0.30 nm, ∫|ψ|² dx = A²σ√π, so A = (πσ²)⁻¹⁄⁴ = 1.37 nm⁻¹⁄², and the probability of finding the particle within 0.30 nm of the centre is erf(1) = 0.843.
3A linear equation fixes the shape and leaves the size blank
The Schrödinger equation iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x² + Vψ has ψ in every term exactly once, so if ψ solves it, so do 3ψ, iψ and Aψ for any complex constant A. Nothing in the dynamics can choose A; the Born rule does. The particle is certainly somewhere, so the density |ψ|² must integrate to 1 over all x. Write ψ = A f(x) with f any convenient solution: then |A|² ∫|f|² dx = 1, which fixes |A| = (∫|f|² dx)⁻¹⁄². The phase is still free, because A and A e(iα) give the same |ψ|² everywhere, so take A real and positive. Two checks come with it. Dimensions: |ψ|² is a probability per metre, so in one dimension ψ carries m⁻¹⁄², and A must supply whatever f leaves over. Time: for a real potential the Schrödinger equation keeps ∫|ψ|² dx constant, as Unit 2 proves from the continuity equation, so a state normalised at t = 0 stays normalised.
4Polynomials in a region: square first, then integrate
Take ψ = A x(a − x) for 0 ≤ x ≤ a and zero outside, a trial state for a particle held between two walls. The integrand is |ψ|² = A² x²(a − x)², so square before integrating; integrating ψ itself answers a different question. Expanding gives A²(a²x² − 2ax³ + x⁴), whose integral from 0 to a is A²a⁵(1/3 − 1/2 + 1/5) = A²a⁵/30. The beta formula ∫₀ᵃ xᵐ(a − x)ⁿ dx = m! n! a(m+n+1)/(m+n+1)! gives the same 2! 2! a⁵/5! = a⁵/30 without the bookkeeping. So A = √(30/a⁵). Check the dimensions: x(a − x) is an area in m², so A must carry m⁻⁵⁄² to leave ψ in m⁻¹⁄², and √(30/a⁵) does. For a = 1.00 nm, A = 5.48 nm⁻⁵⁄² and the peak density |ψ(a/2)|² = 30/(16a) = 1.88 nm⁻¹. That exceeds 1, which is allowed: it is a density, and only its integral over an interval is a probability.
5Gaussians: one integral, and what does not change it
For ψ = A e(−x²/(2σ²)) the density is A² e(−x²/σ²), and the one integral needed is ∫e(−x²/σ²) dx = σ√π over all x, found by squaring the integral and switching to polar coordinates. Then A²σ√π = 1 and A = (πσ²)⁻¹⁄⁴; for σ = 0.50 nm, A = 1.06 nm⁻¹⁄². Two changes leave this A untouched, and seeing why saves work. Moving the centre to x₀ only relabels the integration variable. Multiplying by a phase e(ik₀x), which turns a packet that spreads in place into one that also moves, changes nothing because |e(ik₀x)|² = 1: the density of A e(−(x − x₀)²/(2σ²)) e(ik₀x) is still A² e(−(x − x₀)²/σ²). Watch the widths, though. σ is the width of ψ; |ψ|² is narrower, with standard deviation σ/√2, because squaring doubles the decay rate of the exponential. A polynomial times a Gaussian needs the moments ∫x² e(−x²/σ²) dx = σ³√π/2 and upwards, each obtained from the one before by differentiating with respect to β = 1/σ² and changing the sign.
6Reading a probability off ψ
With ψ normalised, the probability of finding the particle between a and b is ∫ₐᵇ|ψ|² dx, and at this stage that is the only kind of number ψ gives. For the polynomial state above, P(0 ≤ x ≤ c) works out to 10u³ − 15u⁴ + 6u⁵ with u = c/a, so the first quarter of the region holds P = 0.104. A classical particle bouncing between the walls at constant speed spends equal times in equal lengths and would give 0.25; this state has moved probability away from the walls, where it vanishes, into the middle. For the Gaussian, P(|x| ≤ σ) = erf(1) = 0.843, not the familiar 0.683, because σ measures ψ and not |ψ|². If ψ has not been normalised you can still get probabilities: divide the interval integral by the total, P = ∫ₐᵇ|ψ|² dx ÷ ∫|ψ|² dx, and A cancels. Three checks catch most slips. The whole line must give 1, mirror-image intervals of a symmetric state must give equal shares, and no answer may exceed 1.
7Which functions can be normalised at all
Normalisation needs ∫|ψ|² dx to be finite and non-zero. The first condition is the demanding one, and it is usually decided in the tails. If |ψ| falls like C|x|(−p) at large |x|, the tail integral from R to infinity is C²R(1−2p)/(2p − 1), finite only when p exceeds ½. So a 1/x tail normalises, a 1/√x tail fails because its integral grows like ln x without limit, and anything that levels off, p = 0, fails fastest. Exponential and Gaussian tails pass with room to spare. A spike can fail too: near x = 0 a factor |x|(−q) in ψ leaves the integral finite for q below ½ and makes it diverge from q = ½ up, so |x|⁻¹⁄⁴ can be normalised and 1/√|x| cannot. And ψ = 0 everywhere has a finite integral that no A can raise to 1: it describes no particle. Functions that pass make up the square-integrable set, and the next lesson sets this condition beside continuity and single-valuedness in the full list of requirements on ψ.
8Plane waves: useful functions that will not normalise
The plane wave ψ = A e(i(kx − ωt)) is the standard picture of a particle with sharp momentum ħk, and it sits squarely on the failing side: |ψ|² = |A|² everywhere, so the integral over a length L is |A|²L, which grows without bound. No A makes it 1, so no plane wave is a physical state. Two conventions keep it usable. Box normalisation confines x to a length L with periodic ends, sets A = 1/√L, and computes every probability or ratio before letting L grow; one electron of a beam, modelled in a 1.0 μm box, has A = 1.0 × 10³ m⁻¹⁄² and a 1.0% chance of being in any 10 nm stretch. Delta normalisation sets A = 1/√(2π) and asks instead that the overlap of two plane waves be δ(k − k′), a convention the free-particle unit builds properly. Either way the physical state is a wave packet, a superposition of plane waves that does fall off in its tails and does normalise.
Key equations
What each relation says, and the units it is written in.
Normalisation condition
∫ |ψ(x, t)|² dx = 1, integrated over all x
For ψ = A f(x) it reads |A|² ∫|f|² dx = 1. It fixes the size of the constant A that the linear Schrödinger equation leaves free, but not its phase.
|ψ|² in m⁻¹ and dx in m make the integral a pure number, so in one dimension ψ carries m⁻¹⁄².
Probability of an interval
P(a ≤ x ≤ b) = ∫ₐᵇ |ψ(x, t)|² dx
Before normalising, divide by the same integral taken over all x: the ratio is the probability, and A cancels from it.
a and b in m; P is a pure number between 0 and 1 once ψ is normalised.
Gaussian integral
∫ e(−x²/σ²) dx = σ√π, over all x
Gives A = (πσ²)⁻¹⁄⁴ for every Gaussian ψ, unchanged by moving the centre or multiplying by a phase e(ik₀x), since neither alters the integral.
σ in m. The integrand is |ψ|² for ψ ∝ e(−x²/(2σ²)), a density whose standard deviation is σ/√2.
Polynomial integral on 0 ≤ x ≤ a
∫₀ᵃ xᵐ (a − x)ⁿ dx = m! n! a(m+n+1) / (m+n+1)!
Normalises any state of this product form on 0 ≤ x ≤ a without expanding brackets: ψ = A x(a − x) needs A = √(30/a⁵), which carries m⁻⁵⁄².
a in metres; the exponents m and n are whole numbers. For |x(a − x)|², m = n = 2 and the integral is a⁵/30.
Tail test for square-integrability
∫ x(−2p) dx from R to ∞ = R(1−2p) / (2p − 1), finite only for p above ½
A 1/x tail passes; a 1/√x tail diverges like ln x; a plane wave, p = 0, diverges in proportion to the length it covers.
For |ψ| ≈ C|x|(−p) at large |x|: p is a pure number, R in m, C in m(p−1/2).
Box normalisation of a plane wave
ψ(x) = (1/√L) e(ikx) on a length L with periodic ends
Makes a sharp-momentum wave usable: compute probabilities and ratios at finite L, then let L grow. Unit 5 swaps 1/√L for the delta convention 1/√(2π).
L in m, k in m⁻¹; |ψ|² = 1/L at every point of the box, whatever k is.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag A until the top bar stops on the dashed line at 1: at σ = 1.0 nm that is A = 0.75 nm⁻¹⁄², where solid meets dashed. Narrow σ to 0.6 nm and the total falls to 0.60, so A must rise to 0.97. Set w = σ: the window holds 0.843 of the total whatever A or σ.
Live results
- ∫|ψ|² dx OVER ALL x
- 1.772
- A THAT NORMALISES
- 0.751 nm⁻¹⁄²
- ∫|ψ|² dx OVER |x| ≤ w
- 1.494
- P(|x| ≤ w) = WINDOW ÷ ALL
- 0.843
Catch the common trap
The mistake this lesson is built to prevent.
The commonest slip is to normalise ψ instead of |ψ|²: to set ∫ψ dx = 1 as though ψ itself were the probability density. For ψ = A x(a − x) on 0 ≤ x ≤ a this gives A = 6/a³ instead of √(30/a⁵), and the total probability that choice produces, ∫|ψ|² dx = (36/a⁶)(a⁵/30) = 1.2/a, is not even a pure number, let alone 1. The Born rule assigns probability to |ψ|², so it is |ψ|² that must integrate to 1. The dimension check exposes the error at once: ψ must carry m⁻¹⁄², and 6/a³ times an area leaves m⁻¹. For a complex ψ the mistake is worse, because ∫ψ dx can be complex or close to zero: for a moving packet A e(−x²/(2σ²)) e(ik₀x) it shrinks by the factor e(−k₀²σ²/2) as k₀ grows, while the correct normalisation does not depend on k₀ at all.
Test it: what changes?
- Total ∫|ψ|² dx = A²σ√π falls
- Normalising amplitude (πσ²)⁻¹⁄⁴ rises
- Normalised P(|x| ≤ w) rises
- Total ∫|ψ|² dx grows as A²
- Window integral grows as A²
- Ratio P = window ÷ all unchanged
- Normalising A unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
At t = 0 a particle has ψ(x) = A sech(x/b), with b = 0.25 nm and sech u = 1/cosh u. Find A, and the probability of finding the particle within 0.25 nm of the origin.
View solution →
- The density is |ψ|² = A² sech²(x/b). Since d/dx tanh(x/b) = (1/b) sech²(x/b), the antiderivative of the density is A²b tanh(x/b).
- Over all x, tanh runs from −1 to +1, so ∫|ψ|² dx = 2A²b = 1 and A = 1/√(2b) = 1/√(0.50 nm) = 1.41 nm⁻¹⁄². Dimension check: A²b is nm⁻¹ × nm, a pure number, as a total probability must be.
- P(|x| ≤ b) = A²b [tanh(1) − tanh(−1)] = 2A²b tanh(1) = tanh(1) = 0.762.
- Checks: widening the window sends tanh towards 1 and P towards 1, and the answer does not depend on b, as any probability over a window measured in units of the state's own width must not.
Answer. A = 1.41 nm⁻¹⁄²; P(|x| ≤ 0.25 nm) = tanh 1 = 0.762.
MediumExample 2
A particle confined to 0 ≤ x ≤ a, with a = 2.0 nm, has ψ(x) = A x²(a − x) inside and zero outside. Find A and the probability of finding the particle in the left half, and compare with a classical particle crossing at constant speed.
View solution →
- |ψ|² = A² x⁴(a − x)². The beta integral with m = 4 and n = 2 gives ∫₀ᵃ x⁴(a − x)² dx = 4! 2! a⁷/7! = 48a⁷/5040 = a⁷/105.
- So A = √(105/a⁷). With a⁷ = 2.0⁷ nm⁷ = 128 nm⁷, A = √(0.820 nm⁻⁷) = 0.906 nm⁻⁷⁄². Dimension check: x²(a − x) is in m³, and m⁻⁷⁄² × m³ = m⁻¹⁄², as ψ requires.
- Left half: expand to a²x⁴ − 2ax⁵ + x⁶ and integrate from 0 to a/2: a⁷(1/160 − 1/192 + 1/896) = 29a⁷/13440.
- P = (105/a⁷)(29a⁷/13440) = 29/128 = 0.227, so the right half holds 0.773: the x² factor keeps the particle away from the left wall.
- A classical particle crossing at constant speed spends equal times in each half, P = 0.5, so the quantum state is strongly lopsided. Checks: 0.227 + 0.773 = 1, and a cancels from both probabilities.
Answer. A = 0.906 nm⁻⁷⁄²; P(0 ≤ x ≤ 1.0 nm) = 29/128 = 0.227, against 0.5 for the classical particle.
HardExample 3
A particle has ψ(x) = A(1 + x/b) e(−x²/(2b²)) with b = 0.40 nm. Using ∫e(−u²) du = √π over all u, find A and the probability that the particle is found at x above 0, and explain why the cross term matters for one answer and not the other.
View solution →
- |ψ|² = A²(1 + 2x/b + x²/b²) e(−x²/b²). Substitute u = x/b, dx = b du: ∫|ψ|² dx = A²b ∫(1 + 2u + u²) e(−u²) du.
- Over all u: ∫e(−u²) du = √π; the cross term 2u e(−u²) is odd and integrates to zero; ∫u² e(−u²) du = √π/2, by parts. So A²b × 3√π/2 = 1 and A² = 2/(3√π b).
- Numbers: A² = 2/(3 × 1.7725 × 0.40 nm) = 0.940 nm⁻¹, so A = 0.970 nm⁻¹⁄².
- On the half-line u from 0 to ∞ the three pieces are √π/2, ∫2u e(−u²) du = 1, and √π/4. So P(x above 0) = (2/(3√π))(3√π/4 + 1) = ½ + 2/(3√π) = 0.876.
- The cross term is odd, so over the whole line it cancels and leaves the normalisation alone; over a half-line it survives and tilts the probability towards the side where 1 + x/b is large.
- Checks: A²b is nm⁻¹ × nm, a pure number, as a total probability must be; P(x below 0) = ½ − 2/(3√π) = 0.124 and the two add to 1; b cancels from both probabilities; deleting the x/b term makes ψ symmetric and returns ½, as it must.
Answer. A = 0.970 nm⁻¹⁄²; P(x above 0) = ½ + 2/(3√π) = 0.876.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Set ∫|ψ|² dx = 1, never ∫ψ dx = 1.
Write ψ = A f(x), so |A|² ∫|f|² dx = 1 fixes |A|; choose A real and positive. Then P(a ≤ x ≤ b) = ∫ₐᵇ|ψ|² dx, and before normalising divide by the total. The integral must converge: tails falling no faster than |x|⁻¹⁄² fail.
Next: The requirements on ψ(x, t), and the reason for each →