University Physics IV · Foundations of Quantum Mechanics · 5.3

Born's Rule & Probability Current

Everything measurable about a one-dimensional state comes out of one recipe: square the modulus, then integrate over the interval you care about. This lesson makes that recipe honest — what fixes the scale and the units of ψ, what keeps the total at 1 for all time, and how probability gets from one region to another.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Born's rule is the bridge from a complex function to a laboratory number, and it is a postulate rather than a theorem: |ψ(x, t)|² dx is the probability that a position measurement, on a system prepared in ψ, lands between x and x + dx. Two consequences follow at once. Probability is dimensionless and dx is a length, so ψ in one dimension must carry units of one per root metre; and the particle has to be somewhere, so ∫|ψ|² dx = 1 — a condition that spends the free multiplicative constant the linear Schrödinger equation leaves dangling, and that evicts every function which is not square-integrable, plane waves included.

A normalisation imposed at one instant would be worthless if the dynamics could break it, so the third move is a check: differentiate the norm under the integral sign, substitute the time-dependent Schrödinger equation, and watch the potential terms cancel exactly when V is real. What is left is ∂|ψ|²/∂t = −∂j/∂x with j = (ħ/m)Im(ψ* ∂ψ/∂x), a continuity equation: probability is never created or destroyed, only transported, and the total stays pinned at 1 as long as ψ dies off at infinity. The cost is real. ψ itself is unobservable, since a global phase changes nothing measurable; the rule delivers a distribution and never the individual outcome; and j says how an ensemble's probability flows, not how fast one particle rides.

Simple definition: Born's rule says the probability of finding the particle between x and x + dx at time t is |ψ(x, t)|² dx, which makes |ψ|² a probability per unit length and never, on its own, a probability.
Example: For the L = 1.00 nm ground state ψ₁ = √(2/L) sin(πx/L), |ψ₁(L/2)|² = 2.0 nm⁻¹; across a 0.010 nm window at the centre that is a probability of 0.020, and across the whole well it is exactly 1.

Born's rule in one dimensionP(a ≤ x ≤ b) = ∫ₐᵇ |ψ(x, t)|² dx

A probability needs an interval — a density times a length. |ψ|² at a single point is never itself an answer.

ρ = |ψ|² = ψ*ψ, in m⁻¹; ψ in one per root metre; P is a pure number

Normalisation, and the units it hands ψ∫₋∞⁺∞ |ψ(x, t)|² dx = 1

Spends the free constant and fixes ψ's units: 1/√m in one dimension, 1/√(m³) in three.

The Schrödinger equation is linear, so ψ and 2ψ solve it alike; this picks one of them.

Probability current densityj = (ħ/2mi)(ψ* ∂ψ/∂x − ψ ∂ψ*/∂x) = (ħ/m) Im(ψ* ∂ψ/∂x)

Any ψ that is real up to one constant phase gives j = 0 — a bound stationary state stands, it does not flow.

j is real; its sign is the direction of flow; units s⁻¹ in 1D, m⁻² s⁻¹ in 3D

Continuity, and why the norm never drifts∂|ψ|²/∂t + ∂j/∂x = 0dP(a, b)/dt = j(a) − j(b)

Probability inside a region changes only by crossing its edges, so the total over all x is frozen at 1.

Requires a real V. A complex V leaks norm deliberately, to model absorption or decay.

Phase gradient as a local velocityψ = √ρ exp(iS/ħ) → j = ρ (∂S/∂x)/m

Shows what current needs: a phase that varies with x. Flat phase, no flow, however lumpy |ψ|² is.

S(x, t) is the real phase in joule seconds; ∂S/∂x is a local momentum in kg m s⁻¹.

Plane wave: a current with no normalisationψ = A exp[i(kx − ωt)] → ρ = |A|², j = (ħk/m)|A|² = vρ

Flux = density × speed, which is what makes R = j_ref/j_inc and T = jₜᵣ/j_inc meaningful ratios.

∫|A|² dx diverges, so box-normalise with A = 1/√L or normalise to δ(k − k′).

The density is |ψ|², and ψ itself stays invisible

ρ(x, t) = ψ*ψ is real and never negative, whatever ψ is doing in the complex plane, and that is the only reason a squared modulus can serve as a probability density. Multiplying the whole state by a constant phase e^(iα) leaves ρ untouched, and so does the stationary-state factor e^(−iEt/ħ): a single energy eigenstate has a density that does not move at all. What survives squaring is relative phase between two terms, which is why a superposition of two eigenstates carries a cross term 2ψ₁ψ₂cos(ωt) and a single eigenstate carries none. Dimensions follow from the rule itself: probability is a pure number and dx is a length, so ρ has units of m⁻¹ in one dimension and ψ has units of one per root metre — the strangest-looking unit in the course, and a direct consequence of Born's rule.

Normalisation spends the constant the equation leaves free

The time-dependent Schrödinger equation is linear and homogeneous, so if ψ solves it then so does cψ for any complex c. Nothing in the dynamics picks c; Born's rule does, by demanding that the total probability equal 1. Take the infinite-well ground state, ψ = A sin(πx/L) inside the well and zero outside. Then ∫₀ᴸ A²sin²(πx/L) dx = A²L/2, because sin² averages to one half, so A = √(2/L). For L = 1.00 nm that is A = 1.41 nm^(−1/2), or 4.47 × 10⁴ m^(−1/2) in SI. Only the modulus of A is fixed — a leftover phase is free and unmeasurable. And notice what normalisation excludes: any function whose square integral diverges has no c that will work, so it is not a state at all.

Differentiate the norm and the potential cancels

Write ∂ρ/∂t = ψ*(∂ψ/∂t) + ψ(∂ψ*/∂t) and substitute the Schrödinger equation in the form ∂ψ/∂t = (iħ/2m)∂²ψ/∂x² − (i/ħ)Vψ, together with its conjugate. The potential contributes −(i/ħ)V|ψ|² from the first term and +(i/ħ)V*|ψ|² from the second, so the two cancel exactly when V is real. What remains is (iħ/2m)(ψ*ψ″ − ψψ*″), and that is a perfect x-derivative: it equals ∂/∂x[(iħ/2m)(ψ*ψ′ − ψψ*′)] = −∂j/∂x. Hence ∂ρ/∂t + ∂j/∂x = 0. Integrate over the whole line and the right-hand side becomes j(−∞) − j(+∞), which vanishes for any square-integrable ψ, so d/dt∫|ψ|²dx = 0. Normalise once and it stays normalised. Break the assumption on purpose — add an imaginary part, V → V − iΓ/2 — and the norm decays as e^(−Γt/ħ), which is how absorption and decay are modelled.

Reading the current: what it measures, when it vanishes

Because j takes an imaginary part, any ψ that is real up to one overall constant phase gives j = 0 everywhere. The bound states of a well are real sines, so they carry no current: they are standing waves with a frozen density, which is exactly what a stationary state is. Current requires a phase that varies with position. Writing ψ = √ρ e^(iS/ħ) makes this explicit, since then j = ρ(∂S/∂x)/m — a density multiplied by a local velocity built from the phase gradient. The sign of j is the direction of flow, and the integrated form dP(a, b)/dt = j(a) − j(b) is the accounting you actually use: probability inside [a, b] rises only if more flows in at a than out at b. In the figure below, that is why the arrow is longest at the phase where the boundary probability is changing fastest, and zero at the two turning phases.

Plane waves break the rule, and what to do about it

A plane wave ψ = A exp[i(kx − ωt)] has ρ = |A|² everywhere, so ∫ρ dx diverges and no choice of A normalises it. It is not a state; it is a limit that states approach. Its current is finite and useful all the same: j = (ħk/m)|A|² = v|A|², reading as flux equals density times speed. There are two standard repairs. Box normalisation confines the system to length L with periodic boundaries, giving A = 1/√L and a discrete set of k. Delta normalisation keeps the continuum and replaces 1 with a Dirac δ, ∫φ*ₖ φₖ′ dx = 2πδ(k − k′). For scattering problems neither is needed, because reflection and transmission are ratios of currents in which |A|² cancels. Numbers for scale: a 10.0 eV electron has k = 1.62 × 10¹⁰ m⁻¹ and v = 1.88 × 10⁶ m s⁻¹.

In three dimensions, and what the rule will not give you

Nothing structural changes in 3D. The density ρ = |ψ(r, t)|² has units of m⁻³, so ψ carries m^(−3/2); the current becomes a vector J = (ħ/m)Im(ψ*∇ψ) with units m⁻² s⁻¹; and continuity reads ∂ρ/∂t + ∇⋅J = 0, whose integrated form turns the edge terms into a flux through the surface bounding the volume. What Born's rule does not give you is the individual outcome. It is a probability assignment, not a mechanism, and nothing in the deterministic linear Schrödinger equation implies it — which is why it is listed among the postulates. Nor is |ψ|² the whole distribution story: momentum has its own density, obtained from the same ψ by Fourier transform, and the two cannot be read off one classical picture.

Change one variable at a time

Make the relationship visible.

Interactive model

Mixing angle θ: c₁ = cos θ, c₂ = sin θ45 °

Phase φ = (E₂ − E₁)t/ħ60 °

Boundary x_b as a percentage of L50 %

Set the mixing angle to 45° and drag the phase: the hump slides from side to side while the area under the curve never leaves 1, and the arrow is longest exactly when the boundary probability changes fastest, vanishing at φ = 0° and 180°. Take the mixing to 0° or 90° and the current dies.

P LEFT OF x_b0.712

P RIGHT OF x_b0.288

CURRENT j AT x_b0.630 per fs

WEIGHT |c₂|² OF ψ₂0.50

Live interpretationP LEFT OF x_b: 0.712. P RIGHT OF x_b: 0.288. CURRENT j AT x_b: 0.630 per fs. WEIGHT |c₂|² OF ψ₂: 0.50

Catch the common trap

Explain before calculating.

Misconception check

The recurring error is quoting a density as though it were a probability. For the 1.00 nm ground state |ψ₁(L/2)|² = 2/L = 2.0 nm⁻¹, and a student who writes "a 2.0 chance of being at the middle" has produced a probability greater than one that also carries a unit — two impossibilities in a single number. |ψ|² is probability per unit length, and it becomes a probability only after multiplying by an interval: 2.0 nm⁻¹ × 0.010 nm = 0.020. The same slip reappears as "the electron is smeared through the well like a cloud". Born's rule is a statement about repeated measurements on identically prepared systems: every run finds one whole electron at one place, and it is the histogram over many runs that follows |ψ|². The probability current inherits the same warning — j is the rate at which that histogram's weight shifts across a point, not a speed at which anything travels.

An electron sits in the ground state ψ₁ = √(2/L) sin(πx/L) of an infinite well of width L = 1.00 nm. What is the probability of finding it inside a 0.010 nm window centred on the middle of the well?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is confined to an infinite well of width L = 1.00 nm, with ψ(x) = A sin(πx/L) inside and zero outside. Normalise the state, quote A with its unit, and find the probability that the electron is in the left quarter of the well.

Born's rule fixes the scale: ∫₀ᴸ |ψ|² dx = 1, so A² ∫₀ᴸ sin²(πx/L) dx = 1.
Over a whole number of half-periods sin² averages to one half, so ∫₀ᴸ sin²(πx/L) dx = L/2, giving A²L/2 = 1 and A = √(2/L).
With L = 1.00 nm: A = √(2/1.00 nm) = 1.41 nm^(−1/2) = 4.47 × 10⁴ m^(−1/2). The half-power unit is the signature of a one-dimensional wave function.
Probability in the left quarter: P = (2/L)∫₀^(L/4) sin²(πx/L) dx = [x/L − sin(2πx/L)/(2π)]₀^(L/4) = 0.250 − 1/(2π) = 0.250 − 0.159 = 0.091.
Sanity check: a uniform density would give 0.250, and the ground-state density is smallest near the walls, so a number well below a quarter is what you should expect.

AnswerA = 1.41 nm^(−1/2) = 4.47 × 10⁴ m^(−1/2), and P(0 ≤ x ≤ L/4) = 0.091, or 9.1% — far less than the 25% a uniform density would give.

MediumIn the region to the left of a potential step the stationary state is ψ(x) = A e^(ikx) + B e^(−ikx), with |B| = 0.60|A|. Show that the current is uniform, then evaluate it for a 10.0 eV electron with |A|² = 1.0 × 10⁷ m⁻¹, and give the reflection and transmission coefficients.

Differentiate: ψ′ = ik(A e^(ikx) − B e^(−ikx)), so ψ*ψ′ = ik[|A|² − |B|² + (B*A e^(2ikx) − A*B e^(−2ikx))].
The bracketed pair is w − w* = 2i Im(w) with w = B*A e^(2ikx); multiplied by ik it becomes real, so it drops out of the imaginary part. Hence Im(ψ*ψ′) = k(|A|² − |B|²) and j = (ħk/m)(|A|² − |B|²).
The interference term that makes |ψ|² ripple as a standing wave has cancelled in j, so j has no x dependence — as continuity demands, since ∂ρ/∂t = 0 for a stationary state forces dj/dx = 0.
Speed: p = √(2mE) = √(2 × 9.109 × 10⁻³¹ × 1.602 × 10⁻¹⁸) = 1.708 × 10⁻²⁴ kg m s⁻¹, so v = p/m = 1.88 × 10⁶ m s⁻¹ and k = p/ħ = 1.62 × 10¹⁰ m⁻¹.
Currents: j_inc = v|A|² = 1.88 × 10¹³ s⁻¹ and j_ref = −v|B|² = −0.36 j_inc, so j = v(|A|² − |B|²) = 0.64 × 1.88 × 10¹³ = 1.20 × 10¹³ s⁻¹.
Coefficients are ratios of currents: R = |B|²/|A|² = 0.60² = 0.36, and T = 1 − R = 0.64, which is exactly the surviving net current as a fraction of the incident one.

Answerj = +1.20 × 10¹³ s⁻¹, uniform and directed towards the step; R = 0.36 and T = 0.64.

HardAn electron in a 1.00 nm infinite well is prepared as ψ(x,0) = [ψ₁(x) + ψ₂(x)]/√2. Find the probability of finding it in the left half as a function of time, its period, and the peak probability current at the midpoint.

Each eigenstate picks up its own phase: ψ(x, t) = [ψ₁e^{(−iE₁t/ħ)} + ψ₂e^{(−iE₂t/ħ)}]/√2, so ρ = ½(ψ₁² + ψ₂²) + ψ₁ψ₂cos(ωt) with ω = (E₂ − E₁)/ħ.
Energies: E₁ = h²/(8mL²) = (6.626 × 10⁻³⁴)²/(8 × 9.109 × 10⁻³¹ × 10⁻¹⁸) = 6.02 × 10⁻²⁰ J = 0.376 eV, and E₂ = 4E₁, so E₂ − E₁ = 3E₁ = 1.13 eV. Then ω = 1.81 × 10⁻¹⁹/1.055 × 10⁻³⁴ = 1.71 × 10¹⁵ rad s⁻¹ and the period is 2π/ω = 3.67 fs.
Left half: ∫₀^(L/2) ψ₁² dx = ∫₀^(L/2) ψ₂² dx = ½, while the cross term gives ∫₀^(L/2) ψ₁ψ₂ dx = (2/L)∫₀^(L/2) sin(πx/L)sin(2πx/L) dx = 1/π + 1/(3π) = 4/(3π) = 0.424.
So P_left(t) = 0.500 + 0.424 cos(ωt), sloshing between 0.076 and 0.924 — and P_right = 1 − P_left at every instant, because the sloshing never touches the normalisation.
Continuity with j(0) = 0 at the wall gives dP_left/dt = −j(L/2, t), so j(L/2, t) = 0.424 ω sin(ωt) and the peak is 0.424 × 1.71 × 10¹⁵ = 7.27 × 10¹⁴ s⁻¹.
Check it against the current formula directly: j = (4πħ/mL²)c₁c₂ sin(ωt) sin³(πx/L), which at x = L/2 with c₁c₂ = ½ is h/(mL²) = 6.626 × 10⁻³⁴/(9.109 × 10⁻³¹ × 10⁻¹⁸) = 7.27 × 10¹⁴ s⁻¹.

AnswerP_left(t) = 0.500 + 0.424 cos(ωt), oscillating between 0.076 and 0.924 with period 3.67 fs; the midpoint current peaks at 7.27 × 10¹⁴ s⁻¹, that is 0.73 per femtosecond.