University Physics IV · The Schrödinger Equation · 6.2

Normalisation & the Probability Current

Born's rule only pays off if the total probability stays at one. Here you set the entry requirements a function must meet to be a state at all, fix its normalisation constant, and then follow the probability as it moves — because a current, not a promise, is what holds the norm at one.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Born's rule promotes |ψ|² to a probability density, and the promotion comes with a bill. A density must integrate to one, so ψ has to be square-integrable — decaying fast enough that ∫|ψ|²dx converges — and any solution of the Schrödinger equation that fails this is discarded before the physics begins. Normalisation then fixes the multiplicative constant up to a factor e^(iα), and that leftover global phase is pure convention: it cancels out of every |ψ|², every current, every expectation value.

The harder demand is that the norm stay at one at all later times, and here the equation itself supplies the answer. Differentiate ∫|ψ|²dx, substitute iħ∂ψ/∂t = Ĥψ and its conjugate, and what emerges is ∂ρ/∂t + ∂j/∂x = 0 with j = (ħ/m)Im(ψ*ψ′) — a local law of the same form as charge conservation, saying that probability is transported rather than created or destroyed. The cost is a short list of conditions that are easy to forget.

The potential must be real, or the relation gains a source term and the norm drifts. The boundary terms must vanish, which a normalisable state gives for free and a plane wave never does. And ψ must be continuous everywhere, with ψ′ continuous wherever V is finite, or the ψ″ standing in the equation has no meaning.

Simple definition: A wavefunction is admissible when it is single-valued, continuous, and square-integrable, and normalising it means dividing by √(∫|ψ|²dx) so that the probability of finding the particle somewhere is exactly one.
Example: For ψ(x) = A e^(−|x|/L) with L = 0.50 nm, ∫|ψ|²dx = A²L, so A = 1/√L = 1.41 nm^(−1/2); the particle is then found within |x| < L with probability 1 − e⁻² = 0.865.

Admissibility: square-integrability∫|ψ(x, t)|² dx < ∞ over all space

The entry test. A function can solve the equation perfectly and still fail it, in which case it is not a state.

|ψ|² is a probability per unit length, so ψ carries m^(−1/2) in 1-D and m^(−3/2) in 3-D.

The normalisation constantψ_N = ψ / √(∫|ψ|² dx)

Turns any square-integrable solution into one whose total probability is exactly 1, changing no physics on the way.

Determined only up to a factor e^(iα) with α real, since |e^(iα)|² = 1.

Joining conditions at a boundaryΔψ = 0 alwaysΔψ′ = 0 wherever V is finite

Supplies the two matching equations at every step, well edge or spike in a piecewise potential.

At V = −αδ(x), Δψ′ = −(2mα/ħ²)ψ(0); at an infinite wall ψ′ may jump freely.

Probability density and currentρ = |ψ|², j = (ħ/2mi)(ψ*ψ′ − ψψ*′) = (ħ/m) Im(ψ*ψ′)

For a plane wave j = (ħk/m)|A|² = vρ, the classical flux; for any real ψ, j = 0 at every point.

ρ in m⁻¹, j in s⁻¹ — a probability per second crossing a point, signed by direction of flow.

Local conservation of probability∂ρ/∂t + ∂j/∂x = 0 ⟹ dP[a, b]/dt = j(a) − j(b)

Probability lost from a region is exactly what crossed its ends — which is what makes R + T = 1 true.

Holds pointwise for real V. In a stationary state ∂ρ/∂t = 0, so j takes the same value at every x.

Where the norm can died/dt ∫|ψ|² dx = −[j] evaluated from −∞ to +∞ + (2/ħ)∫ Im(V) |ψ|² dx

Names the only two leaks: flux escaping at infinity, or an absorptive imaginary part in the potential.

Both terms vanish for a normalisable state in a real potential. Im V = −Γ/2 gives τ = ħ/Γ.

Not every solution of the equation is a state

The Schrödinger equation is a linear PDE, and like any such equation it has far more solutions than physics wants. Born's rule is the filter. If |ψ(x, t)|²dx is the probability of finding the particle in dx, then ∫|ψ|²dx over all space must converge — ψ must be square-integrable. e^(−|x|/L) qualifies; e^(+x²/L²) and the constant function do not, and neither is a state however neatly it solves the equation. Two further requirements come from the equation rather than from Born. ψ must be single-valued, or |ψ|² is not even a function of position. And ψ must be continuous everywhere, with ψ′ continuous wherever V is finite, because ψ″ appears in the equation: a kink in ψ makes ψ″ a delta function, and the only thing that can balance a delta on one side is a delta on the other. Where V is not finite the licence is withdrawn — at the walls of an infinite square well, ψ′ jumps, and it is allowed to.

Fixing the constant, and the phase you can never see

Square-integrability makes normalisation possible; it does not perform it. If ∫|ψ|²dx = I, then ψ_N = ψ/√I integrates to 1. Take the trial function ψ = A x(a − x) on 0 < x < a, zero outside. Then ∫₀ᵃ x²(a − x)²dx = a⁵/3 − a⁵/2 + a⁵/5 = a⁵/30, so A = √(30/a⁵); with a = 1.00 nm that is A = 5.48 nm^(−5/2). Watch the units as a check: in one dimension |ψ|² is a probability per unit length, so ψ carries m^(−1/2), and in three dimensions m^(−3/2). Normalisation leaves exactly one thing undetermined, an overall factor e^(iα) with α real. That global phase cancels out of |ψ|², out of j, and out of every expectation value, so it is not a physical degree of freedom — two states differing only by it are the same state. A relative phase is a different animal: in ψ = c₁ψ₁ + c₂ψ₂ the difference arg c₂ − arg c₁ survives in the cross term and is measurable.

Differentiate the norm and a current falls out

Normalising at t = 0 is worth nothing unless the norm stays at 1, so test it. With ρ = ψ*ψ, ∂ρ/∂t = ψ*ψ̇ + ψψ̇*, and the equation supplies both pieces: ψ̇ = (iħ/2m)ψ″ − (i/ħ)Vψ, and its conjugate ψ̇* = −(iħ/2m)ψ*″ + (i/ħ)V*ψ*. Substituting gives ∂ρ/∂t = (iħ/2m)(ψ*ψ″ − ψψ*″) + (i/ħ)(V* − V)|ψ|². Take V real and the second term dies outright. The first is already a perfect derivative, because ψ*ψ″ − ψψ*″ = ∂ₓ(ψ*ψ′ − ψψ*′). So ∂ρ/∂t + ∂j/∂x = 0 with j = (ħ/2mi)(ψ*ψ′ − ψψ*′) = (ħ/m)Im(ψ*ψ′), the last form following because ψ*ψ′ − ψψ*′ = 2i Im(ψ*ψ′). Nothing was assumed about the shape of V or the form of ψ. This is a local conservation law of the same type as charge conservation in electromagnetism: probability moves through the intervening space rather than disappearing here and reappearing there.

Reading the current off a state

j has units of probability per second — a rate at which probability crosses a point — and its sign gives the direction. Three readings are worth committing to memory. For a plane wave ψ = A e^{(i(kx − ωt)}), ψ*ψ′ = ik|A|², so j = (ħk/m)|A|² = vρ: density times velocity, exactly the classical flux. For any real ψ, ψ*ψ′ is real, so Im(ψ*ψ′) = 0 and j vanishes identically — every non-degenerate bound state of a real one-dimensional potential can be chosen real, so bound states carry no current, though they are certainly not at rest, since ⟨p⟩ = 0 while ⟨p²⟩ > 0. And in any stationary state ρ is time-independent, so ∂ρ/∂t = 0 forces dj/dx = 0: the current is the same number at every x. That last fact is the whole basis of scattering coefficients, R = |j_refl|/j_inc and T = jₜᵣₐₙₛ/j_inc, and it is why R + T = 1 is a statement about currents rather than about amplitudes.

The two things the norm needs to survive

Integrate the continuity relation over all space and you get dN/dt = −[j] from −∞ to +∞ plus (2/ħ)∫Im(V)|ψ|²dx, with N = ∫|ψ|²dx. Two conditions have to hold for N to stay at 1. First, the boundary terms must vanish: a normalisable state takes ψ and ψ′ to zero at infinity and j goes with them, but nothing about a raw solution of the equation guarantees this in advance — it is a condition you impose, not one you inherit. Second, V must be real, or Im(V) acts as a source or a sink. Physicists exploit that deliberately. Setting V = V₀ − iΓ/2 gives dN/dt = −(Γ/ħ)N, so N(t) = e^(−t/τ) with τ = ħ/Γ: a state that leaks away with a lifetime. For τ = 1.6 ns, Γ = ħ/τ = 4.1 × 10⁻⁷ eV, a natural linewidth near 100 MHz. This is a modelling device, not new physics — the missing probability has gone into channels the Hamiltonian omits, and the price is that Ĥ is no longer Hermitian.

Plane waves, and the three ways to live with them

ψ = Ae^(ikx) has |ψ|² = |A|² everywhere, so ∫|ψ|²dx = |A|² times the length of all space. The momentum eigenstates therefore sit outside the space of physical states, and there are three standard repairs. Box normalisation: confine the particle to a length L with periodic boundaries, take A = 1/√L, allow only k = 2πn/L, and let L → ∞ at the end of the calculation. Delta normalisation: keep the continuum and accept ∫ψ*_(k′)ψₖ dx = 2πδ(k − k′), so that ψₖ = e^(ikx)/√(2π) is orthonormal to a distribution rather than to a Kronecker δ, and sums over states become integrals. Or build a packet: superpose a spread of k with a square-integrable weight, and the result is normalisable, with a finite Δx and Δp. For scattering none of this is required, because a ratio of currents stays finite even when neither state is normalisable — which is why transmission is always quoted as a current ratio.

Change one variable at a time

Make the relationship visible.

Interactive model

Reflected amplitude r = |B|/|A|0.60

Wavenumber k4.0 nm⁻¹

Slide k and the fringes crowd together while the current line does not move at all — interference changes where probability sits, not how much flows. Then push r to 1: the fringes reach full contrast, ψ becomes 2cos(kx), which is real, and the current drops to zero.

FRINGE MAX (1 + r)²2.56

FRINGE MIN (1 − r)²0.16

CURRENT j / v = 1 − r²0.64

FRINGE SPACING π / k0.79 nm

Live interpretationFRINGE MAX (1 + r)²: 2.56. FRINGE MIN (1 − r)²: 0.16. CURRENT j / v = 1 − r²: 0.64. FRINGE SPACING π / k: 0.79 nm

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to think a plane wave can be normalised if only you choose A small enough. It cannot. For ψ = Ae^(ikx), ∫|ψ|²dx = |A|² times the length of all space, and the divergence lives in the extent of the integral, not in the size of the constant: shrinking A by a factor of 10⁶ still leaves infinity, and the only A giving a finite answer is A = 0, which is no particle at all. The point is not bookkeeping. A momentum eigenstate has Δx infinite and genuinely does not describe a localisable particle; it is a limit, useful only as a basis element — box-normalised, δ-normalised, or superposed into a packet. The same trap catches ψ = A, the "free particle at rest", and every non-decaying branch of a solution, including the growing exponential you must discard outside a finite well.

A particle in an infinite square well of width a is in its ground state ψ(x) = √(2/a) sin(πx/a), which is real. What is the probability current inside the well, and what does that tell you about the particle?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA particle is described by the triangular wavefunction ψ(x) = A(1 − |x|/L) for |x| ≤ L, and ψ = 0 outside, with L = 2.0 nm. Find A, then find the probability of locating the particle within |x| < L/2.

Normalisation demands ∫|ψ|²dx = 1. The function is even, so ∫|ψ|²dx = 2A²∫₀^L (1 − x/L)² dx.
Substituting u = 1 − x/L turns that integral into L∫₀¹ u² du = L/3, so ∫|ψ|²dx = 2A²L/3.
Setting it to 1 gives A = √(3/2L) = √(3 / 4.0 nm) = 0.87 nm^(−1/2). The half-power unit is the check: |ψ|² is a probability per unit length, so ψ must carry m^(−1/2).
For |x| < L/2 the same substitution runs u from 1/2 to 1: P = 2A²L∫_{1/2}^{1} u² du = 2A²L(1 − 1/8)/3 = 7A²L/12.
Insert A² = 3/2L: P = (7/12)(3/2) = 7/8 = 0.875. The constant cancels, as it must — a probability cannot depend on the units L was quoted in.

AnswerA = √(3/2L) = 0.87 nm^(−1/2), and P(|x| < 1.0 nm) = 7/8 = 0.875. The outer halves of the triangle occupy half the width but hold only one eighth of the probability.

MediumAn electron beam of kinetic energy 4.0 eV meets a potential step and is partly reflected, so that in the incident region ψ(x) = A e^(ikx) + B e^(−ikx) with |B| = 0.60|A|. Find j(x), show it does not depend on x, and evaluate it in terms of |A|².

j = (ħ/m)Im(ψ*ψ′). Here ψ′ = ik(Ae^(ikx) − Be^(−ikx)) and ψ* = A*e^(−ikx) + B*e^(ikx).
Multiplying out, ψ*ψ′ = ik[(|A|² − |B|²) + (AB*e^(2ikx) − A*Be^(−2ikx))]. The bracketed pair equals 2i Im(AB*e^(2ikx)), so multiplying it by i makes it real and it drops out of the imaginary part: Im(ψ*ψ′) = k(|A|² − |B|²).
So j = (ħk/m)(|A|² − |B|²), a constant — no x survives. That had to happen: the state is stationary, so ∂ρ/∂t = 0 and continuity forces dj/dx = 0.
The density is not constant: ρ = |A|² + |B|² + 2Re(AB*e^(2ikx)) carries fringes of contrast 2|A||B|. Interference redistributes where probability sits without transporting any of it.
Numbers: k = √(2mₑE)/ħ = √(2 × 9.11 × 10⁻³¹ × 6.41 × 10⁻¹⁹)/1.055 × 10⁻³⁴ = 1.02 × 10¹⁰ m⁻¹, so ħk/mₑ = 1.19 × 10⁶ m s⁻¹. With |B|² = 0.36|A|², j = 0.64 × 1.19 × 10⁶ |A|² = 7.6 × 10⁵ |A|² s⁻¹ when |A|² is in m⁻¹.
Read as a ratio, the same current gives R = |B|²/|A|² = 0.36 and T = 1 − R = 0.64.

Answerj = (ħk/m)(|A|² − |B|²) = 7.6 × 10⁵ |A|² s⁻¹, independent of x. R = 0.36 and T = 0.64. The standing-wave fringes in ρ carry none of the current, which is why R and T are defined from j and never from |ψ|².

HardA metastable state is modelled by a complex potential V = V₀ − iΓ/2, with V₀ real and Γ > 0. Starting from the continuity relation derived without assuming V real, find how the norm evolves, identify the lifetime, evaluate Γ and Im V for a state that lives 1.6 ns, and say what has been given up.

Keeping the term that was discarded for real V: ∂ρ/∂t + ∂j/∂x = (i/ħ)(V* − V)ρ = (2/ħ)Im(V)ρ.
Integrate over all space. The current term becomes the boundary values of −j at ±∞, which vanish for a square-integrable ψ, leaving dN/dt = (2/ħ)∫Im(V)ρ dx with N = ∫ρ dx.
Im(V) = −Γ/2 is a constant, so it comes outside the integral: dN/dt = −(Γ/ħ)N.
Hence N(t) = N(0)e^(−Γt/ħ) = e^(−t/τ) with τ = ħ/Γ. The norm is not conserved — that is the purpose of the model, not a mistake in it.
For τ = 1.6 ns: Γ = ħ/τ = 1.055 × 10⁻³⁴ / 1.6 × 10⁻⁹ = 6.6 × 10⁻²⁶ J = 4.1 × 10⁻⁷ eV, so Im V = −Γ/2 = −2.1 × 10⁻⁷ eV. As a frequency, Γ/h = 1.0 × 10⁸ Hz — the natural linewidth, near 100 MHz, of a transition with that lifetime.
What has been surrendered is hermiticity. With a complex V the Hamiltonian is not Hermitian, its eigenvalues become complex, E₀ − iΓ/2, and probability leaves the description. The particle has not vanished; it has gone into channels the Hamiltonian never modelled.

AnswerN(t) = e^(−t/τ) with τ = ħ/Γ. A 1.6 ns lifetime needs Γ = 4.1 × 10⁻⁷ eV and Im V = −2.1 × 10⁻⁷ eV, a linewidth of about 100 MHz. The cost is a non-Hermitian Ĥ: real potentials conserve the norm, complex ones absorb it.