Born Rule and Conservation of Normalisation

Derivation

Born Rule and Conservation of Normalisation

D-138 Home PU-202 Threads chance · matter · waves Depends on The Time-Dependent Schrödinger Equation, Probability Current and the Continuity Equation

Statement

For a single particle described by a wavefunction ψ(x, t) obeying the time-dependent Schrödinger equation with a real potential, the quantity ρ = |ψ|² is a probability density, and its space integral N(t) = ∫|ψ|² dx is independent of time: dN/dt = 0. Consequently a state normalised so that N(0) = 1 remains normalised, N(t) = 1, for all t.

Why it matters

The Born rule — that |ψ(x)|² dx is the probability of finding the particle in [x, x+dx] — is only meaningful if those probabilities sum to one and stay summing to one as the state evolves. If normalisation drifted, the particle would spontaneously appear or vanish, and predicted probabilities would be uninterpretable.

This result is what certifies Schrödinger dynamics as a legitimate probability-conserving law. It ties the abstract structure (Hermiticity of the Hamiltonian) to a concrete accounting statement (a continuity equation), and it is the wave-mechanics face of unitary time evolution.

Assumptions

The Born rule: ρ = |ψ|² is the probability density.Without it there is nothing to normalise; the theorem below would be a statement about an integral with no physical meaning.

ψ obeys the time-dependent Schrödinger equation.The specific form of ∂ψ/∂t is what makes the potential terms cancel; a different evolution law need not conserve N.

The potential V(x, t) is real (the Hamiltonian is Hermitian).A complex V leaves an uncancelled source term and N decays or grows — this is exactly how absorbing/optical potentials model particle loss.

ψ is square-integrable and ψ, ∂ψ/∂x → 0 as x → ±∞.The boundary flux would not vanish; probability could leak in from infinity and dN/dt ≠ 0 despite real V. Plane waves (not normalisable) sit outside the theorem.

Differentiation under the integral sign is valid (dominated convergence).Exchanging d/dt and ∫dx could fail for pathological states, invalidating the very first line of the proof.

Derivation

ρ(x, t) = ψ*ψ = |ψ|², N(t) = ∫_−∞^∞ ρ dx

Definition of the probability density (Born rule) and the total normalisation. A

dN/dt = ∫_−∞^∞ ∂ρ/∂t dx

Differentiate under the integral. Legal because the limits are constant and ρ is smooth and integrably dominated (dominated convergence). C

∂ρ/∂t = ψ* (∂ψ/∂t) + ψ (∂ψ*/∂t)

Product rule on ψ*ψ. Purely algebraic. A

∂ψ/∂t = (iℏ/2m) ∂²ψ/∂x² − (i/ℏ) Vψ

Rearrange the TDSE iℏ ∂ψ/∂t = −(ℏ²/2m)∂²ψ/∂x² + Vψ, dividing by iℏ. A

∂ψ*/∂t = −(iℏ/2m) ∂²ψ*/∂x² + (i/ℏ) Vψ*

Complex-conjugate step 4. Here V* = V is used explicitly — a real potential. This is the step that fails for a complex V. C

∂ρ/∂t = (iℏ/2m) (ψ* ∂²ψ/∂x² − ψ ∂²ψ*/∂x²)

Insert steps 4 and 5 into step 3. The two ±(i/ℏ)V|ψ|² terms are equal and opposite, so the potential cancels exactly. B

ψ* ∂²ψ/∂x² − ψ ∂²ψ*/∂x² = ∂/∂x (ψ* ∂ψ/∂x − ψ ∂ψ*/∂x)

The bracket is a perfect x-derivative: expanding the right side, the cross terms ∂ψ*/∂x·∂ψ/∂x cancel, leaving the left side. C

∂ρ/∂t = − ∂j/∂x, j = (ℏ/2mi)(ψ* ∂ψ/∂x − ψ ∂ψ*/∂x)

Combine steps 6 and 7 and identify the probability current j (the prior result). This is the continuity equation ∂ρ/∂t + ∂j/∂x = 0. A

dN/dt = −∫_−∞^∞ ∂j/∂x dx = −[ j(∞,t) − j(−∞,t) ]

Substitute step 8 into step 2 and apply the fundamental theorem of calculus. A

ψ, ∂ψ/∂x → 0 ⇒ j(±∞,t) = 0 ⇒ dN/dt = 0

Square-integrability forces ψ and its slope to vanish at infinity, so the boundary current is zero. C

N(t) = N(0); choose N(0) = 1 ⇒ N(t) = 1 ∀ t

A vanishing derivative means N is constant. Fixing the free multiplicative constant of ψ at one instant fixes it forever. A

Result

d/dt ∫_−∞^∞ |ψ(x, t)|² dx = 0 ⇒ ∫_−∞^∞ |ψ|² dx = 1 (∀ t)

Reading. The total probability of finding the particle somewhere is conserved. Locally, probability can flow — the density at a point rises or falls — but only by an equal and opposite flux j through neighbouring regions, never by creation or destruction. Normalisation set once holds forever.

Units check. N is dimensionless. In 1D [ρ] = |ψ|² = m⁻¹ so that ρ dx is a pure number; [j] = (ℏ/m)·|ψ|²·m⁻¹ = (m² s⁻¹)(m⁻²) = s⁻¹. Then [∂j/∂x] = m⁻¹s⁻¹ = [∂ρ/∂t], and [dN/dt] = s⁻¹, equal on both sides — consistently zero.

Limiting cases

Stationary state ψ = φ(x) e^−iEt/ℏ: ρ = |φ|² is time-independent, so ∂ρ/∂t = 0 pointwise — conservation holds trivially and locally.
Real wavefunction (e.g. a bound-state amplitude, box eigenfunctions): ψ* ∂ψ/∂x is real, so j = 0 everywhere; the density is frozen and N is constant with no flow at all.
Free particle V = 0: still real, so N conserved even as a wave packet spreads — broadening lowers the peak but the area stays one.
Three dimensions: the same proof gives ∂ρ/∂t + ∇·j = 0; the divergence theorem turns the volume integral into a surface flux at infinity, which vanishes for normalisable ψ.
Slowly varying V(t): time-dependence of a real potential is irrelevant — the cancellation in step 6 uses only reality, not stationarity.

Breaks when

Complex (absorbing) potential V = V₀ − iΓ/2: step 5 no longer conjugates cleanly; an extra source term −(Γ/ℏ)ρ survives and dN/dt = −(Γ/ℏ)N < 0. Norm decays as e^−Γt/ℏ — this is the point of optical potentials modelling capture/decay.
Non-normalisable states (plane waves, unbounded scattering states): j(±∞) ≠ 0, so step 10 fails. These need box normalisation or wave-packet superposition before probabilities make sense.
Particle number not fixed (relativistic pair creation, second quantisation): a single-particle |ψ|² is not the conserved quantity; charge or particle-number current replaces it, and the Klein–Gordon "density" is not even positive-definite.
Open systems / non-Hermitian effective dynamics (Lindblad master equations, measurement collapse): probability is redistributed to an environment or to unobserved branches, and N for the tracked wavefunction is not preserved.

Failure modes

Dropping the reality of V: writing ∂ψ*/∂t with V instead of V*, so the potential appears to cancel even for complex V — hiding the decay term.
Sign error in the conjugate equation: forgetting that conjugation flips i → −i in both the ∂²/∂x² and V terms; only then do the potentials cancel and the current emerge.
Assuming j = 0 in general because "probability is conserved". Conservation is global; the local current is generally non-zero (only real ψ gives j = 0).
Confusing ∂ρ/∂t = 0 with dN/dt = 0: the density can slosh in time while the integral is fixed; only stationary states have both.
Keeping a boundary term: applying the theorem to a plane wave and concluding norm is conserved, ignoring that j(±∞) ≠ 0 there.
Normalising once and forgetting: re-normalising a numerically evolved state "because it drifted" — drift signals a broken integrator (non-unitary step), not new physics.

Discussion

The heart of the proof is a cancellation: the kinetic term of the Hamiltonian generates the current j (a divergence, hence a boundary effect), while the potential term contributes ±(i/ℏ)V|ψ|² which cancels iff V is real. This is no accident: reality of V is exactly Hermiticity of the Hamiltonian, Ĥ = Ĥ^†. Conservation of normalisation is therefore the wave-mechanics shadow of a deeper statement — that time evolution U(t) = e^−iĤt/ℏ is unitary, preserving the inner product ⟨ψ|ψ⟩ and with it all of quantum probability.

The result also fixes what kind of law the Schrödinger equation is. It is first order in time precisely so that specifying ψ at one instant determines it forever; the continuity equation guarantees that this deterministic flow of the amplitude never spoils the interpretation of |ψ|² as a conserved probability. The particle is never created or destroyed — it is merely transported, its probability fluid stirred but not spilled.

Structurally, the pairing "conserved density + its current obeying a continuity equation" recurs throughout physics: electric charge with ρ_q, J, energy with the Poynting vector, particle number in fluids. Here the conserved charge is probability itself. By Noether's theorem the underlying symmetry is the global U(1) phase invariance ψ → e^iαψ of the Schrödinger Lagrangian: the same freedom that makes the overall phase of a state unobservable is what makes total probability conserved. In relativistic theory this promotes to conservation of charge, since the naive probability density loses positivity.

Common misconceptions. "Normalisation is just a choice we impose once" — true that we choose N(0)=1, but it is a theorem, not a further choice, that it stays one. "Conservation means nothing moves" — no; the density flows via j, only the total is fixed. "A spreading packet loses norm" — the peak falls but the width grows so the area is invariant. "It only works for stationary states" — it holds for every solution with real V.

Worked examples

Example 1 — Ground state of an infinite square well: real amplitude, zero current, trivially conserved.

ψ₁(x) = A sin(πx/L), 0 ≤ x ≤ L, ψ = 0 outside

The ground-state spatial amplitude; A to be fixed by normalisation. A

∫₀^L A² sin²(πx/L) dx = A² (L/2) = 1 ⇒ A = √(2/L)

Use ∫₀^L sin²(πx/L)dx = L/2; solve for A. A

j = (ℏ/2mi)(ψ* ψ′ − ψ ψ*′) = 0 (ψ real)

ψ* = ψ real, so the bracket is ψψ′ − ψψ′ = 0. B

L = 1.0 nm: A = √(2 / 1.0×10⁻⁹ m) = 4.47×10⁴ m^−1/2

Plug the number after the symbolic result. Units: m^−1/2, correct for a 1D amplitude. A

A = √(2/L) = 4.47×10⁴ m^−1/2, j = 0, N = 1

Reading. A real eigenfunction carries no probability current, so the density is static and N stays 1 with nothing flowing. Units check. A² = 2/L = m⁻¹, and A²·(L/2) is dimensionless as required.

Example 2 — Free Gaussian wave packet: the peak falls and the packet spreads, yet N = 1 at every time.

|ψ(x,t)|² = (1 / √(2π) σ_t) exp(−x² / 2σ_t²), σ_t = σ₀√(1 + (ℏt / 2mσ₀²)²)

Standard free-particle spreading of a minimum-uncertainty Gaussian; width grows with t. B

N(t) = ∫_−∞^∞ |ψ|² dx = (1/√(2π)σ_t) · √(2π)σ_t = 1

Gaussian integral ∫ e^−x²/2σ²dx = √(2π)σ; the σ_t-factors cancel for every t. A

Electron: m = 9.11×10⁻³¹ kg, σ₀ = 1.0 nm; spread time τ = 2mσ₀²/ℏ

Define the characteristic spreading time to put numbers on σ_t. A

τ = 2(9.11×10⁻³¹)(1.0×10⁻⁹)² / (1.055×10⁻³⁴) = 1.7×10⁻¹⁴ s

At t = τ, σ_t = √2 σ₀ = 1.41 nm; peak height falls by 1/√2. A

t = 0: peak = 1/(√(2π)·10⁻⁹) = 3.99×10⁸ m⁻¹; t = τ: peak = 2.82×10⁸ m⁻¹

Area = peak × effective width stays fixed: 3.99×10⁸ · √(2π)·10⁻⁹ = 1 and likewise at τ. B

N(0) = N(τ) = 1 (peak: 3.99×10⁸ → 2.82×10⁸ m⁻¹, width: 1.0 → 1.41 nm)

Reading. The packet visibly delocalises — lower and wider — yet the enclosed probability is exactly 1 throughout, the concrete meaning of conserved normalisation. Units check. [peak] = m⁻¹, [σ_t] = m, product dimensionless; [τ] = kg·m² / (J·s) = s.

Problems

Show that any real normalisable wavefunction has zero probability current everywhere, and hence trivially conserves N. Comment on why a stationary state ψ=φe^−iEt/ℏ with complex φ can still have j=0.
Solution
For real ψ, ψ*=ψ, so j=(ℏ/2mi)(ψψ′−ψψ′)=0 identically. Then ∂ρ/∂t=−∂j/∂x=0, density static, N constant. For a stationary state, |ψ|²=|φ|² is time-independent so N is conserved regardless of φ being complex; j is generally non-zero (e.g. a travelling stationary scattering state), but for a bound state whose φ can be chosen real, j=0. So real-valued amplitude is sufficient but not necessary for conservation.
An absorbing potential V = V₀ − iΓ/2 (Γ>0 real) models decay. Derive dN/dt and solve for N(t); find the mean lifetime for Γ = 6.6×10⁻¹⁶ J (about 4 keV).
Solution
Repeat the proof; the potential terms now give (1/iℏ)|ψ|²(V−V*) = (2/ℏ)Im(V)|ψ|² = −(Γ/ℏ)ρ. Thus ∂ρ/∂t = −∂j/∂x −(Γ/ℏ)ρ; integrating and dropping boundary flux, dN/dt = −(Γ/ℏ)N, so N(t)=N₀e^−Γt/ℏ. Mean lifetime τ=ℏ/Γ = 1.055×10⁻³⁴/6.6×10⁻¹⁶ = 1.6×10⁻¹⁹ s. Probability leaks out at rate Γ/ℏ.
Normalise ψ(x) = A e^−|x|/a on (−∞,∞) and evaluate A for a = 0.53 Å = 5.3×10⁻¹¹ m (the Bohr radius).
Solution
∫|ψ|²dx = A²∫_−∞^∞e^−2|x|/adx = A²·2∫₀^∞e^−2x/adx = A²·2·(a/2)=A²a = 1. So A = 1/√a = 1/√(5.3×10⁻¹¹) = 1.37×10⁵ m^−1/2. (Units: A²a = m⁻¹·m dimensionless, correct.)
For a plane wave ψ = A e^i(kx−ωt) compute ρ and j, and show j = ρv with v = ℏk/m. Why does the normalisation theorem not apply?
Solution
ρ=|A|² (constant). ∂ψ/∂x=ikψ, so ψ*∂ψ/∂x=ik|A|² and j=(ℏ/2mi)(ik|A|²−(−ik)|A|²)=(ℏ/2mi)(2ik|A|²)=(ℏk/m)|A|². With v=ℏk/m this is j=ρv — a probability fluid of density ρ drifting at the group/particle velocity. The theorem fails because a plane wave is not square-integrable: ∫|A|²dx diverges and j(±∞)=ρv≠0, so the boundary term in step 9 does not vanish. Box or wave-packet normalisation is needed.
Two orthonormal real stationary states φ₁,φ₂ with energies E₁,E₂ form ψ=c₁φ₁e^{−iE₁t/ℏ}+c₂φ₂e^{−iE₂t/ℏ}. Show N is constant while the local density oscillates at the Bohr frequency ω=(E₂−E₁)/ℏ. Take c₁=c₂=1/√2.
Solution
N=∫|ψ|²dx=|c₁|²+|c₂|²+2Re[c₁c₂*e^{−i(E₁−E₂)t/ℏ}]∫φ₁φ₂dx. Orthogonality makes the integral 0, so N=|c₁|²+|c₂|²=½+½=1, time-independent. But the local density |ψ(x,t)|²=½φ₁²+½φ₂²+φ₁φ₂cos(ωt) with ω=(E₂−E₁)/ℏ: the cross term sloshes probability back and forth (a classic dipole oscillation) while the total stays pinned at 1 — conservation is global, not local.