University Physics V · Particle in a Box · 5.4

The Sine Eigenbasis: Orthonormality & Completeness

Two integrals do the whole job. One says the sine states are mutually perpendicular unit vectors; the other says nothing has been left off the list. Together they let you stop doing calculus on wavefunctions and start doing linear algebra on coefficients — and they tell you exactly what a truncated sum still gets wrong.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Orthonormality and completeness are two different claims, and the box lets you check both. Orthogonality is not a coincidence of sines: Ĥ is self-adjoint on the Dirichlet domain, so (Eₙ − Eₘ)⟨m|n⟩ = 0 forces distinct levels apart, and the integral (2/L)∫₀ᴸ sin(mπx/L) sin(nπx/L) dx = δₘₙ is the audit rather than the reason — the √(2/L) is simply what normalisation leaves over. Completeness is the harder half and it does not follow: delete |2⟩ and the surviving set is still perfectly orthonormal.

What supplies it is Sturm–Liouville theory — the Dirichlet Laplacian on a bounded interval has compact resolvent, so its eigenvectors span the whole of L²([0, L]) — and the statement is Σₙ|n⟩⟨n| = 1, whose position-basis form is the closure relation (2/L)Σₙ sin(nπx/L) sin(nπx′/L) = δ(x − x′). Together they collapse a state into a column of numbers: cₙ = ⟨n|ψ⟩, one integral each, with Parseval guaranteeing Σ|cₙ|² = ⟨ψ|ψ⟩ = 1, so the coefficients are already the Born probabilities P(Eₙ). The cost hides in the word converges.

It means in the L² norm, an integral of squared error — not at any given point. Every partial sum obeys ψ(0) = ψ(L) = 0 whether ψ does or not, and near a jump the sum overshoots by 8.95% of it forever, the ear narrowing as 1/N without ever getting shorter.

Simple definition: The sine eigenbasis is the orthonormal set |n⟩ = √(2/L) sin(nπx/L), n = 1, 2, 3, …, whose overlaps obey ⟨m|n⟩ = δₘₙ and whose projectors sum to the identity on L²([0, L]), so every state in the box is one series Σcₙ|n⟩ with cₙ = ⟨n|ψ⟩.
Example: For ψ = √(30/L⁵)⋅x(L − x), symmetry kills every even n and cₙ = 4√60/(n³π³): c₁ = 0.99928 and c₃ = 0.03701, so two terms already hold 99.99% of the norm while Σ|cₙ|² = 1 exactly.

The normalised eigenvectors⟨x|n⟩ = ψₙ(x) = √(2/L) sin(nπx/L), n = 1, 2, 3, …

The Dirichlet condition already chose the sine. Normalisation is the one constant left, and it fixes it up to a phase.

ψₙ carries m^(−1/2); √(2/L) = 4.47 × 10⁴ m^(−1/2) at L = 1.00 nm. Zero outside [0, L].

Orthonormality⟨m|n⟩ = (2/L) ∫₀ᴸ sin(mπx/L) sin(nπx/L) dx = δₘₙ

Turns projection into one integral per level: no simultaneous equations, no matrix to invert, one coefficient at a time.

Dimensionless. Product-to-sum leaves cosines spanning whole numbers of periods, which integrate to zero unless m = n.

Resolution of the identityΣₙ |n⟩⟨n| = 1 ⇔ (2/L) Σₙ sin(nπx/L) sin(nπx′/L) = δ(x − x′)

This is completeness, and it licenses inserting Σ|n⟩⟨n| anywhere inside a bracket. Orthonormality alone never gives it.

Sum over n = 1 to ∞. Both sides of the closure relation carry m⁻¹, and the delta lives on the open interval (0, L).

Expansion coefficientcₙ = ⟨n|ψ⟩ = √(2/L) ∫₀ᴸ sin(nπx/L) ψ(x) dx

One number per level: the state stops being a function of x and becomes a column vector you can hand to NumPy.

ψ in m^(−1/2) makes cₙ dimensionless. It is exactly the Fourier sine coefficient of ψ on [0, L].

Parseval and the Born rule⟨φ|ψ⟩ = Σₙ bₙ* cₙ, ⟨ψ|ψ⟩ = Σₙ |cₙ|² = 1, P(Eₙ) = |cₙ|²

Normalisation becomes a convergent series, and 1 − Σ(n ≤ N)|cₙ|² is the exact fraction of the state a truncation discarded.

bₙ = ⟨n|φ⟩, every term dimensionless, Eₙ = n²π²ħ²/(2mL²). The basis change is unitary, so brackets survive it.

Convergence in the norm only‖ψ − S_N‖ → 0, yet max|ψ − S_N| → 0.0895 × jump

Names the price of completeness: mean-square agreement never promises agreement at a point, and Gibbs is the standing proof.

S_N is the partial sum to n = N. The overshoot sits within about L/(2N) of a jump in the odd 2L-periodic extension of ψ.

Orthogonality is forced before it is computed

Take Ĥ = −(ħ²/2m) d²/dx² on the Dirichlet domain — functions in H²([0, L]) with ψ(0) = ψ(L) = 0. Integrating by parts twice, ⟨m|Ĥψₙ⟩ − ⟨Ĥψₘ|ψₙ⟩ equals the Wronskian boundary term −(ħ²/2m)[ψₘ*ψₙ′ − ψₘ*′ψₙ]₀ᴸ, and the boundary condition kills it at both ends. The left side is (Eₙ − Eₘ)⟨m|n⟩ because every Eₙ is real, and the 1D box has no degeneracy, so E₁, E₂, E₃, … are distinct and every off-diagonal overlap must vanish before you integrate anything. Now integrate anyway, because that is what catches a wrong convention. Product-to-sum gives sin A sin B = ½[cos(A − B) − cos(A + B)], so for m ≠ n the integrand is a difference of cosines at (m − n)π/L and (m + n)π/L, each completing a whole number of periods on [0, L]; both integrate to zero. For m = n the first cosine is cos 0 = 1 and contributes L/2, so (2/L)(L/2) = 1.

Normalisation is the only freedom the walls left

The eigenvalue problem fixes the shape and not the size: any multiple of sin(nπx/L) still solves it and still vanishes at both walls. Born's rule fixes the size. Since ∫₀ᴸ sin²(nπx/L) dx = L/2 for every n — sin² averages to ½ over a whole number of half-periods, and n only changes how many — the constant is √(2/L), the same for every level. Units follow: |ψ|² is a probability per unit length, so ψ carries m^(−1/2). At L = 1.00 nm that is √(2/L) = 4.47 × 10⁴ m^(−1/2), and every |ψₙ|² peaks at 2/L = 2.00 nm⁻¹ whatever n is; raising n moves the antinodes around without raising the peak. One freedom survives normalisation: e^(iθ)|n⟩ is equally normalised, so the ket is a ray. Choosing the real sine that is positive just inside x = 0 is a convention, and it is the convention that keeps every cₙ real when ψ is real — say it out loud before comparing signs with anyone else's code.

Completeness is a second theorem, not a corollary

Orthonormality never implies completeness. Throw away |2⟩: the surviving set still satisfies ⟨m|n⟩ = δₘₙ exactly, while ψ₂ itself now expands to the zero vector and Parseval fails on it. The missing input is spectral. On a bounded interval, −d²/dx² with Dirichlet conditions is a regular Sturm–Liouville operator with compact resolvent, so its eigenvectors form an orthonormal basis of the whole of L²([0, L]). In Dirac notation that is Σₙ|n⟩⟨n| = 1; in the position basis it is the closure relation (2/L)Σₙ sin(nπx/L) sin(nπx′/L) = δ(x − x′), a distribution, so read it only under an integral. Numerically the two halves come apart. Build S with Sₙⱼ = √(2Δx/L) sin(nπxⱼ/L) for n ≤ N on a grid of J points: S @ S.T ≈ I_N tests orthonormality of the states you kept, while S.T @ S ≈ I_J tests completeness and only approaches the identity as N approaches J. Truncation breaks completeness first, and never breaks orthonormality at all.

Projection, and parity that halves the integrals

Because Σₙ|n⟩⟨n| = 1, writing |ψ⟩ = Σcₙ|n⟩ and hitting it with ⟨m| returns cₘ = ⟨m|ψ⟩ at once: orthonormality has already decoupled the equations, so there is no linear system to solve. In the position basis that projection is the Fourier sine coefficient, cₙ = √(2/L)∫₀ᴸ sin(nπx/L) ψ(x) dx. Use symmetry before calculus. The basis has definite parity about the midpoint, ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x), so a state symmetric about L/2 has zero overlap with every even n and an antisymmetric one with every odd n. For the normalised parabola ψ = √(30/L⁵) x(L − x) that deletes half the integrals unseen, and the survivors give cₙ = 4√60/(n³π³): c₁ = 0.99928, c₃ = 0.03701, c₅ = 0.00799. The state is 99.86% ground state, which is why a parabola is such a good freehand sketch of ψ₁ — and why c₂ = 0 exactly, not merely small.

Parseval turns the norm into probabilities

Insert the identity into the norm: ⟨ψ|ψ⟩ = ⟨ψ|(Σₙ|n⟩⟨n|)|ψ⟩ = Σₙ⟨ψ|n⟩⟨n|ψ⟩ = Σₙ|cₙ|². Normalisation stops being an integral and becomes a convergent series, and the same move on two states gives ⟨φ|ψ⟩ = Σₙbₙ*cₙ — the change of basis is unitary, so it preserves every bracket, not merely lengths. The physics arrives when the Born rule is read in the energy basis: P(Eₙ) = |cₙ|², so Parseval is literally the statement that the probabilities of the possible energy outcomes sum to one, and ⟨H⟩ = Σₙ|cₙ|²Eₙ. It is also your numerical error bar. Truncating at N discards exactly 1 − Σ(n ≤ N)|cₙ|² of the norm — and therefore of the probability — which for the parabola is 1.4 × 10⁻³ at N = 1 and 7.5 × 10⁻⁵ at N = 3. Print that residual in any box code you write; it is the one convergence diagnostic that costs nothing.

Decay rate is a boundary question, and Gibbs is the bill

How fast cₙ falls is set by how well ψ imitates the basis at the walls, because every ψₙ vanishes there and so does every ψₙ″. A state matching the first condition but not the second — the parabola, whose second derivative is −2A everywhere — gives cₙ ∝ n⁻³ and |cₙ|² ∝ n⁻⁶. A state failing even the first — the flat ψ = 1/√L, a perfectly good vector of L²([0, L]) though nowhere near the domain of Ĥ — gives only cₙ = 2√2/(nπ) for odd n, so |cₙ|² ∝ n⁻² and eight terms retain just (8/π²)(1 + 1/9 + … + 1/225) = 0.9747 of the norm, against the parabola's 0.99993 with two. The mismatch surfaces as Gibbs. The odd 2L-periodic extension of the flat state is a square wave jumping by 2 at each wall, and the partial sum overshoots by 0.0895 of that jump: the eight-term sum reaches 1.1803/√L at x = L/16. Keep sixteen terms and the peak moves to x = L/32 and drops only to 1.1793/√L.

Change one variable at a time

Make the relationship visible.

Interactive model

Odd harmonics kept M3

Probe position x₀/L0.10

Park the probe at 0.10L and drag M from 1 to 5: the reading climbs through 0.99 at M = 3, then overshoots to 1.18 — an 18% ear that more terms only sharpen. Move the probe to 0.50L and the same sweep closes on 1 like 1/(πM).

HIGHEST n KEPT5

CAPTURED NORM Σ|cₙ|²0.9331

PROBE VALUE × √L0.991

POINTWISE ERROR0.009

Live interpretationHIGHEST n KEPT: 5. CAPTURED NORM Σ|cₙ|²: 0.9331. PROBE VALUE × √L: 0.991. POINTWISE ERROR: 0.009

Catch the common trap

Explain before calculating.

Misconception check

“The sine basis is complete” gets read as “the series equals ψ at every point”. It does not. Completeness is a statement in the norm: ‖ψ − S_N‖² = ∫₀ᴸ|ψ − S_N|² dx → 0, an integral of squared error — and an integral can be driven to zero while the error at particular points stays large, provided the bad region shrinks fast enough. That is exactly what happens. Every partial sum obeys S_N(0) = S_N(L) = 0 whether ψ does or not, and near a jump the overshoot keeps 8.95% of the jump height forever while its width falls as L/(2N), so its share of the norm dies as 1/N and convergence is untouched. So never read a truncated expansion as a pointwise approximation near a wall or a discontinuity, and stop asking the limit to agree with ψ at isolated points: in L² two functions differing on a set of measure zero are the same vector, and those endpoints were never part of what the vector claimed.

The flat state ψ(x) = 1/√L on (0, L) has sine coefficients cₙ = 2√2/(nπ) for odd n and 0 for even n, and Parseval closes exactly: Σ|cₙ|² = (8/π²)(π²/8) = 1. Which statement about this expansion is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFor |n⟩ = √(2/L) sin(nπx/L) on [0, L], show by direct integration that ⟨1|3⟩ = 0 and ⟨3|3⟩ = 1. Then, for L = 1.00 nm, give the numerical value of the normalisation constant and the peak probability density of |3⟩.

Product-to-sum: sin A sin B = ½[cos(A − B) − cos(A + B)]. With A = πx/L and B = 3πx/L, sin(πx/L) sin(3πx/L) = ½[cos(2πx/L) − cos(4πx/L)].
∫₀ᴸ cos(2πx/L) dx = (L/2π)[sin(2πx/L)]₀ᴸ = 0, and the 4π term vanishes the same way: both cosines complete a whole number of periods on [0, L]. So ⟨1|3⟩ = (2/L)(0) = 0.
That zero was compulsory. E₁ ≠ E₃ and Ĥ is self-adjoint on the Dirichlet domain, so the two eigenvectors had to be orthogonal; the integral only confirms the convention has not been mangled.
For m = n = 3 the same identity gives sin²(3πx/L) = ½[1 − cos(6πx/L)]; the cosine again integrates to zero, leaving ∫₀ᴸ sin²(3πx/L) dx = L/2, so ⟨3|3⟩ = (2/L)(L/2) = 1 — the √(2/L) exists precisely to cancel that L/2.
With L = 1.00 nm = 1.00 × 10⁻⁹ m, √(2/L) = √(2.00 × 10⁹ m⁻¹) = 4.47 × 10⁴ m^(−1/2), and |ψ₃|² peaks at 2/L = 2.00 × 10⁹ m⁻¹ = 2.00 nm⁻¹ at each of its three antinodes.

Answer⟨1|3⟩ = 0 and ⟨3|3⟩ = 1. At L = 1.00 nm the normalisation constant is 4.47 × 10⁴ m^(−1/2) and |ψ₃|² peaks at 2.00 nm⁻¹.

MediumA particle in a box of width L is prepared in ψ(x) = A x(L − x). Normalise it, find cₙ = ⟨n|ψ⟩ for every n, state P(E₁) and P(E₂), and say how much of the norm survives truncation at n = 3.

Normalise: ∫₀ᴸ x²(L − x)² dx = L⁵/30, so A = √(30/L⁵) and ψ carries m^(−1/2) as it must.
Use parity first. ψ is symmetric about L/2 and ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x), so every even-n integrand is antisymmetric about the midpoint and cₙ = 0 for all even n — half the integrals gone before any calculus.
For the rest, ∫₀ᴸ x(L − x) sin(nπx/L) dx = 2L³(1 − (−1)ⁿ)/(n³π³) = 4L³/(n³π³) for odd n. Hence cₙ = √(2/L)⋅√(30/L⁵)⋅4L³/(n³π³) = 4√60/(n³π³), independent of L as a dimensionless coefficient must be.
Numbers: 4√60 = 30.984 and π³ = 31.006, so c₁ = 0.99928, c₃ = c₁/27 = 0.03701, c₅ = c₁/125 = 0.00799.
P(E₁) = c₁² = 0.99856 and P(E₂) = 0 exactly, not merely small. Keeping n ≤ 3 retains c₁² + c₃² = 0.99856 + 0.00137 = 0.99992, so 7.5 × 10⁻⁵ of the norm — and of the probability — has been thrown away.
Parseval closes exactly: Σ over odd n of 960/(n⁶π⁶) = (960/π⁶)(63/64)(π⁶/945) = 1, with no remainder to hide an error in.

Answercₙ = 4√60/(n³π³) for odd n and 0 for even n; P(E₁) = 0.9986 and P(E₂) = 0; truncating at n = 3 keeps 0.99992 of the norm.

HardThe same box is prepared in the flat state ψ(x) = 1/√L on (0, L). Find cₙ, verify Parseval, compute the norm retained by n ≤ 15, evaluate the partial sum at x = L/16, and find ⟨H⟩.

cₙ = ⟨n|ψ⟩ = √(2/L)(1/√L)∫₀ᴸ sin(nπx/L) dx = (√2/L)(L/nπ)(1 − (−1)ⁿ) = 2√2/(nπ) for odd n and 0 for even n.
Parseval: Σ over odd n of 8/(n²π²) = (8/π²)(π²/8) = 1 exactly. The flat state is a legitimate vector of L²([0, L]) even though it violates ψ(0) = ψ(L) = 0 — that condition defines the domain of Ĥ, not membership of the Hilbert space.
Norm retained by n ≤ 15, which is eight odd terms: (8/π²)(1 + 1/9 + 1/25 + 1/49 + 1/81 + 1/121 + 1/169 + 1/225) = 0.81057 × 1.20249 = 0.97470. Because cₙ falls only as 1/n, 2.5% of the state still sits above n = 15.
The eight-term sum is S₈(x) = (4/(π√L)) Σ over odd n ≤ 15 of sin(nπx/L)/n, and S₈′ ∝ sin(16πx/L)/sin(πx/L) first vanishes at x = L/16, so that is where the peak sits. Summing sin(11.25n°)/n for n = 1, 3, …, 15 gives 0.92699, and 4/π times that is 1.1803.
So S₈(L/16) = 1.1803/√L, an 18.0% overshoot above the target 1/√L. Sixteen terms move the peak to x = L/32 and lower it only to 1.1793/√L: the ear narrows as L/(2N) and never shortens, converging on 0.0895 × the jump of 2.
⟨H⟩ = Σ|cₙ|²Eₙ = Σ over odd n of (8/n²π²)(n²E₁) = (8E₁/π²) Σ over odd n of 1, which diverges. Expandability is a Hilbert-space fact; a finite ⟨H⟩ is a domain fact, and this state has the first without the second.

Answercₙ = 2√2/(nπ) for odd n, 0 for even n; Parseval closes exactly; n ≤ 15 retains 0.9747; S₈(L/16) = 1.1803/√L, an 18% ear; ⟨H⟩ diverges.