Orthogonality and completeness of the sine basis
Perpendicular axes for functions, with none missing.
Every expansion, probability and time evolution in the rest of this unit leans on two facts set out here. Orthogonality turns each coefficient into a single integral. Completeness promises that the sines leave nothing out, in the precise sense that the squared error integrates to zero, which is weaker than matching at every point.
Build the model
Orthogonality makes each coefficient one integral; completeness says none is missing.
1The big idea
The stationary states of the infinite well are more than a list of solutions: they are a set of perpendicular axes for functions on [0, L], with ∫₀ᴸ f*g dx playing the dot product. Orthogonality, ∫₀ᴸ ψₘψₙ dx = δₘₙ, follows from a product-to-sum identity, but its real source is the equation: (kₙ² − kₘ²) times the overlap equals a boundary term, and the walls, where every state vanishes, set that term to zero. Two states of different energy cannot overlap, and the same proof serves every one-dimensional well in the course. Completeness is a separate claim, that no axis is missing: for every square-integrable f the squared coefficients add up to ∫|f|² dx. Drop the even-n states and the rest are still orthonormal but no longer complete. For the sines, completeness is Fourier's theorem applied to the odd extension of f, which is why functions that do not vanish at the walls can still be built. The convergence is in mean square, the integral of |f − SN|² going to zero, and that is what physics needs, because every prediction is an integral of |ψ|². It is not pointwise: beside a jump the partial sums overshoot by about 18% however many terms are kept, and at the walls they give 0 whatever f is. Nor does it reach derivatives: the flat function's series converges while its energy diverges.
2In one sentence
The well's sine states are orthonormal, ∫₀ᴸ ψₘψₙ dx = δₘₙ, and complete: every square-integrable f on [0, L] is the mean-square limit of Σ cₙψₙ, with cₙ = ∫₀ᴸ ψₙ f dx.
Example. The normalised parabola √(30/L⁵) x(L − x) has cₙ = 8√15/(nπ)³ for odd n and 0 for even n: ψ₁ alone carries |c₁|² = 99.86% of its squared norm, ψ₃ another 0.137%, and the full set exactly 100%.
3One identity turns the overlap into two cosines
Integrate the product of two states: (2/L)∫₀ᴸ sin(mπx/L) sin(nπx/L) dx. The identity sin a sin b = ½[cos(a − b) − cos(a + b)] turns the product into cos((m − n)πx/L) and cos((m + n)πx/L). Any cos(jπx/L) with j a nonzero integer integrates over [0, L] to (L/(jπ)) sin(jπ) = 0, so for m ≠ n both terms vanish. For m = n the first cosine is cos 0 = 1, leaving (2/L)(½)(L) = 1. So ∫₀ᴸ ψₘψₙ dx = δₘₙ, and the dimensions agree: each ψ has dimensions of length(−1/2), the product length⁻¹, and dx restores a pure number. Symmetry explains half the pairs at a glance. ψ₁ is even about x = L/2 and ψ₂ odd, so their product is odd and cancels, positive on the left half and negative on the right. ψ₁ and ψ₃ are both even, their product is positive in the outer thirds and negative in the middle one, and only the identity shows those areas balance. The interval matters too: over the left half alone, ∫₀(L/2) ψ₁ψ₂ dx = 4/(3π) = 0.424.
4The walls, not the sine function, make the overlap vanish
A second proof never uses the formula for ψ. Each state obeys ψₙ″ = −kₙ²ψₙ, with kₙ² proportional to Eₙ. Multiply the equation for ψₘ by ψₙ, the one for ψₙ by ψₘ, and subtract: ψₙψₘ″ − ψₘψₙ″ = (kₙ² − kₘ²)ψₘψₙ. The left side is the derivative of ψₙψₘ′ − ψₘψₙ′, so over [a, b] the overlap times kₙ² − kₘ² equals that bracket evaluated between the ends. Across the whole well the bracket vanishes at both walls, where every state is zero, and with different energies the overlap must be zero. That is the real reason: boundary conditions plus distinct eigenvalues. A potential V(x) cancels in the subtraction, so the same lines, with the ends taken at ±∞ where bound states vanish, prove orthogonality in the finite well and the oscillator, where no identity helps; and bound states in one dimension are never degenerate, so no pair escapes. The argument even predicts the half-well value. At x = L/2, ψ₂ = 0, ψ₁ = √(2/L) and ψ₂′ = −(2π/L)√(2/L), so the bracket is 4π/L², and dividing by k₂² − k₁² = 3π²/L² gives 4/(3π).
5The overlap integral works as a dot product
In ordinary space a vector v has components vᵢ = êᵢ ⋅ v along orthonormal unit vectors, and |v|² is the sum of their squares. Functions on [0, L] behave the same way once ∫₀ᴸ f*g dx plays the dot product. Orthonormality makes each coefficient a single projection, cₙ = ∫₀ᴸ ψₙ f dx, with no conjugate needed because ψₙ is real, fixed without reference to any other term; UQMI-06.06 puts that to work on given states. Here the payoff is geometric. Keep the first N terms, SN = Σ cₙψₙ, and orthogonality gives ∫|f − SN|² dx = ∫|f|² dx − Σ|cₙ|² through n = N. The error is read off the coefficients, and no other N coefficients do better, just as the shortest path from a point to a plane is the perpendicular. The left side is never negative, so Σ|cₙ|² ≤ ∫|f|² dx for any orthonormal set: Bessel's inequality. For the flat function 1/√L, c₁ = 2√2/π = 0.900, so ψ₁ alone leaves 1 − 0.811 = 0.189 unaccounted for, and S₁ has squared norm ∫|S₁|² dx = 0.811, not 1: a truncated series is not a normalised state until rescaled.
6Completeness is a separate claim: no axis is missing
Orthogonality says the axes are perpendicular; completeness says there are enough of them. The set is complete if Bessel's inequality becomes Parseval's identity, Σ|cₙ|² = ∫|f|² dx, for every square-integrable f, which is the same as the truncation error falling to zero. The two properties are independent. Discard the even-n states and the rest are just as orthonormal, but every odd-n sine is symmetric about x = L/2, and so is every sum of them. A function antisymmetric about the centre, such as x − L/2, overlaps none of them: all its coefficients vanish while its squared norm, L³/12, does not. Bessel holds and Parseval fails. For the full set, completeness is Fourier's theorem: extend f to [−L, 0] as an odd function, repeat with period 2L, and the sine series converges to that extension in mean square. That is how sines, each zero at the walls, build a function that is not. Square-integrable is a real condition: a spike δ(x − x₀) has cₙ = √(2/L) sin(nπx₀/L), whose squares never die away, and Σ|cₙ|² diverges.
7Mean-square convergence leaves an 18% spike at every jump
The flat function 1/√L jumps to zero at each wall, and its series shows what mean-square convergence does and does not promise. Its coefficients are 2√2/(nπ) for odd n, so the error after keeping n = 1 to N is 1 − (8/π²)Σ 1/n² over odd n, close to 4/(π²(N + 1)): 0.099 through n = 3, 0.020 through n = 19, 0.0041 through n = 99. It goes to zero, as completeness requires. Point by point the story differs. At x = 0 and x = L every partial sum is exactly 0, the midpoint of the odd extension's jump from −1/√L to +1/√L. Just inside, SN climbs past f to a first maximum at x = L/(N + 1), of height 1.200, 1.180 and 1.179 times 1/√L for N = 3, 19 and 99: the Gibbs overshoot, which tends to (2/π)Si(π) = 1.179, not to 1, and never dies away. What shrinks is its width, and an error of fixed height over a width of order L/(N + 1) adds only of order 1/(N + 1) to the integral. Probabilities over intervals converge; a value read at or beside a jump from a truncated series cannot be trusted.
8Smoothness sets the price, and energy is the stricter test
How fast cₙ falls is fixed by the first derivative of the odd, 2L-periodic extension that jumps. The flat function itself jumps, and |cₙ| ∝ 1/n. A tent rising linearly to a peak at L/2 is continuous with a kinked slope, and |cₙ| ∝ 1/n². The parabola x(L − x) has a continuous slope and a jump only in its second derivative, and |cₙ| ∝ 1/n³. So the error left after n = N falls as 1/N, 1/N³ and 1/N⁵: carrying ten times as many terms cuts it by 10 for the flat function, about 1000 for the tent and about 10⁵ for the parabola. Energy asks more. The derivatives of the states are cosines, orthogonal too, so a normalised state vanishing at the walls has energy Σ|cₙ|²n²E₁: (10/π²)E₁ = 1.013 E₁ for the parabola, (12/π²)E₁ = 1.216 E₁ for the tent. For the flat function every term is 8E₁/π² and the sum diverges, the infinite cost UQMI-06.01 found for a nonzero value at a wall. The classical string clamped at both ends has these same sines as its modes: plucked at its midpoint, it takes the tent's shape, its harmonics fall as 1/n², and its energy is finite.
Key equations
What each relation says, and the units it is written in.
Orthonormality of the well's stationary states
∫₀ᴸ ψₘ(x) ψₙ(x) dx = δₘₙ
Lets one integral isolate each coefficient of an expansion, independent of every other term. It needs the whole well: over part of it, pairs such as ψ₁ and ψ₂ overlap.
ψₙ = √(2/L) sin(nπx/L) has units m(−1/2) for L in metres; δₘₙ is 1 when m = n and 0 otherwise, a pure number like the integral.
Orthogonality from the Schrödinger equation
(kₙ² − kₘ²) ∫ₐᵇ ψₘψₙ dx = [ψₙψₘ′ − ψₘψₙ′] evaluated from x = a to x = b
Distinct energies plus a bracket that vanishes at both ends force zero overlap without trigonometry; V(x) cancels, so with ends at ±∞ the proof also serves the finite well and oscillator.
kₙ, in m⁻¹, is the wavenumber in ψₙ″ = −kₙ²ψₙ, so kₙ² is proportional to Eₙ; the bracket, in m⁻², is zero wherever both ψ vanish.
Parseval's identity, the test of completeness
Σ |cₙ|² (n = 1 to ∞) = ∫₀ᴸ |f(x)|² dx
Holds for every square-integrable f exactly when the set is complete. Delete ψ₂ from the set and it fails for f = ψ₂, whose coefficients on every remaining state are zero.
cₙ = ∫₀ᴸ ψₙ f dx, dimensionless for f in m(−1/2); a normalised f makes both sides 1. f need not vanish at the walls.
Mean-square error of an N-term sum
∫₀ᴸ |f − SN|² dx = ∫₀ᴸ |f|² dx − Σ |cₙ|² (n = 1 to N)
Reads the error off the coefficients without integrating the partial sum. The right side can only fall as N grows; completeness is the claim that it reaches zero.
SN = Σ cₙψₙ through n = N. Uses orthonormality alone, so it holds for any orthonormal set; no other N coefficients do better.
Coefficient decay is set by the first jump
|cₙ| ∝ n(−j−1), where the jth derivative of the odd 2L-periodic extension of f is the first to jump
Prices a truncation: a mean-square error below 0.1% takes 203 nonzero terms for the flat function, 3 for the tent and 2 for the parabola.
j = 0 for the flat function (|cₙ| ∝ 1/n), 1 for a tent peaked at L/2 (1/n²), 2 for x(L − x) (1/n³).
Gibbs overshoot beside a jump
max SN → (2/π) Si(π) f₀ = 1.179 f₀, reached at x = L/(N + 1)
The spike narrows as 1/(N + 1) but never dies away, so the pointwise error stays near 18% while the mean-square error goes to zero. At the wall itself every SN is 0.
Flat f of height f₀, in m(−1/2); N is the highest odd n kept. Si(π) = ∫ (sin t/t) dt from 0 to π = 1.852.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At w = 0, step N from 3 to 39: the dot only drops from 1.200 to 1.179 f₀ while the error falls from 9.9% to 1.0%. Set w = 0.10 and the spike fades as N passes about 10. At w = 0 the energy of SN grows in step with N; for any nonzero w it heads for a finite limit.
Live results
- ODD HARMONICS KEPT
- 5
- SHARE OF ∫|f|² IN ψ₁
- 81.06 %
- |cN| ÷ |c₁|, LAST TERM
- 0.1111
- FLAT-f ERROR ≈ 4/(π²(N + 1))
- 4.05 %
Catch the common trap
The mistake this lesson is built to prevent.
Completeness gets read as equality at every point, and that reading misleads in two opposite directions. One student reasons that every ψₙ vanishes at the walls, so no sum of them can equal a function that does not, and decides the sine basis cannot hold such a state. Another expects the spike beside a jump to die away once enough terms are kept. Completeness claims neither. It claims that ∫₀ᴸ |f − SN|² dx goes to zero, and for the flat function 1/√L it does, as 4/(π²(N + 1)), while SN(0) stays exactly 0 and the spike settles at 1.179/√L, only narrowing. A disagreement confined to a width that shrinks to nothing cannot change an integral, and every probability quantum mechanics predicts is an integral of |ψ|² over some interval, so for every such prediction the series and the function agree.
Test it: what changes?
- Mean-square error falls, roughly as 4/(π²(N + 1))
- Peak of SN tends to 1.179 f₀, not f₀
- First peak moves in to x = L/(N + 1)
- SN stays exactly 0 at each wall
- Coefficients decay as 1/n², not 1/n
- Overshoot beside each wall fades at large N
- Energy of SN tends to a finite limit
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A well has width L = 0.80 nm. Using sin a sin b = ½[cos(a − b) − cos(a + b)], evaluate ∫ψ₂ψ₃ dx and ∫ψ₃ψ₃ dx over the whole well, then ∫ψ₂ψ₃ dx over its left half, 0 ≤ x ≤ L/2, alone. Check the dimensions of each answer.
View solution →
- Dimensions first: ψₙ = √(2/L) sin(nπx/L) is in nm(−1/2), a product of two is in nm⁻¹, and dx in nm leaves a pure number. Every answer must be dimensionless, so L = 0.80 nm has to cancel.
- Product to sum: sin(2πx/L) sin(3πx/L) = ½[cos(πx/L) − cos(5πx/L)], and sin²(3πx/L) = ½[1 − cos(6πx/L)]. The prefactor (2/L)(½) = 1/L multiplies each integral.
- Whole well: ∫₀ᴸ cos(jπx/L) dx = (L/(jπ)) sin(jπ) = 0 for j = 1, 5 and 6. So ∫ψ₂ψ₃ dx = (1/L)(0 − 0) = 0, while ∫ψ₃ψ₃ dx = (1/L)(L − 0) = 1.
- Left half: the cosine integrals now stop at x = L/2, giving (L/π) sin(π/2) = L/π and (L/(5π)) sin(5π/2) = L/(5π). So ∫ψ₂ψ₃ dx = (1/L)(L/π − L/(5π)) = 4/(5π) = 0.255, with L gone as the dimensions demanded.
- Cross-check with the boundary term: (k₃² − k₂²)∫ψ₂ψ₃ dx over [0, L/2] equals ψ₃ψ₂′ − ψ₂ψ₃′ at x = L/2, since both states vanish at x = 0. There ψ₂ = 0, ψ₃ = −√(2/L) and ψ₂′ = −(2π/L)√(2/L), so the bracket is 4π/L²; dividing by 5π²/L² gives 4/(5π) again.
Answer. Over the whole well ∫ψ₂ψ₃ dx = 0 and ∫ψ₃ψ₃ dx = 1; over the left half ∫ψ₂ψ₃ dx = 4/(5π) = 0.255, dimensionless and the same for every L.
MediumExample 2
Test completeness on the flat function f(x) = 1/√L inside the well, which does not vanish at the walls. Find its coefficients, confirm Parseval's identity using Σ 1/n² over odd n = π²/8, find the mean-square error after n = 1, 3 and 5, and find how many nonzero terms bring that error below 1.0%.
View solution →
- cₙ = ∫₀ᴸ √(2/L) sin(nπx/L)(1/√L) dx = (√2/L)(L/(nπ))(1 − cos nπ). The bracket is 2 for odd n and 0 for even n, so cₙ = 2√2/(nπ) for odd n: c₁ = 0.900, c₃ = 0.300, c₅ = 0.180, all dimensionless as they must be for a normalised f.
- Parseval: Σ|cₙ|² = (8/π²) Σ 1/n² over odd n = (8/π²)(π²/8) = 1, equal to ∫₀ᴸ |f|² dx = L × (1/L) = 1. No norm is lost, although every ψₙ is zero at the walls where f is not.
- After three terms: |c₁|² + |c₃|² + |c₅|² = 0.8106 + 0.0901 + 0.0324 = 0.9331, so ∫|f − S₅|² dx = 1 − 0.9331 = 0.0669, a mean-square error of 6.7%.
- The tail: each odd n above N stands for an interval of width 2 in n, so Σ 1/n² over odd n beyond N ≈ ∫ dn/(2n²) from N + 1 to ∞ = 1/(2(N + 1)), and the error is about (8/π²)/(2(N + 1)) = 4/(π²(N + 1)). At N = 5 that gives 0.0675, against the exact 0.0669.
- Below 1.0% needs N + 1 to exceed 400/π² = 40.5, and N + 1 is even, so N = 41. Exact sums confirm it: 1.013% through n = 39, 0.965% through n = 41. The odd n from 1 to 41 are 21 terms.
Answer. cₙ = 2√2/(nπ) for odd n and Σ|cₙ|² = 1; the error is 6.7% after n = 1, 3, 5 and first drops below 1.0% at n = 41, 21 nonzero terms, since it shrinks only as 4/(π²(N + 1)).
HardExample 3
Keep the flat function's series through odd n = N: SN(x) = (4/(π√L)) Σ sin(nπx/L)/n over odd n ≤ N. Locate the first maximum of SN just inside the left wall, find its height for N = 9 and N = 39 and its limit as N → ∞, and reconcile a peak that does not shrink with a mean-square error that does.
View solution →
- Put θ = πx/L and K = (N + 1)/2 terms. Then dSN/dθ = (4/(π√L)) Σ cos((2k − 1)θ) for k = 1 to K. Multiplying that sum by 2 sin θ telescopes, since 2 sin θ cos((2k − 1)θ) = sin(2kθ) − sin((2k − 2)θ), so the sum is sin(2Kθ)/(2 sin θ).
- The slope first vanishes where 2Kθ = π, at θ = π/(2K), which is x = L/(2K) = L/(N + 1): x = 0.100L for N = 9 and 0.025L for N = 39. The peak moves toward the wall as 1/(N + 1).
- Heights: evaluating (4/π) Σ sin(nπ/(N + 1))/n over odd n ≤ N gives SN√L = 1.1823 for N = 9 and 1.1792 for N = 39, both about 18% above f = 1/√L.
- The limit: with tₖ = (2k − 1)π/(2K), spaced by π/K, the same sum is (2/π) Σ (sin tₖ/tₖ)(π/K), a midpoint Riemann sum for (2/π)∫ (sin t/t) dt from 0 to π = (2/π)(1.8519) = 1.1790. The overshoot tends to 0.179/√L, 8.9% of the odd extension's jump of 2/√L, and never dies away however large N is.
- Reconcile: the mean-square error, 1 − (8/π²) Σ 1/n² over odd n ≤ N, is 0.0404 for N = 9 and 0.0101 for N = 39, four times smaller for four times the harmonics. The spike keeps its height, but its width is of order L/(N + 1), so its share of ∫|f − SN|² dx falls as 1/(N + 1) along with the rest.
Answer. The first maximum sits at x = L/(N + 1), with height 1.182/√L for N = 9 and 1.179/√L for N = 39, tending to (2/π)Si(π)/√L = 1.179/√L; the mean-square error still falls from 0.0404 to 0.0101, because the spike narrows rather than drops.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Orthogonality isolates each cₙ; completeness makes Σ|cₙ|² equal ∫|f|² dx.
Find each coefficient as cₙ = ∫₀ᴸ ψₙ f dx and test Parseval's identity. Convergence is in mean square, ∫|f − SN|² dx → 0, so interval probabilities agree; values at or beside a jump do not, since SN is 0 at the walls and overshoots by about 18%.