Completeness: expanding any state on an eigenbasis
One integral per coefficient; nothing left over, in mean square.
Orthogonality told you the eigenfunctions do not overlap; completeness says they leave nothing out. Take that on trust and the payoff is large: each coefficient costs one integral, normalisation moves onto the coefficients, and an operator acting on ψ merely rescales them. The care goes into what the equals sign in ψ = Σ cₙψₙ promises.
Build the model
Orthonormality finds each coefficient; completeness promises nothing is left over.
1The big idea
Orthogonality, proved in the previous lesson, says the eigenfunctions of a Hermitian operator do not overlap; it does not say they are enough. Completeness is that second claim: every square-integrable ψ on the problem's region can be written as ψ = Σ cₙψₙ. It is the infinite-dimensional meaning of a basis, and this course asserts it rather than proving it, because the proof belongs to Sturm–Liouville and spectral theory. Granted completeness, orthonormality does all the work. Each coefficient is one overlap integral, cₙ = ∫ ψₙ*ψ dx, with no simultaneous equations to solve; the norm moves onto the coefficients, Σ |cₙ|² = ∫ |ψ|² dx; and an operator applied to ψ only multiplies each cₙ by its eigenvalue. Three limits keep the statement honest. The sum converges in mean square, ∫ |ψ − SN|² dx → 0, which lets it miss ψ at a wall for ever and overshoot beside a jump by a height that never shrinks. Letting an operator act term by term needs more than a finite Σ |cₙ|²: it needs Σ |aₙcₙ|² to converge, which is the operator's domain showing up in the coefficients. And an operator such as p̂ on the whole line has no normalisable eigenfunctions at all, so the sum becomes an integral over a continuous spectrum, the Fourier transform, named here and put to work in Unit 5.
2In one sentence
An orthonormal set of eigenfunctions ψₙ is complete when every square-integrable ψ equals Σ cₙψₙ with cₙ = ∫ ψₙ*ψ dx, in the sense that the partial sums SN satisfy ∫ |ψ − SN|² dx → 0 as N → ∞.
Example. In a box of width a, ψ = √(16/(5a)) sin³(πx/a) is exactly (3ψ₁ − ψ₃)/√10 on the eigenbasis ψₙ = √(2/a) sin(nπx/a): c₁ = 0.949, c₃ = −0.316, every other cₙ = 0, and the weights 0.900 + 0.100 add to 1.
3Orthonormality turns each coefficient into one integral
Suppose ψ = Σ cₙψₙ, where the ψₙ are eigenfunctions of a Hermitian Â, normalised and orthogonalised so that ∫ ψₘ*ψₙ dx = δₘₙ (UQMI-03.06). Multiply both sides by ψₘ* and integrate over the region. The integral may be taken inside the sum because the series converges in mean square, and then every term with n ≠ m vanishes: cₘ = ∫ ψₘ*ψ dx. It is how a vector's x-component is its dot product with the unit vector along x, found without knowing any other component. In a basis that is not orthogonal, every coefficient depends on all the others, and N of them cost N simultaneous equations. The conjugate belongs on the basis function. Put it on ψ instead and the integral returns cₘ*; leave it off a complex ψₘ, as on a ring where ψₘ = e(2πimx/L)/√L, and the integral returns the coefficient of the opposite momentum. Dimensions check the formula: ψₘ* and ψ each carry m⁻¹⁄² and dx carries m, so cₘ is a pure number. Often no integral is needed: sin³θ = (3 sin θ − sin 3θ)/4 expands sin³(πx/a) on two box functions by inspection.
4The norm passes to the coefficients
Put the expansion into ∫ |ψ|² dx and use orthonormality once more: the double sum Σₘ Σₙ cₘ*cₙ ∫ ψₘ*ψₙ dx collapses to Σ |cₙ|². This is Parseval's relation, and for a normalised ψ it reads Σ |cₙ|² = 1, so a state built from eigenfunctions is normalised by adjusting coefficients, not by integrating. The relative phases of the cₙ do not enter. In a box, (ψ₁ + ψ₂)/√2 and (ψ₁ − ψ₂)/√2 both have weights ½ and ½, yet their densities lean towards opposite walls, because the cross term ±ψ₁ψ₂ in the density flips sign between them, and ψ₁ψ₂ is positive in the left half and negative in the right. The weights record how much of each eigenfunction ψ contains, and nothing about where the particle sits. Unit 4 reads |cₙ|² as the probability that a measurement of A returns aₙ (UQMI-04.03), and adds as a postulate that the result aₙ leaves the system described by ψₙ (UQMI-04.04); whether anything physically collapses is an interpretive question the course flags without settling. Here Σ |cₙ|² = 1 is simply a theorem about integrals.
5Complete is a property of the whole set
Orthonormality can be checked pair by pair; completeness cannot, because it claims nothing is missing. For any orthonormal set, ∫ |ψ − SN|² dx = ∫ |ψ|² dx − Σ |cₙ|² over n ≤ N can never be negative, which gives Bessel's inequality, Σ |cₙ|² ≤ ∫ |ψ|² dx. The set is complete when equality holds for every square-integrable ψ, or equivalently when the only function orthogonal to every ψₙ is zero. Remove ψ₂ from the box sines and what remains is still orthonormal, yet ψ₂ itself has every coefficient zero: Σ |cₙ|² = 0 against a norm of 1. A subtler hole is a missing partner in a degenerate level. On a ring of circumference L, the normalised constant and cosines cos(2πnx/L) are orthonormal eigenfunctions of the free Ĥ, but sin(2πx/L) is orthogonal to all of them, so each level's sine must be included. For the Hamiltonians this course solves, Sturm–Liouville and spectral theory supply completeness; here it is taken on trust and tested through Parseval. The classical ancestor is the plucked string, expanded on the same sines as the box, whose energy is a sum over modes with no cross terms.
6Mean square, not point by point
Integrate |ψ − Σ bₙψₙ|² for any trial coefficients bₙ with n ≤ N and the result is ∫ |ψ|² dx − Σ |cₙ|² + Σ |bₙ − cₙ|². The Fourier coefficients are therefore the best choice for every N, and the error they leave is the one completeness drives to zero. For the flat state ψ = 1/√a in a box, cₙ = 2√2/(nπ) for odd n and zero for even n. The errors after N = 1, 3 and 9 are 0.189, 0.099 and 0.040, and at large N the error falls as 4/(π²N). Point by point the story differs. Every ψₙ vanishes at the walls, so SN(0) = 0 for every N while ψ(0⁺) = 1/√a. Beside the wall the partial sums overshoot to about 1.18/√a, and more terms squeeze the overshoot towards the wall without lowering it. That is the Gibbs phenomenon; at a jump inside the region the sums converge to the midpoint of the jump. None of this shows in the error, because a strip of vanishing width carries vanishing area. So the equals sign in ψ = Σ cₙψₙ means equality in mean square: functions differing only at isolated points give the same value for every integral, and describe the same state.
7An operator acts term by term, inside its domain
For a finite sum, linearity gives Â Σ cₙψₙ = Σ aₙcₙψₙ at once: on its own eigenbasis an operator only rescales coefficients. For an infinite sum the same line holds exactly when Σ |aₙcₙ|² converges, which is the statement that Âψ is square-integrable. The domain on which  was proved Hermitian (UQMI-03.05) therefore reappears as a rate of decay of the cₙ, and smoothness sets that rate. In a box, a jump in ψ, including a nonzero value at a wall, gives cₙ ∝ 1/n, a corner gives 1/n², and a ψ smooth inside that vanishes at both walls gives 1/n³ or faster. Since Eₙ ∝ n², Ĥψ exists only in the last case. The flat state shows what is at stake. Inside the box −(ħ²/2m)ψ″ = 0, so a careless ∫ ψ*Ĥψ dx gives zero, yet Σ Eₙ|cₙ|² over odd n ≤ N is (8/π²)E₁ times the number of terms: 4.05E₁ at N = 9 and 40.5E₁ at N = 99, growing without bound. The two disagree because ψ does not vanish at the walls, so it lies outside the domain where Ĥ is Hermitian and term-by-term action is licensed. A ψ with a jump costs unbounded kinetic energy.
8When the spectrum is continuous, the sum becomes an integral
On the whole line, p̂ = −iħ d/dx has an eigenfunction e(ipx/ħ)/√(2πħ) for every real p, and none can be normalised: |e(ipx/ħ)|² = 1 at every x, so its norm integral diverges. The label n becomes the continuous p, the sum becomes an integral, and Fourier's trick becomes the Fourier transform φ(p) = (2πħ)⁻¹⁄² ∫ e(−ipx/ħ) ψ(x) dx, with ψ(x) = (2πħ)⁻¹⁄² ∫ e(ipx/ħ) φ(p) dp and ∫ |φ|² dp = ∫ |ψ|² dx in place of Parseval. Orthonormality becomes δ-normalisation, handled in UQMI-05.03, and the transform is put to work in UQMI-05.05. A limiting case joins the two pictures. On a ring the momenta 2πnħ/L are discrete, spaced by 2πħ/L; let L grow and the spacing closes, the sum over n turns into (L/2πħ) ∫ dp, and the ring's expansion becomes the line's. For ψ ∝ e(−x²/(2b²)) with b = 0.10 nm, the transform is φ(p) ∝ e(−p²b²/(2ħ²)), a Gaussian whose width in the same sense is ħ/b = 1.05 × 10⁻²⁴ kg m s⁻¹: narrowing ψ widens φ, as any Fourier pair must.
Key equations
What each relation says, and the units it is written in.
Expansion on a complete eigenbasis
ψ(x) = Σₙ cₙ ψₙ(x), with Âψₙ = aₙψₙ
Completeness is the claim that such cₙ exist for every square-integrable ψ, even one outside Â's domain. The course asserts it rather than proving it.
ψ and ψₙ in m⁻¹⁄² in one dimension, cₙ dimensionless and in general complex. Every member of a degenerate level is included.
Fourier's trick
cₙ = ∫ ψₙ*(x) ψ(x) dx
Multiply the expansion by ψₘ* and integrate: orthonormality kills every term except n = m. One integral per coefficient, no simultaneous equations.
The conjugate sits on the basis function. Units m⁻¹⁄² × m⁻¹⁄² × m, so cₙ is a pure number.
Parseval: the norm moves onto the coefficients
Σₙ |cₙ|² = ∫ |ψ|² dx
Normalise from the coefficients instead of an integral. A total below 1 for a normalised ψ means the set is missing members.
Both sides dimensionless, and equal to 1 for a normalised ψ. The relative phases of the cₙ drop out of the sum.
Mean-square error of a truncated sum
∫ |ψ − SN|² dx = ∫ |ψ|² dx − Σ |cₙ|², summed over n ≤ N
Completeness means exactly that this tends to zero. Being an integral, it says nothing about SN at a single point: a wall or a jump can be missed for ever.
SN = Σ cₙψₙ over n ≤ N. Dimensionless. No other choice of the first N coefficients leaves a smaller error.
An operator acts term by term
Âψ = Σₙ aₙ cₙ ψₙ, valid when Σ |aₙ cₙ|² converges
On its own eigenbasis  only rescales coefficients. In a box, Ĥψ is square-integrable only if |cₙ| falls faster than n⁻⁵⁄².
aₙ carries the units of Â. The convergence condition is Â's domain, read off the coefficients.
Continuous spectrum: the sum becomes an integral
φ(p) = (2πħ)⁻¹⁄² ∫ e(−ipx/ħ) ψ(x) dx
Fourier's trick with the sum over n replaced by an integral over p, and Σ |cₙ|² by ∫ |φ|² dp. Named here; UQMI-05.05 puts it to work.
p in kg m s⁻¹, φ in (kg m s⁻¹)⁻¹⁄². Each e(ipx/ħ)/√(2πħ) is a p̂ eigenfunction that cannot be normalised.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Set x₀ = 0.50a: the even-n bars vanish. Raise N to 12 and the error falls to 0.051, yet SN still overshoots beside each jump and the dot sits at 0.49 of ψ's height, the jump's midpoint. EN|cN|²/E₁ never dies away: 2.64 at N = 3, still 2.15 at N = 11, so the energy sum diverges.
Live results
- GROUND WEIGHT |c₁|²
- 0.504
- LAST WEIGHT KEPT |cN|²
- 0.1012
- ENERGY TERM EN|cN|²/E₁
- 0.911
- HEIGHT OF ψ
- 1.83 a⁻¹⁄²
Catch the common trap
The mistake this lesson is built to prevent.
Students read ψ = Σ cₙψₙ as an identity at every point. Summing the box sines for the flat state ψ = 1/√a, they find SN(0) = 0 at every N and conclude either that the sines cannot represent a function that does not vanish at the walls, or that enough terms will eventually close the gap. Both conclusions are wrong. The first fails Parseval's test: with cₙ = 2√2/(nπ) for odd n, Σ |cₙ|² = (8/π²)(1 + 1/9 + 1/25 + …) = (8/π²)(π²/8) = 1, so nothing is missing from the set. The second fails at the wall itself: every ψₙ vanishes at x = 0, so no finite sum is ever nonzero there. What completeness promises is ∫ |ψ − SN|² dx → 0, and a wall value, or an overshoot squeezed into a strip that narrows as N grows, encloses no area. Since every quantity the physics computes from ψ is an integral, the expansion is exact in the only sense that matters.
Test it: what changes?
- Mean-square error never increases
- Coefficients already kept do not change
- Overshoot beside each jump persists
- Height 1/√w of ψ rises
- Ground weight |c₁|² falls
- Even-n weights stay zero
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron on a ring of circumference L = 1.2 nm has the p̂ eigenbasis uₙ = e(2πinx/L)/√L, n = 0, ±1, ±2, …, with eigenvalues pₙ = 2πnħ/L. It is prepared in ψ = A[2cos(2πx/L) + i sin(4πx/L)]. Expand ψ on the uₙ, fix A from the coefficients rather than by integrating |ψ|², confirm one coefficient by Fourier's trick, apply p̂ term by term, and list the momenta present.
View solution →
- Rewrite with Euler's formula: 2cos(2πx/L) = e(2πix/L) + e(−2πix/L), and i sin(4πx/L) = ½[e(4πix/L) − e(−4πix/L)]. Each exponential is √L times one uₙ, so ψ = A√L[u₁ + u₋₁ + ½u₂ − ½u₋₂].
- Parseval: Σ |cₙ|² = A²L(1 + 1 + ¼ + ¼) = (5/2)A²L = 1, so A = √(2/(5L)) = 0.577 nm⁻¹⁄². The coefficients are c₁ = c₋₁ = √(2/5) = 0.632 and c₂ = −c₋₂ = 1/√10 = 0.316, pure numbers as they must be.
- Fourier's trick for c₋₂, conjugate on the basis function: c₋₂ = ∫₀ᴸ (e(4πix/L)/√L) ψ dx. Over a full lap only the e(−4πix/L) part of ψ survives, giving (A/√L)(−½)L = −½A√L = −0.316. Dropping the conjugate would pick out the e(4πix/L) part and return +0.316, the coefficient of the opposite momentum.
- Direct check: the two terms of ψ are real and imaginary respectively, so |ψ|² = A²[4cos²(2πx/L) + sin²(4πx/L)], whose integral round the ring is A²(2L + L/2) = (5/2)A²L, just what Parseval gave.
- Term by term, p̂ψ = Σ pₙcₙuₙ = (2πħ/L)A√L[u₁ − u₋₁ + u₂ + u₋₂] = (2πħ/L)A[2i sin(2πx/L) + 2cos(4πx/L)]. Differentiating directly, −iħ dψ/dx gives the same function, as it must for a finite sum.
- The momenta present are ±2πħ/L = ±5.52 × 10⁻²⁵ kg m s⁻¹ and ±4πħ/L = ±1.10 × 10⁻²⁴ kg m s⁻¹, with weights 0.40, 0.40, 0.10 and 0.10. Limiting case: drop the sine term and ψ becomes the standing wave √(2/L) cos(2πx/L), an equal mix of ±2πħ/L, just as a standing wave on a hoop is two waves running opposite ways.
Answer. A = √(2/(5L)) = 0.577 nm⁻¹⁄²; c₁ = c₋₁ = 0.632, c₂ = 0.316, c₋₂ = −0.316, weights summing to 1. Momenta ±5.52 × 10⁻²⁵ and ±1.10 × 10⁻²⁴ kg m s⁻¹.
MediumExample 2
Take as supplied data the box eigenbasis ψₙ = √(2/a) sin(nπx/a) on 0 < x < a, with a = 0.80 nm. The ramp ψ = √(3/a³) x is normalised on 0 < x < a but jumps to zero at x = a. Find cₙ by Fourier's trick, confirm Σ |cₙ|² = 1 using Σ 1/n² = π²/6, find the mean-square error of SN for N = 1, 3 and 10, and compare SN with ψ just inside x = a.
View solution →
- cₙ = √(2/a)√(3/a³) ∫₀ᵃ x sin(nπx/a) dx. Integrating by parts, the boundary term is −(a²/(nπ)) cos(nπ) = (−1)(n+1) a²/(nπ), and what remains is an integral of cos(nπx/a) over whole half-periods, which is zero.
- So cₙ = (√6/a²)(−1)(n+1) a²/(nπ) = (−1)(n+1)√6/(nπ): c₁ = 0.780, c₂ = −0.390, c₃ = 0.260. The width cancels, as it must: cₙ is dimensionless, and a only sets the scale of x.
- Parseval: Σ |cₙ|² = (6/π²) Σ 1/n² = (6/π²)(π²/6) = 1. The sines pass the completeness test for this ψ even though ψ does not vanish at x = a.
- Mean-square error, 1 − (6/π²) Σ 1/n² over n ≤ N. N = 1: 1 − 0.608 = 0.392. N = 3: 1 − 0.608 × 1.361 = 0.173. N = 10: 1 − 0.608 × 1.550 = 0.058. For large N the error is close to 6/(π²N), so pushing it below 0.01 takes about 60 terms.
- Point by point: every ψₙ vanishes at x = a, so SN(a) = 0 at every N, while ψ just inside the wall is √(3/a) = 1.94 nm⁻¹⁄². Beside the wall SN overshoots to about 1.18 times that before plunging to zero, and more terms only narrow the overshoot. Mean-square convergence holds; pointwise convergence at the wall never does.
- Domain: cₙ ∝ 1/n is the signature of the jump, and Σ Eₙ|cₙ|² ∝ Σ n² × (1/n²) diverges. The ramp is square-integrable and fully expandable, but it lies outside the domain of Ĥ and has no finite energy.
Answer. cₙ = (−1)(n+1)√6/(nπ): 0.780, −0.390, 0.260, …, with Σ |cₙ|² = 1. Errors 0.392, 0.173 and 0.058 for N = 1, 3 and 10, approaching 6/(π²N) at large N, while SN(a) = 0 at every N against ψ = 1.94 nm⁻¹⁄² just inside the wall.
HardExample 3
An electron is in a box of width a = 0.50 nm, with the supplied eigenbasis ψₙ = √(2/a) sin(nπx/a) and Eₙ = n²E₁. It is prepared in ψ = √(2/a)|sin(2πx/a)|, which has exactly the probability density of ψ₂. Find its coefficients, check Parseval to four non-zero terms, compare Σ Eₙ|cₙ|² with E₂, and decide whether Ĥ may be applied to this ψ term by term.
View solution →
- Symmetry first. ψ is even about the centre x = a/2, while every ψₙ with even n is odd about it, so every even-n coefficient vanishes, c₂ included: ∫ ψ₂ψ dx = (2/a) ∫ sin(2πx/a)|sin(2πx/a)| dx = 0, positive over the left half and negative over the right. Same density as ψ₂, zero overlap with it.
- For odd n, double the left half and use product-to-sum: cₙ = (4/a) ∫₀(a/2) sin(2πx/a) sin(nπx/a) dx = 8 sin(nπ/2)/(π(4 − n²)). So c₁ = 8/(3π) = 0.849, c₃ = 8/(5π) = 0.509, c₅ = −8/(21π) = −0.121 and c₇ = 8/(45π) = 0.0566.
- Parseval: the weights are 0.7205, 0.2594, 0.0147 and 0.0032, totalling 0.9978, and the higher terms supply the last 0.002. The ground state carries only 72%: the corner at x = a/2 needs ψ₃ and above.
- Energy. With E₁ = π²ħ²/(2ma²) = 1.504 eV, the direct integral (ħ²/2m) ∫ |ψ′|² dx is 4E₁ = 6.02 eV, the same as E₂, because |ψ′| = |ψ₂′| everywhere but one point. From the coefficients, Σ Eₙ|cₙ|²/E₁ = 0.72 + 2.33 + 0.37 + 0.16 + … reaches 3.67 at n = 9, 3.85 at n = 21 and 3.97 at n = 101, creeping towards 4 because its terms fall only as 1/n².
- Term by term: Eₙcₙ/E₁ = 8n² sin(nπ/2)/(π(4 − n²)) tends to ±8/π = ±2.55 instead of decaying, so Σ |Eₙcₙ|² diverges and Ĥψ is not square-integrable. Directly, ψ′ jumps by (4π/a)√(2/a) = 50.3 nm⁻³⁄² at x = a/2, so ψ″ holds a δ function there: ψ is outside the domain of Ĥ, and Σ Eₙcₙψₙ converges to no function.
- The energy stays finite because that delta sits where ψ(a/2) = 0 and adds nothing to ∫ ψ*Ĥψ dx. The ladder reads: Σ |cₙ|² converges, so ψ is a state; Σ Eₙ|cₙ|² converges, so its energy is finite; Σ Eₙ²|cₙ|² diverges, so Ĥψ does not exist.
Answer. cₙ = 8 sin(nπ/2)/(π(4 − n²)) for odd n, zero for even n: 0.849, 0.509, −0.121, 0.057, weights totalling 0.998. Σ Eₙ|cₙ|² → 4E₁ = 6.02 eV = E₂ though c₂ = 0, and Ĥ cannot act term by term.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Fourier's trick finds each cₙ; Parseval checks nothing is missing.
Take cₙ = ∫ψₙ*ψ dx with the conjugate on the basis function, and check Σ|cₙ|² = ∫|ψ|² dx. The sum converges in mean square, so walls and jumps may be missed point by point. Apply  term by term only when Σ|aₙcₙ|² converges.
Next: Expectation values, variance, and standard deviation →