Complete Orthonormal Bases

Abstract vectors are beautiful and useless. To compute anything — a probability, an overlap, a time evolution — you must turn a state into numbers, and the machine that does it is a complete orthonormal basis: a set of mutually perpendicular unit vectors that reaches every direction of Hilbert space. Read a state against such a set and it becomes a column of components; read an operator against it and it becomes a matrix. Quantum mechanics turns into linear algebra you can put on a computer. This article pays two debts the corpus has been carrying. Stationary States and Separation of Variables expanded every initial condition as $\sum_n c_n\psi_n$ and promised that the guarantee behind the move — completeness — lived in the Formalism chapter. And both Inner Product and Norm of States and Hermitian Operators and Observables deferred the manufacturing question: given raw, non-orthogonal vectors, how do you build an orthonormal set? Here you get the expansion theorem with its coefficient formula $c_n = \langle n|\psi\rangle$ , the Gram–Schmidt algorithm that manufactures orthonormality to order, and Bessel's inequality — the accounting rule that tells you, term by term, how much of a state your basis has captured, with equality (Parseval) as the certificate of completeness. The payoff generalizes far past this chapter: every representation you will use — position, momentum, energy, angular momentum — is a choice of complete orthonormal set, and "change of representation" is re-reading one vector against new axes. Learn the machinery once, abstractly, and every concrete basis in the subject arrives pre-assembled.

Key takeaways

An orthonormal set satisfies $\langle m|n\rangle = \delta_{mn}$ : mutually perpendicular, unit length. Complete means it spans the space — no state is left outside, so expansions miss nothing.
The expansion theorem: every state is $|\psi\rangle = \sum_n c_n|n\rangle$ with coefficients $c_n = \langle n|\psi\rangle$ , and the expansion is unique. This is the abstract engine behind the Fourier trick of Infinite Square Well.
Gram–Schmidt manufactures orthonormal sets from raw vectors: subtract the projections onto what you already have, normalize, repeat. Applied to $\{1, x, x^2\}$ it produces the Legendre family.
Bessel's inequality $\sum_n |c_n|^2 \leq \langle\psi|\psi\rangle$ holds for any orthonormal set; equality for every state — Parseval — is the exact criterion of completeness.
For a normalized state the captured norm $\sum |c_n|^2$ climbs monotonically toward 1: a flat state in the box basis reaches 81% with one term, 96% with five, and needs terms through $n = 41$ for 99%.
Key idea: a complete orthonormal basis converts abstract states into unique coefficient lists $c_n = \langle n|\psi\rangle$ — orthonormality makes the components clean, completeness makes them the whole story, and Parseval's equality is the test that certifies it.

The core insight

A complete orthonormal basis converts every state into a unique list of components $c_n = \langle n|\psi\rangle$ : orthonormality extracts each coefficient without contamination, completeness guarantees nothing is left over, and Bessel-to-Parseval accounting measures exactly how much of the state any partial basis has captured.

Where this lives today

"Basis-set convergence" is a line item in every computational-chemistry paper: molecular energies are computed in finite orthonormal-ish basis sets (STO-3G up to cc-pV5Z ladders), and the march toward the complete-basis-set limit is precisely a Bessel staircase — practitioners extrapolate the deficit just as Example 2's audit does. Plane-wave density-functional codes expose the same physics as a single dial, the energy cutoff, which is nothing but "how far up the momentum basis do we truncate." Quantum computing reads this article twice. State tomography reconstructs an unknown $|\psi\rangle$ by measuring components against complete sets of bases; and every quantum algorithm's promise of speed hinges on which basis a state is sparse in — the quantum Fourier transform is a change of complete orthonormal basis executed in hardware. Meanwhile tensor-network methods compress many-body states by finding, via singular value decompositions (Gram–Schmidt's industrial descendant), the small orthonormal set that captures 99.99% of a state's norm — Bessel's deficit, minimized by machine, a million times a second.

The physical picture

Stand a vector in ordinary space next to three perpendicular unit axes. Its shadows on the axes — three numbers — describe it completely: perpendicularity means each shadow is measured without disturbing the others, and having all three axes means nothing escapes the description. A complete orthonormal basis is this picture promoted to Hilbert space, where the axes may be infinite in number and each is itself a quantum state.

Figure

Components are shadows — orthonormality keeps them clean, completeness catches them all

Two independent properties do the work, and it pays to keep them separate. Orthonormality is clean bookkeeping: $\langle m|n\rangle = \delta_{mn}$ means projecting onto one axis extracts that component and nothing else. Completeness is coverage: the axes span the whole space, so the component list is the entire state. Either can hold without the other — one unit vector in a plane is orthonormal but incomplete; three coplanar unit vectors at 120° cover the plane but over-count it. Only the combination gives the two-way dictionary between states and coefficient lists. The physics reading is the one Stationary States and Separation of Variables ran on credit: a Hamiltonian's eigenfunctions form exactly such a set (supplied by Hermitian Operators and Observables, its degenerate eigenspaces organized by this article's Gram–Schmidt), so any initial wavefunction expands over stationary states and the whole future is a list of rotating phases. Choosing a basis is choosing a language — position for potentials, momentum for kinetic energy, energy for time; the state is the language-independent thing the coefficients describe.

Orthonormality and completeness, stated precisely

The two defining properties of the sets this article studies:

\langle m|n\rangle = \delta_{mn} \qquad (1)

where: $\{|n\rangle\}$ — the candidate basis vectors; $\delta_{mn}$ — the Kronecker delta, $1$ for $m = n$ , else $0$ . The diagonal says each vector is normalized; the off-diagonal says they are mutually orthogonal — in Inner Product and Norm of States' language, perfectly distinguishable in one shot. Completeness is the coverage clause: the set is complete if every state in the Hilbert space can be written as a (possibly infinite) superposition of its members — no vector is orthogonal to all of them, nothing hides outside the span. For infinite sums, "can be written" means the partial sums converge in norm to the state, the convergence sense the completeness axiom of Hilbert Space and State Vectors was built to guarantee. How completeness upgrades to a statement about operators — the resolution of the identity $\sum_n |n\rangle\langle n| = \hat 1$ — is the next serial's business, Completeness Relation and Resolution of Identity; this article works at the level of states and coefficients, where the theorems live.

Derivation AThe expansion theorem — extracting coefficients and proving them unique

We derive the master formula of representation theory: if $\{|n\rangle\}$ is a complete orthonormal set, every state is $|\psi\rangle = \sum_n c_n|n\rangle$ with $c_n = \langle n|\psi\rangle$ , and no other coefficient list works. The strategy is to assume an expansion exists (that is what completeness grants) and let orthonormality interrogate it one axis at a time; uniqueness then falls out of the same move applied to the difference of two candidate expansions. Watch how each property carries exactly one load — completeness grants existence, orthonormality grants the formula and the uniqueness.

Step 1. Existence, from completeness. The set spans the space, so some coefficients $c_n$ exist with

|\psi\rangle = \sum_n c_n\,|n\rangle

Why this is not trivial: for infinite sets "spans" means the partial sums converge in norm — the guarantee purchased by the completeness axiom of the Hilbert space itself.

Step 2. Extraction, from orthonormality. Take the inner product of both sides with a chosen basis bra $\langle m|$ :

\langle m|\psi\rangle = \sum_n c_n\,\langle m|n\rangle = \sum_n c_n\,\delta_{mn} = c_m

The Kronecker delta kills every term but one: the coefficient along $|m\rangle$ is just the overlap $\langle m|\psi\rangle$ . Why this is the abstract Fourier trick: replace kets by box eigenfunctions and the inner product by its $L^2$ integral, and this line becomes the coefficient integral performed concretely in Infinite Square Well.

Step 3. Uniqueness. Suppose two lists both work: $|\psi\rangle = \sum_n c_n|n\rangle = \sum_n c_n'|n\rangle$ . Subtract and project with $\langle m|$ :

0 = \Big\langle m\Big|\sum_n (c_n - c_n')\,\Big|n\Big\rangle = c_m - c_m'

for every $m$ . The lists are identical: a state has exactly one address in a given basis. Why physics needs this: if two coefficient lists could describe one state, probabilities read from coefficients would be ambiguous — the Born-rule machinery two serials ahead leans on this uniqueness silently. *Verified: numerically on a random complex 4-dimensional state against a random unitary basis — the extracted coefficients rebuild the state to $10^{-12}$ , and re-extracting from the rebuilt state returns the identical list.*

With the expansion theorem in hand, inner products and norms become component arithmetic. For two states with components $b_n = \langle n|\phi\rangle$ and $c_n = \langle n|\psi\rangle$ :

\langle\phi|\psi\rangle = \sum_n b_n^*\,c_n, \qquad \langle\psi|\psi\rangle = \sum_n |c_n|^2 \qquad (2)

where: the star implements the bra side's conjugation (Dirac Bra-Ket Notation's antilinearity). The second equality is Parseval's identity — length survives being read against any complete orthonormal frame. Hilbert Space and State Vectors proved this as "Pythagoras in Hilbert space" and exhibited it on its tent state; what this article adds below is the incomplete case, where equality degrades to an inequality that measures what the basis misses.

Manufacturing bases: Gram–Schmidt

Nature rarely hands you orthonormal vectors. Degenerate eigenspaces (Hermitian Operators and Observables) arrive as unorganized frames; approximation methods start from convenient trial functions. Both Inner Product and Norm of States and the hermiticity article deferred the systematic fix to this serial. Here it is.

Go deeper · Derivation B · 1 minGram–Schmidt — subtract the shadows, normalize, repeat

Derivation BGram–Schmidt — subtract the shadows, normalize, repeat

We build, from any linearly independent list $|v_1\rangle, |v_2\rangle, \dots$ , an orthonormal list $|e_1\rangle, |e_2\rangle, \dots$ spanning the same space at every stage. The idea is geometric and single-minded: each new vector is guilty of overlapping the axes already built, so subtract those shadows (the projections), keep the perpendicular remainder, and scale it to unit length. We prove the remainder is orthogonal to everything before it and can never vanish — the two facts that make the algorithm an algorithm rather than a hope.

Step 1. Initialize. Normalize the first vector:

|e_1\rangle = \frac{|v_1\rangle}{\sqrt{\langle v_1|v_1\rangle}}

Step 2. Induct. Given orthonormal $|e_1\rangle, \dots, |e_{k-1}\rangle$ spanning the same space as $|v_1\rangle, \dots, |v_{k-1}\rangle$ , form the remainder of the next raw vector:

|w_k\rangle = |v_k\rangle - \sum_{j=1}^{k-1} \langle e_j|v_k\rangle\,|e_j\rangle

Why each subtracted term is a shadow: $\langle e_j|v_k\rangle|e_j\rangle$ is precisely the component of $|v_k\rangle$ along the finished axis $|e_j\rangle$ — the expansion theorem, used as a scalpel.

Step 3. Check orthogonality. Project the remainder onto any finished axis $|e_i\rangle$ ( $i < k$ ):

\langle e_i|w_k\rangle = \langle e_i|v_k\rangle - \sum_{j=1}^{k-1} \langle e_j|v_k\rangle\,\langle e_i|e_j\rangle = \langle e_i|v_k\rangle - \langle e_i|v_k\rangle = 0

Orthonormality of the finished axes collapses the sum to the single $j = i$ term, which cancels the first term exactly. The remainder is perpendicular to the whole finished set.

Step 4. Check survival, then normalize. If $|w_k\rangle$ were zero, $|v_k\rangle$ would be a combination of $|e_1\rangle, \dots, |e_{k-1}\rangle$ — i.e. of $|v_1\rangle, \dots, |v_{k-1}\rangle$ — contradicting linear independence. So $\langle w_k|w_k\rangle > 0$ and $|e_k\rangle = |w_k\rangle/\sqrt{\langle w_k|w_k\rangle}$ is legal. Each stage spans the same space as the raw vectors it consumed, so the finished set inherits the raw set's coverage: feed it a complete raw family and it returns a complete orthonormal basis. Why this closes the degenerate-eigenspace loophole: inside an eigenspace every vector is an eigenvector, and Gram–Schmidt never leaves the span — the organized frame it returns is still made of eigenvectors. *Verified: sympy runs the algorithm on $\{1, x, x^2\}$ over $L^2[-1,1]$ and returns exactly $1/\sqrt2$ , $\sqrt{3/2}\,x$ , $\sqrt{45/8}\,(x^2 - \tfrac13)$ with the Gram matrix equal to the identity; each output matches the normalized Legendre polynomial $\sqrt{(2n+1)/2}\,P_n(x)$ .*

The accounting rule: Bessel, then Parseval

How much of a state has your basis captured after $N$ terms? There is an exact ledger.

Go deeper · Derivation C · 1 minBessel's inequality — and Parseval as the certificate of completeness

Derivation CBessel's inequality — and Parseval as the certificate of completeness

We prove that for any orthonormal set — complete or not — the squared components never overdraw the state: $\sum_n |c_n|^2 \leq \langle\psi|\psi\rangle$ , with the deficit equal to the squared length of what the set misses. The strategy is to name the miss: subtract from $|\psi\rangle$ its shadow on the span, observe that the remainder is orthogonal to the shadow, and let Pythagoras do the accounting. Completeness is then readable off the ledger — the set is complete precisely when the deficit vanishes for every state, which is Parseval's equality promoted to a test.

Step 1. Split the state against the (possibly partial) set $\{|n\rangle\}_{n \leq N}$ :

|\psi\rangle = \underbrace{\sum_{n=1}^{N} c_n|n\rangle}_{|\psi_\parallel\rangle} + |r\rangle, \qquad c_n = \langle n|\psi\rangle

Step 2. Show the remainder is orthogonal to every axis used:

\langle m|r\rangle = \langle m|\psi\rangle - \sum_{n=1}^{N} c_n\,\delta_{mn} = c_m - c_m = 0 \quad (m \leq N)

Why this is Gram–Schmidt's Step 3 in new clothing: subtracting the shadow always leaves a perpendicular remainder — the same geometry powers both derivations.

Step 3. Apply Pythagoras (the norm additivity of orthogonal pieces, from Inner Product and Norm of States):

\langle\psi|\psi\rangle = \sum_{n=1}^{N} |c_n|^2 + \langle r|r\rangle \;\geq\; \sum_{n=1}^{N} |c_n|^2

Bessel's inequality, with the deficit identified: $\langle r|r\rangle$ is exactly the squared norm your truncated basis fails to reach. The partial sums increase monotonically (each new term adds $|c_n|^2 \geq 0$ ) and are bounded above — so they always converge, whether or not the set is complete.

Step 4. Read off the completeness test. If the set is complete, the remainder can be driven to zero in norm and $\sum_n |c_n|^2 = \langle\psi|\psi\rangle$ — Parseval. Conversely, if equality holds for every state, no state has a nonzero perpendicular remainder, so nothing lies outside the span: the set is complete. Equality-for-all is not a consequence of completeness but equivalent to it — the working criterion used to certify basis sets in practice. Verified: numerically — partial sums for a random $\mathbb{C}^4$ state against a random orthonormal set climb monotonically, stay below the squared norm, and hit it exactly at four of four terms; a deliberately incomplete two-vector set stalls strictly below. The box-basis exhibit of the worked example reproduces the same staircase against an infinite basis.

Figure

The Bessel staircase — a flat state's norm captured term by term in the box basis

Limiting cases

case	result	meaning
complete set, any state	$\sum_n \lvert c_n\rvert^2 = \langle\psi\vert\psi\rangle$	Parseval: the coefficient list is the whole state
incomplete set	$\sum_n \lvert c_n\rvert^2 < \langle\psi\vert\psi\rangle$ for some state	Bessel deficit = squared norm of the missed remainder
finite dimension $N$	any $N$ orthonormal vectors are complete	counting suffices; convergence questions vanish
degenerate eigenspaces	orthonormal frame per eigenspace via Gram–Schmidt	the supply of eigenbases stays organized (Hermitian Operators and Observables)
continuous spectrum	sums → integrals, $\delta_{mn} \to \delta(x - x')$	delta-normalized "bases"; owned by Continuous Spectra and Delta Normalization
mixed spectrum	discrete sum + continuum integral together	e.g. finite well: bound states alone are not complete

Worked examples

Example 1 — manufacturing the Legendre basis (confidence builder). Orthonormalize the raw list $\{1, x, x^2\}$ on $L^2[-1, 1]$ (real inner product $\langle f|g\rangle = \int_{-1}^{1} f g\,\mathrm{d}x$ ).

Step 1. First axis. $\langle 1|1\rangle = 2$ , so $e_0 = 1/\sqrt{2}$ .

Step 2. Second axis. The shadow of $x$ on $e_0$ is $\langle e_0|x\rangle e_0 = 0$ by parity — odd against even. The remainder is $x$ itself; normalizing with $\langle x|x\rangle = 2/3$ gives $e_1 = \sqrt{3/2}\,x$ .

Step 3. Third axis. Shadows of $x^2$ : on $e_0$ , $\langle e_0|x^2\rangle e_0 = \tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt2\,}{3}\cdot\tfrac{1}{\sqrt2}$ — cleaner via the raw overlap $\langle 1|x^2\rangle = 2/3$ , giving remainder $w = x^2 - \tfrac13$ ; on $e_1$ the shadow vanishes by parity. Normalizing with $\langle w|w\rangle = 8/45$ :

e_2 = \sqrt{\tfrac{45}{8}}\,\Big(x^2 - \tfrac13\Big)

Step 4. Recognize the family. Each output is the normalized Legendre polynomial $\sqrt{(2n+1)/2}\,P_n(x)$ : Gram–Schmidt, fed powers of $x$ , manufactures the Legendre basis — the same family that will organize angular problems later in the track. Orthonormality checks exactly: the $3 \times 3$ Gram matrix is the identity.

Example 2 — the flat state in the box, audited (exam level). An electron in an infinite square well of width $L = 1$ nm ( $E_1 = 0.376$ eV) is prepared in the flat state $\psi = 1/\sqrt{L}$ on $(0, L)$ . Expand it over the box eigenbasis $\phi_n = \sqrt{2/L}\sin(n\pi x/L)$ and audit the expansion.

Step 1. Coefficients. $c_n = \int_0^L \phi_n\psi\,\mathrm{d}x = \tfrac{\sqrt2}{n\pi}(1 - \cos n\pi)$ : zero for even $n$ (the flat state is symmetric about the center; even- $n$ modes are odd there), and

c_n = \frac{2\sqrt2}{n\pi} \quad (n \text{ odd}) \qquad\Rightarrow\qquad |c_n|^2 = \frac{8}{\pi^2 n^2}

Step 2. Certify with Parseval. $\sum_{\text{odd}} 8/\pi^2 n^2 = (8/\pi^2)(\pi^2/8) = \textbf{1}$ exactly — the basis is complete and the state is fully captured. If an energy measurement follows, the probabilities are the $|c_n|^2$ : $P(E_1) = 8/\pi^2 = \textbf{81.1\%}$ , $P(E_3) = \textbf{9.0\%}$ , $P(E_5) = \textbf{3.2\%}$ (the probability reading itself is Born Rule and Measurement Probabilities' law, two serials ahead).

Step 3. Audit the truncation — the Bessel staircase. Captured norm: 0.8106 ( $n \leq 1$ ), 0.9006 ( $n \leq 3$ ), 0.9331 ( $n \leq 5$ ), 0.9496 ( $n \leq 7$ ), 0.9596 ( $n \leq 9$ ); reaching 99% takes every odd term through $n = \textbf{41}$ . The staircase climbs monotonically, as Bessel demands — but slowly, because rebuilding the jumps at the walls from sines that vanish there is expensive.

Step 4. One honest flag. Norm convergence is not everything: each odd term contributes the same $8/\pi^2\,E_1 \approx 0.81\,E_1$ to the energy sum $\sum |c_n|^2 E_n$ (50 terms in: $\approx 40.5\,E_1$ , still climbing), which diverges — the flat state has no finite mean energy. Norms forgive wall discontinuities; energies do not. The tent state of Hilbert Space and State Vectors showed the mild version (slow convergence from a kink); the full story of which functions are physically admissible at all is Admissibility Conditions on Wavefunctions'. (Every number in both examples — the Legendre normalizations, the Gram matrix, $c_n$ , the exact Parseval sum, the staircase values, $n = 41$ , the linear energy growth — is recomputed in this article's verification script.)

Origins · 1 minFriedrich Wilhelm Bessel and how this was discovered

Friedrich Wilhelm Bessel

1784–1846 · Königsberg Observatory

The inequality that audits every quantum basis was an astronomer's bookkeeping rule. Bessel — a self-taught calculator who left a merchant's apprenticeship at 20 after recomputing the orbit of Halley's comet, and who ran the Königsberg observatory from 25 — spent his life extracting maximal information from imperfect data: he was the first human to measure the distance to a star (61 Cygni, 1838, by parallax of a third of an arcsecond). Decomposing periodic astronomical records into trigonometric components, he proved in 1828 the accounting rule now bearing his name: the squared components extracted against an orthogonal family can never exceed the total — measurement against clean axes never overdraws the data. He could not have known the rule would become quantum mechanics' completeness auditor, or that equality — proved for general orthonormal systems by Parseval's heirs and enshrined in the Riesz–Fischer theorem the Hilbert Space and State Vectors portrait celebrates — would certify the eigenbases of atoms. His functions, meanwhile, ride along every cylindrical quantum problem. An astronomer who never wrote a word of quantum theory supplied its most-used inequality after Cauchy–Schwarz.

History

1807

— Fourier claims arbitrary functions expand over sinusoids; the coefficient-extraction integral is born with the claim.

1828

— Bessel proves his inequality for trigonometric components of astronomical data: extracted energy never exceeds the total.

1896

— Parseval's name attaches to the equality case, a century after his formal memoir on series.

1907

— Riesz and Fischer prove L² complete: coefficient lists and states are in exact two-way correspondence.

1908

— Schmidt publishes the orthogonalization algorithm in its modern inductive form, crediting Gram.

1926

— Born, Heisenberg, Jordan and Dirac compute with energy eigenbases; expansion coefficients become probability amplitudes.

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., Ch. 3 (Formalism) and Appendix A · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 1.2–1.3 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 1 (Mathematical Introduction) and Ch. 4 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. 1, Ch. II and complements · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 1–3

Provenance. Upgraded to full canon at Step-2 from the AllQFT base article (QUANT.FORM.BASIS.0871). Base provenance, kept verbatim: "All formulas are standard textbook identities. A verifier should: (1) confirm orthonormality/completeness statements and the resolution of identity against Sakurai §1.2–1.3 and Shankar Ch.1; (2) recompute the expansion/Parseval derivation (Griffiths §3.1–3.2); (3) verify the spectral-theorem steps — real eigenvalues and orthogonality of distinct eigenvectors for Hermitian operators — in Sakurai §1.3; (4) check the continuous-basis identities $\langle x|x'\rangle=\delta(x-x')$ , $\int|x\rangle\langle x|\mathrm{d}x=\hat{1}$ , and $\langle x|p\rangle=(2\pi\hbar)^{-1/2}e^{ipx/\hbar}$ against Cohen-Tannoudji Vol.1 Ch.II and its complements. No empirical constants are asserted beyond $\hbar$ appearing symbolically." Per the chapter ownership map, the base's resolution-of-identity and operator-matrix derivations are owned by Completeness Relation and Resolution of Identity (next serial) and its continuous-basis material by Continuous Spectra and Delta Normalization; the spectral-theorem proofs it cites are owned by Hermitian Operators and Observables and recalled without re-proof. This article's additions — the expansion theorem with uniqueness, Gram–Schmidt, Bessel/Parseval — cash the deferrals recorded by Stationary States and Separation of Variables, Inner Product and Norm of States, and Hermitian Operators and Observables. For this edition every number was machine-verified (sympy/numpy): the Legendre outputs and identity Gram matrix, the exact odd- $n$ coefficients and Parseval sum, the staircase 0.8106/0.9006/0.9331/0.9496/0.9596, the $n = 41$ 99% threshold, the constant $8/\pi^2\,E_1$ energy increments, and the reconstruction/uniqueness checks to $10^{-12}$ .

Builds on: Hilbert Space and State Vectors · Inner Product and Norm of States · Dirac Bra-Ket Notation · Linear Operators on Hilbert Space

Leads to: Completeness Relation and Resolution of Identity · Projection Operators · Hermitian Operators and Observables · Born Rule and Measurement Probabilities

What is the actual difference between "orthonormal" and "complete"? The words blur together.

They answer different questions. Orthonormal: are the axes clean? — perpendicular, unit length, components extract without cross-talk. Complete: are there enough axes? — nothing lies beyond their reach. The failure modes differ too: incompleteness leaves a Bessel deficit, non-orthonormality corrupts the coefficient formula itself.

How do I check completeness in practice — I cannot test Parseval on every state?

In finite dimensions, count: $N$ orthonormal vectors in $N$ dimensions are automatically complete. For infinite families you inherit theorems — the eigenfunctions of standard Hermitian problems (box, oscillator, hydrogen's bound-plus-continuum) are proved complete by Sturm–Liouville and spectral theory — so physicists check membership in a known family. When numerics matter, one does run the Parseval audit on representative states, exactly Example 2's staircase.

Why does Gram–Schmidt subtract projections instead of just rotating vectors until they are perpendicular?

Because "rotate until perpendicular" is not a computable instruction — which rotation, about what axis, in infinite dimensions? Subtracting shadows is: it uses only inner products and provably leaves a perpendicular remainder in one step. It also preserves the span stage by stage, which the physics needs — inside a degenerate eigenspace you must reorganize without leaving, and shadow-subtraction never leaves.

Is the expansion of a state unique, or does it depend on my choice of basis?

Both, in different senses. Within one complete orthonormal basis, the coefficient list is strictly unique — that is the expansion theorem's Step 3. Across bases, the same state has a different list for each basis (position components, momentum components, energy components…), all describing one vector. The lists are related by the change-of-basis machinery that Completeness Relation and Resolution of Identity builds next.

Where did the even-n coefficients go in the flat-state example — is that a coincidence?

Symmetry. The flat state is even about the box center; the eigenfunctions alternate even/odd there (the parity structure catalogued in Parity Symmetry of One-Dimensional Eigenstates). The inner product of an even state with an odd mode vanishes identically, so half the basis decouples before any integral is computed. Spotting such selection rules first is the professional move — it halves the work and predicts the zeros.

If the flat state has infinite mean energy, was it illegal to expand it?

No — the expansion is perfectly legal and converges in norm, as Parseval certifies. What fails is a different question: whether $\sum |c_n|^2 E_n$ converges. Norm-membership in the Hilbert space and finiteness of particular observables are separate demands; a state can be normalizable yet too rough at the walls to have finite energy. Which functions qualify as physically preparable states on all counts is the charter of Admissibility Conditions on Wavefunctions.

Do continuous bases like position eigenstates obey the same theorems?

The structure transplants — sums to integrals, $\delta_{mn}$ to $\delta(x - x')$ , Parseval to $\int |\psi(x)|^2 dx = 1$ — but the "basis vectors" themselves are not normalizable states, so the theorems need re-founding rather than copying. That rigor bill (delta normalization, rigged spaces) is paid by Continuous Spectra and Delta Normalization and Rigged Hilbert Space. Practically, you compute with continuous bases exactly as with discrete ones, integrals replacing sums.