Hilbert Spaces and Orthonormal Bases

Hilbert space is the daily working language of quantitative science. Every quantum computation runs in one: a register of qubits is a unit vector in $\mathbb{C}^{2^N}$, gates are unitaries preserving the inner product, and an outcome's probability is a squared coefficient $|c_n|^2$ — Parseval as the statement that the probabilities sum to one. In signal processing the discrete Fourier and wavelet transforms are changes of orthonormal basis, and Parseval's relation ("energy in time equals energy in frequency") is what lets JPEG and MP3 discard small coefficients while controlling the norm error. Spectral and finite-element PDE solvers are orthogonal projection onto a finite basis, the residual bounded exactly by the Bessel deficit. Quantum chemistry expands molecular orbitals in finite basis sets and lives or dies by how completely they span the true $L^2$ space — the "basis-set incompleteness error" that benchmark calculations chase toward the complete-basis limit. And an active frontier is rigged Hilbert space, the extension that makes Dirac's continuous "bases" $|x\rangle$, with $\langle x|x'\rangle=\delta(x-x')$ and $\int\mathrm{d}x\,|x\rangle\langle x|=\mathbb{1}$, fully rigorous.

Think of a Hilbert space as $\mathbb{C}^n$ carried to $n\to\infty$, but done with care. Choose an orthonormal basis and the components of a vector are just inner products, $c_n=\langle e_n,f\rangle$; the schoolbook fact that "the squared length is the sum of the squared components" survives intact as Parseval's relation, provided completeness is there to make the infinite sums land somewhere. The first figure keeps honest books on that transfer: for a concrete vector it shows how much of the squared length a truncated basis captures (Bessel), and how a complete basis fills the budget exactly (Parseval).

Completeness of the space is felt best through its failure. Picture a family of continuous ramps that steepen around a single point until they mimic a jump: each ramp is continuous and the terms crowd together in mean-square (they are Cauchy), yet the vector they converge toward is a discontinuous step that no longer belongs to the space of continuous functions. That space has a hole exactly where the step should sit; passing to $L^2$ fills every such hole, just as passing from the rationals to the reals adds the missing limit points. The second figure makes that hole visible with a real convergence rate.

The starting datum is the inner product and the length it induces; a Hilbert space carries the physics convention throughout, linear in the second argument and conjugate-linear in the first. Everything geometric — projection, orthogonality, the reconstruction of a vector from its coefficients — is downstream of these two objects.

where: $\langle f,g\rangle$ — Hermitian inner product, linear in $g$, conjugate-linear in $f$; $\lVert f\rVert$ — the norm, the length induced by the inner product. Reading: the inner product measures overlap, and a vector's self-overlap is the square of its length. The one axiom that promotes an inner-product space to a Hilbert space is completeness. It is a statement not about any single vector but about the space as a whole: the limit that a Cauchy sequence is straining toward must exist and must live inside the space.

where: $(f_n)$ — a sequence of vectors in $\mathcal{H}$; $f$ — its limit, required to belong to $\mathcal{H}$. Reading: no Cauchy sequence may point at a hole; if the terms bunch up, something real is there to catch them. The Cauchy–Schwarz inequality holds in any inner-product space, before completeness enters. It bounds an overlap by the product of lengths and is the fact that makes the cosine legal, the triangle inequality true, and the inner product continuous.

where: $f,g$ — any two vectors. Reading: you cannot project more of $f$ onto $g$ than the whole of $f$; equality only when the two are parallel. A family of vectors is orthonormal when distinct members are perpendicular and each has unit length; its members' overlaps with a target vector are the generalized Fourier coefficients.

where: $\{e_n\}$ — an orthonormal family; $\delta_{mn}$ — Kronecker delta, $1$ if $m=n$ and $0$ otherwise; $c_n$ — the $n$-th coefficient of $f$. Reading: an orthonormal basis is a coordinate frame in which the metric is the identity, so overlaps become conjugated coordinate sums. For any orthonormal set the coefficients can never store more squared length than the vector has — Bessel's inequality. The gap between the two sides is the squared length of the part of $f$ the set cannot represent.

where: $\sum_n|c_n|^2$ — total squared weight captured by the set; $\lVert f\rVert^2$ — the full squared length. Reading: a truncated basis always loses something, and Bessel bounds exactly how much. When the orthonormal set is complete — maximal, so that no nonzero vector is orthogonal to all of it — Bessel's inequality becomes Parseval's equality and the vector is fully reconstructed by its coefficients. These two lines are the operational heart of the subject.

where: the sum in $f=\sum_n c_n e_n$ converges in norm (mean-square), not necessarily pointwise; $d_n=\langle e_n,g\rangle$. Reading: expand, and lengths and overlaps are computed coordinate by coordinate exactly as in $\mathbb{C}^n$. Two structural relations close the circle. The resolution of the identity repackages completeness as an operator statement, and the Riesz representation theorem identifies every continuous linear functional with a vector — the rigorous basis of Dirac's bra–ket notation.

where: $\mathbb{1}$ — the identity operator; $e_n\langle e_n,\cdot\rangle$ — the orthogonal projector onto the line through $e_n$; $\phi$ — a bounded linear functional; $g_\phi$ — its representing vector. Reading: inserting $\mathbb{1}=\sum_n|e_n\rangle\langle e_n|$ between operators is the completeness relation of quantum mechanics, and every bra is a vector.

Hilbert space specializes to familiar settings according to dimension and the choice of measure; each row is a fully valid instance a physicist meets in practice.

1 · Parseval on the sawtooth and the Basel problem (fully worked)

On $L^2[-\pi,\pi]$ use the orthonormal sine family $u_n(x)=\sin(nx)/\sqrt{\pi}$ to expand $f(x)=x$, then apply Parseval's relation to evaluate $\sum_{n=1}^\infty 1/n^2$. This is Euler's Basel sum, produced here purely by completeness.

Step 1. Compute the coefficients $a_n=\langle u_n,f\rangle=\frac{1}{\sqrt{\pi}}\int_{-\pi}^{\pi}x\sin(nx)\,\mathrm{d}x$. The integrand is even, and integrating by parts gives $\int_0^\pi x\sin(nx)\,\mathrm{d}x=\pi(-1)^{n+1}/n$, so $\int_{-\pi}^{\pi}x\sin(nx)\,\mathrm{d}x=2\pi(-1)^{n+1}/n$. Hence $a_n=\dfrac{2\sqrt{\pi}\,(-1)^{n+1}}{n}$ and $|a_n|^2=\mathbf{4\pi/n^2}$.

Step 2. Compute the squared norm directly: $\lVert f\rVert^2=\int_{-\pi}^{\pi}x^2\,\mathrm{d}x=\big[\tfrac13x^3\big]_{-\pi}^{\pi}=\mathbf{2\pi^3/3}$. Numerically this is $20.6709$, the total squared length that Parseval must account for.

Step 3. Parseval's relation says the squared norm equals the sum of the squared coefficients: $\lVert f\rVert^2=\sum_{n=1}^\infty|a_n|^2$, i.e. $\dfrac{2\pi^3}{3}=\sum_{n=1}^\infty\dfrac{4\pi}{n^2}=4\pi\sum_{n=1}^\infty\dfrac{1}{n^2}$. Solve for the sum: $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}=\frac{2\pi^3/3}{4\pi}=\boxed{\dfrac{\pi^2}{6}}\approx1.64493$.

Step 4. The point is where this number came from: not from clever series manipulation but from completeness. Parseval is an equality only because the sine family is a complete orthonormal basis; a single missing mode would replace $\pi^2/6$ by something smaller. The Basel sum $\sum 1/n^2=\pi^2/6$ is Parseval's relation for the sawtooth, exact because the basis is complete.

2 · The Bessel deficit of a truncated basis (confidence builder)

Work the figure's example in full. Take $f=(2,3,6)$ in $\mathbb{C}^3$ with the standard orthonormal basis $\{e_1,e_2,e_3\}$. Keep only the incomplete set $\{e_1,e_2\}$, find the Bessel deficit, then restore completeness and confirm Parseval.

Step 1. The coefficients are $c_n=\langle e_n,f\rangle$, so $c_1=2$, $c_2=3$, $c_3=6$. The full squared length is $\lVert f\rVert^2=2^2+3^2+6^2=4+9+36=\mathbf{49}$, so $\lVert f\rVert=7$.

Step 2. With the truncated set $\{e_1,e_2\}$ the partial sum is $S_2=2e_1+3e_2=(2,3,0)$, capturing $|c_1|^2+|c_2|^2=4+9=\mathbf{13}$. Bessel's inequality holds with room to spare: $13\le49$. The residual is $f-S_2=(0,0,6)$, whose squared length is the Bessel deficit $=36$ — precisely $|c_3|^2$, the direction $\{e_1,e_2\}$ cannot represent.

Step 3. Restore the missing element. The set $\{e_1,e_2,e_3\}$ is complete, so Parseval must hold: $\sum_n|c_n|^2=4+9+36=49=\lVert f\rVert^2$. The deficit has vanished. A truncated orthonormal set obeys Bessel (13 ≤ 49) with a deficit equal to the discarded squared components; only the complete basis achieves Parseval equality (49 = 49).

Hilbert space is one of the rare abstractions that were needed before they were named: analysts were manipulating infinite orthonormal expansions for a century before anyone axiomatized the space in which they live.

Texts. Stone & Goldbart, Mathematics for Physics, Ch. 2 (Function Spaces) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 5 · Hassani, Mathematical Physics, Chs. 5–6 (Hilbert spaces, generalized functions) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 5 · Kreyszig, Introductory Functional Analysis with Applications, Ch. 3 · Reed & Simon, Methods of Modern Mathematical Physics I: Functional Analysis. The key derivations rest on three checkpoints: Cauchy–Schwarz follows from the optimal-$\lambda$ / perfect-square method; Bessel and Parseval follow from expanding $\lVert f-S_N\rVert^2=\lVert f\rVert^2-\sum|c_n|^2$; and the Riesz–Fischer argument turns on the Cauchy estimate $\lVert S_M-S_N\rVert^2=\sum_{N+1}^M|c_n|^2$. The worked numbers are: sawtooth coefficients $|a_n|^2=4\pi/n^2$ with $\lVert x\rVert^2=2\pi^3/3$ giving $\sum 1/n^2=\pi^2/6=1.64493$; finite example $f=(2,3,6)$ with $\lVert f\rVert^2=49$, two-term partial sum $13$, and deficit $36$. The Fourier example uses the orthonormal family $u_n=\sin(nx)/\sqrt{\pi}$ on $[-\pi,\pi]$.

Prerequisites. Inner-Product Spaces and Norms · Orthogonality and the Gram–Schmidt Process · Vector Spaces and Subspaces · Linear Independence, Basis, and Dimension. Leads to. Adjoint and Hermitian Operators · Diagonalization and the Spectral Decomposition · Linear Operators and Matrix Representations · Change of Basis and Similarity Transformations.

Ask the tutor: Want to see how Parseval turns the sawtooth into $\pi^2/6$, or why the completeness relation $\sum|n\rangle\langle n|=\mathbb{1}$ is the move behind nearly every quantum-mechanics calculation? Ask and we will work it line by line.