Hilbert Spaces

A Hilbert space is an inner-product space that is complete in the norm induced by its inner product. It combines linear superposition and orthogonality with a guarantee that norm-convergent approximation processes do not leave the space.

Finite-dimensional inner-product spaces are automatically Hilbert spaces. The word complete becomes substantive in infinite dimensions, where wave functions, infinite series, operator domains, and limiting procedures enter.

This page introduces the general structure. Detailed treatments of L2 Spaces, Orthonormal Bases, and Separable Hilbert Spaces have their own canonical pages.

Definition

Let $\mathcal H$ be a complex vector space with inner product

$$\langle\phi\vert\psi\rangle,$$

conjugate-linear in $\phi$ and linear in $\psi$. The inner product induces the norm and metric

$$\begin{aligned} \lVert\psi\rVert &= \sqrt{\langle\psi\vert\psi\rangle},\\ d(\phi,\psi) &= \lVert\phi-\psi\rVert. \end{aligned}$$

A sequence $(\psi_n)$ is Cauchy when

$$\begin{gathered} \forall\epsilon>0,\ \exists N\text{ such that}\\ m,n\ge N \quad\Longrightarrow\quad \lVert\psi_m-\psi_n\rVert<\epsilon. \end{gathered}$$

The space is complete when every Cauchy sequence has a limit in $\mathcal H$:

$$\exists\psi\in\mathcal H \quad\text{such that}\quad \lVert\psi_n-\psi\rVert\longrightarrow0.$$

Completeness concerns the norm topology. It does not say that every pointwise-convergent function sequence converges in norm, or that every formally written vector belongs to the space.

Why Completeness Matters

Many useful vectors are not specified in one step. They arise as limits of:

• partial sums of basis expansions;
• variational approximations;
• wave-packet regularizations;
• sequences of numerical truncations;
• solutions constructed by successive approximation;
• limits of observables or time-evolution schemes.

If the space is incomplete, an internally Cauchy approximation can converge to an object missing from the declared state space. Completeness ensures that all information needed to define the limit remains inside the space.

It also supports central Hilbert-space theorems:

• square-summable orthonormal expansions converge;
• closed subspaces have nearest-point orthogonal projections;
• bounded linear functionals have vector representatives;
• bounded operators form a complete normed operator space.

Completeness is therefore not an ornamental axiom. It closes the space under the limiting operations used throughout analysis.

An Explicit Incomplete Inner-Product Space

Let $c_{00}$ be the vector space of complex sequences with finite support:

$$\begin{aligned} c_{00} &= \bigl\lbrace x=(x_n)_{n\ge1}:\\ &\qquad \lvert\operatorname{supp}x\rvert<\infty \bigr\rbrace. \end{aligned}$$

Give it the inner product

$$\langle x\vert y\rangle =\sum_{n=1}^{\infty}x_n^*y_n.$$

The sum is finite for vectors in $c_{00}$. Let $e_n$ be the sequence with a one in position $n$, and define

$$x^{(N)} =\sum_{n=1}^{N}\frac{1}{n}e_n.$$

For $M>N$,

$$\left\lVert x^{(M)}-x^{(N)} \right\rVert^2 = \sum_{n=N+1}^{M}\frac{1}{n^2}.$$

The tail tends to zero because $\sum_n n^{-2}$ converges, so $(x^{(N)})$ is Cauchy. Its natural limit is

$$x= \left( 1,\frac12,\frac13,\ldots \right),$$

which has finite squared norm but does not have finite support. Thus $x\notin c_{00}$, and $c_{00}$ is not complete.

The completion of $c_{00}$ is $\ell^2$, the space obtained by adjoining all square-summable limits. This example isolates exactly what completeness adds.

Completion

Every inner-product space $V$ has a Hilbert-space completion $\overline V$. Informally, one adjoins limits for all Cauchy sequences. More precisely, one can:

1. take the set of Cauchy sequences in $V$;
2. identify two sequences whose difference converges to zero;
3. define vector operations and the inner product on equivalence classes.

The original space embeds isometrically and densely into $\overline V$. The completion is unique up to a unique isometric isomorphism that fixes the embedded copy of $V$.

Completion changes the space, not the norm formula on the original vectors. For example,

$$\overline{c_{00}}=\ell^2$$

in the sequence norm, while suitable spaces of test functions complete to $L^2$ spaces in the integral norm.

Fundamental Geometry

All inner-product spaces satisfy the Cauchy–Schwarz inequality,

$$\left\lvert \langle\phi\vert\psi\rangle \right\rvert \le \lVert\phi\rVert\, \lVert\psi\rVert,$$

and therefore the triangle inequality. Orthogonal vectors, for which $\langle\phi\vert\psi\rangle=0$, satisfy the Pythagorean identity

$$\lVert\phi+\psi\rVert^2 =\lVert\phi\rVert^2+\lVert\psi\rVert^2.$$

The norm obeys the parallelogram law:

$$\lVert\phi+\psi\rVert^2 +\lVert\phi-\psi\rVert^2 =2\lVert\phi\rVert^2 +2\lVert\psi\rVert^2.$$

Conversely, a norm comes from an inner product precisely when it satisfies the parallelogram law. With the convention used here, the complex polarization identity is

$$\begin{aligned} \langle\phi\vert\psi\rangle &= \frac14 \bigl( \lVert\phi+\psi\rVert^2 -\lVert\phi-\psi\rVert^2 \bigr)\\ &\quad -\frac{i}{4} \bigl( \lVert\phi+i\psi\rVert^2 -\lVert\phi-i\psi\rVert^2 \bigr). \end{aligned}$$

Thus the inner product and its induced norm determine one another.

Standard Examples

Finite coordinate spaces. The space $\mathbb C^d$ with

$$\langle x\vert y\rangle =\sum_{j=1}^{d}x_j^*y_j$$

is complete. Every finite-dimensional complex Hilbert space is unitarily isomorphic to one of these.

Square-summable sequences. The space

$$\ell^2(\mathbb N) = \left\{ x=(x_n): \sum_{n=1}^{\infty}\lvert x_n\rvert^2<\infty \right\}$$

has inner product

$$\langle x\vert y\rangle =\sum_{n=1}^{\infty}x_n^*y_n.$$

It is the coordinate model for every countably infinite-dimensional separable Hilbert space.

Square-integrable functions. For a measure space $(X,\mu)$,

$$L^2(X,\mu) = \left\{ \psi: \int_X\lvert\psi\rvert^2\,d\mu<\infty \right\}$$

after identifying functions equal almost everywhere. The inner product is

$$\langle\phi\vert\psi\rangle = \int_X\phi^*\psi\,d\mu.$$

Closed subspaces. Every closed linear subspace of a Hilbert space is a Hilbert space with the inherited inner product. A nonclosed subspace is incomplete in that norm.

Direct sums and Hilbert-space tensor products provide further examples. In infinite dimensions their definitions include convergence or completion conditions that are invisible in finite-dimensional notation.

Closed Subspaces and Orthogonal Complements

For a subset $M\subseteq\mathcal H$, define

$$\begin{aligned} M^\perp &= \bigl\lbrace \psi\in\mathcal H: \langle m\vert\psi\rangle=0\\ &\qquad \text{for every }m\in M \bigr\rbrace. \end{aligned}$$

The orthogonal complement is always a closed subspace. One has

$$M^{\perp\perp}=\overline{\operatorname{span}M}.$$

If $M$ is itself a closed linear subspace, the projection theorem gives the orthogonal direct sum

$$\mathcal H=M\mathbin{\oplus_\perp}M^\perp.$$

Every $\psi\in\mathcal H$ has a unique decomposition

$$\psi=P_M\psi+(I-P_M)\psi,$$

where $P_M\psi\in M$ and $(I-P_M)\psi\in M^\perp$. The retained component is the unique best approximation:

$$\lVert\psi-P_M\psi\rVert = \inf_{m\in M}\lVert\psi-m\rVert.$$

Closedness is essential. If a subspace omits one of its norm limits, the nearest candidate may lie outside it. The finite-dimensional construction is developed in Projectors.

Orthonormal Families and Hilbert Bases

An orthonormal family $\{e_\alpha\}$ satisfies

$$\langle e_\alpha\vert e_\beta\rangle =\delta_{\alpha\beta}.$$

It is complete when

$$\overline{\operatorname{span}\{e_\alpha\}} =\mathcal H,$$

or equivalently when the only vector orthogonal to every $e_\alpha$ is zero. Such a family is called an orthonormal basis or Hilbert basis.

This is not generally a Hamel basis. Infinite Hilbert expansions are limits of finite sums, not finite algebraic combinations. In a separable Hilbert space, a complete orthonormal basis can be indexed by $n=1,2,\ldots$, and

$$\begin{aligned} \psi &= \sum_{n=1}^{\infty} \langle e_n\vert\psi\rangle e_n,\\ \lVert\psi\rVert^2 &= \sum_{n=1}^{\infty} \left\lvert \langle e_n\vert\psi\rangle \right\rvert^2. \end{aligned}$$

The first series converges in Hilbert-space norm. Pointwise convergence of a wavefunction representation is a separate question. See Completeness and Orthonormal Bases for Bessel’s inequality, Parseval’s identity, and convergence modes.

Separability and the Sequence Model

A Hilbert space is separable when it has a countable dense subset. For Hilbert spaces, this is equivalent to possessing a finite or countably infinite complete orthonormal basis.

If $\mathcal H$ has countable complete orthonormal basis $\{e_n\}$, the coordinate map

$$\begin{aligned} U:\mathcal H&\longrightarrow\ell^2,\\ U\psi &= \bigl( \langle e_1\vert\psi\rangle, \langle e_2\vert\psi\rangle, \ldots \bigr) \end{aligned}$$

is unitary. Hence all countably infinite-dimensional separable Hilbert spaces are abstractly unitarily isomorphic to $\ell^2$.

This does not make all quantum models physically identical. The choice of operators, domains, preferred representations, tensor-factor structure, and symmetries carries additional information not contained in the abstract Hilbert space alone.

Continuous position variables are compatible with separability. A space such as $L^2(\mathbb R)$ has a countable Hilbert basis even though position has a continuous range and generalized position eigenvectors are not ordinary $L^2$ vectors.

Riesz Representation for Functionals

Let $F:\mathcal H\to\mathbb C$ be a bounded linear functional. The Riesz representation theorem states that there is a unique $\phi_F\in\mathcal H$ such that

$$F(\psi) = \langle\phi_F\vert\psi\rangle$$

for every $\psi\in\mathcal H$. Moreover,

$$\lVert F\rVert = \lVert\phi_F\rVert.$$

This theorem identifies the continuous dual of a Hilbert space with the space itself, conjugate-linearly in the representing vector. It is the rigorous basis for converting kets to continuous bras.

The word bounded cannot be omitted. Discontinuous algebraic linear functionals are not represented by Hilbert-space vectors.

Vectors and Representations

A Hilbert-space vector is not the same object as one coordinate representation. The same vector may appear as:

• a coefficient sequence in an orthonormal basis;
• a square-integrable position wavefunction;
• a square-integrable momentum wavefunction;
• components relative to a finite truncation.

A unitary representation map preserves inner products and norms. Equality of vectors therefore means equality in the Hilbert-space sense, not necessarily pointwise equality of chosen function representatives.

In $L^2$, functions that differ on a measure-zero set define the same vector. Likewise, a generalized position ket $\lvert x\rangle$ is not an ordinary normalizable vector merely because it is useful notation in a distributional representation.

Bounded and Unbounded Operators

A bounded linear operator $A:\mathcal H\to\mathcal H$ satisfies

$$\lVert A\psi\rVert \le C\lVert\psi\rVert$$

for some finite $C$ and every $\psi$. Bounded operators are continuous and extend uniquely from a dense subspace to the whole Hilbert space.

Important quantum operators can be unbounded. An unbounded operator is specified by both a rule and a domain:

$$A:\mathcal D(A)\subseteq\mathcal H \longrightarrow\mathcal H.$$

Products, adjoints, commutators, and exponentials then require domain analysis. A formula that is valid for finite matrices may be meaningless if the relevant vector is outside an operator domain.

See Bounded Operators, Unbounded Operators, and Domains of Operators for the canonical treatments.

Worked Example: Even Coordinates in Sequence Space

Let $\mathcal H=\ell^2(\mathbb N)$ and define

$$M= \left\{ x\in\ell^2: x_{2k-1}=0 \text{ for every }k \right\}.$$

This is the closed subspace of sequences supported on even indices. For $x=(x_1,x_2,x_3,x_4,\ldots)$, define

$$P_{\mathrm{even}}x = (0,x_2,0,x_4,\ldots).$$

Then

$$P_{\mathrm{even}}^2 =P_{\mathrm{even}}, \qquad P_{\mathrm{even}}^\dagger =P_{\mathrm{even}}.$$

The complementary projector retains the odd coordinates. For every $x\in\ell^2$,

$$\begin{aligned} \lVert x\rVert^2 &= \lVert P_{\mathrm{even}}x\rVert^2\\ &\quad+ \lVert(I-P_{\mathrm{even}})x\rVert^2. \end{aligned}$$

This is an infinite-dimensional orthogonal decomposition with exactly the same geometry as a finite coordinate projection. The analytic content is that both subspaces are closed and both projected sequences remain in $\ell^2$.

Truncations and Convergence

Let $\{e_n\}$ be a complete orthonormal basis and define finite-rank projectors

$$P_N = \sum_{n=1}^{N} \lvert e_n\rangle\langle e_n\rvert.$$

For each fixed $\psi$ with coefficients $c_n=\langle e_n\vert\psi\rangle$,

$$\begin{aligned} \lVert(I-P_N)\psi\rVert^2 &= \sum_{n>N}\lvert c_n\rvert^2\\ &\longrightarrow0. \end{aligned}$$

Thus $P_N\to I$ strongly: convergence holds after applying the operators to each fixed vector. But in an infinite-dimensional space,

$$\lVert I-P_N\rVert_{\mathrm{op}}=1$$

for every $N$, because $I-P_N$ fixes any normalized basis vector beyond the truncation. The convergence is not in operator norm.

This distinction matters in numerical quantum mechanics. A truncation may converge for every fixed state while failing to approximate the identity uniformly over all normalized states. For unbounded operators, controlling only the Hilbert norm may also be insufficient; domain-adapted or graph norms can be necessary.

Finite versus Infinite Dimension

Several finite-dimensional facts require qualification:

• an infinite-dimensional subspace need not be closed;
• a linear operator need not be bounded or defined everywhere;
• a bounded closed set need not be compact;
• a self-adjoint operator need not have an eigenbasis of normalizable vectors;
• spectra may contain continuous parts;
• a bounded operator need not have a finite trace;
• strong convergence need not imply operator-norm convergence;
• generalized eigenvectors may live outside the Hilbert space.

The overview Finite- versus Infinite-Dimensional Quantum Mechanics tracks the corresponding physics implications.

Boundary with Quantum Postulates

Hilbert-space mathematics supplies vectors, inner products, subspaces, operators, and convergence. It does not by itself say:

• which rays or density operators represent physical preparations;
• how probabilities are assigned;
• which self-adjoint operator represents a laboratory quantity;
• how states evolve;
• how composite-system factors are chosen.

Those are quantum-theoretic postulates and modeling choices. See State Vectors for the first physical use of Hilbert-space vectors.

Common Mistakes

• Defining a Hilbert space as merely a vector space with an inner product.
• Confusing norm convergence with pointwise convergence.
• Assuming every dense subspace is complete.
• Calling an infinite orthonormal Hilbert basis a Hamel basis.
• Omitting closure in the statement $M^{\perp\perp}=\overline{\operatorname{span}M}$.
• Projecting onto a nonclosed subspace and assuming a nearest point exists.
• Treating every algebraic functional as a continuous bra.
• Assuming every infinite-dimensional operator is bounded.
• Ignoring the domain of an unbounded operator.
• Interpreting strong convergence as operator-norm convergence.
• Treating generalized eigenvectors as normalizable Hilbert-space vectors.
• Inferring the physical postulates of quantum mechanics from completeness.

Exercises

1. In $c_{00}$ with the $\ell^2$ norm, let

   $$x^{(N)} =\sum_{n=1}^{N}\frac{1}{n}e_n.$$

   Prove directly that $(x^{(N)})$ is Cauchy and explain why it has no limit in $c_{00}$.

Solution

For $M>N$,

$$\begin{aligned} \left\lVert x^{(M)}-x^{(N)} \right\rVert^2 &= \sum_{n=N+1}^{M}\frac{1}{n^2} \\ &\le \sum_{n=N+1}^{\infty}\frac{1}{n^2}. \end{aligned}$$

The tail of a convergent positive series tends to zero, so the sequence is Cauchy. If it converged in $c_{00}$ to $y$, continuity of each coordinate functional would give $y_n=1/n$ for every $n$. That sequence has infinitely many nonzero entries, contradicting $y\in c_{00}$. The limit exists only after completing the space to $\ell^2$.

2. Let $M$ be a linear subspace of a Hilbert space $\mathcal H$. Prove that $M$, with the inherited norm, is complete if and only if $M$ is closed in $\mathcal H$.

Solution

Assume first that $M$ is closed. Any Cauchy sequence in $M$ is Cauchy in $\mathcal H$, so it converges to some $x\in\mathcal H$. Closedness implies $x\in M$, hence $M$ is complete.

Conversely, assume $M$ is complete and let $(x_n)$ be a sequence in $M$ converging in $\mathcal H$ to $x$. The sequence is Cauchy in the inherited norm. Completeness of $M$ gives a limit $y\in M$. Limits in a metric space are unique, so $x=y\in M$. Therefore $M$ contains all of its limits and is closed.

3. On $\ell^2$, let $x=(x_1,x_2,x_3,x_4,\ldots)$ and define

   $$Px =(0,x_2,0,x_4,\ldots).$$

   Show that $P$ is an orthogonal projector, determine its range and kernel, and compute its operator norm.

Solution

Applying $P$ twice changes nothing, so $P^2=P$. For $x,y\in\ell^2$,

$$\begin{aligned} \langle x\vert Py\rangle &= \sum_{k=1}^{\infty} x_{2k}^*y_{2k}\\ &= \langle Px\vert y\rangle, \end{aligned}$$

so $P^\dagger=P$. Its range is the closed subspace of even-supported sequences, and its kernel is the closed subspace of odd-supported sequences.

Because deleting coordinates cannot increase the norm,

$$\lVert Px\rVert\le\lVert x\rVert.$$

Equality holds for every nonzero even-supported sequence, so $\lVert P\rVert_{\mathrm{op}}=1$.

4. Let $\{e_n\}$ be a complete orthonormal basis and

   $$P_N =\sum_{n=1}^{N} \lvert e_n\rangle\langle e_n\rvert.$$

   Prove that $P_N\psi\to\psi$ for every fixed $\psi$, but $\lVert I-P_N\rVert_{\mathrm{op}}=1$ for every $N$.

Solution

Write

$$\psi = \sum_{n=1}^{\infty}c_ne_n, \qquad \sum_{n=1}^{\infty}\lvert c_n\rvert^2 <\infty.$$

Then

$$\lVert\psi-P_N\psi\rVert^2 = \sum_{n>N}\lvert c_n\rvert^2 \longrightarrow0.$$

This proves strong convergence. Since $I-P_N$ is an orthogonal projector, its operator norm is at most one. On the normalized vector $e_{N+1}$,

$$(I-P_N)e_{N+1}=e_{N+1},$$

so its norm is at least one. Therefore

$$\lVert I-P_N\rVert_{\mathrm{op}}=1$$

for every finite $N$.

References

• J. B. Conway, A Course in Functional Analysis, 2nd ed., Springer, 1990.
• E. Kreyszig, Introductory Functional Analysis with Applications, Wiley, 1978.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• L. Debnath and P. Mikusiński, Introduction to Hilbert Spaces with Applications, 3rd ed., Elsevier, 2005.