Hilbert Space as State Space

A Hilbert space supplies a complete linear space of quantum state representatives, together with the inner product needed to compare preparations and calculate probabilities. Completeness ensures that convergent approximation procedures have a state-space limit. It does not select the observables, Hamiltonian, or admissible preparations: these are additional parts of a quantum model. This page treats the usual complex Hilbert-space formulation; pure states are its rays, while mixed states require density operators or, more generally, positive functionals on an observable algebra.

Required background. State Vectors supplies inner products, normalization, and linear combinations of representatives.

Completeness and quantum approximations

A complex Hilbert space $\mathcal H$ is an inner-product space complete in the norm $\|\psi\|=\sqrt{\langle\psi,\psi\rangle}$. Completeness means that every norm-Cauchy sequence has a norm limit in $\mathcal H$. This is stronger than requiring finite linear combinations to exist.

For example, let $c_{00}$ be the space of complex sequences with finite support, with

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

The vectors $x^{(N)}=(2^{-1},2^{-2},\ldots,2^{-N},0,\ldots)$ are Cauchy because

$$\|x^{(M)}-x^{(N)}\|^2 =\sum_{n=N+1}^{M}4^{-n}\longrightarrow0.$$

Their limit has infinitely many nonzero coordinates and is absent from $c_{00}$. Completing this space gives $\ell^2(\mathbb N)$. A finite expansion can therefore be an excellent computational representative without the set of finite expansions being a complete state space.

This distinction matters whenever a basis truncation is removed. If $\{e_n\}$ is a countable orthonormal basis and $\psi\in\mathcal H$, then

$$\psi_N=\sum_{n=1}^{N}\langle e_n,\psi\rangle e_n, \qquad \|\psi-\psi_N\|^2 =\sum_{n>N}|\langle e_n,\psi\rangle|^2\longrightarrow0.$$

Completeness and the basis expansion theorem justify the limit. They do not give a rate without information about the coefficients. The orthonormal-basis and completion results are developed in Teschl, Chapter 1; the estimates below explain their quantum significance.

Norm convergence controls bounded experiments

An effect is a bounded operator $E$ with $0\leq E\leq I$. In a normalized vector $\psi$, its event probability is $p_\psi(E)=\langle\psi,E\psi\rangle$. For normalized $\phi,\psi$,

$$\begin{aligned} |p_\psi(E)-p_\phi(E)| &=|\langle\psi-\phi,E\psi\rangle +\langle\phi,E(\psi-\phi)\rangle|\\ &\leq2\|E\|\,\|\psi-\phi\|. \end{aligned}$$

Thus norm approximation of normalized representatives controls every bounded event probability uniformly. The estimate need not be optimal to be useful: a bound on the state error produces a bound on measurement error without diagonalizing each effect.

The same argument applies to any bounded observable, with $\|E\|$ replaced by its operator norm. It does not control an unbounded energy or position expectation. A small norm component can lie at arbitrarily high energy; Infinite-Dimensional Caveats gives an explicit example.

Normalization requires a further step in truncated calculations. If $\psi_N\to\psi$ and $\|\psi\|=1$, then $\|\psi_N\|\to1$. For sufficiently large $N$, the normalized vector $\widehat\psi_N=\psi_N/\|\psi_N\|$ is defined and also converges to $\psi$. A projection onto too small a trial space may be zero, so normalization cannot be performed before checking the denominator.

Representatives, states, and observable structure

Multiplying a nonzero vector by a nonzero complex scalar leaves its pure-state ray unchanged. The exact equivalence with normalized phase classes and rank-one projections belongs to Physical States as Rays. Here the important distinction is between the linear arena and the physical state space: the zero vector belongs to $\mathcal H$ but represents no normalized preparation, and adding rays is not intrinsically defined.

Complex scalars permit relative phases and unitary dynamics in the conventional formulation. The Hilbert-space axioms alone do not prove that physical theories must use the complex rather than real or quaternionic field. Choosing complex Hilbert spaces is part of the framework being studied, not a consequence of completeness.

Nor does an abstract Hilbert space determine a system. An infinite-dimensional separable Hilbert space is unitarily isomorphic to $\ell^2(\mathbb N)$, but this does not make a free particle and a harmonic oscillator the same model. Their Hamiltonians have different spectral properties. An equivalence of models must intertwine the relevant operators and their domains, not merely provide a unitary correspondence of bare spaces.

Spaces used by elementary models

A qubit uses $\mathbb C^2$, and a fixed spin-$j$ degree of freedom uses $\mathbb C^{2j+1}$. A spinless particle on the line uses $L^2(\mathbb R,dx)$; in three spatial dimensions it uses $L^2(\mathbb R^3,d^3x)$. These spaces describe different kinds of coordinates even when their abstract Hilbert-space structure is easily classified.

An $L^2$ vector is an equivalence class of functions equal almost everywhere. Evaluation at a single point is not a well-defined bounded functional on that space. Changing a representative at one point changes no state. Probabilities for measurable regions, by contrast, are well defined:

$$p_\psi(\Omega)=\int_\Omega|\psi(x)|^2\,dx.$$

The Hamiltonian then specifies an operator on a domain. On the line, the free Hamiltonian has domain $H^2(\mathbb R)$, a proper dense subspace of $L^2(\mathbb R)$. A vector outside that domain is still a legitimate Hilbert-space state; it need not possess a norm derivative under the unitary evolution at time zero. See Stone’s Theorem for the exact distinction.

Exercise: a controlled truncation

Let $\psi=\sqrt3\sum_{n\geq1}2^{-n}e_n$. Compute the omitted probability after retaining the first $N$ basis vectors, and the overlap with the normalized truncation.

Solution

The state is normalized since $3\sum_{n\geq1}4^{-n}=1$. If $P_N$ projects onto the first $N$ basis vectors, then

$$\|(I-P_N)\psi\|^2=4^{-N}, \qquad \|P_N\psi\|^2=1-4^{-N}.$$

For $\widehat\psi_N=P_N\psi/\sqrt{1-4^{-N}}$,

$$|\langle\widehat\psi_N,\psi\rangle|^2=1-4^{-N}.$$

This verifies convergence of the preparation and quantifies its lost tail. It gives no energy-error estimate until the Hamiltonian and its growth on $e_n$ are specified.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — Hilbert-space quantum models and the separation of states from observables.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapter 1 — completeness, orthonormal expansions, and Hilbert-space operators.