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Separability Assumptions

A Hilbert space is separable when it has a countable dense subset, equivalently a finite or countably infinite orthonormal basis. This hypothesis makes basis expansions and many representation arguments manageable. It does not require discrete measurement outcomes, finite dimension, or absence of entanglement. Standard single-particle spaces such as L2(Rd)L^2(\mathbb R^d) are separable even though position has a continuous spectrum.

Required background. Hilbert Space as State Space supplies dense spans, completion, and orthonormal expansions.

A subset D⊆HD\subseteq\mathcal H is dense if every vector can be approximated arbitrarily closely in norm by elements of DD. If a countable orthonormal basis {en}\{e_n\} exists, the finite linear combinations with coefficients in Q+iQ\mathbb Q+i\mathbb Q form a countable dense set. Countability follows from taking a countable union of finite products; density follows by first truncating a basis expansion and then approximating its finitely many coefficients.

Conversely, starting from a countable dense sequence and applying Gram–Schmidt, while discarding vectors already in the previously constructed span, produces a finite or countable orthonormal family with dense span. This is a Hilbert basis: infinite expansions converge in norm. It is not generally a Hamel basis, whose vectors would have to express every state as a finite sum.

Every orthonormal set in a separable Hilbert space is countable. Distinct unit orthogonal vectors are a distance 2\sqrt2 apart. Their open balls of radius 1/31/3 are disjoint, and each contains a point from a fixed countable dense set. Distinct balls require distinct points, which prevents an uncountable orthonormal set.

These arguments identify exactly what the assumption buys. In an infinite-dimensional separable space, choosing an orthonormal basis gives a unitary map

H⟶ℓ2(N),ψ⟼(⟨en,ψ⟩)n≥1.\mathcal H\longrightarrow\ell^2(\mathbb N), \qquad \psi\longmapsto(\langle e_n,\psi\rangle)_{n\geq1}.

The map identifies Hilbert-space structures; it does not identify the operators specifying different physical systems.

On L2(R,dx)L^2(\mathbb R,dx), step functions supported on bounded intervals with rational endpoints and rational complex values form a countable dense family. Density follows from approximation by simple functions and regularity of Lebesgue measure. Thus this position-space Hilbert space is separable.

Nevertheless the position operator has spectrum R\mathbb R. A position ket is a generalized object, not a unit vector in L2(R)L^2(\mathbb R). The formal equation ⟨x∣y⟩=δ(x−y)\langle x|y\rangle=\delta(x-y) therefore does not exhibit an uncountable orthonormal Hilbert-space basis. Dirac delta normalization is distributional and is not the Kronecker normalization of actual vectors.

Compare this with ℓ2(R)\ell^2(\mathbb R) using counting measure. There is now a unit vector exe_x for every real label xx, supported at that one label, with ⟨ex,ey⟩=δxy\langle e_x,e_y\rangle=\delta_{xy}. This is a nonseparable Hilbert space. The distinction is the measure and vector structure, not the fact that both labels range over the real numbers.

The Unbounded Spectral Theorem explains how a separable Hilbert space supports continuous spectral measures without normalizable eigenvectors for every spectral value.

Individual vectors still have countable support

Section titled “Individual vectors still have countable support”

Suppose {eα}α∈I\{e_\alpha\}_{\alpha\in I} is an orthonormal basis, allowing an uncountable index set. For any fixed ψ\psi, only countably many coefficients ⟨eα,ψ⟩\langle e_\alpha,\psi\rangle can be nonzero.

To prove this, for each positive integer mm define

Im={α:∣⟨eα,ψ⟩∣≥1/m}.I_m=\{\alpha:|\langle e_\alpha,\psi\rangle|\geq1/m\}.

Bessel’s inequality bounds ∣Im∣|I_m| by m2∥ψ∥2m^2\|\psi\|^2, so each ImI_m is finite. Every nonzero coefficient belongs to at least one ImI_m, making the support a countable union of finite sets.

This does not make the entire space separable. The countable support can depend on the vector; there need not be one countable basis subset supporting all vectors simultaneously. Similarly, every trace-class density operator has separable support, since its nonzero spectral subspaces span at most a countable-dimensional closed subspace. A family of density operators can still collectively explore a nonseparable space.

Strongly continuous dynamics has separable cyclic orbits

Section titled “Strongly continuous dynamics has separable cyclic orbits”

Let U(t)U(t) be a strongly continuous unitary group and fix ψ\psi. Define

Hψ=span⁡‾{U(t)ψ:t∈R}.\mathcal H_\psi =\overline{\operatorname{span}}\{U(t)\psi:t\in\mathbb R\}.

Strong continuity permits every U(t)ψU(t)\psi to be approximated by U(qn)ψU(q_n)\psi with rational qn→tq_n\to t. Consequently Hψ\mathcal H_\psi is already the closed span of the countable rational-time orbit and is separable. It is invariant under the group by the group law and continuity.

This result is useful even when an ambient representation is nonseparable: one initial vector under one strongly continuous time evolution generates a separable invariant subspace. It does not imply that adding arbitrary observables or uncountably many preparation sectors leaves that subspace invariant.

Separability should therefore be stated where a theorem uses it, rather than promoted to a universal conclusion about every algebraic quantum representation. In finite-degree Schrödinger models it is standard; in representation theory and infinite-system constructions one must inspect the particular space being used.

Explain why a Bell state in C2⊗C2\mathbb C^2\otimes\mathbb C^2 does not contradict separability of its Hilbert space, and construct a countable dense subset of that space.

Solution

The Hilbert space is isomorphic to C4\mathbb C^4. Vectors whose four coordinates lie in Q+iQ\mathbb Q+i\mathbb Q form a countable dense set. The Bell vector (∣00⟩+∣11⟩)/2(|00\rangle+|11\rangle)/\sqrt2 is entangled because it is not a product vector. Hilbert-space separability is a topological countability property; state separability is a convex or tensor-product property. They answer different questions.