Separable Hilbert Spaces

A Hilbert space is separable if it contains a countable dense subset. In Hilbert-space practice, this is equivalent to having a finite or countably infinite complete orthonormal basis.

Most Hilbert spaces used in ordinary quantum mechanics are separable. This fact is one reason countable basis expansions, truncations, and numerical approximations are so effective, even when the system has continuous position variables.

Dense Subsets

A subset $D\subset\mathcal H$ is dense if every vector in $\mathcal H$ can be approximated arbitrarily well by vectors from $D$:

$$\forall \psi\in\mathcal H,\quad \forall \epsilon>0,\quad \exists d\in D \quad \text{such that} \quad \lVert\psi-d\rVert\lt\epsilon.$$

The word “dense” refers to the Hilbert-space norm. It does not mean that every vector literally belongs to $D$.

A Hilbert space is separable if some dense subset can be listed as

$$d_1,d_2,d_3,\ldots.$$

Finite-dimensional Hilbert spaces are automatically separable.

Countable Orthonormal Bases

For Hilbert spaces, separability is equivalent to the existence of a countable complete orthonormal basis:

$$\{e_1,e_2,e_3,\ldots\}.$$

If such a basis exists, every vector has an expansion

$$\psi = \sum_{n=1}^{\infty} c_n e_n, \qquad c_n=\langle e_n\vert\psi\rangle,$$

with convergence in norm.

Conversely, if a Hilbert space has a countable dense subset, one can construct a countable complete orthonormal system by applying a Gram–Schmidt-type procedure to a carefully chosen independent sequence. The theorem is standard functional analysis; the practical lesson is that countable approximation and countable orthonormal expansions are the same separability idea.

The expansion side is developed in Completeness and Orthonormal Bases.

Countable Does Not Mean Few States

Separability does not mean the Hilbert space contains only countably many states. Even $\mathbb C^2$ contains uncountably many normalized rays.

The point is that a countable set is enough to approximate every state. For example, in $\mathbb C^2$, vectors with rational real and imaginary parts form a countable dense subset after normalization where appropriate.

In an infinite-dimensional separable Hilbert space with orthonormal basis $\{e_n\}$, finite linear combinations

$$\sum_{n=1}^{N} q_n e_n$$

with rational real and imaginary parts $q_n$ form a countable dense subset.

Standard Examples

The following Hilbert spaces are separable:

• $\mathbb C^n$ for finite $n$;
• $\ell^2(\mathbb N)$, the space of square-summable sequences;
• $L^2([a,b])$ and $L^2(\mathbb R^d)$ with the usual Lebesgue measure;
• Hilbert spaces obtained from standard bound-state bases such as the harmonic oscillator eigenfunctions;
• finite tensor products of separable Hilbert spaces.

The separability of $L^2(\mathbb R^d)$ can feel surprising because $\mathbb R^d$ has continuum many points. The reason is that $L^2$ functions are controlled by integral norm, and simple functions with rational coefficients on rational boxes form a countable dense subset.

Continuous Spectra Are Compatible with Separability

Separability does not forbid continuous spectra. The Hilbert space $L^2(\mathbb R)$ is separable, but the position and momentum observables have continuous spectra in the usual wave-mechanics setting.

The resolution is that position kets $\lvert x\rangle$ and momentum kets $\lvert p\rangle$ are generalized eigenvectors, not members of a countable orthonormal basis of normalizable states. A separable Hilbert space can have a countable complete orthonormal basis and still support observables whose spectral description involves integrals over continuous variables.

For the practical representation language, see Position and Momentum Representations.

Tensor Products

Finite tensor products of separable Hilbert spaces are separable. If $\{e_m\}$ is a countable orthonormal basis for $\mathcal H_A$ and $\{f_n\}$ is one for $\mathcal H_B$, then

$$\{e_m\otimes f_n:m,n\in\mathbb N\}$$

is countable and forms an orthonormal basis for $\mathcal H_A\otimes\mathcal H_B$ after completion.

This is one reason ordinary two-particle wave mechanics can still be handled with countable basis expansions even though the configuration space is continuous. The finite-dimensional tensor-product construction is Tensor Products.

Nonseparable Hilbert Spaces

Nonseparable Hilbert spaces exist. A simple example is $\ell^2(I)$ for an uncountable index set $I$, with orthonormal vectors $\{e_i:i\in I\}$. Any dense subset must be large enough to approximate all those mutually orthogonal directions, so no countable dense subset exists.

Such spaces are not the default setting for elementary wave mechanics or ordinary finite-particle quantum systems. They can appear in more specialized mathematical contexts, so one should not state separability as automatic without hypotheses.

Common Mistakes

• Confusing a separable Hilbert space with a separable mixed state in entanglement theory.
• Thinking separability means there are only countably many physical states.
• Thinking $L^2(\mathbb R)$ is nonseparable because $\mathbb R$ is uncountable.
• Confusing a countable orthonormal basis with the continuum of position labels.
• Assuming separability rules out continuous spectra.
• Forgetting the completion step when forming infinite-dimensional tensor products.
• Treating nonseparable Hilbert spaces as impossible rather than merely nonstandard for introductory quantum mechanics.

Cross-Links

• Hilbert Spaces
• Completeness and Orthonormal Bases
• L2 Spaces
• Position and Momentum Representations
• Finite-Dimensional Hilbert Spaces
• Tensor Products
• Finite vs Infinite-Dimensional Quantum Mechanics

References

• P. R. Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, 2nd ed., Chelsea, 1957.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• G. B. Folland, Real Analysis, 2nd ed., Wiley, 1999.

Exercises

1. Why is every finite-dimensional Hilbert space separable?

Solution

Choose a finite basis. Finite linear combinations with rational real and imaginary coefficients form a countable dense subset. Equivalently, a finite orthonormal basis is already a countable complete orthonormal basis.

2. Explain why $\ell^2(\mathbb N)$ is separable.

Solution

The standard sequence vectors $e_n$ form a countable complete orthonormal basis. Finite linear combinations of the $e_n$ with rational complex coefficients form a countable dense subset.

3. Does separability rule out a continuous position spectrum?

Solution

No. The space $L^2(\mathbb R)$ is separable, but the position operator has continuous spectrum in the usual representation. The position labels refer to generalized eigenvectors, not to a countable orthonormal basis of normalizable states.

4. Give a simple example of a nonseparable Hilbert space.

Solution

Let $I$ be an uncountable set. The space $\ell^2(I)$ has an orthonormal family $\{e_i:i\in I\}$ with uncountably many mutually orthogonal unit vectors. No countable subset can be dense, so the Hilbert space is nonseparable.