Separable and Entangled Mixed States

In infinite dimensions, a separable normal state is a trace-norm limit of finite convex mixtures of product states. Requiring one exact finite or countable ensemble would give a smaller class in general. The canonical Separable Mixed States article develops the physical distinction between product, separable, and entangled states; this page supplies the topological qualification.

Helpful background. Product States establishes normal product states and their continuity. Trace Norm and Operator Norm supplies the operational topology. Convex State Space supplies convexity and closure.

Separability uses the closed convex hull

Let both factors be separable Hilbert spaces. Let $\mathcal P$ denote the set of pure product projectors. Define

$$\operatorname{Sep}(A:B) =\overline{\operatorname{conv}\mathcal P}^{\,\|\cdot\|_1}.$$

A normal state outside this set is entangled. The definition includes every finite mixture, every trace-norm convergent countable mixture, and every limit of such preparations. It keeps the subsystem split explicit.

Allowing mixed product factors gives the same closed convex hull. Each local density operator has a countable spectral decomposition; taking products and finite approximations expresses a mixed product as a trace-norm limit of pure product mixtures.

In finite dimension $d=d_Ad_B$, the pure product projectors form a compact set. Their convex hull is compact, so closure adds nothing. Carathéodory’s theorem gives a representation using at most $d^2$ pure product projectors, since trace-one Hermitian matrices have real affine dimension $d^2-1$. There is no dimension-independent finite bound to carry into the infinite-dimensional setting.

Continuous ensembles are ordinary probability mixtures

Let $\nu$ be a Borel probability measure on the pair of local pure-state spaces, represented by rank-one projectors $P,Q$. The integral

$$\rho=\int P\otimes Q\,d\nu(P,Q)$$

exists as a Bochner integral in the trace-class Banach space. The integrand is measurable and has trace norm one. It is positive and has trace one after integration.

Such a barycenter is separable: approximate the integrable function in mean trace norm by simple functions, with product-state values, to obtain finite convex mixtures converging to it. Conversely, every separable normal state admits a probability-measure representation of this form. That converse is a nontrivial barycentric theorem, proved by Holevo, Shirokov, and Werner; it is not an application of finite-dimensional Carathéodory with the dimension silently omitted.

A purely atomic measure gives a countable ensemble. General probability measures need not be atomic, and a continuous ensemble cannot always be replaced by a countable one yielding exactly the same density operator.

Why countable decomposability is stronger

If

$$\rho=\sum_kp_k|a_k,b_k\rangle\langle a_k,b_k|, \qquad p_k>0,$$

then any one term gives a nonzero product projector dominated by $\rho$:

$$\rho\geq p_k|a_k,b_k\rangle\langle a_k,b_k|.$$

The same necessary condition holds for countable mixtures of mixed product states after refining their spectral decompositions. Some separable states dominate no nonzero product rank-one operator. They therefore cannot have such an ensemble, despite being limits of finite mixtures.

Holevo, Shirokov, and Werner construct explicit examples from correlated translations of two vectors on the circle whose Fourier coefficients are all nonzero. Their theorem establishes non-countable decomposability; merely displaying a continuous integral would not establish it, because the same state might otherwise have an unrelated discrete ensemble.

The distinction is between exact representation and arbitrarily accurate approximation. It does not obstruct preparation to any prescribed trace-norm accuracy by finite mixtures.

Phase randomization gives a calculable barycenter

For an illustration of a continuous mixture, let $\mathcal H_A=\mathcal H_B=\ell^2(\mathbb N_0)$ and take unit vectors $a=\sum_ma_m|m\rangle$, $b=\sum_nb_n|n\rangle$. Set $U_\theta|n\rangle=e^{in\theta}|n\rangle$ and average the product state of $U_\theta a$ and $U_\theta b$ uniformly over $\theta\in[0,2\pi]$.

The resulting matrix elements are

$$\langle m,n|\rho|p,r\rangle =a_mb_n\overline{a_p b_r}\, \delta_{m+n,p+r}.$$

This follows by integrating $e^{i(m+n-p-r)\theta}$. Common phase randomization removes coherence between different total-number sectors, while retaining possible coherence inside each sector. The resulting state is separable by its construction even though it need not be diagonal in a product basis.

The integrand is trace-norm continuous in $\theta$: the unitary orbit is norm continuous on each fixed vector, and rank-one projectors inherit that continuity. Finite Riemann sums therefore converge in trace norm. This example establishes the approximation mechanism; no non-countable-decomposability claim is needed for it.

Every entangled normal state has a bounded witness

The separable set is closed and convex in the real Banach space of self-adjoint trace-class operators. If $\rho$ lies outside it, the Hahn–Banach separation theorem gives a continuous real functional and a strict separating threshold.

The duality between trace class and bounded operators writes that functional as $\sigma\mapsto\operatorname{Tr}(H\sigma)$ for bounded self-adjoint $H$. Subtracting the threshold times $I$ produces $W$ with

$$\operatorname{Tr}(W\sigma)\geq0 \quad(\sigma\in\operatorname{Sep}), \qquad \operatorname{Tr}(W\rho)<0.$$

This proves existence of a bounded entanglement witness. It does not promise a simple formula, efficient measurement, or a finite number of settings. For pure states, the largest-Schmidt-coefficient construction gives an explicit witness.

Exercises

1. Explain why an entangled state has a trace-norm neighborhood containing only entangled states.

Solution

The separable set is closed, so its complement is open. Quantitatively, a bounded witness with $\operatorname{Tr}(W\rho)=-\eta<0$ stays negative whenever $\|\rho-\rho'\|_1<\eta/\|W\|$. Trace duality bounds the change in its expectation.

2. Does an ensemble consisting entirely of entangled pure states prove that its average is entangled?

Solution

No. Separability asks whether the density operator belongs to the closed convex hull of product projectors, regardless of one displayed ensemble. Different ensembles can have the same barycenter. A witness or another valid criterion must rule out all separable representations.

References

• Holevo, Alexander S., Maksim E. Shirokov, and Reinhard F. Werner, “Separability and Entanglement-Breaking in Infinite Dimensions,” 2005, arXiv:quant-ph/0504204 — closed-convex separability, probability-measure representations, and non-countably-decomposable examples. Paper.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, second revised and expanded edition, De Gruyter, 2019 — infinite-dimensional state spaces and entanglement. Publisher.
• Watrous, John, The Theory of Quantum Information, Cambridge University Press, 2018 — finite-dimensional convex geometry and entanglement witnesses. Author’s text.