Product vectors, separable states and entangled states of a two-factor composite

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $H,K$ be nonzero complex Hilbert spaces (Hilbert space) and let $H\widehat{\otimes }K$ be their completed tensor product with inclusion $\kappa$ (Hilbert tensor construction: inner product, completion and product bases). Density operators are the self-adjoint positive trace-class operators of trace one (Density operators and scalar spectral probability clause (a), Trace class operator), and for $u,u^{\mathrm{\prime }}\in H$, $v,v^{\mathrm{\prime }}\in K$ the rank-one operator $T_{u\otimes v,u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}}$ is the one of Rank-one operator calculus.

(a) Product vectors and pure product states. A vector $\mathrm{\Psi }\in H\widehat{\otimes }K$ is a product vector when $\mathrm{\Psi }=\kappa (u\otimes v)$ for some $u\in H$, $v\in K$. A pure product state of the composite is a density operator of the form $T_{\mathrm{\Psi },\mathrm{\Psi }},\mathrm{\Psi }=\kappa (u\otimes v),\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }v\mathrm{\parallel }=1.$ A vector is non-product when it is not a product vector.

(b) Separable states. Write $\mathcal{P}$ for the set of all pure product states, a subset of the set $D(H\widehat{\otimes }K)$ of density operators. The separable states are the elements of the trace-norm closed convex hull $\mathcal{S}(H,K):=\stackrel{}{\mathrm{conv}}\mathcal{P}\subseteq D(H\widehat{\otimes }K),$ where $\mathrm{conv}$ denotes the set of finite convex combinations and the closure is taken in the trace norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$ of Trace class is a two sided Banach operator ideal. A state $\rho \in D(H\widehat{\otimes }K)$ is entangled (with respect to the declared splitting $H,K$) when $\rho \mathrm{\notin }\mathcal{S}(H,K)$.

(c) Conventions and elementary consequences of the form of the definition. The set $\mathrm{conv}\mathcal{P}$ consists exactly of the finite sums ${\sum }_{j=1}^n{w}_j{T}_{{\mathrm{\Psi }}_j,{\mathrm{\Psi }}_j}$ with $n\ge 1$, weights $w_j\ge 0$, ${\sum }_j{w}_j=1$ and unit product vectors ${\mathrm{\Psi }}_j$: such sums are convex combinations of density operators and hence density operators (The state space is convex, closed under countable mixtures, and complete in the trace norm clause (a)), without any orthogonality requirement on the product vectors. Every element of $\mathrm{conv}\mathcal{P}$ is a density operator, and $\mathcal{S}(H,K)$ is a convex subset of $D(H\widehat{\otimes }K)$: if $\rho ,\sigma$ are limits of convex combinations ${\rho }_n\to \rho$, ${\sigma }_n\to \sigma$ and $t\in [0,1]$, then $t\rho +(1-t)\sigma$ is the trace-norm limit of the convex combinations $t{\rho }_n+(1-t){\sigma }_n$. Separability of a state is a property of the declared splitting $(H,K)$; no assertion is made here about its behaviour under a different factorization of a larger space.

(d) Finite dimension. When $H$ and $K$ are finite dimensional, $\mathcal{P}$ is compact and $\mathrm{conv}\mathcal{P}$ is already closed in the trace norm (it is a compact, hence closed, convex set in the finite-dimensional space of endomorphisms of $H\widehat{\otimes }K$), so $\mathcal{S}(H,K)=\mathrm{conv}\mathcal{P}$ and no closure convention is needed. In the general case the closure in clause (b) is declared explicitly and is part of the definition.

(e) Pure-state specialization and status. A pure state $T_{\mathrm{\Psi },\mathrm{\Psi }}$ with $\mathrm{\parallel }\mathrm{\Psi }\mathrm{\parallel }=1$ is separable in the precise trace-norm closed-convex-hull sense of clause (b) exactly when $\mathrm{\Psi }$ is a product vector of clause (a); otherwise it is entangled. This is the proved mathematical specialization of Pure-state separability is equivalent to being a product: a Schmidt overlap witness, whose locally defined $\mathcal{P},\mathcal{S}$ are exactly those above. Clauses (a)-(e) fix mathematical vocabulary and adopt no physical assumption, dynamics, measurement or probability law. Whether a laboratory preparation realizes a separable or entangled state for the declared splitting remains a tested identification with uncertainty, not a consequence about nature of this definition.

• Why the closure is explicit. In infinite dimension the convex hull of the pure product states need not be trace-norm closed, and different authors complete it differently. Clause (b) declares the trace-norm closure, which is the convention used by this category; the mathematical results of the page are stated for this declared class.

• Entanglement is relative to a splitting. The same vector of a three-factor space may be a product vector with respect to one grouping of the factors and non-product with respect to another, so a separability or entanglement statement without the declared pair $(H,K)$ is incomplete; the definition above always carries that pair.

• Finite-dimensional closed convex hull. Suppose $H$ and $K$ have complex dimensions $h,k\ge 1$, put $L=H\widehat{\otimes }K$ and $N=hk$, and let $W=\{T\in \mathcal{B}(L):T=T^{\ast }\}$, regarded as a real vector space. In orthonormal coordinates, $W$ has real dimension $d=N^2$: there are $N$ real diagonal entries and $N(N-1)\mathrm{/}2$ free complex entries above the diagonal. The product of the unit spheres of $H$ and $K$ is a closed bounded subset of ${\mathbb{R}}^{2h+2k}$, hence compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. The map $(u,v)\mapsto \kappa (u\otimes v)$ is continuous: bilinearity and the tensor norm formula give $\mathrm{\parallel }\kappa (u\otimes v)-\kappa (u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }})\mathrm{\parallel }\le \mathrm{\parallel }u-u^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel }$. The map $q\mapsto T_{q,q}$ is trace-norm continuous by Rank-one operator calculus clause (f). Therefore $\mathcal{P}$ is compact by The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset.

  Every finite convex combination of more than $d+1$ members $P_1,\dots ,P_m\in \mathcal{P}$ with positive weights $w_i$ can be reduced. The vectors $(P_i,1)$ in $W\times \mathbb{R}$, a real vector space of dimension $d+1$, are linearly dependent; hence there are real $c_i$, not all zero, with ${\sum }_i{c}_i{P}_i=0$ and ${\sum }_i{c}_i=0$. The $c_i$ have both signs. Set $t={\min }_{c_i>0}{w}_i\mathrm{/}{c}_i$ and $w_i^{\mathrm{\prime }}=w_i-t{c}_i$. Then every $w_i^{\mathrm{\prime }}\ge 0$, ${\sum }_i{w}_i^{\mathrm{\prime }}=1$, ${\sum }_i{w}_i^{\mathrm{\prime }}{P}_i={\sum }_i{w}_i{P}_i$, and at least one $w_i^{\mathrm{\prime }}$ is zero. Removing zero weights and repeating leaves at most $d+1$ terms. Since $\mathcal{P}$ is nonempty, shorter combinations can be padded with zero weights. Thus $\mathrm{conv}\mathcal{P}$ is the image of ${\mathcal{P}}^{d+1}\times {\mathrm{\Delta }}_d$ under the continuous map $((P_i),(w_i))\mapsto {\sum }_i{w}_i{P}_i$, where ${\mathrm{\Delta }}_d=\{(w_1,\dots ,w_{d+1}):w_i\ge 0,{\sum }_i{w}_i=1\}$. The trace-norm and Euclidean coordinate topologies on $W$ agree. In an orthonormal basis $(g_i{)}_{i=1}^N$ of $L$, write $T={\sum }_{i,j}{t}_{ij}{T}_{g_i,g_j}$ and use the Euclidean coordinate norm $\mathrm{\parallel }T{\mathrm{\parallel }}_2:=({\sum }_{i,j}\mathrm{\mid }{t}_{ij}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. Each coordinate satisfies $\mathrm{\mid }{t}_{ij}\mathrm{\mid }=\mathrm{\mid }⟨T{g}_j,g_i⟩\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }\le \mathrm{\parallel }T{\mathrm{\parallel }}_1$; conversely, $\mathrm{\parallel }T{\mathrm{\parallel }}_1\le {\sum }_{i,j}\mathrm{\mid }{t}_{ij}\mathrm{\mid }\mathrm{\parallel }{T}_{g_i,g_j}{\mathrm{\parallel }}_1={\sum }_{i,j}\mathrm{\mid }{t}_{ij}\mathrm{\mid }\le N\mathrm{\parallel }T{\mathrm{\parallel }}_2$, since the matrix units have trace norm one by Rank-one operator calculus and there are $N^2$ coordinates. Thus $\mathrm{\parallel }T{\mathrm{\parallel }}_2\le N\mathrm{\parallel }T{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }T{\mathrm{\parallel }}_1\le N\mathrm{\parallel }T{\mathrm{\parallel }}_2$, proving equality of the topologies. Hence in finite coordinates $\mathcal{P}$ is closed and bounded because it is compact; the simplex ${\mathrm{\Delta }}_d$ is also closed and bounded. Therefore ${\mathcal{P}}^{d+1}\times {\mathrm{\Delta }}_d$ is closed and bounded in a Euclidean space, hence compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Its image is compact and therefore closed in the trace-norm space. Finally, $\mathrm{conv}\mathcal{P}\subseteq D(L)$ by The state space is convex, closed under countable mixtures, and complete in the trace norm clause (a), and its trace-norm closure remains in $D(L)$ by that lemma's clause (d). This proves the finite-dimensional assertion of clause (d); clause (b)'s trace-norm closure remains the declared convention in every dimension.