Reduced preparations of a declared composite: the subsystem states of the framework

Adopt the declared composite model of Composite systems of the framework: product bases, local readouts, local effects and local instruments with subsystems $A,B$, assigned spaces $H,K$ and composite space $H\widehat{\otimes }K$, the state postulate P1 (State and composition postulate P1: Hilbert spaces, preparations and composite systems) and the state space $D(H\widehat{\otimes }K)$ of the framework (The state space of the framework: preparations, pure rays and density operators). Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), the hypothesis of the cited partial-trace and tensor construction results. The following vocabulary is used by this page.

(a) Reduced preparations. For a preparation $\rho \in D(H\widehat{\otimes }K)$ of the declared composite, the reduced preparation of subsystem $A$ is ${\rho }_A:={\mathrm{Tr}}_K\rho \in D(H),$ the image of $\rho$ under the partial trace ${\mathrm{Tr}}_K$ of Separable partial trace and its basic properties clause (a), characterized by $\mathrm{tr}({\rho }_{A}X)=\mathrm{tr}(\rho (X\otimes I_K))\text{for every bounded }X\in \mathcal{B}(H);$ the reduced preparation of subsystem $B$ is ${\rho }_B:={\mathrm{Tr}}_H\rho \in D(K)$, characterized by $\mathrm{tr}({\rho }_{B}Y)=\mathrm{tr}(\rho (I_H\otimes Y))$ for every bounded $Y\in \mathcal{B}(K)$. By clause (b) of Separable partial trace and its basic properties the partial trace maps density operators to density operators and is trace preserving, so ${\rho }_A$ and ${\rho }_B$ are states of the respective subsystems in the sense of the state space of the framework.

(b) Affineness and continuity. The maps $\rho \mapsto {\rho }_A$ and $\rho \mapsto {\rho }_B$ are affine and continuous in the trace norm: ${\mathrm{Tr}}_K$ is linear and $\mathrm{\parallel }{\mathrm{Tr}}_{K}T{\mathrm{\parallel }}_1\le \mathrm{\parallel }T{\mathrm{\parallel }}_1$ by clause (a) of Separable partial trace and its basic properties, with the trace norm of Trace class is a two sided Banach operator ideal. In particular the reduction of a statistical mixture ${\sum }_n{w}_n{\rho }_n\in D(H\widehat{\otimes }K)$ of composite preparations is the corresponding mixture ${\sum }_n{w}_n{\rho }_{A,n}$ of the reduced preparations.

(c) Product preparations reduce to their factors. If $\mathrm{\Psi }=\kappa (u\otimes v)$ with $\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }v\mathrm{\parallel }=1$ and $\rho =T_{\mathrm{\Psi },\mathrm{\Psi }}$ is a product preparation (Composite systems of the framework: product bases, local readouts, local effects and local instruments clause (b), Hilbert tensor construction: inner product, completion and product bases), then $T_{u\otimes v,u\otimes v}=T_{u,u}\otimes T_{v,v}$ as bounded operators on $H\widehat{\otimes }K$, because both sides agree on every elementary tensor $\kappa (x\otimes y)$ in the sense of Bounded tensor maps: extension, norm estimate, adjoint and commutation clause (a) and $\kappa [H\odot K]$ is dense (Hilbert tensor construction: inner product, completion and product bases clause (c)); consequently ${\rho }_A=T_{u,u},{\rho }_B=T_{v,v}$ by clause (c) of Separable partial trace and its basic properties. Thus a product preparation is exactly a preparation whose reductions reconstruct it, $\rho ={\rho }_A\otimes {\rho }_B$, and whose reductions are pure.

(d) Correlated and uncorrelated preparations. A composite preparation $\rho$ is called uncorrelated when $\rho ={\rho }_A\otimes {\rho }_B$ and correlated otherwise; a pure preparation $T_{\mathrm{\Psi },\mathrm{\Psi }}$ is called a product preparation when $\mathrm{\Psi }$ is a product vector and non-product otherwise (Product vectors, separable states and entangled states of a two-factor composite clauses (a), (b)). These are descriptive names for the displayed algebraic conditions; the statistical reading of the distinction, including the failure of factorization of local outcome laws for non-product pure preparations, is a theorem of this page and is not assumed here.

(e) Status. Clauses (a)-(d) fix vocabulary: the subsystem states attached to a composite preparation by the partial trace. The reduction is a mathematical operation on the model description. It is not a dynamics, not an interaction, not a measurement and not a physical discarding of subsystem $B$; no disturbance of either subsystem and no empirical fact is asserted, and the question whether a laboratory procedure prepares a state whose reduction is a prescribed state is a tested identification with uncertainty under P7. The physical reading of clause (a) — that the reduced preparation reproduces every local outcome law of subsystem $A$ — is a theorem of this page and is not assumed here.

• Reverse product criteria. To justify the reverse implications in clause (c), take a unit vector $\mathrm{\Psi }$ with Schmidt expansion $\mathrm{\Psi }={\sum }_r{s}_r\kappa (v_r\otimes u_r)$, where $s_r>0$, ${\sum }_r{s}_r^2=1$, and the two factor families are orthonormal. The proof of Infinite Schmidt decomposition and the pure product criterion steps 3.1 and 6.1 identifies the eigenvalues of ${\rho }_A={\mathrm{Tr}}_K{T}_{\mathrm{\Psi },\mathrm{\Psi }}$ as $s_r^2$ and proves $\mathrm{tr}({\rho }_A^2)={\sum }_r{s}_r^4=1$ exactly when there is one Schmidt coefficient, exactly the product-vector case. The partial-trace formula in Separable partial trace and its basic properties clause (c), applied to finite Schmidt truncations, gives the finite reduction sums. The rank-one map is trace-norm continuous and the partial trace is trace-norm contractive, so passage to the trace-norm limit gives ${\rho }_B={\sum }_r{s}_r^2{T}_{u_r,u_r}$; hence its purity is the same and purity of either reduction implies that $\mathrm{\Psi }$ is a product vector. Clause (c)'s partial-trace calculation proves the converse.

  The other reverse criterion, $\rho ={\rho }_A\otimes {\rho }_B$ implying a product vector, follows from the same expansion. For each $r$, put $q_r=\kappa (v_r\otimes u_r)$. The expansion and orthonormality give $⟨\rho q_r,q_r⟩=s_r^2$ and $⟨{\rho }_A{v}_r,v_r⟩=⟨{\rho }_B{u}_r,u_r⟩=s_r^2$. The bounded tensor calculus then gives $⟨({\rho }_A\otimes {\rho }_B)q_r,q_r⟩=s_r^4$. Equality of the composite operators forces $s_r^2=s_r^4$ for each positive coefficient. Since ${\sum }_r{s}_r^2=1$, exactly one coefficient is present. Conversely clause (c)'s product calculation gives $\rho ={\rho }_A\otimes {\rho }_B$ for a product vector.