Separable partial trace and its basic properties

[The rank-one values] For $T_{u\otimes v,u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}}$ and bounded $A$, $T(A\otimes I)=T_{u\otimes v,(A\otimes I{)}^{\ast }(u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }})}=T_{u\otimes v,A^{\ast }{u}^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}}$ by [F1], whose trace is $⟨u\otimes v,A^{\ast }{u}^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}⟩=⟨u,A^{\ast }{u}^{\mathrm{\prime }}⟩⟨v,v^{\mathrm{\prime }}⟩=⟨Au,u^{\mathrm{\prime }}⟩⟨v,v^{\mathrm{\prime }}⟩$. On the other hand $\mathrm{tr}((⟨v,v^{\mathrm{\prime }}⟩T_{u,u^{\mathrm{\prime }}})A)=⟨v,v^{\mathrm{\prime }}⟩\mathrm{tr}(T_{u,u^{\mathrm{\prime }}}A)=⟨v,v^{\mathrm{\prime }}⟩⟨Au,u^{\mathrm{\prime }}⟩$ by [F1]. Hence the assignment ${\mathrm{Tr}}_K{T}_{u\otimes v,u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}}:=⟨v,v^{\mathrm{\prime }}⟩T_{u,u^{\mathrm{\prime }}}$ satisfies the defining identity on rank-one tensors, and $\mathrm{\parallel }{\mathrm{Tr}}_K{T}_{u\otimes v,u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}}{\mathrm{\parallel }}_1\le \mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }{v}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }=\mathrm{\parallel }u\otimes v\mathrm{\parallel }\mathrm{\parallel }{u}^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}\mathrm{\parallel }$ by [F1].

[Arbitrary rank-one vectors] Fix $w,z\in H\widehat{\otimes }K$. By density and Countable Choice choose sequences of finite algebraic tensors converging to these two vectors; collect their countably many second-factor vectors and let $K_0$ be their closed linear span. Finite rational complex combinations form an explicit dense sequence in $K_0$, so [F4] provides a finite/countable orthonormal basis $(f_b)$ of $K_0$. The maps taking ${\sum }_i{u}_i\otimes v_i$ to ${\sum }_i⟨v_i,f_b⟩u_i$ have norm at most one by orthogonal coordinates and extend to the completion. Their values $w_b,z_b\in H$ satisfy $w={\sum }_b{w}_b\otimes f_b$, $z={\sum }_b{z}_b\otimes f_b$ and ${\sum }_b\mathrm{\parallel }{w}_b{\mathrm{\parallel }}^2=\mathrm{\parallel }w{\mathrm{\parallel }}^2$, ${\sum }_b\mathrm{\parallel }{z}_b{\mathrm{\parallel }}^2=\mathrm{\parallel }z{\mathrm{\parallel }}^2$: the finite-coordinate projections are contractions, converge on algebraic tensors in $H\odot K_0$, and hence converge on its closure, which contains $w,z$. Define $Q(w,z):={\sum }_b{T}_{w_b,z_b}$. By [F1], [F2] and Cauchy–Schwarz, this converges in trace norm and $\mathrm{\parallel }Q(w,z){\mathrm{\parallel }}_1\le {\sum }_b\mathrm{\parallel }{w}_b\mathrm{\parallel }\mathrm{\parallel }{z}_b\mathrm{\parallel }\le \mathrm{\parallel }w\mathrm{\parallel }\mathrm{\parallel }z\mathrm{\parallel }$. Finite-coordinate truncations and step 1.1 give $\mathrm{tr}(Q(w,z)A)=⟨(A\otimes I)w,z⟩=\mathrm{tr}(T_{w,z}(A\otimes I))$; rank-one trace-norm continuity and bounded tensor maps [F4] pass this identity to limits. The identity uniquely determines $Q(w,z)$ by [F3], so it is independent of the supports, basis and approximants, is sesquilinear in $w,z$, and agrees with the elementary assignment.

[Extension to trace class] For a nuclear representation $T={\sum }_j{T}_{w_j,z_j}$, define ${\mathrm{Tr}}_{K}T:={\sum }_{j}Q(w_j,z_j)$. Step 2.1 and [F2] give trace-norm convergence and the bound $\mathrm{\parallel }{\mathrm{Tr}}_{K}T{\mathrm{\parallel }}_1\le \inf {\sum }_j\mathrm{\parallel }{w}_j\mathrm{\parallel }\mathrm{\parallel }{z}_j\mathrm{\parallel }=\mathrm{\parallel }T{\mathrm{\parallel }}_1$. The defining trace pairing follows by summing that of step 2.1, using trace-norm continuity and $\mathrm{\parallel }A\otimes I\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }$. If two nuclear representations are used, both images have the same pairings against every bounded $A$, and their difference vanishes by [F3]. Thus the definition is independent of representation. The same uniqueness proves linearity, since the sum and scalar multiple of images satisfy the pairing for the sum and scalar multiple of inputs. This proves (a) and the elementary formula (c).

[Positivity and trace preservation] Step 2.1 gives $Q(w,w)={\sum }_b{T}_{w_b,w_b}$, which is self-adjoint positive since $⟨T_{w_b,w_b}x,x⟩=\mathrm{\mid }⟨x,w_b⟩{\mathrm{\mid }}^2\ge 0$ and trace-norm convergence implies operator-norm convergence. Every self-adjoint positive trace-class input of positive trace becomes a density operator after division by its trace, so it has the positive spectral rank-one expansion of Density operators and scalar spectral probability; a zero-trace self-adjoint positive trace-class input is zero by Trace of a positive operator is the sum of its eigenvalues. The bounded map of step 3.1 carries the partial sums of this expansion to positive sums of the corresponding $Q(w,w)$ and hence carries its limit to a self-adjoint positive operator. Taking $A=I_H$ in the defining pairing gives $\mathrm{tr}{\mathrm{Tr}}_{K}T=\mathrm{tr}T$. Hence density operators map to density operators.

[Uniqueness] If two linear maps ${\mathrm{Tr}}_K,{\mathrm{Tr}}_K^{\mathrm{\prime }}$ satisfy the defining identity for all $T$ and $A$, then for fixed $T$ the difference $S={\mathrm{Tr}}_{K}T-{\mathrm{Tr}}_K^{\mathrm{\prime }}T$ satisfies $\mathrm{tr}(SA)=0$ for all bounded $A$, so $S=0$ by [F3]. Hence the map is unique and, in particular, independent of any basis; the symmetrically defined ${\mathrm{Tr}}_H$ has the stated identity.