adjoint operator of the partial trace map

Question

Could someone explain to me, what is the adjoint map of the partial trace map the (tensored with the identity map), or why does the following equality hold? $Tr(C_A\cdot Tr_{B} D_{AB})=Tr((C_A\otimes Id_B)\cdot D_{AB})$? I think I can see it intuitively from wikipedia, but mathematically can someone prove it?

Answer 1

If you write $D=\sum_j X_j\otimes Y_j$, with $X_j\in A$, $Y_j\in B$, then $$ \text{Tr}_B(D)=\sum_j\text{Tr}(Y_j)X_j, $$ and $$ \text{Tr}(C\cdot\text{Tr}(D))=\text{Tr}(\sum_j \text{Tr}(Y_j)CX_j)=\sum_j\text{Tr}(CX_j)\text{Tr}(Y_j). $$ Also, $$ \text{Tr}((C\otimes I)D)=\text{Tr}(\sum_j CX_j\otimes Y_j)=\sum_j\text{Tr}(CX_j)\text{Tr}(Y_j). $$

Answer 2

A bit late, but the formula you wrote above can be seen as the coordinate-free definition of the partial trace. Namely, the partial trace $tr_W$ is the unique operator, that for all $M \in V\otimes W$ and $N \in V$ fulfills that

$ tr( M_{VW} ⋅ N_{V} \otimes I_W) = tr( tr_W(M_{VW}) \cdot N_V)$.

Because $\langle A,B \rangle = tr(A^\dagger B)$ is the Hilber-Schmidt inner product, the partial trace can be seen as the the adjoint to the map $V \to V \otimes I$.