How do I make sense of
$$
tr\Big(\rho(X\otimes I)\Big)=\sum_{a,b,a',b'} \rho_{ab,a'b'}X_{a,a'}\delta_{b,b'}
$$
If this does not involve any tensor products it would be a simple elementwise matrix multiplication.
If I consider sample of smaller matrices I would be able to probabily verify this. But is there any way I can see through this expression for trace which involves tensor products ?
Original Context from my Reference
Refer to Page 18
where $\rho^A$ and $\rho^{AB}$ are positive definite matrices and $X$ is hermitian.
$$(\rho^A X)_{ab}=\sum_{a'} \rho^A_{aa'}X_{a'b}\implies tr(\rho^A X)=\sum_{a}(\rho^A X)_{aa}=\sum_{aa'} \rho^A_{aa'}X_{a'a}$$
How does $tr(\rho^A X)=\sum_{aa'} \rho^A_{aa'}X_{aa'}$ instead of $\sum_{aa'} \rho^A_{aa'}X_{a'a}$ ?
Best Answer
One trick that I often find to be helpful is to begin by seeing what happens when all matrices involved in the computation are expressed in the form of a tensor product. In this case, this would mean replacing $A \otimes B$. From there, we can extend our observation to the general case either by noting that the expression is linear or by expressing $\rho$ as a linear combination of matrices of the form $A \otimes B$. For instance, if we let $E_{ij}$ denote the matrix with a $1$ as its $i,j$ entry and zeros elsewhere, then we can always write $\rho = \sum_{i,j} E_{ij} \otimes \sigma_{ij}$, where $\sigma_{ij}$ is the $i,j$ "block-entry" of $\rho$.
For this particular problem, if we write $\rho = \sum_{i,j} E_{ij} \otimes \sigma_{ij}$, then we find that $$ \begin{align} \operatorname{tr}(\rho(X \otimes I)) & = \operatorname{tr}\left(\sum_{i,j} (E_{ij} \otimes \sigma_{ij})(X \otimes I)\right) \\ & = \sum_{i,j} \operatorname{tr}[(E_{ij} \otimes \sigma_{ij})(X \otimes I)] \\ & = \sum_{i,j} \operatorname{tr}[(E_{ij} X) \otimes \sigma_{ij}] \\ & = \sum_{i,j}\operatorname{tr}(E_{ij}X) \operatorname{tr}(\sigma_{ij}) = \sum_{i,j}x_{ji} \operatorname{tr}(\sigma_{ij}). \end{align} $$
I suspect that $\rho$ is indexed such that $$ \rho = \sum_{a,b,a',b'} \rho_{ab,a'b'} E_{ab}\otimes E_{a'b'}. $$ Equivalently, if the tensor product of matrices is expressed as a single matrix using the Kronecker product, then $\rho_{ab,a'b'}$ is the $b,b'$-entry of the "block-entry" $\sigma_{a,a'}$ referenced in my first paragraph.
With that established, my simplification above can be rewritten as $$ \begin{align} \operatorname{tr}(\rho(X \otimes I)) & = \sum_{a,a'}X_{a',a} \operatorname{tr}(\sigma_{a,a'}) \\ & = \sum_{a,a'}X_{a',a} \sum_{b,b'} \sigma_{a,a'}[b,b'] \delta_{b,b'} \\ & = \sum_{a,a'}X_{a',a} \sum_{b,b'} \rho_{ab,a'b'} \delta_{b,b'} = \sum_{a,b,a',b'}X_{a',a} \rho_{ab,a'b'} \delta_{b,b'}. \end{align} $$ We could also have arrived at this formula directly from the expansion $$ \rho = \sum_{a,b,a',b'} \rho_{ab,a'b'} E_{ab}\otimes E_{a'b'}. $$ Indeed, we have $$ \begin{align} \operatorname{tr}(\rho(X \otimes I)) & = \operatorname{tr}\left(\sum_{a,b,a',b'} (\rho_{ab,a'b'} E_{aa'} \otimes E_{bb'})(X \otimes I)\right) \\ & = \sum_{a,b,a',b'} \rho_{ab,a'b'} \operatorname{tr}[(E_{aa'}\otimes E_{bb'})(X \otimes I)] \\ & = \sum_{a,b,a',b'} \rho_{ab,a'b'} \operatorname{tr}(E_{aa'}X)\operatorname{tr}(E_{bb'}) = \sum_{a,b,a',b'} \rho_{ab,a'b'} X_{a',a} \delta_{b,b'}. \end{align} $$