# Trace of the quantum map $A^n_m (\rho) = \sum_{ij} | i…i \rangle^n \langle i…i|^m \rho | j…j \rangle^m \langle j…j|^n$

We define some quantum map $$A^n_m (\rho)$$ and let it act on density matrix $$\rho$$:

$$A^n_m (\rho) = \sum_{ij} | i...i \rangle^n \langle i...i|^m \rho | j...j \rangle^m \langle j...j|^n.$$

Taking the trace of this should result in the following:

$$Tr(A^n_m (\rho)) = \sum_i \langle i...i|^m \rho | i...i \rangle^m.$$

I would greatly appreciate it if someone could explain me where this step comes from. For context see arXiv:1704.08668 (https://arxiv.org/abs/1704.08668) page 4.

• Those are $i$'s on the right in the 2nd formula, not $j$'s. Is it more clear then? If not, why not -- what is your definition of the trace? – Norbert Schuch Sep 5 at 14:13
• And can you explain your title? – Norbert Schuch Sep 5 at 14:21
• @NorbertSchuch Hi Norbert Schuch, I'll change the title to something more general. I initially did not give it much thought. You are right about there being i's in the second line of equations. I think the problem is indeed with my definition of the trace. As far as I know it simply sums over the diagonal components. – zef wolffs Sep 5 at 14:56
• @NorbertSchuch is the following correct: – zef wolffs Sep 5 at 15:01
• What following? – Norbert Schuch Sep 5 at 17:00

Given a matrix $$B$$ acting on an $$n$$-partite system, that is, a matrix/tensor with $$2n$$ indices $$B_{i_1\cdots i_n,j_1\cdots j_n}$$, its trace is defined as $$\mathrm{Tr} B \equiv \sum_{k_1\cdots k_n} B_{k_1\cdots k_n,k_1\cdots k_n}.$$ This is nothing but a rewriting of the usual formula $$\operatorname{Tr}A=\sum_i A_{i,i}$$ when there are $$2n$$ indices instead of just two.
$$\mathrm{Tr}(A^n_m (\rho)) = \sum_{k_1\cdots k_n}\langle k_1\cdots k_n| \left( \sum_{ij} | i...i \rangle^n \langle i...i|^m \rho | j...j \rangle^m \langle j...j|^n \right) |k_1\cdots k_n\rangle\\ =\sum_{\ell}\langle \underbrace{\ell\cdots \ell}_m|\rho|\underbrace{\ell\cdots \ell}_m\rangle,$$ where we used $$\langle k_1\cdots k_n|\underbrace{i\cdots i}_n\rangle=\delta_{k_1,i}\cdots \delta_{k_n,i}.$$