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.

  • $\begingroup$ 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? $\endgroup$ – Norbert Schuch Sep 5 at 14:13
  • $\begingroup$ And can you explain your title? $\endgroup$ – Norbert Schuch Sep 5 at 14:21
  • $\begingroup$ @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. $\endgroup$ – zef wolffs Sep 5 at 14:56
  • $\begingroup$ @NorbertSchuch is the following correct: $\endgroup$ – zef wolffs Sep 5 at 15:01
  • $\begingroup$ What following? $\endgroup$ – 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.

Applying this to the map under consideration you get

$$ \mathrm{Tr}(A^n_m (\rho)) = \sum_{k_1\cdots k_n}\langle k_1\cdots k_n| \left( \sum_{ij} | i...i \rangle^n \langle i...i|^m \rho | j...j \rangle^m \langle j...j|^n \right) |k_1\cdots k_n\rangle\\ =\sum_{\ell}\langle \underbrace{\ell\cdots \ell}_m|\rho|\underbrace{\ell\cdots \ell}_m\rangle,$$ where we used $$\langle k_1\cdots k_n|\underbrace{i\cdots i}_n\rangle=\delta_{k_1,i}\cdots \delta_{k_n,i}.$$


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