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I have come accross a formula that puzzles me a bit in the proof of lemma 23 (page 32) of this paper.

The authors start from a (translationally-invariant) matrix product state: $$\lvert\psi\rangle := \sum_{i_1, \ldots, i_n}\textrm{tr}(A_{i_1} \ldots A_{i_n})\lvert i_1 \ldots i_n\rangle$$

where the $A_i$ are matrices of dimension $D$ satisfying $\sum_{i}A_i^{\dagger}A_i = \sum_{i}A_iA_i^{\dagger} = \mathbf{1}_{D \times D}$ (that is, they constitute Kraus operators for a unital channel). Then, they claim that the reduced state of the first $l$ qudits looks like:

$$\rho_l = \sum_{i_1, \ldots, i_l}\textrm{tr}\left(A_{i_1} \ldots A_{i_l}\frac{\mathbf{1}_{D \times D}}{D}A_{j_1}^{\dagger} \ldots A_{j_l}^{\dagger}\right)\lvert i_1 \rangle\langle j_1\rvert \otimes \ldots \otimes \lvert i_l \rangle\langle j_l \rvert$$

Yet, the first expression that looks obvious to me would rather be: $$\rho_l = \sum_{\substack{i_1, \ldots, i_l\\j_1, \ldots, j_l}}\left(\sum_{i_{l + 1}, \ldots, i_n}\textrm{tr}\left(A_{i_1}\ldots A_{i_{l}}A_{i_{l + 1}}\ldots A_{i_n}\right)\textrm{tr}\left(A_{i_n}^{\dagger}\ldots A_{i_{l + 1}}^{\dagger}A_{j_l}^{\dagger}\ldots A_{j_1}^{\dagger}\right)\right)\lvert i_1 \rangle\langle j_1\rvert \otimes \ldots \otimes \lvert i_l \rangle\langle j_l \rvert$$

It looks as if by some magic the authors converted the sum of product of traces above to the trace of a product! Any idea where that would come from? At first, I thought it had to do with the $A$ being randomized (see paragraph on top of page 32), but by giving it a closer look, it seems the statements of lemma 23 do not include an expectation value with respect to the Haar measure, as opposed e.g to lemma 24, so I am not sure what to think.

EDIT: Corrected some typos in the indices in the second equation for $\rho_l$.

EDIT 2: What about this simple example? Set $D = 2, A_0 = \begin{pmatrix} \sqrt{1 - \gamma} & 0\\ 0 & 0 \end{pmatrix}, A_1 = \begin{pmatrix} \sqrt{\gamma} & 0\\ 0 & 1 \end{pmatrix}$ with $0 < \gamma < 1$ Clearly $A_0^{\dagger}A_0 + A_1^{\dagger}A_1 = A_0A_0^{\dagger} + A_1A_1^{\dagger} = \mathbf{1}_{2 \times 2}$, but \begin{align} |\psi\rangle_{12} & = (1 - \gamma)|00\rangle + \sqrt{\gamma(1 - \gamma)}|01\rangle + \sqrt{\gamma(1 - \gamma)}|10\rangle + (1 + \gamma)|11\rangle\\ \rho_1 & := \textrm{tr}_2\left[|\psi\rangle\langle\psi|\right] = (1 - \gamma)|0\rangle\langle0| + 2\sqrt{\gamma(1 - \gamma)}|0\rangle\langle 1| + 2\sqrt{\gamma(1 - \gamma)}|1\rangle\langle 0| + (1 + \gamma)|1\rangle\langle 1| \end{align} while $$ \sum_{i_1, j_1}\textrm{tr}\left[A_{i_1}\frac{\mathbf{1}_{2 \times 2}}{2}A_{j_1}^{\dagger}\right]|i_1\rangle\langle j_1| = \frac{1}{2}\left[(1 - \gamma)|0\rangle\langle0| + \sqrt{\gamma(1 - \gamma)}|0\rangle\langle 1| + \sqrt{\gamma(1 - \gamma)}|1\rangle\langle 0| + (1 + \gamma)|1\rangle\langle 1|\right] $$ There is a discrepancy in the off-diagonal terms.

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  • $\begingroup$ The j summation indices should be ordered descendingly. Other than that, it seems fine. In your formula, on the other hand, some $i$s should be $j$s. Note that the trace is also just a sum: Did you try to write out your sum including the trace and use the conditions on $\sum A_i A_i^\dagger$ etc.? -- Or, more elegantly, use $\mathrm{tr}(A\otimes B)=\mathrm{tr}(A)\,\mathrm{tr}(B)$. $\endgroup$ Commented Nov 28, 2019 at 10:03
  • $\begingroup$ Yes, I did try to use the trace-preserving and unital assumptions, rewriting the trace as the sum, but without success for now. I have added what I think may be a counterexample to the question; please let me know whether it makes sense or I missed the point! $\endgroup$ Commented Nov 28, 2019 at 17:05

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