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We define \begin{equation} \sum_{i=1}^{4^l} P_i \otimes P_i, \sum_{i=1}^{4^m} Q_i \otimes Q_i, \end{equation} where $P_l$ is the $n$ qubit Pauli string and $Q_m$ is the $m$ qubit Pauli string.

Does the following equality hold? \begin{equation} \sum_{i=1}^{4^l} P_i \otimes P_i \otimes \sum_{i=1}^{4^m} Q_i \otimes Q_i = \sum_{i=1}^{4^{l+m}} O_i \otimes O_i, \end{equation} where $O_i$ is the $l+m$ qubit Pauli string.

Would the summation be the same although the elements of the summation are different? In fact, they should be the same.

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It doesn't hold. You can consider the simple example where $l=1$ and $m=1$ as a counter-example.

$$\sum_i P_i \otimes P_i = II + XX + YY + ZZ$$ $$\sum_i Q_i \otimes Q_i = II + XX + YY + ZZ$$

Then,

\begin{align} \sum_i P_i \otimes P_i \sum_i Q_i \otimes Q_i &= IIII + IIXX + IIYY + IIZZ \\ &+ XXII + XXXX + XXYY + XXZZ \\ &+ YYII + YYXX + YYYY + YYZZ \\ &+ ZZII + ZZXX + ZZYY + ZZZZ \\ &\neq \sum_i O_i \otimes O_i \end{align}

What you have in the RHS are $4^{l+m}$ Pauli strings of a set of $4^{2(l+m)}$ elements, since they are defined as $O_i \otimes O_i$, then their length is $2(l+m)$. Notice that these Pauli strings are symmetric with respect to the middle. However, from the LHS you don't have this symmetry.

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  • $\begingroup$ Would the summation be the same although the elements of the summation are different? $\endgroup$ Commented Feb 9, 2023 at 4:08
  • $\begingroup$ @Michael.Andy notice what happens when you swap the second and third paulis in each string - i think that answers your question. $\endgroup$
    – forky40
    Commented Feb 9, 2023 at 20:02

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