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Define the operator on a qudit system as \begin{align} o &= \sum_{s, s^\prime=1}^d o_{s,s^\prime}\vert s\rangle\langle s \vert \otimes \vert s^\prime\rangle \langle s^\prime \vert. \tag{1} \end{align} Then I want to compute the composite operator of $N$ qudit systems, which is \begin{align} O &= o^{\otimes N} = \sum_{\textbf{s}, \textbf{s}^\prime} O_{\textbf{s}, \textbf{s}^\prime} \vert \textbf{s} \rangle \langle \textbf{s} \vert \otimes \vert \textbf{s}^\prime \rangle \langle \textbf{s}^\prime \vert, \tag{2} \end{align} where \begin{align} \vert \textbf{s} \rangle = \bigotimes_{i=1}^N \vert s_i \rangle, \vert \textbf{s}^\prime \rangle = \bigotimes_{i=1}^N \vert s_i^\prime \rangle \tag{3} \end{align} are the computational basis states and $i$ is the index of the qudit system.

I derive a result that is different from eq. (2). How to obtain eq. (2)? This problem originates in eq. (30) of this paper

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It would probably be easier to see how they derived (2) by defining a single label for each pair: $t=(s,s^\prime)$, where there are $d^2$ possible values of $t$. Then we write $$o=\sum_t o_t |t\rangle\langle t|$$ and $$O=\sum_{\mathbf{t}} O_{\mathbf{t}} |\mathbf{t}\rangle\langle \mathbf{t}|.$$ Can you see how these are obtained?

I can write \begin{align} o^{\otimes N}&=\sum_{t_1,t_2,\cdots,t_N}o_{t_1}o_{t_2}\cdots o_{t_N}|t_1\rangle\langle t_1|\otimes |t_2\rangle\langle t_2|\otimes\cdots\otimes |t_N\rangle\langle t_N|\\ &=\sum_{\mathbf{t}}o_{t_1}o_{t_2}\cdots o_{t_N}|\mathbf{t}\rangle\langle \mathbf{t}| \end{align} to identify $O_{\mathbf{t}}=o_{t_1}o_{t_2}\cdots o_{t_N}$.

Once you can do this calculation with a single index $t$, it is straightforward to do with two indices $s$ and $s^\prime$. Or, we can just take our result, notice that $|\mathbf{t}\rangle\langle \mathbf{t}|=|\mathbf{s}\rangle\langle \mathbf{s}|\otimes |\mathbf{s^\prime}\rangle\langle \mathbf{s^\prime}|$, and we're done.

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  • $\begingroup$ Thank you QM. I derive $|\mathbf{s}\rangle\langle \mathbf{s}|\otimes |\mathbf{s^\prime}\rangle\langle \mathbf{s^\prime}| = \vert \mathbf{s} \mathbf{s}^\prime \rangle \langle \mathbf{s} \mathbf{s}^\prime \vert = \vert s_1,...s_N, s_1^\prime,...s_N^\prime \rangle \langle s_1,...s_N, s_1^\prime,...s_N^\prime \vert \neq \vert t_1 \rangle \langle t_1 \vert \otimes ... \otimes \vert t_N \rangle \langle t_N \vert = \vert \mathbf{t} \rangle \langle \mathbf{t} \vert$. Anything wrong? $\endgroup$ May 28, 2023 at 1:54
  • $\begingroup$ @Michael.Andy it's just about the labels. $|s_1,s_N,s_1^\prime,s_N^\prime\rangle \langle s_1,s_N,s_1^\prime,s_N^\prime|$ is the same as $|s_1\rangle\langle s_1|\otimes |s_N\rangle\langle s_N|\otimes\rangle\langle s_1^\prime\rangle\langle s_1^\prime|\otimes |s_N^\prime \rangle\langle s_N^\prime|$ to me. And you can order the labels of $\mathbf{t}=(s_1,\cdots,s_N,s_1^\prime,\cdots,s_N^\prime)$ if it makes life easier. The point is to give you the method; you can use any convention you want for ordering the labels $\endgroup$ May 28, 2023 at 15:19

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