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


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.