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$\newcommand{\bra}[1]{\left<#1\right|} \newcommand{\ket}[1]{\left|#1\right>}$Here is what I tried:

Given that we have two projectors: $$ A = \sum_i \ket{i} \bra{i}, \hspace{2em} B = \sum_j \ket{j} \bra{j} $$ The goal is to prove that: $$ A \otimes B = \sum_k \ket{k} \bra{k}. \tag1\label1 $$ Plugging into \eqref{1}, we get: $$ A \otimes B = \left( \sum_i \ket{i} \bra{i} \right) \otimes \left( \sum_j \ket{j} \bra{j} \right) = \sum_{i,j} \ket{i} \bra{i} \otimes \ket{j} \bra{j} \tag{2}\label{2} $$ I'm not sure how to proceed from \eqref{2}. It would be convenient if for every $\ket{i}$ and $\ket{j}$ there is a $\ket{k}$ for which the following identity is true: $$ \ket{k} \bra{k} = \ket{i} \bra{i} \otimes \ket{j} \bra{j} \tag{3}\label{3} $$ This would prove \eqref{1} immediately. Is \eqref{3} true though? If yes, why? If not, how else can we proceed to prove \eqref{1}?

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    $\begingroup$ it's just a redefinition. More precisely you should write $|k\rangle=|i,j\rangle$. The only identity you are using is $|i\rangle\!\langle i|\otimes\lvert j\rangle\!\langle j\rvert=\lvert i,j\rangle\!\langle i,j\rvert$ $\endgroup$
    – glS
    Aug 28, 2020 at 22:35
  • $\begingroup$ Thanks, that answers my question. I tried proving $|i\rangle\!\langle i|\otimes\lvert j\rangle\!\langle j\rvert=\lvert i,j\rangle\!\langle i,j\rvert$ and managed to convince myself of its truth by expanding the terms into matrices/vectors and see what gets multiplied by what, but I'm wondering if there's a "nicer" way to do this. I tried looking at the definitions in Nielsen & Chuang (10th edition) page 73 but couldn't find anything useful, which is weird because the book usually introduces the necessary identities prior to the exercises. $\endgroup$
    – Attila Kun
    Aug 29, 2020 at 3:20

2 Answers 2

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I wouldn't approach the problem the way that you have. Instead, I'd take the definition of what it means to be a projector: $P$ is a projector if and only if $P^2=P$ and $P=P^\dagger$.

So, let's take $$ P=P_A\otimes P_B. $$ We can calculate $P^\dagger=P_A^\dagger\otimes P_B^\dagger=P_A\otimes P_B=P$, which follows from the assumption that $P_A$ and $P_B$ are projectors.

Similarly, $$ P^2=P_A^2\otimes P_B^2=P_A\otimes P_B=P. $$ You're done!

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$\begin{aligned} {[A \otimes B](|a\rangle \otimes|b\rangle) } &=\left[\sum_i|i\rangle\langle i|\otimes \sum_j| j\rangle\langle j|\right](|a\rangle \otimes|b\rangle) \\ &=\left[\sum_i|i\rangle\langle i \mid a\rangle\right] \otimes\left[\sum_i|j\rangle\langle j \mid b\rangle\right] \\ &=\sum_{i, j}\langle i | a\rangle\langle j \mid b\rangle(|i\rangle \otimes|j\rangle) \\ &=\sum_{i, j}\langle i j \mid a b\rangle|i j\rangle \\ &=\left[\sum_{i, j}|i j\rangle\langle i j|\right]|a\rangle \otimes|b\rangle \end{aligned}$

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    – Community Bot
    Nov 9, 2022 at 21:15

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