# Nielsen & Chuang Exercise 2.32: Show that the tensor product of two projectors is a projector

$$\newcommand{\bra}{\left<#1\right|} \newcommand{\ket}{\left|#1\right>}$$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}?

• it's just a redefinition. More precisely you should write $|k\rangle=|i,j\rangle$. The only identity you are using is $|i\rangle\!\langle i|\otimes\lvert j\rangle\!\langle j\rvert=\lvert i,j\rangle\!\langle i,j\rvert$
– glS
Aug 28 '20 at 22:35
• 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. Aug 29 '20 at 3:20

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!