Quantum Computing Asked on June 24, 2021
$newcommand{bra}[1]{left<#1right|}
newcommand{ket}[1]{left|#1right>}$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:
$$
A otimes B = sum_k ket{k} bra{k}. tag1label1
$$
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}?
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 $$ P=P_Aotimes P_B. $$ We can calculate $P^dagger=P_A^daggerotimes P_B^dagger=P_Aotimes P_B=P$, which follows from the assumption that $P_A$ and $P_B$ are projectors.
Similarly, $$ P^2=P_A^2otimes P_B^2=P_Aotimes P_B=P. $$ You're done!
Correct answer by DaftWullie on June 24, 2021
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