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I am reading Introduction to Quantum Computing by Kaye, Laflamme, and Mosca. As an exercise, they write

"Prove that if the operators $P_{i}$ satisfy $P_{i}^{*}=P_{i}$ and $P_{i}^{2}=P_{i}$ , then $P_{i}P_{j}=0$ for $i\ne j$.''

In the context of this problem, it has been assumed that $I=\sum_{i=1}^{n} P_{i}$, where I suppose that $n$ could be infinite. I have shown that this is true in the trivial case $n=2$, but the general case has been eluding me. How should I attack this?

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For each $j$, $$P_j=P_jIP_j=P_j\left(\sum_{k=1}^n P_k\right)P_j=\sum_{k=1}^nP_jP_kP_j=P_j+\sum_{k\neq j}P_jP_kP_j,$$ so $\sum\limits_{k\neq j}P_jP_kP_j=0$. For each $i\neq j$, $P_jP_iP_j=(P_iP_j)^*P_iP_j$ is a positive operator, and a sum of positive operators is positive, so $-P_jP_iP_j=\sum\limits_{k\neq i,j}P_jP_kP_j$ is also positive. This is only possible if $P_jP_iP_j=0$. Since $\|P_iP_j\|^2=\|(P_iP_j)^*P_iP_j\|=\|P_jP_iP_j\|$, it follows that $P_iP_j=0$.

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Because of the properties you state, $\|P_{j}x\|^{2}=(x,P_{j}x)=(P_{j}x,x)$. Therefore, $$ \|x\|^{2} = (\sum_{j}P_{j}x,x)= \sum_{j}\|P_{j}x\|^{2}. $$ Apply this identity to $x=P_{k}y$, and use the fact that $P_{k}^{2}=P_{k}$: $$ \|P_{k}y\|^{2} = \sum_{j\ne k}\|P_{j}P_{k}y\|^{2}+\|P_{k}y\|^{2}. $$ The only way this can happen is $P_{j}P_{k}y=0$ for all $j \ne k$.

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Here is a slight variant of Jonas’ argument.


Assume that $ p_{1},\ldots,p_{n} $ are projection elements of a unital $ C^{*} $-algebra $ A $, where $ n \in \Bbb{N}_{\geq 2} $, such that $$ \sum_{k = 1}^{n} p_{k} = 1_{A}. $$ Choose distinct $ i,j \in [n] $, where $ [n] \stackrel{\text{df}}{=} \Bbb{N}_{\leq n} $. Then \begin{align} p_{i} & = p_{i} 1_{A} \\ & = p_{i} \sum_{k \in [n]} p_{k} \\ & = \sum_{k \in [n]} p_{i} p_{k} \\ & = p_{i}^{2} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} \\ & = p_{i} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k}. \end{align} It follows that $$ p_{i} p_{j} p_{j}^{*} = p_{i} p_{j} = - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} = - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*}, $$ and consequently, $$ (\spadesuit) \qquad (p_{i} p_{j}) (p_{i} p_{j})^{*} = p_{i} p_{j} p_{j}^{*} p_{i}^{*} = - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*} p_{i}^{*} = - \sum_{k \in [n] \setminus \{ i,j \}} (p_{i} p_{k}) (p_{i} p_{k})^{*}. $$ On the extreme left of $ (\spadesuit) $, we have a positive element, while on the extreme right of $ (\spadesuit) $, we have a negative element. This can only mean that both extremes are zero, so $ (p_{i} p_{j}) (p_{i} p_{j})^{*} = 0_{A} $. Hence, $$ \| p_{i} p_{j} \|_{A}^{2} = \| (p_{i} p_{j}) (p_{i} p_{j})^{*} \|_{A} = \| 0_{A} \|_{A} = 0, $$ or equivalently, $ p_{i} p_{j} = 0_{A} $.

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For all $i,j$ you have $P_i+P_j\le\sum_k P_k\le I$, hence $P_i\le I-P_j$.

Multiplying by $P_j$ on left and right on LHS and RHS you then get $$P_j P_i P_j\le P_j(I-P_j)P_j=0,$$ hence $P_j P_i P_j=0$, which implies $P_j P_i P_j=(P_j P_i)(P_j P_i)^\dagger=0$ and thus $P_i P_j=P_j P_i=0$.


You can also prove the other direction: if $P_i P_j=0$ for all $i\neq j$ then $\sum_k P_k$ is a projector, as $$\left(\sum_k P_k\right)^2=\sum_k P_k + \sum_{i<j}(P_i P_j + P_j P_i)=\sum_k P_k.$$

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