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I would like to complete the following exercise:

Prove that if the operators $P_i$ satisfy $P_i^{\dagger}$ = $P_i$ and $P_i^2$ = $P_i$, then $P_iP_j=0$ for all $i\neq j$.

From $P_i^2 = P_i$ I conclude that $P_i(P_i - \mathbb{I}) = 0$, where $\mathbb{I}$ is the identity matrix. So $P_i$ is either $0$ or $\mathbb{I}$ but this is wrong because I think there should be more such operators than those two trivial ones...

Could you please point out my mistake and provide the legit proof for the excercise?

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    $\begingroup$ It does not follow from $P_i(P_i-I)=0$ that $P_i$ is 0 or I. I can be written as $\sum_j P_j$ for some orthonormal set of projectors. So if you subtract $P_i$ from that sum you are just left with a sum of projectors orthogonal to $P_i$. In addition, I think there must be more information in the exercise you are doing than you are giving us because I could pick two projectors like $|0><0|$ and $\frac{1}{2}(|0>+|1>)(<0|+<1|)$ which do not satisfy the equation you gave but also are not ruled out by anything in the information you gave in your question. $\endgroup$
    – alanf
    Commented Apr 28, 2014 at 11:10
  • $\begingroup$ I provided you with the full content of the exercise. Notice that your reasoning goes in the opposite direction of the implication. First we assume that the set of operators we are dealing with is the one containing operators that satisfy given rules and for them we prove. $\endgroup$
    – Ghostwriter
    Commented Apr 28, 2014 at 11:24
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    $\begingroup$ The operators I have given are Hermitian and satisfy $P=P^2$. Since they do not satisfy $P_iP_j=0$ there must be more to the question than you are telling us. What other information have you been given about the $P_i$? $\endgroup$
    – alanf
    Commented Apr 28, 2014 at 11:28
  • $\begingroup$ Sorry, I misunderstood you, you are right with your examples. I uploaded the original source. What is the problem then? $\endgroup$
    – Ghostwriter
    Commented Apr 28, 2014 at 11:56
  • $\begingroup$ The excercise does not provide enough information to be solved. Something along the lines of $$ \sum_i P_i = 1$$ is missing. $\endgroup$
    – Neuneck
    Commented Apr 28, 2014 at 12:16

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