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Suppose I have an $n$-qubit Hermitian operator $U$ such that $U^2=I$.

The projection operators with eigenvalue $+1$ and $−1$ are $P_+$ and $P_-$.

How can I prove that $P_+=\frac{1}{2}(1+U)$ and $P_-=\frac{1}{2}(1-U)$?

I think $\sum P=1$ but I have no idea to get this conclusion.

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First, we can start with $U = P_+ - P_-$, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If $U^2 = I$, that means $I = (P_+ - P_-)(P_+ - P_-) = P_+P_+ - P_+P_- - P_-P_+ + P_-P_-$. Since the two projectors correspond to orthogonal eigenspaces, operating on themselves doesn't change them and one operating on the other gets a zero, so we get $P_+ + P_- = I$. With $P_+ + P_- = I$ and $P_+ - P_- = U$, we can add the two equations to get $P_+ = \frac{1}{2}(I + U)$ or subtract the $U$ from the $I$ to get $P_- = \frac{1}{2}(I - U)$.

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  • $\begingroup$ That makes sense! Thank you $\endgroup$
    – user13341
    Dec 17 '20 at 1:58
  • $\begingroup$ In the first line you're assuming that the projectors are rank one but the OP says that $U$ is an $n$-qubit operator. In particular if $n>1$ then they cannot be rank $1$ as $U$ needs full rank to square to the identity. $\endgroup$
    – Rammus
    Dec 17 '20 at 12:47
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    $\begingroup$ Fixed, doesn't change the rest sans clarifying "eigenspaces" rather than "eigenstates". $\endgroup$ Dec 17 '20 at 14:25

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