Quantum Computing Asked on January 5, 2021

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.

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 $P_- = frac{1}{2}(I - U)$.

Correct answer by Joseph Geipel on January 5, 2021

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