Martin Muñoz
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I'm studying for my Quantum Computing exam. It's at 2 PM EST today. If anyone can give me a nudge in the right direction before then that would be excellent!
Problem:
Assume the operators P_i satisfy:
Attempt:
This seemed really obvious to me intuitively but I've been struggling with a proof.
First I wrote P_i = \textbf{1} - \sum_{i \ne m}{P_m} and P_j = \textbf{1} - \sum_{j \ne n}{P_n}. I then tried to apply the first to the second, but it got messy and I couldn't get anywhere.
Then I tried to create some identities to see if that would make things more clear:
P_i P_j = P_i^\dagger P_j^\dagger = (P_j P_i)^\dagger
and
P_i P_j = P_i^2 P_j^2 = P_i P_i P_j P_j = P_i (P_j P_i)^\dagger P_j
but all I could think of. :(
Problem:
Assume the operators P_i satisfy:
- \textbf{1} = \sum_i{P_i}
- P_i^{\dagger} = P_i
- P_j^2 = P_j.
Attempt:
This seemed really obvious to me intuitively but I've been struggling with a proof.
First I wrote P_i = \textbf{1} - \sum_{i \ne m}{P_m} and P_j = \textbf{1} - \sum_{j \ne n}{P_n}. I then tried to apply the first to the second, but it got messy and I couldn't get anywhere.
Then I tried to create some identities to see if that would make things more clear:
P_i P_j = P_i^\dagger P_j^\dagger = (P_j P_i)^\dagger
and
P_i P_j = P_i^2 P_j^2 = P_i P_i P_j P_j = P_i (P_j P_i)^\dagger P_j
but all I could think of. :(