- #1

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**Problem:**

Assume the operators [tex]P_i[/tex] satisfy:

- [tex]\textbf{1} = \sum_i{P_i}[/tex]
- [tex]P_i^{\dagger} = P_i[/tex]
- [tex]P_j^2 = P_j[/tex].

**Attempt:**

This seemed really obvious to me intuitively but I've been struggling with a proof.

First I wrote [tex]P_i = \textbf{1} - \sum_{i \ne m}{P_m}[/tex] and [tex]P_j = \textbf{1} - \sum_{j \ne n}{P_n}[/tex]. 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:

[tex]P_i P_j = P_i^\dagger P_j^\dagger = (P_j P_i)^\dagger[/tex]

and

[tex]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[/tex]

but all I could think of. :(