- #1
Martin Muñoz
- 5
- 0
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 [tex]P_i[/tex] satisfy:
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. :(
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. :(