- #1
McLaren Rulez
- 292
- 3
Hi,
I finished reading about the Schmidt decomposition from Preskill's notes today. I understand and follow his derivation but it still seems completely non intuitive to me. We have
[tex]
\mid\psi\rangle_{AB}=\sum_{i,u}a_{iu}\mid i\rangle_{A}\mid u\rangle_{B}=\sum_{i}\mid i\rangle_{A}\mid\tilde{i}\rangle_{B}
[/tex]
where we have
[tex]
\mid\tilde{i}\rangle_{B}=\sum_{i,u}a_{iu}\mid u\rangle_{B}
[/tex]
At this stage, the system B is represented by [itex]\mid\tilde{i}\rangle_{B}[/itex] which are not orthonormal. Then, one chooses a basis for the first subsystem such that the partial trace [itex]\rho_{a}=Tr_{B}\rho_{AB}[/itex] is diagonal. Somehow, this automatically makes the vectors of system B orthonormal. I can see that it's true but I have no clue why. If anyone has some intuition for what is going on, I'd be very grateful.
Thank you.
I finished reading about the Schmidt decomposition from Preskill's notes today. I understand and follow his derivation but it still seems completely non intuitive to me. We have
[tex]
\mid\psi\rangle_{AB}=\sum_{i,u}a_{iu}\mid i\rangle_{A}\mid u\rangle_{B}=\sum_{i}\mid i\rangle_{A}\mid\tilde{i}\rangle_{B}
[/tex]
where we have
[tex]
\mid\tilde{i}\rangle_{B}=\sum_{i,u}a_{iu}\mid u\rangle_{B}
[/tex]
At this stage, the system B is represented by [itex]\mid\tilde{i}\rangle_{B}[/itex] which are not orthonormal. Then, one chooses a basis for the first subsystem such that the partial trace [itex]\rho_{a}=Tr_{B}\rho_{AB}[/itex] is diagonal. Somehow, this automatically makes the vectors of system B orthonormal. I can see that it's true but I have no clue why. If anyone has some intuition for what is going on, I'd be very grateful.
Thank you.