Normal trace is equivalent to the sum of the eigenvalues (or diagonal elements) of a matrix. Partial trace acts only on part of the system, so for a density matrix.. say it's a pure state but entangled,
[tex]\rho_{AB}=\frac{1}{2}(|01\rangle +|10\rangle )(\langle 01|+\langle 10 |)[/tex]
The partial trace over subsystem B gives the reduced density matrix [tex]\rho_A[/tex], so [tex]Tr_B(\rho_{AB})=\rho_A[/tex]
So
[tex]\rho_A=_B\langle 0 |\rho_{AB}|0\rangle_B +_B\langle 1 |\rho_{AB}|1\rangle_B[/tex]
[tex]\rho_A=|1\rangle \langle 1 | + |0\rangle \langle 0 |[/tex]
My question is how do I prove that
[tex]Tr_B (\rho \sigma) = Tr_B (\sigma \rho)[/tex]
where [tex]\rho[/tex] and [tex]\sigma[/tex] are both density matrices of a system AB.
Thanks!