Is the partial trace $\mathrm{Tr}_B(\rho)$ equal to $\sum_k \mathrm{Tr}[(\sigma_k\otimes I)^\dagger \rho]\sigma_k$?

Question

Assume a composite quantum systes with state $|\psi_{AB}\rangle$ or better $\rho=|\psi_{AB}\rangle\langle\psi_{AB}|$. I want to know the state of system $A$ only, i.e. $\rho_A$.

Is there any difference if I trace out system $B$, i.e. $\rho_A=Tr_B\rho$ compared to building up $\rho_A$ from projections on the Pauli operators, i.e. $\displaystyle\rho_A=\sum_{k=1,x,y,z}Tr\big((\sigma_k\otimes 1\big)^\dagger \rho)\sigma_k$.

Some numerics indicate, that this is the same...

Answer 1

They are exactly the same. Remember that you can write $$ \rho=\sum_{i,j}\rho_{ij}\sigma_i\otimes\sigma_j. $$ If you take the partial trace, you have $$ \rho_A=\sum_{i,j}\rho_{ij}\sigma_i \text{Tr}(\sigma_j). $$ $\text{Tr}(\sigma_j)=0$ unless $j=0$, i.e. the identity operator. Thus, we can write $$ \rho_A=\sum_i2\rho_{i0}\sigma_i, $$ and of course we can calculate $$ 2\rho_{i0}=\text{Tr}((\sigma_i\otimes I)\rho). $$