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

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...

• – glS
Aug 7 '20 at 12:40

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 $$2\rho_{i0}=\text{Tr}((\sigma_i\otimes I)\rho).$$
• +1 thanks. Funny I always used a kind of non-square matrix representation of a super operator $\hat P$ to do the trace-out job: $\rho_A= mat( \hat P \circ vec (\rho))$, it rather sums up amplitudes than settings things to zero... Aug 7 '20 at 10:18
• @user1936752 I evaluated the right hand side, expressing $\rho$ using the first equation. The only set of Pauli’s with non-zero trace is $I\otimes I$. Aug 8 '20 at 7:04