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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...

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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 $$ 2\rho_{i0}=\text{Tr}((\sigma_i\otimes I)\rho). $$

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  • $\begingroup$ +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... $\endgroup$ – draks ... Aug 7 at 10:18
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    $\begingroup$ That's certainly a useful way for computers to handle it, or if you want to deal with noisy operations e.g. via the Lindblad equation. But it's not generally the first choice for this sort of problem as converting backwards and forwards risks introducing extra work/mistakes (in my view). $\endgroup$ – DaftWullie Aug 7 at 10:44
  • $\begingroup$ Sorry if I'm missing something but how did you arrive at the last line? $\endgroup$ – user1936752 Aug 8 at 0:36
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    $\begingroup$ @user1936752 I evaluated the right hand side, expressing $\rho$ using the first equation. The only set of Pauli’s with non-zero trace is $I\otimes I$. $\endgroup$ – DaftWullie Aug 8 at 7:04
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    $\begingroup$ Although there may be an issue with factor of 2.... $\endgroup$ – DaftWullie Aug 8 at 7:05

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