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It is known that density matrix $\rho$ is hermitian. How do I prove that for a bipartite system $AB$, reduced density matrix of $A$, $\rho_A = Tr_B\{\rho_{AB}\}$, is hermitian, given that $\rho_{AB}$ is hermitian as well?

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This can be worked out from the definition itself.

The partial trace $tr_B$ is defined as the linear extension of the mapping $$tr_B : S \otimes T \rightarrow tr(T)S$$ for any matrix $S$ on $H_A$ and $T$ on $H_B$.

Let ${|a_i\rangle}$ be a basis of $H_A$, and ${|b_i\rangle}$ be a basis of $H_B$. Any density matrix $\rho_{AB}$ on $H_A \otimes H_B$ can then be decomposed as $\rho_{AB} = \sum_{ijkl} m_{ijkl}|a_i\rangle \langle a_j| \otimes |b_k\rangle \langle b_l|$.

We know that $\rho_{AB}^\dagger = \rho_{AB}$.

This implies $m_{ijkl} = m^*_{jilk}$.

The partial trace then reads $\rho_A = tr_B \rho_{AB} = \sum_{ijkl} m_{ijkl}|a_i\rangle \langle a_j| \langle b_l|b_k\rangle$.

Since $\langle b_l| b_k \rangle = \langle b_k| b_l\rangle$, ensuring $m_{ijkl} = m^*_{jilk}$ should imply that $\rho_A = \rho^\dagger_A$.

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