# How do I prove that reduced density matrix is Hermitian?

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?

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 $$\rho_A = \rho^\dagger_A$$.

• @Neijal Kanderbalt, is there anything in my answer that confuses you? How can I clarify? – exp ikx Sep 29 '19 at 13:28