# Is the partial trace of a mixed state always mixed? If not, are there natural examples where the partial trace of a mixed state is a pure state?

I know that the partial trace of a pure entangled state must be mixed and that of a product pure state must be pure; but I couldn't find an answer to my above question.

$$\rho = \frac{1}{2} \lvert 0 \rangle_{_A} \lvert 0 \rangle_{_B} \langle 0 \rvert_{_A} \langle 0 \rvert_{_B} + \frac{1}{2} \lvert 0 \rangle_{_A} \lvert 1 \rangle_{_B} \langle 0 \rvert_{_A} \langle 1 \rvert_{_B}$$ traces out to \begin{align} \rho_{_A} &= \lvert 0 \rangle_{_A} \langle 0 \rvert_{_A} \\ \rho_{_B} &= \frac{1}{2} \lvert 0 \rangle_{_A} \langle 0 \rvert_{_A} + \frac{1}{2} \lvert 1 \rangle_{_A} \langle 1 \rvert_{_A} . \end{align}
Since all the above density operators are diagonal, it is easy to see that $\rho$ and $\rho_{_B}$ are mixed states while $\rho_{_A}$ is a pure state.
• Makes sense. $\rho$ is a product state of a mixed ($\rho_B$) and a pure ($\rho_A$) state. Tracing out the pure state gives the mixed state. Dec 8 '17 at 0:19