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.


2 Answers 2


Start with a pure qubit and a mixed qubit, then trace out the mixed qubit. The overall state goes from mixed to pure.


Writing out Craig's answer explicitly (I personally find explicit examples helpful):

$$ \rho = \frac{1}{2} \lvert 0 \rangle_{_A} \lvert 0 \rangle_{_B} \langle 0 \rvert_{_A} \langle 0 \rvert_{_B} + \frac{1}{2} \lvert 0 \rangle_{_A} \lvert 1 \rangle_{_B} \langle 0 \rvert_{_A} \langle 1 \rvert_{_B} $$ 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.

  • $\begingroup$ 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. $\endgroup$ Dec 8, 2017 at 0:19
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    $\begingroup$ A follow-up question. Is there a mixed state which is not a product of a mixed and a pure state (so, it could be a product of 2 mixed states or it could not be a product state at all) such that it has a pure partial trace? $\endgroup$ Dec 8, 2017 at 0:22

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