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If we have the following two states \begin{equation} |\psi\rangle_1 = \frac{1}{\sqrt{2}}|0\rangle_A|0\rangle_B + \frac{1}{\sqrt{2}} |1\rangle_A |1\rangle_B \end{equation} \begin{equation} |\psi\rangle_2 = \frac{1}{\sqrt{2}}|0\rangle_A|0\rangle_B - \frac{1}{\sqrt{2}} |1\rangle_A |1\rangle_B \end{equation} How do you mix them with the same proportion to create a mixed state? What would be the resulting density operator?

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You can prepare the mixed state as follows. Flip a perfect coin. If it comes up heads, prepare $|\psi\rangle_1$, otherwise prepare $|\psi\rangle_2$. Finally, forget the result of flipping the coin.

The corresponding density operator is

$$ \rho = \frac{1}{2} |\psi\rangle_1\langle\psi|_1 + \frac{1}{2} |\psi\rangle_2\langle\psi|_2. $$

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Another thing that you can do to prepare your state into a mixed state is to simply measure your pure state.

Either it will be in $|\psi\rangle_1$ or $|\psi\rangle_2$ with 50/50 probability, which (after collapse) will not have any "special quantum weirdness" which would be observed via interference and can only be described with mixed states (classical probability).

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