Suppose that we have two entangled particles A and B with pure state vector
$|\psi\rangle=a|0\rangle_A |1\rangle_B + b|1\rangle_A |0\rangle_B \hspace{1cm}(1)$
When we take the partial trace over the degrees of freedom of one of the two entangled particles, lets say B, we get the reduced density matrix for the A, which is identical to a mixed state density matrix. But is the state of the particle A a real mixed state? i.e. is its state some definite state, $|0\rangle_A$ or $|1\rangle_A$, but we simply don't know which one? Wouldn't that mean that the whole system is in $|0\rangle_A |1\rangle_B$ or $|1\rangle_A |0\rangle_B$, and not in the superposition (1)?