How lack of information changes state? Suppose two orthonormal states $|\psi_0\rangle$ and $|\psi_1\rangle$ are kept in two separate boxes. If we know which box shares which state, the state of the composite system is $|\psi_0\rangle |\psi_1\rangle$. But if this information is not available what is the state of the composite system ? 
 A: If you have incomplete information about your system, then you can no longer consider the system to be in a pure state, and you need to describe the resulting mixed state as a density matrix that's a convex combination of the full-information states
$$
\rho_{01}=|\psi_0\rangle\langle\psi_0|\otimes|\psi_1\rangle\langle\psi_1|
$$
and
$$
\rho_{10}=|\psi_1\rangle\langle\psi_1|\otimes|\psi_0\rangle\langle\psi_0|.
$$
How you mix them will depend on how much information you have about where each state went, and if you really have no information at all then you should use the even-weights mixture
$$
\rho=\frac12\bigg(
|\psi_0\rangle\langle\psi_0|\otimes|\psi_1\rangle\langle\psi_1|
+
|\psi_1\rangle\langle\psi_1|\otimes|\psi_0\rangle\langle\psi_0|
\bigg).
$$
It's important to note, on the other hand, that this is rather different to coherent superpositions of the form $|\psi\rangle = \frac{1}{\sqrt{2}}\bigg( |\psi_0\rangle|\psi_1\rangle + |\psi_1\rangle |\psi_0\rangle \bigg)$, where the information isn't just unavailable - it just doesn't exist.
