Consider an EPR situation in which there are two particles, a and b, of which state is given by
$\Psi = \frac{1}{\sqrt2}(|1\rangle|0\rangle + |0\rangle|1\rangle)$,
where $|0\rangle$ and $|1\rangle$ are two eigenstates of an observable O on single-particle Hilbert space (such as z-spin).
Now, while a and b are not in pure states, each of them may be in a mixed state. For instance, one could say that a is in a mixed state represented by the following reduced density matrix:
$\frac{1}{2}(|0\rangle\langle0|+|1\rangle\langle1|)$.
Typically, a mixed state is interpreted as representing our ignorance. On this view, $\frac{1}{2}(|0\rangle\langle0|+|1\rangle\langle1|)$ would correspond to a state in which a is either in $|1\rangle$ or $|0\rangle$ but we simply do not know which state a is in.
This interpretation, however, would be equivalent to the refuted position (attributed to EPR) that the states of a and b were already determined even before measurement, and we simply do not have access to those states. Since this 'local hidden variable' view was refuted, the above reduced density matrix cannot be interpreted as implying our ignorance.
Then how should we interpret the reduced density matrix?
The only reasonable move I can think of is to say that while a density matrix normally represents our ignorance, a reduced density matrix doesn't. This, however, seems very unsatisfying.
So, what's the physical meaning of the reduced density matrix in the EPR case?
Best Answer
I don't know if this will answer your question, but maybe it will help. Consider the case in which we have one spin-1/2 particle in a pure state \begin{align} |+\rangle=\frac{1}{2}\left(|0\rangle + |1\rangle\right) \end{align} with $S_z|0\rangle = -1/2$, $S_z|1\rangle = +1/2$. While the outcome of the measurement of $S_z$ in this state is random, we know that this state is an eigenstate of $S_x$, so if we told ahead of time that our spin is prepared in state $|+\rangle$, we know that we will always obtain the value $1/2$ if we measure $S_x$. This is true for any pure state $|\psi\rangle$ of a spin-1/2 system, i.e., we can always find a direction $\hat{\mathbf{n}}$ such that $|\psi\rangle$ is an eigenstate of $\mathbf{S}\cdot\mathbf{n}$.
In contrast, if our spin is in a mixed state—for example the mixed state obtained by tracing out the other spin in a Bell state—then the result of a measurement along any axis is random. In other words, given the complete quantum state of the spin, there is no measurement of which is guaranteed. I think this is a better way of stating the sense in which a mixed-state density matrix encodes "ignorance."
If we accept the interpretation you describe, where the mixed state $\rho = \frac{1}{2}\left(|0\rangle\langle0|+|1\rangle\langle1|\right)$ means our system is either in $|0\rangle$ or $|1\rangle$, this means that our state is definitely not in either of the eigenstates $|-\rangle$, $|+\rangle$ of $S_x$. But this leads us to a contradiction, because we can also write our state as $\rho = \frac{1}{2}\left(|-\rangle\langle -|+|+\rangle\langle+|\right)$, and this should mean that the system is either in $|-\rangle$ or $|+\rangle$. So I think the stated interpretation is incorrect, as is any conclusion you might draw from it.