If we consider a bipartie system that is entangled,
$$|\psi^{AB}\rangle=\frac{|{\uparrow_z\downarrow_z}\rangle-|{\downarrow_z\uparrow_z}\rangle}{\sqrt{2}} $$
And we want to know the probability distribution of an experiment that involves only system A, we can use the reduced density matrix:
$$\rho^A=\frac12 |{\uparrow_z}\rangle\langle{\uparrow_z}|+\frac12|{\downarrow_z}\rangle\langle{\downarrow_z}|$$
Which results from "ignoring" the rest of the bipartite system.
I'm struggling to make sense of this state. The density matrix describes a mixed ensemble, which may contain an even mixture of ups and down in any axis with equal probability. But here we have a single electron for example which can't be a statistical mixture.
Normally if I want to describe a quantum state, I think of an ensemble of states that are all prepared exactly the same and then I measure different operators on the states.But when I want to use this way of thinking for the state considered here, someone might say: well, if you entangle a lot of particles and you collect always the second particle and use these as your ensemble, you got a statistical mixture of pure states.
-> But this can't be true since none of the electrons is in a pure state?
Also, if someone else measures the spin of the other particles and therefore makes the entangled state collapse, now my ensemble "really is a mixture" of pure states, but I can't know that until I recieve the information from him.
So my two questions are:
Is there a difference between my ensemble before the other person has measured his particles and after the measurement?
How can the state of system A of the entanglement be a mixed state if it is only one particle? Is the description by the reduced density matrix incomplete?
