Suppose I have a system composed of two subsystems (each is a 2-state system). I understand, that there exist two types of such systems: separable, and entangled. A separable system can be written as $$|\psi\rangle = (a|0\rangle_A+b|1\rangle_A)\otimes(c|0\rangle_B-d|1\rangle_B)$$ Then I measure the subsystem $A$ and find out that it is in state $|0\rangle$. After the measurement the system becomes
$$|\psi'\rangle = |0\rangle_A\otimes(c|0\rangle_B-d|1\rangle_B)$$ So, after the measurement of subsystem $A$, I do not get any information about the state of subsystem $B$. An entangled state would be $$|\phi\rangle = \frac{1}{\sqrt 2}(|0\rangle_A\otimes|0\rangle_B +|1\rangle_A\otimes|1\rangle_B)$$ In this case, if I measure the subsystem $A$ and get $|0\rangle_A$ then I can be 100% sure, that $B$ is in state $|0\rangle_B$, and the wavefunction of the full system collapses into $|\phi'\rangle = |0\rangle_A\otimes|0\rangle_B$. Now my questions are:
- As far as I know, state $|\psi\rangle$ is a superposition of the four basis vectors ($|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$), but it is considered a pure state.
- In contrary, $|\psi'\rangle$ is said to be a classical statistical mixture, so it is not a pure state. Why? Isn't it also a simple superposition of the basis vectors?
- Why is it called a classical statistical mixture?
- The state $|\phi'\rangle$ is also a pure state. Does this mean that entangled subsystems can not be in a mixed state?
- How can I decide if a state is a mixture, or a pure state?