Is it possible to differentiate between pure states and mixed ones in the laboratory? I am wondering if it is possible to differentiate between pure states and mixed ones in the laboratory?
For example consider a quantum system define relative to the orthonormal basis {up,down}. We can build a mixed state which is a combination of 2 or more pure states. Then we measured the probability of the system to be equal to $p$ for the system to be in the 'up' state, and equals to $1-p$ for the system to be in the 'down' state.
However, we can build a pure state with the same probabilities as above. So, how can we differentiate the both situations?
 A: Yes, you can differentiate (assuming you have many many copies of the identically produced systems) between a pure state and a mixed state. As you notice, you cannot differentiate using a single operator as you can imagine a pure state having the same probability distribution over the spectrum of that operator. However, if you start doing measurements with another operator that doesn't commute with the first one, you'd get different probability distributions for a mixed state and a pure state because the mixed state would lack interference.
For example, a mixed state of equal probabilities for spin up state and spin down state in $z$ direction would give you equal probabilities also for spin up and spin down measurements in $x$ directions. However, a pure state such as $\frac{1}{\sqrt{2}} \big(\vert \uparrow_z~\rangle + \vert \downarrow_z~\rangle \big)$ would not.
A: Well, a pure state is an eigenstate of some operator. So if you can measure that observable you can readily distinguish between pure and mixed states.
For example with spin halves, there will be some polarisation angle along which the pure state is an eigenstate and you’ll always get the same outcome. But for a mixed state there’s no such direction.
