# Trying to understand mixed states

I took a basic quantum chemistry course (McQuarrie's "Quantum Chemistry"), but it never dealt with mixed states -- only pure states (or if it did, we never got to it in class).

So I'm trying to understand them on my own. Consider a situation where Bob is in the lab and flips a coin. If it is heads, he prepares the system into pure state $|\psi_1\rangle$. If the coin is tails, he prepares the system into pure state $|\psi_2\rangle$. Now he invites Dave into the room. Bob knows which way the coin landed, but Dave doesn't. All Dave knows is that the system is either in the pure state $|\psi_1\rangle$ or $|\psi_2\rangle$, each with 50% probability.

...so could you say the system is in a pure state to Bob and a mixed state to Dave? Or am I way off base here?

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Yes, you are absolutely correct. – Mark Mitchison May 22 '13 at 10:08

"Purity" of "mixedness" (if you permit these words) is a property of the system and not the observer. A system is said to be in a pure state if it is in one of the allowed states $|\psi_i\rangle$, $i = 1 \ldots n$ or in a linear superposition $|\phi\rangle = \sum_{i=1}^{i=n}\alpha_i|\psi_i\rangle$ of such states. If it is in a mixed state, then such a representation is not possible and we have to express it as a density matrix $\hat{\rho} = \sum_{i=1}^{i=n}\beta_i|\psi_i\rangle\langle\psi_i|$.
What is the difference? The expectation value of an observable $\hat{A}$ is $\langle\phi|\hat{A}|\phi\rangle$ for a pure state and $Tr(\hat{\rho}\hat{A})$. Polarized light is an example of a system being in pure state, while light from an incandescent bulb is in a mixed state.