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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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    $\begingroup$ Yes, you are absolutely correct. $\endgroup$
    – Mark Mitchison
    Commented May 22, 2013 at 10:08
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70436/2451 , physics.stackexchange.com/q/121337/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Aug 14, 2016 at 19:09
  • $\begingroup$ I just wanted to clarify that, if they repeat the experiment, then yes, they sample from a mixed state, which can always be viewed as a classical sampling of pure quantum states. Generally, a mixed state is a "convex mixture" of pure states, meaning $\langle A \rangle=$tr$(A \rho)$ = $\sum_i \lambda_i \,\langle \psi_i | A |\psi_i \rangle$, and the mixed state is the sum $\rho = \sum_i \lambda_i \,|\psi_i \rangle\langle\psi_i|$, which applies to the scenario you describe, thermal states, etc. $\endgroup$
    – just a phase
    Commented Dec 9, 2023 at 21:05

1 Answer 1

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"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.

You can refer to section 42 of Schiff's book for a crisp explanation of the concepts and their analogues in classical mechanics.

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    $\begingroup$ This is just wrong, purity and mixedness is frequently a property of the observer. To pretend otherwise is just going to confuse the OP, whose statement is completely correct. The only time mixedness is a fundamental property is when one considers one part of an entangled system. $\endgroup$
    – Mark Mitchison
    Commented May 22, 2013 at 10:08
  • $\begingroup$ @MarkMitchison, thank you for pointing it out. I didn't know that purity and "mixedness" depend on the observer. Can you illustrate the point with an example? It would help if you could use the one I referred in my answer - polarized light versus that from an incandescent bulb. $\endgroup$
    – Amey Joshi
    Commented May 22, 2013 at 14:49
  • $\begingroup$ Well, the example given by the OP is the canonical one, so you should make sure you understand that first. I think that the question of "why thermal states are mixed" is too deep to be answered in a comment, but you should ask it as a separate question (if you cannot find the answer elsewhere on Physics.SE). It is a good one and I'm sure it will attract some interesting answers. $\endgroup$
    – Mark Mitchison
    Commented May 22, 2013 at 16:25

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