Why is this the right form for the density operator in "classical" mixture cases? It is well known that the set of density operators $\{\rho\}$ for a quantum theory form a convex set. As I have seen them defined, we simply say that a state corresponds to some linear operator $\rho$ which is self-adjoint, positive semidefinite, and of unit trace.
Now it is often taken in books as obvious that, for a situation (I'm struggling to make this precise which is part of the question I suppose) in which there is some "classical" uncertainty about the state of some quantum system which may be, with probability $p_n$, in some pure state $|\psi_n\rangle \langle \psi_n|$, we must represent the state of the system via a weighted sum of these pure states. More precisely, we take
$$\rho = \sum_n p_n |\psi_n\rangle \langle \psi_n|, \sum p_n = 1, p_n \in [0,1]   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  (A)$$
Is there any reason other than "experiment confirms this choice" that we might take this? Obviously this also depends on exactly which physical contexts imply this choice for $\rho$. For instance, if I enter a laboratory and am unsure if a spin has been prepared up or down but I know it is in one or the other, is it correct for me to assign $\rho = \frac{1}{2}|+\rangle \langle +|+\frac{1}{2}|-\rangle \langle -|$? It seems absurd since this cannot reproduce the statistics of measurements of the actual pure state.
Thus to rephrase and try to be more precise, my questions are:
(1) When is (A) the correct choice for $\rho$?
(2) Can we prove that it is the correct choice without appealing to experiment? NB that I understand that this $\rho$ is a valid state since $\{\rho\}$ forms a convex set; thus my question here is not "why is it a valid state" but rather "why is it the valid state for this physical situation"?
 A: The heuristic I've seen before is to stipulate that there should be two kinds of averages,

*

*a coherent (quantum) one in which we take expectation values as $\langle \psi | \hat{A} | \psi \rangle$, and


*an incoherent (classical) one in which we take averages over a probability distribution in the usual way, i.e., $\bar{a} = \sum_n p_n a_n$.
Then, if you have a system that could be in multiple quantum states $|\psi_n\rangle$---and you don't know which one but you can assign a probability distribution $p_n$ to them---the average is given by the "double" average
$$
\overline{\langle \hat{A} \rangle} =
\sum_n p_n \langle\psi_n|\hat{A}|\psi_n\rangle\,.
$$
From here, it's a matter of rearranging things. First, choose any orthonormal basis $|m\rangle$, and resolve the identity in this expression as
$$
\overline{\langle \hat{A} \rangle} =
\sum_n p_n \langle\psi_n|\hat{A}|\psi_n\rangle
=
\sum_n p_n \langle\psi_n|\hat{A}\left(
\sum_m |m\rangle\langle m |
\right)|\psi_n\rangle\,.
$$
Rearranging, we get
$$
\overline{\langle \hat{A} \rangle}
=
\sum_m\sum_n p_n \langle\psi_n|\hat{A}
 |m\rangle
\langle m |\psi_n\rangle
=
\sum_m\sum_n p_n \langle m |\psi_n\rangle\langle\psi_n|\hat{A}
 |m\rangle
\,.
$$
Finally, push the sum over $n$ through to get
$$
\overline{\langle \hat{A} \rangle}
=
\sum_m \langle m |\left(
\sum_n p_n|\psi_n\rangle\langle\psi_n|\right)\hat{A}
 |m\rangle
=\operatorname{Tr}\left(\rho\hat{A}\right)\,,
$$
where
$$
\rho = \sum_n p_n|\psi_n\rangle\langle\psi_n|\,.
$$
A: Short answers : yes.
I guess that your confusion comes from the different natures of randomness.

*

*Quantum randomness : as you raised it, quantum systems possess an intrinsic randomness, absent from classiscal systems. Indeed, even when there is no uncertainty about the state of the considered system, that is when we know its state is $|\psi\rangle$, hence a pure state density matrix $\hat{\rho} = |\psi\rangle\langle\psi|$, it is still described by a propability distribution carried by $\psi$.


*Classical randomness : this type of randomness corresponds to a mere statistical mixture (of pure states), hence the form $\hat{\rho} = \sum_np_n|\psi_n\rangle\langle\psi_n|$. In other words, when doing measurements, you pick a state $|\psi_n\rangle$ in the same manner you would pick a given ball from an urn. Nevertheless, another specificity of quantum statistical mixtures over classical ones in that case is the superposition principle; indeed, the quantum system is in every state of the mixture at the same time, hence the linear combination structure, such that the individual (entangled) states cannot be separated as the balls in the urn could be $-$ even by measurements, since the wavefunction collapse "destroys" the other states of the mixture.


*"Bayesian randomness" : the last type of randomness, the one modelled by the $\frac{1}{2}$-weights in $\hat{\rho} = \frac{1}{2}|+\rangle\langle+| + \frac{1}{2}|-\rangle\langle-|$, doesn't describe the system as such but your guess about it, because of your lack of knowledge when entering the room, in the same way as the a priori distribution of bayesian theory, which you will update afterwards, thanks the additional informations due to measurements.
