In textbooks, it is sometimes written that a mixed state can be represented as mixture of $N$ (I assume here $N<+\infty$) quantum pure states $|\psi_i\rangle$ with classical probabilities $p_i$:
$$\rho = \sum_{i=1}^N p_i |\psi_i \rangle \langle \psi_i| \tag{1}\:.$$
Above $p_i \in (0,1]$ and $\sum_i p_i =1$ and a do not necessarily assume that $\langle \psi_i|\psi_j\rangle =0$ if $i\neq j$ but I require that $\langle\psi_i |\psi_i\rangle =1$ so that $\rho \geq 0$ and $tr(\rho)=1$. (There is another procedure to obtain mixed states using a partial trace on a composite system, but I am not interested on this here).
I am not sure that it makes any sense to distinguish between classical probabilities embodied in the coefficients $p_i$ and quantum probabilities included in the pure states $|\psi_i\rangle$ representing the quantum part of the state. This is because, given $\rho$ as an operator, there is no way to uniquely extract the numbers $p_i$ and the states $|\psi_i\rangle$.
I mean, since $\rho = \rho^\dagger$ and $\rho$ is compact, it is always possible, for instance, to decompose it on a basis of its eigenvectors (and there are many different decompositions leading to the same $\rho$ whenever $\rho$ has degenerate eigenspaces). Using non orthogonal decompositions many other possibilities arise.
$$\rho = \sum_{j=1}^M q_j |\phi_j\rangle \langle \phi_j|\tag{2}$$
where again $q_j \in (0,1]$ and $\sum_j q_j =1$ and now $\langle \phi_i|\phi_j\rangle =\delta_{ij}$. I do not think there is a physical way to decide, a posteriori, through suitable measurements of observables if $\rho$ has been constructed as the incoherent superposition (1) or as the incoherent superposition (2). The mixed state has no memory of the procedure used to construct it.
To pass from (1) to (2) one has, in a sense, to mix (apparently) classical and quantum probabilities.
So I do not think that it is physically correct to associate a classical part and a quantum part to a mixed state, since there is no a unique physical way to extract them from it.
Perhaps my impression is simply based on a too naively theoretical interpretation of the formalism.
I would like to know your opinions about this issue.
Best Answer
Yes, the density matrix reconciles all quantum aspects of the probabilities with the classical aspect of the probabilities so that these two "parts" can no longer be separated in any invariant way.
As the OP states in the discussion, the same density matrix may be prepared in numerous ways. One of them may look more "classical" – e.g. the method following the simple diagonalization from equation 1 – and another one may look more quantum, depending on states that are not orthogonal and/or that interfere with each other – like equations 2.
But all predictions may be written in terms of the density matrix. For example, the probability that we will observe the property given by the projection operator $P_B$ is $$ {\rm Prob}_B = {\rm Tr}(\rho P_B) $$ So whatever procedure produced $P_B$ will always yield the same probabilities for anything.
Unlike other users, I do think that this observation by the OP has a nontrivial content, at least at the philosophical level. In a sense, it implies that the density matrix with its probabilistic interpretation should be interpreted exactly in the same way as the phase space distribution function in statistical physics – and the "quantum portion" of the probabilities inevitably arise out of this generalization because the matrices don't commute with each other.
Another way to phrase the same interpretation: In classical physics, everyone agrees that we may have an incomplete knowledge about a physical system and use the phase space probability distribution to quantify that. Now, if we also agree that probabilities of different, mutually excluding states (eigenstates of the density matrix) may be calculated as eigenvalues of the density matrix, and if we assume that there is a smooth formula for probabilities of some properties, then it also follows that even pure states – whose density matrices have eigenvalues $1,0,0,0,\dots$ – must imply probabilistic predictions for most quantities. Except for observables' or matrices' nonzero commutator, the interference-related quantum probabilities are no different and no "weirder" than the classical probabilities related to the incomplete knowledge.