If a wave function represents knowledge, what does a density matrix mean, then?

Question

I'm curious about this. I've heard of interpretations of quantum theory in which the wave function $\psi$ is taken as representing knowledge, or information (e.g. "Quantum Bayesianism"), about the physical parameters of a particle like its position and momentum, and thus things like the famous "wave function collapse" are taken as simply an update of knowledge due to new information received about the system. Moreover, with the Bayesian interpretation you can say a prior wave function before a measurement could also incorporate assumed knowledge as opposed to that actually received from the environment.

However, what I don't get is this. I also notice there's a thing called a "mixed state" or "density matrix" where you end up with a classical (i.e. not complex number-based, or norm-1 real, not norm-2 complex, probability) probability distribution among a wide panoply of wave functions, representing uncertain knowledge in what wave function it actually is, such as in the case of describing one particle of an entangled system (which has a composite wave function $\psi(x_1, x_2)$ and so the wave function $\psi_1(x_1)$ depends on specifically what value attains for $x_2$ which is unknown, and conversely), or where that a system is interacting with the macroscopic environment, or where there is uncertainty in how the system was prepared.

Yet it seems if we interpret the first wave function as knowledge, then the density matrix becomes some kind of redundant-seeming "knowledge about knowledge", which makes no sense. Does this mean the knowledge interpretation of a wave function itself is problematic, or is there some way to reconcile these two and explain why the density matrix is not redundant? It makes sense to consider it non-redundant in a "realist" interpretation where that $\psi$ is a physical wave or something like that, since then the DM straightfowardly becomes lack of knowledge about what the real physical wave is, but not where $\psi$ itself is still knowledge.

So what's up with this, anyways?

Answer 1

Quoting from the abstract of

Peres, Asher. "What is a state vector?." American Journal of Physics 52.7 (1984): 644-650.

"a state vector represents a procedure for preparing or testing one or more physical systems. "

In this sense a pure state thus represents a "perfect" preparation procedure, where all information about the state is known. In this same sense a mixed state is an imperfect preparation procedure, where some information about the state is not known. This imperfect knowledge is modelled as a statistical distribution of perfect procedures.

Answer 2

A wave function tells you the probability distribution of a particle. As you say, this probability distribution may be influenced by external factors, such as the presence of other particles or uncertainty about the nature of the system.

A mixed state $\Psi$ is a probability distribution on the state space. The density matrix tells you how likely it is to be in some state. Knowledge of a mixed state is not the same as knowledge of a state; you may be confounding a mixed state with its expected value $\mathbb E[\Psi]$, which is a pure state with the property that for all measurable sets $A$, $$(\mathbb E(\Psi))(A) = \mathbb E[\mathbb P(A)]$$ where the expected value on the right is taken over the state space w.r.t. the probability measure $\Psi$.