What is the actual meaning of the density operator?

Firstly, what is a state?

A state gives you the complete description of a system. Let's label the state of a system $\lvert \psi \rangle$. This is a normalised state vector which belongs in the vector space of states. Keep in mind that we are talking about the full state; I haven't decomposed it into basis states, and I will not. This is not what the density matrix is all about.

The state vector description is a powerful one, but it is not the most general. There are some quantum experiments for which no single state vector can give a complete description. These are experiments that have additional randomness or uncertainty, which might mean that either state $\lvert \psi_1⟩$ or $\lvert \psi_2⟩$ is prepared. These additional randomness or uncertainties arise from imperfect devices used in experiments, which inevitably introduce this classical randomness, or they could arise from correlations of states due to quantum entanglement.

In this case, then, it is convenient to introduce the density matrix formalism. Since in quantum mechanics all we calculate are expectation values, how would you go about calculating the expectation value of an experiment where in addition to having intrinsic quantum mechanical randomness you also have this classical randomness arising from imperfections in your experiment?

Recall that $$Tr(\lvert\phi_1\rangle\langle\phi_2\rvert)=Tr(\lvert\phi_1\rangle\otimes\langle\phi_2\rvert)=\langle\phi_1\mid\phi_2\rangle,$$ and $$\hat{O}\circ(\lvert\phi_1\rangle\langle\phi_2\rvert)=(\hat{O}\lvert\phi_1\rangle)\otimes\langle\phi_2\rvert$$

Now, using the linearity of the trace, we can compute the expectation value as:

$$ \langle \hat{O} \rangle = p_1\langle \psi_1 \lvert \hat{O} \lvert \psi_1 \rangle + p_2\langle \psi_2 \lvert \hat{O} \lvert \psi_2 \rangle$$ $${} = p_1Tr(\hat{O} \lvert \psi_1 \rangle \langle \psi_1 \lvert) + p_2Tr(\hat{O} \lvert \psi_2 \rangle \langle \psi_2 \lvert) $$ $${} =Tr(\hat{O} (p_1 \lvert \psi_1 \rangle \langle \psi_1 \lvert) + p_2 \lvert \psi_2 \rangle \langle \psi_2 \lvert)) = Tr(\hat{O} \rho)$$

where $p_1$ and $p_2$ are the corresponding classical probabilities of each state being prepared, and $\rho$ is what we call the density matrix (aka density operator): it contains all the information needed to calculate any expectation value for the experiment.

So your suggestion 1 is correct, but suggestion 2 is not, as this is not a superposition. The system is definitely in one state; we just don't know which one due to a classical probability.

Check the form of your density operator it should be something like $$\rho=\sum_i p_i |\psi_i\rangle\langle\psi_i|$$

I find the most intuitive way to think of the density matrix as follows. Consider an experimenter in his lab who has a "machine" which produces the quantum state $|\psi\rangle$ but the machine doesn't work perfectly and produces some other states which we don't want. Then to carry out predictions about these states we need something which captures our ignorance of the states which we have i.e we could not use state vectors as we do not have pure states but an ensemble or mixture of states. The density matrix is then what we use. Your suggestion 1.

As an aside the density operator can also describe pure states.

Your suggestion number two is incorrect as superpositions and ensembles are physically quite different things. Consider the pure state $$|\psi\rangle=\sum_i \sqrt{p_i}\ |\psi_i\rangle$$ Note that the weights are the square root of the probability which relates to the born rule.

The density operator of a quantum system summarises the expectation values of the observables of that system alone. If you have two systems $S_1$ and $S_2$ that are entangled with one another, then there is no pure Schrodinger picture state for either of the entangled systems. However, there are observables of $S_1$ (or $S_2$) and those observables have expectation values. The standard way of writing expectation values in terms of the pure state is given by $\langle\psi|\hat{A}|\psi\rangle$ for some arbitrary observable $\hat{A}$, but this can also be written as $\text{tr}(|\psi\rangle\langle\psi|\hat{A})$. So then the question is whether there is an operator on the Hilbert space of $S_1$ alone that would give you the right expectation values using the second formula. It turns out that the correct choice is given by the partial trace since the partial trace is the only choice consistent with the Born rule, see "Quantum Computation and Quantum Information" by M.A. Nielsen, I.L. Chuang, p. 107.

You ask:

Does this mean the system is in exactly one of the $|\psi_i\rangle$ states and we don't know in which state it is in and we just know the probabilities?

No. If $S_1$ and $S_2$ can be manipulated jointly in a coherent process then they can undergo quantum interference. During quantum interference, the square magnitude of probability amplitudes do not in general obey the calculus of probabilities, see

• Machines, Logic and Quantum Physics. D. Deutsch et al. Bull. Symb. Log. 6 no. 3, 265 (2000), arXiv:math/9911150.

Your next question is:

Or it is in a superposition of these k states with probabilities being interpreted as weights?

This is also wrong because the density matrix of a system is an instrumental description of what you would get by doing measurements on that system alone. There is not necessarily anything wrong with using such a description if you know its limitations.

I should also add that, contrary to some of the answers above, the density matrix of a system in general does not provide a complete description of the system. A complete description of a system will include information about what systems it is entangled with and how it is entangled, which is not given by the system's density matrix. Such information is given by a system's Heisenberg picture observables, see:

• Information Flow in Entangled Quantum Systems. D. Deutsch and P. Hayden. Proc. R. Soc. Lond. A 456 no.1999, 1759 (2000), arXiv:quant-ph/9906007.