quantum mechanics - What is the actual meaning of the density operator?

I am not able to understand the definition of the density operator. I know that if $V$ is a vector space and if I have $k$ states belonging to this vector space, say $|\psi_{i}\rangle$ for $1\le i\le k$, and the probability of the system being in $|\psi_{i}\rangle$ is $p_i$, then the density operator for the system is given by $$\rho=\sum_{i=1}^{i=k}p_i|\psi_i\rangle\langle\psi_i |. $$ Now what I am unable to understand is

1. 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 probability?


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

Answer

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 which 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?

Well, the expectation value would be:

$$ \langle \hat{O} \rangle = p_1\langle \psi_1 \lvert \hat{O} \lvert \psi_1 \rangle + p_2\langle \psi_2 \lvert \hat{O} \lvert \psi_2 \rangle $$

where $p_1$ and $p_2$ are the corresponding classical probabilities of each state being prepared. All we need to do know is to use some trace properties to construct a more general and more compact form of the above equation, which will finally be the density matrix:

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

Remember, we can introduce the trace here because the trace of a scalar is the scalar itself, so there is no ambiguity in what we are doing here. In fact it's a mathematical trick which greatly helps in our derivation.

Now, using the cyclic invariance and linearity properties of the trace:

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

where $\rho$ is what we call the density matrix. The density operator 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.