Why is the state of a quantum system called "Density $\textbf{Operator}$"?

Question

In quantum mechanics, a $d$-dimensional pure state is represented by a vector belonging to a $d$-dimensional Hilbert space $\mathcal{H}^d$. A mixed state is represented by a density matrix $\rho \in \mathcal{H}^{d} \otimes {\mathcal{H}^{d*}}$. Where * denotes the dual.

Operators act on the state $\rho$. Why is $\rho$ called an operator when it is the object on which operators are acting? What am I missing?

Does this have something to do with the $C^*$ algebraic formulation of quantum mechanics? If so please suggest some books on $C^*$ algebraic formulation of quantum mechanics.

Answer 1

The terminology can be confusing because in the phrases "state vector", "pure state", and "mixed state", the word "state" is being used in different ways.

The terms pure state and mixed state do not directly refer to a state vector in the Hilbert space, but to our state of knowledge about the physical system. If we say a system is in a pure state, that means that we can represent the system as a single state vector in Hilbert space.$^\star$ If we say the system is in a mixed state, it means that we do not know specifically what state vector represents the system; we can only assign probabilities for the system to be represented by any given state vector.

Pure states can be represented by a state vector in the Hilbert space. Mixed states cannot. On the other hand, while only a pure state can be represented by a state vector, both pure and mixed states can be represented by the density matrix.

The density matrix is an operator, not a state vector, by definition. A ket is a map that takes a bra and produces a complex number. A linear operator is a map that takes one bra and one ket and produces a complex number, and is a bilinear function of the bra and ket. The density matrix is the latter kind of mathematical object. This is clear from the usual way to represent the density matrix as an expansion over state vectors$^\dagger$ \begin{equation} \hat{\rho} = \sum_a p_a |\Psi_a\rangle \langle \Psi_a| \end{equation} since given a bra $\langle b |$ and a ket $| k \rangle$, we see that $\langle b | \hat{\rho} | k \rangle$ is a complex number, and $\hat{\rho}$ is also bilinear function of the bra and ket.

The definitions of pure and mixed states have nothing in particular to do with $C^\star$-algebras, which are a way of formalizing the idea of an algebra of observables. You can understand the density matrix without that language.

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$^\star$If the system is in a pure state, sometimes people will also refer to the state vector itself as the pure state. That's fine so long as the meaning is understood. In this answer, I want to try to clearly distinguish the concepts of a pure state, mixed state, and state vector, so I will intentionally avoid saying this.

$^\dagger$ Note this is not a definition of the density matrix, for example see this answer from Emilio Pisanty

Answer 2

Respectfully, this is actually a deeper question than it may seem at first. From a purely mathematical standpoint, the answer is pretty trivial: we call the density operator an "operator" because, well, it is formally a linear operator on the Hilbert space.

But conceptually, it is not so easy to explain in words what this operator actually does to vectors in the Hilbert space. IMO, of all the operators that come up in QM, only unitary operators have a simple conceptual interpretation in terms of their actual action on state vectors - because only unitary operators take state vectors to state vectors.

Density operators are not unitary, so it's tricky to explain in words what they actually do to an input vector. They just formally give expectation values of observables $A$ via the rather abstract and unintuitive formula $\langle A \rangle = \mathrm{Tr}(\rho A)$.

But it's similarly difficult to explain what an observable operator $A$ "does" as an operator. It only has another rather formal interpretation in terms of its eigenvalues and eigenvectors. (And in reality, only the eigenvectors are actually important.)

So yes, the actual "operation" of the density operator (i.e. the nature of its mapping from vectors to vectors) is difficult to explain conceptually. But the same is true of observable operators as well. Only unitary operators have a straightforward conceptual interpretation as operators, i.e. linear maps between vectors in the Hilbert space. We just have to accept that all other operators on the Hilbert have fairly abstract and formal mathematical meanings.