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.

  • $\begingroup$ Hi Abhishek Banerjee. Welcome to Phys.SE. Are you also asking why it is called a density matrix? $\endgroup$
    – Qmechanic
    Apr 15, 2022 at 14:11
  • $\begingroup$ No, I think I understand why it's a density $\textbf{matrix}$. My confusion is as follows. Is it called an operator just because it has the form of a matrix, which are representations of operators when a basis is chosen? Or does it actually behave like an operator between two vector spaces? $\endgroup$
    – junfan02
    Apr 15, 2022 at 14:13

1 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.

$^\star$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

  • $\begingroup$ I understand why we need a density matrix, my question is about why is the density matrix being called an operator on the same Hilbert space. Does it take vectors(pure states) to pure states? The definition of the operator that I am implicitly assuming here is it is a linear map between two vector spaces. $\endgroup$
    – junfan02
    Apr 15, 2022 at 14:18
  • $\begingroup$ @AbhishekBanerjee That's addressed in the second to last paragraph of my answer. It indeed is a linear map that maps kets to kets, or bras to bras. $\endgroup$
    – Andrew
    Apr 15, 2022 at 14:19
  • $\begingroup$ Is it called an operator because it looks like the representation of an operator(matrix)? $\endgroup$
    – junfan02
    Apr 15, 2022 at 14:20
  • $\begingroup$ @AbhishekBanerjee The definition of a linear operator in quantum mechanics is that it is a linear map from kets to kets. Or, equivalently, it maps a bra and a ket to a complex number. The density matrix satisfies this definition. $\endgroup$
    – Andrew
    Apr 15, 2022 at 14:21
  • $\begingroup$ That clears it up, thank you $\endgroup$
    – junfan02
    Apr 15, 2022 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.