I was reading Valter Moretti's book on Spectral Theory and Quantum Mechanics, and saw 2 definitions of a quantum state:

1.Let $\mathcal{H}$ be a Hilbert space. A positive, trace-class linear map $\rho:\mathcal{H} \to \mathcal{H}$ with unit trace is called a $\mathbf{state}$.

2.Let $\mathfrak{A}$ be the $C^{*}$-algebra with unit of the quantum theory. Then a $\mathbf{state}$ is a positive linear functional $\omega:\mathfrak{A} \to \mathbb{C}$, which is normalized, i.e. \begin{equation} \omega(a^{*}a) \geq 0 \; \; \forall a \in \mathfrak{A}, \; \; \omega(\mathbb{I})=1. \end{equation}

Are these two notions equivalent? The passage $2 \to 1$ I think should be done via GNS representation, but it's not entirely clear to me how. Conversely, how does one go $1 \to 2$?

I do understand that in QFT, the algebraic approach is more general, but in QM my intuition tells they should be equivalent cause of Stone-von-Neumann, however I haven't seen an explicit proof.


2 Answers 2


If there is a representation $\pi$ of $\mathfrak{A}$ on $\mathcal{H}$, then any density matrix (or "state in Hilbert space") $\rho$ induces a state in the algebraic sense by $$ A\mapsto \mathrm{Tr}_\mathcal{H}(\rho \pi(A)).$$

Note that this is physically the expectation value of $\pi(A)$ with respect to the state $\rho$. The algebraic states are meant to be exactly this - maps on the algebra that assign expectation values to operators.

Conversely, the GNS construction constructs for every algebraic state $\omega$ a Hilbert space $\mathcal{H}_\omega$ and a vector $\psi_\omega$ such that $\rho_\omega : v\mapsto \langle v, \psi_\omega\rangle \psi_\omega $ is a corresponding state in Hilbert space.

When the algebra has only a single irreducible unitary representation $\mathcal{H}$, then we indeed get a bijection between the algebraic states and the density matrices on $\mathcal{H}$: The algebraic states are the convex hull of the pure states, and in this case all pure states are vectors in $\mathcal{H}$, and the convex hull of the pure state $\rho$s are themselves density matrices on $\mathcal{H}$.

When the algebra has more than one irreducible unitary representation, this is no longer necessarily true, and you cannot fix a single $\mathcal{H}$ on which the density matrices are equivalent to all algebraic states.

  • $\begingroup$ Thanks for your answer! I understand the first part. Could you please elaborate on the second part? How to see that the thus defined $\rho_\omega$ is positive,trace class with unit trace? Could you please write down the concrete bijection? $\endgroup$
    – ProphetX
    Commented Mar 25, 2023 at 14:41
  • 1
    $\begingroup$ @ProphetX 1. $\rho_\omega$ is just the projector onto the vector $\psi_\omega$. 2. The bijection is $\omega\mapsto \rho_\omega$ on the pure states and then just extends linearly to all states, that's what I mean when I talk about the convex hull. $\endgroup$
    – ACuriousMind
    Commented Mar 25, 2023 at 16:12
  • $\begingroup$ ah, right, actually for every pure state, the density matrix is a projector into it, so we basically constructed the projector into that pure state $\psi_\omega$, which is given by the GNS representation. Then we linearly extend this to obtain it for the mixed states. thanks for the clarification. $\endgroup$
    – ProphetX
    Commented Mar 25, 2023 at 17:19
  • $\begingroup$ However there is an important fact which deserves to be stressed. $B({\cal H})$ is a C* algebra in its own right and there are algebraic states on it that are not representable as density matrices on ${\cal H}$ when that Hilbert space is not finite dimensional. Algebraic states are more than normal states. Normal states are here the algebraic ones which are also strongly continuous. $\endgroup$ Commented Sep 18, 2023 at 22:13

To add to the existent answer: In case where $\mathfrak A = \mathcal B(H)$, where $H$ is a finite-dimensional Hilbert space, the two notions are equivalent.

Indeed, every density matrix $\rho$, i.e. positive semi-definite operator of unit trace, gives rise to a state in the algebraic sense via $$\omega_\rho (A):=\mathrm{Tr}\,\rho\,A \quad \forall A\in \mathcal B(H) \quad . \tag 1 $$ It is easy to check that this indeed is a state according to definition 2. in the question.

Conversely, every state in the algebraic sense is of the form $(1)$ for some density matrix $\rho$. To see this, note that we can define an inner product on $\mathcal B(H)$ (seen as a Hilbert space) via $\langle A,B\rangle:=\mathrm{Tr}\,A^\dagger\, B$ for $A,B\in \mathcal B(H)$. Now the Riesz representation theorem shows that every linear functional $\phi$ over $\mathcal B(H)$ is of the form $\phi(\cdot)=\langle \rho_\phi,\cdot\rangle$ for some $\rho_\phi \in \mathcal B(H)$.

Positivity and normalization of algebraic states in turn imply that for each $\omega$ there exists a unique density matrix $\rho_\omega$ such that

$$\omega(A)=\mathrm{Tr}\,\rho_\omega\,A \quad \forall A\in \mathcal B(H) \quad .\tag 2$$

  • 2
    $\begingroup$ Thank you very much! Neatly written answer. This clarifies the finite-dimensional case. $\endgroup$
    – ProphetX
    Commented Mar 25, 2023 at 14:56

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