4
$\begingroup$

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.

$\endgroup$

2 Answers 2

7
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
5
$\begingroup$

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

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

$\endgroup$
1
  • 2
    $\begingroup$ Thank you very much! Neatly written answer. This clarifies the finite-dimensional case. $\endgroup$
    – ProphetX
    Commented Mar 25, 2023 at 14:56

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.