# States in algebraic QFT, and non-diagonal matrix elements

Let $\mathcal A$ be a $C^*$ algebra in the sense of Haag and Kastler. We define a state as an element of the space dual to $\mathcal A$, that is, $$\phi\colon\mathcal A\to\mathbb C$$

This definition of a state is supposed to generalise the standard definition (as an element of a Hilbert space over which the algebra $\mathcal A$ acts on). I believe the correct correspondence is $$\phi(A)\overset{\text{identify}}\sim \langle \phi|A|\phi\rangle$$ where the r.h.s. corresponds to the usual "matrix notation" of standard quantum mechanics, with $A\in\mathcal A$ an arbitrary operator.

I guess I missed something about this definition, because I really don't know how to represent a non-diagonal matrix element of the form $$\langle\phi|A|\psi\rangle$$ where $\phi,\psi$ are two distinct states. If my identification above is correct, what should we write in the l.h.s. of $$\square\overset{\text{identify}}\sim \langle \phi|A|\psi\rangle\,?$$

• First states are positive normalized functionals $\phi$. The matrix elements you consider are dedined by fixing two observables $C$ and $D$ and are maps $A \mapsto \phi(C^*AD)$. Oct 31, 2017 at 20:40
• @ValterMoretti, what is the intuition on this? Why taking two kets $|\phi\rangle,|\psi\rangle$ and forming $\langle \phi | A| \psi\rangle$ corresponds to maps $A\mapsto \phi(C^\ast A D)$ with $C,D$ fixed?
– Gold
Oct 31, 2017 at 20:49
• Look at the proof of GNS theorem and you will see that this is the correct point of view. Now I am too tired to write down an extended answer. Oct 31, 2017 at 21:06
• @ValterMoretti I believe I got the point by looking at the GNS construction. I posted one answer with my conclusion. If I got it incorrectly, please let me know.
– Gold
Oct 31, 2017 at 21:45
• @user1620696 you got the point indeed. Nov 1, 2017 at 6:00

This is a tentative answer based on @ValterMoretti's comment and the GNS theorem following these notes. If anything is found to be wrong because I misunderstood it, please let me know.

Let $\mathcal{A}$ be the $C^\ast$-algebra and $\phi : \mathcal{A}\to\mathbb{C}$ a state. Endow the algebra it with the semi-inner product $\langle x,y\rangle = \phi(x^\ast y)$.

Consider then $N_\phi = \{x\in \mathcal{A} : \langle x,x\rangle = 0\}$. The GNS construction is based on noting that $N_\phi$ is a closed left ideal so that on the quotient $\mathcal{A}/N_\phi$ one gets a well-defined inner product descending from $\langle \cdot,\cdot\rangle$. More precisely let $\Pi : \mathcal{A}\to \mathcal{A}/N_\phi$ be the canonical surjective projection, namely $\Pi(x)=x+N_\phi$. Define then

$$\langle \Pi(x),\Pi(y)\rangle=\langle x,y\rangle$$

It turns out that $\langle \Pi(x),\Pi(x)\rangle= 0$ if and only if $\Pi(x) = 0$. Indeed by definition the condition is equivalent to $\langle x,x\rangle = 0$ so that $x\in N_\phi$ and thus $\Pi(x) = 0$ as wanted.

Define $\mathcal{H}_\phi$ the completion of the inner product space $\mathcal{A}/N_\phi$ so that it is a Hilbert space. We represent $\mathcal{A}$ in $\mathcal{H}_\phi$ by

$$\pi_\phi : \mathcal{A}\to \mathcal{A}(\mathcal{H}_\phi)$$

given the action on the elements of $\mathcal{A}/N_\phi$ by $\pi_\phi(a)(x+N_\phi)=ax+N_\phi$.

Remember now that the elements of $\mathcal{H}_\phi$ are to be the usual state vectors. They are thus the elements of $\mathcal{A}/N_\phi$ plus the limits of Cauchy sequences.

Let then $a\in \mathcal{A}$, so that we want to define its matrix elements.

Let $\Phi,\Psi\in\mathcal{A}/N_\phi$ be two state vectors so that $\Phi=\Pi(x)$ and $\Psi=\Pi(y)$. We then have

$$\langle \Phi,\pi_{\phi}(a)\Psi\rangle=\langle \Pi(x),\pi_{\phi}(a)\Pi(y)\rangle=\langle \Pi(x),\Pi(ay)\rangle=\phi(x^\ast ay).$$

Thus we see that the matrix elements as done in QM are obtained in the above form from elements of $\mathcal{A}$ which descend to states.

I believe that to generalize the whole of $\mathcal{H}_\phi$ one will finaly just need to argue with the limits of Cauchy sequences.