## Tuesday, August 10, 2010

### KMS states and symmetries.

I'm still looking at Quantum Statistical Mechanical systems, which I'm treating as a C* algebra $\mathcal{A}$ together with a time evolution $\sigma_t$. Particularly, I'm interested in its equilibrium states, which we can characterize via the KMS condition (recall that a state on a QSM ($\mathcal{A}$,$\sigma_t$) is a positive [or zero] linear functional $\phi : \mathcal{A} \to \mathbb{C}$ with $\phi(1)=1$, or has operator norm 1 if $\mathcal{A}$ has no unit.)

A state $\phi$ satisfies the KMS condition at inverse temperature $\beta$ if for all $a,b \in \mathcal{A}$ we can find a function $F_{a,b} (z)$ with the following properties:
1. Its holomorphic on the strip $\{z \in \mathbb{C} : 0 < \Im(z) < \beta \}$

2. continuous on boundary and bounded

3. $F_{a,b}(t) = \phi(a \sigma_t(b))$ and $F_{a,b}(t+i\beta) = \phi(\sigma_t(b) a)$

4. for all $t \in \mathbb{R}$
We call such states $KMS_{\beta}$ states. We also have $KMS_{\infty}$ states, but rather than define them on half-plane (we call those ground states), we use a stronger condition (For reasons I neither know nor understand at this point): A state $\phi$ is a $KMS_{\infty}$ state if we have $KMS_{\beta}$ states $phi_{\beta}$ such that, for sufficiently large $\beta$, $\phi(a) = \lim_{\beta \to \infty} \phi_\beta (a)$ for all $a \in \mathcal{A}$. Mathematically, all I should need is the definitions, just to understand them and go from there. But I'm not quite satisfied with them. I'm wishing I took a course in statistical thermodynamics/mechanics at this point, as I have little intuition for how these equilibrium states relate to our phase transitions of the system (indeed, I'm still not sure what "phase transition" means for a C* dynamical system.) The $KMS_{\beta}$ states (we denote the set of them by $\Sigma_{\beta}$) are the equilibrium states [states that we can actually measure] for "temperatures" of $1/\beta$. At 0 temperature we have the $KMS_{\infty}$ states. Now, if $\phi_1$ and $\phi_2$ are two $KMS_{\beta}$ states, its not too hard to see that any linear combination of the two will also be, and from that its not too hard to see that $\Sigma_{\beta}$ is a convex set (there's a path between any two $KMS_{\beta}$ states). Connes goes further, and states that its compact under the weak topology (pointwise convergence?) and that its a Choquet Simplex. And I have no idea what that is. But for this blog's purposes, I don't think I need to. The important bit is that $KMS_{\beta}$ states are a convex set. Its not too hard to see that $\Sigma_{\infty}$ is a convex set too. Convex sets have extremal points (points that aren't in a line segment joining two other points.) We'll denote the set of extremal $KMS_{\beta}$ states by $\xi_{\beta}$. Connes has a proposition stating that the extremal $KMS_{\beta}$ states are the same as the states that have a factorial GNS representation (while I have an idea what a GNS representation is - the image of the state under a homomorphism to a Hilbert space, where the homomorphism has certain properties and is constructed in a certain way - I have no idea what it means for one to be factorial.) R. Haag in Local Quantum Physics has a similar statement, that $\xi_{\beta}$ is the same as the set of $KMS_{\beta}$ states that are primary. But I don't know what that means either.

So what's the structure and intuition behind $\xi_{\beta}$? How should I think of it? If I'm thinking just in terms of a physical thermodynamic system, what do the extremal equilibrium states represent? My guess is states of matter, or pure phase states (considering that a change in a state of matter is a "phase transition"). So does that mean there are "impure phase states", id est, states in a mixture of phases (so that $\Sigma_{\beta}$ is non-trivial). Back to the QSM: If we're working in a system with only one phase transition (I'm not sure what a phase transition means in this context) does that mean that there are only two extremal $KMS_{\beta}$ states?

Apart from the equilibrium states, we're also interested in symmetries of the QSM system. Rather frustratingly, I cannot talk much about the physics of QSM symmetries (but I resolve to be able to soon! as I do with statistical thermodynamics!), so I must stick to the mathematics. We're interested in both automorphisms (symmetries of the whole system) and endomorphisms (symmetries of only a part of the system....I think.) In either case, they must be compatible [commute] with our time evolution, id est, if g is our auto/endo-morphism, then $g \sigma_t = \sigma_t g$ for all t. for automorphism g, we say $g \in Aut(\mathcal{A}, \sigma_t)$.

A subgroup $G \subset Aut(\mathcal{A}, \sigma_t)$ we call a group of symmetries by automorphisms. It acts on KMS states via pullback, id est, $g^{*}(\phi) (a) = \phi ( g (a) )$. Connes included a lemma stating that inner automorphisms (from a unitary state $u \in \mathcal{A}$, inner automorphisms acts by $a \mapsto u a u^{-1}$) act trivially on KMS states. Actually, the lemma stated something else, but that was in the proof. The proof essentially uses an analytic continuation of the map $t \mapsto \sigma_t(a)$ to the whole complex plane (I don't know how that's possible) and uses the fact that KMS states have the property that $\phi(b \sigma_{i \beta} (a)) = \phi (ab)$. It then follows that $\phi(u a u^{-1}) = \phi(a)$, and the action is trivial.

An endomorphism $\rho$ also has its pullback, $\rho^{*} (\phi) (a) = \frac{\phi(\rho(a))}{\phi(e)}$, assuming that $\phi(e) \neq 1$.
$e=\phi(1)$ (assuming $\mathcal{A}$ has unity, of course) and its not too hard to see that e is an idempotent ($e^2 = e$, because $e^2 = \phi(1)^2 = \phi (1^2) = e$). Inner endomorphisms also act trivially, by a similar proof.

Things I don't know this post:

Choquet Simplex
factorial GNS representation
primary KMS state.
The analytic continuation used in the proof of the triviality of the action of a group of inner automorphism.
Connes also talks a bit about multiplier algebras and essential ideas - I may look into those tomorrow, but as best I can tell I won't be needing those ideas any time soon. Marcolli doesn't mention them at all, for instance.
Connes also mentions that we can pushforward KSM states: I largely understood what I read there [or I think I did], but I choose not to write about it, largely because of time, and largely because I don't yet see how it relates.