# States

Definition (State, State Space)

A normalized positive linear functional on a C*-algebra is called a

state.

Thestate space$S(A)$ of a C*-algebra $A$ is the set of all its states.

Note that $S(A)$ is a subset of the closed unit ball in the dual space of $A$.

A linear functional $ω$ on a C*-algebra is a state if and only if $ω(1)=1=∥ω∥$.

The state space of a C*-algebra is convex and weak* compact.

Proof Let $A$ be a C*-algebra and let $S(A)$ be its state space. First, we show convexity. Let $ω_{0},ω_{1}$ be states on $A$ and let $t∈(0,1)$. Consider the convex combination $ω=(1−t)ω_{0}+tω_{1}$. Clearly, $ω$ is linear and $ω(1)=1$. By the triangle inequality, $∥ω∥≤1$. It follows from the lemma above that $ω$ lies in $S(A)$. This proves that $S(A)$ is convex.

Next we show weak* compactness. Since $S(A)$ is contained in the closed unit ball in the dual of $A$, which is weak* compact by the Banach–Alaoglu Theorem, it will suffice to show that $S(A)$ is weak* closed. Let $(ω_{i})$ be a net of states that weak* converges to some bounded linear functional $ω$ on $A$. This means that $ω_{i}(x)→ω(x)$ for every $x∈A$. For all $i$ we have $ω_{i}(x)≥0$ for $x≥0$ and $ω_{i}(1)=1$; hence $ω(x)≥0$ for $x≥0$ and $ω(1)=1$. Thus, $ω$ is again a state. This shows that the state space is weak* closed, completing the proof. $□$

TODO: state space is nonempty

Definition (Pure State)

We say that a state of a C*-algebra $A$ is

pureif it is an extreme point of the state space $S(A)$.

The set of pure states of $A$ is denoted by $P(A)$.