A state $\Psi$ on a $C^*$ -algebra $A$ is a positive linear functional $\Psi\colon A \to \Cset$ , $\Psi(a^*a) \geq 0$ for all $a \in A$ , with unit norm. The norm of a positive linear functional is defined by \begin{equation} \norm{\Psi} = \sup_{a \in A}\{|\Psi(a)| : \norm{a}\leq 1\}. \end{equation}For a unital $C^*$ -algebra, $\norm{\Psi} = \Psi(\identity)$ .

The space of states is a convex set. Let $\Psi_1$ and $\Psi_2$ be states, then the convex combination \begin{equation} \lambda\Psi_1+(1-\lambda)\Psi_2, \quad \lambda \in [0,1], \end{equation}is also a state.

A state is pure if it is not a convex combination of two other states. Pure states are the extreme points of the convex set of states. A pure state on a commutative $C^*$ -algebra is equivalent to a character.

A state is called a tracial state if it is also a trace.

When a $C^*$ -algebra is represented on a Hilbert space $\hilbert$ , every unit vector $\psi \in \hilbert$ determines a (not necessarily pure) state in the form of an expectation value, \begin{equation} \Psi(a) = \langle\psi, a\psi\rangle. \end{equation}In physics, it is common to refer to such states by their vector $\psi$ rather than the linear functional $\Psi$ . The converse is not always true; not every state need be given by an expectation value. For example, delta functions (which are distributions not functions) give pure states on $C_0(X)$ , but they do not correspond to any vector in a Hilbert space (such a vector would not be square-integrable).

Bibliography

1

G. Murphy, $C^*$ -Algebras and Operator Theory.
Academic Press, 1990.