Let me start with some objects. Consider the $\mathrm{C}^*$-algebra $A$ defined by:
$$A=M_1(\mathbb{C})\oplus M_2(\mathbb{C})\subset B(\mathbb{C}^3).$$
Let $x=\mathbb{C}^3$ be given by $(e_1+e_2)/\sqrt{2}\,$ (the one dimensional factor acts on $e_1$).
We have a vector state given by $\rho_x(f)=\langle x,f(x)\rangle$, and
$$\rho_x\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1+\frac{1}{2}c_{11}.$$
If I read something about quantum mechanics for such a state, as it is is given by a unit vector, it is called a pure state.
However this contradicts the answer to this question… and when I pick up Murphy he says that a pure state is such that if $\rho_0\leq \rho_x$ is a positive linear functional, then $\rho_0=t\rho_x$ for $t\in[0,1]$.
However for $\rho_x$ we have the linear functional:
$$\rho_0\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1$$
is a bounded linear functional such that $\rho_x-\rho_{0}$ is a positive linear functional — half the state given by the vector $e_2$, not $t\rho_x$… and so Murphy would say $\rho_x$ is not pure.
Can you help me with my confusion?
Best Answer
If $\pi :A\to B(H)$ is any non-degenerate representation of the C*-algebra $A$, and if $x$ is a unit vector in $H$, then the state $$ \varphi (a) = \langle x,\pi(a)x\rangle $$ is pure iff the cyclic space $$ [\pi (A)x] = \overline{\text{span}}\{\pi (a) x: a\in A\} $$ is irreducible for the action of $A$ there.
This is because the representation of $A$ on $[\pi (A)x]$ is equivalent to the GNS representation associated to $\varphi $, and it is well known that a GNS representation is irreducible iff the state is pure.
In the example given, one can show that $[Ax]=\mathbb C^3$, which is not irreducible, so $\rho _x$ is not pure.
On the other hand, as noted by @MaoWao, if $A$ is an irreducible algebra of operators on $H$, such as $B(H)$ itself, $[\pi (A)x] = H$, for every unit vector $x$, whence every vector state is pure.