TITLE: A question about pure state
QUESTION [15 upvotes]: For every unit vector $x$ in a Hilbert space $H$,let $F_x$ be the linear functional on $\mathcal B(H)$ (bounded linear operators) defined by $F_x(T)=(Tx,x)$. Prove that each $F_x$ is pure state and $\mathcal B(H)$ has pure states which are not like this.

REPLY [10 votes]: Here is another way to see that not all pure states on $B(H)$ are equal to some $F_x$ (the second part of your question). Recall the following theorem:

Theorem. Let $a$ be a positive element of a non-zero $C^*$-algebra $A$. Then there exists a pure state $\rho$ of $A$ such that $\lVert a \rVert = \rho(a)$.
Proof. See e.g. Gerard J. Murphy, $C^*$-algebras and operator theory, theorem 5.1.11.

Now assume that $H$ is infinite-dimensional and let $\{y_n\}_{n=1}^\infty$ be an orthonormal sequence in $H$. Let $T \in B(H)$ be the diagonal operator given by $y_n \mapsto \frac{n - 1}{n} y_n$ (and $z \mapsto 0$ if $\{z\} \perp \{y_n\}_{n=1}^\infty$). Note that $T$ is positive with $\lVert T \rVert = 1$, but it does not preserve the norm of any non-zero vector. We apply the above theorem to the positive operator $T^2$, so we find a pure state $\rho$ satisfying $\rho(T^2) = \lVert T^2\rVert = 1$. But for any unit vector $x\in H$ we have
$$ F_x(T^2) = \langle T^2x,x\rangle = \langle Tx,Tx\rangle = \lVert Tx\rVert^2 < 1, $$
since we established that $T$ does not preserve the norm of any non-zero vector. Consequently, $\rho$ is not of the form $F_x$.