A question about pure state 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.
 A: One can give a simple cardinality argument: let $H$ be a Hilbert space with an orthonormal basis $(e_i)_{i\in \Gamma}$. Let $\ell_\infty(\Gamma)$ act on $H$ via diagonal operators wrt to the chosen basis. Then $\ell_\infty=C(\beta \Gamma)$ has $2^{2^{|\Gamma|}}$ (Pospíšil) characters which correspond to points in $\beta \Gamma$. Characters (multiplicative functionals on commutative C*-algebras) are pure. Use Krein-Milman to extend them to pure states on $\mathscr{B}(H)$. Consequently, we have $2^{2^{|\Gamma|}}$ pure states on $\mathscr{B}(H)$ but only at most $2^{|\Gamma|}$ vectors in $H$ (and these correspond to vector states). 
A: Consider an orthonormal basis $\{x_j\}$ of $H$ with $x_1=x$, and consider $\{E_{kj}\}$ the corresponding matrix units (we don't really need matrix units, just the projection onto the span of $x$, but it might help understand). Assume that you can write $F_x=\alpha f + (1-\alpha)g$, for some $\alpha\in[0,1]$ and states $f,g$. Then, since $0\leq E_{11}\leq I$,
$$
1=\langle E_{11}x,x\rangle = F_x(E_{11})=\alpha f(E_{11}) + (1-\alpha) g(E_{11})\leq
\alpha+1-\alpha = 1.
$$
So $\alpha f(E_{11}) + (1-\alpha) g(E_{11})=1$. But as $f(E_{11})\leq1$, $g(E_{11})\leq1$, we conclude that $f(E_{11})=g(E_{11})=1$. In particular, $f(I-E_{11})=0$. Then, for any $T\in B(H)$,
$$
0\leq|f(T(I-E_{11}))|\leq f(T^*T)^{1/2} f((I-E_{11})^2)^{1/2}=f(T^*T)^{1/2}f(I-E_{11})^{1/2}=0.
$$
Thus $f(T)=f(T\,E_{11})$ for all $T$. Taking adjoints, $f(T)=f(E_{11}T)$. But then $f(T)=f(TE_{11})=f(E_{11}TE_{11})$. As $E_{11}TE_{11}=\langle Tx,x\rangle\,E_{11}=F_x(T)E_{11}$,
$$
f(T)=f(E_{11}TE_{11})=F_x(T)\,f(E_{11})=F_x(T).
$$
Similarly with $g$, so $F_x$ is an extreme point. 
Edit: thanks to Matthew for noting that my argument in the last paragraph from the previous version was wrong. Here is the new argument.
Let $\pi_0:B(H)\to B(H)/K(H)$ be the quotient map onto the Calkin algebra $C(H)$. Let $\pi_1:C(H)\to B(K)$ be an irreducible representation of $C(H)$. Then 
$$
\pi=\pi_1\circ\pi_0:B(H)\to B(K)
$$
is an irreducible representation. Using the correspondence between irreps and pure states, there exists a pure state $\varphi$ on $B(H)$ such that its GNS representation is unitarily equivalent to $\pi$. But this tells us that $\varphi(T)=0$ for all $T\in K(H)$, and so $\varphi$ cannot be a point state.
A: 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 $F_x$.
