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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.

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How are pure states defined in your textbook? –  Raskolnikov Nov 1 '11 at 11:20
It is a extreme point of state space. –  Strongart Nov 2 '11 at 10:54

2 Answers 2

up vote 7 down vote accepted

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.

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Nice answer!The matrix units is good! –  Strongart Feb 12 '12 at 10:47
In the final paragraph, I'm worried that to get all states, you'd want to take weak$^*$ limits in $\mathcal{B}(H)^*$, and that there is no reason why this should preserve normality? Or did I make a mistake? –  Matthew Daws Mar 28 '12 at 13:25
Looks like you are right, Matthew, thank you. I'll try to think about it later today. –  Martin Argerami Mar 28 '12 at 17:09
It's fixed now. Thanks again, Matthew! –  Martin Argerami Mar 28 '12 at 18:28
+1 Nice fix! (Stuff to get to word limit) –  Matthew Daws Mar 28 '12 at 19:14

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).

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this is simple modulo very strong results –  Norbert Dec 5 '13 at 15:00
Norbert; one can get rid of Krein-Milman here, and also the fact that $|\beta \Gamma| = 2^{2^{|\Gamma|}}$ is not terribly difficult once you know how to construct big independent families in the power-set of $\Gamma$. –  Tomek Kania Dec 5 '13 at 16:49
and what about $\ell_\infty=C(\beta\Gamma)$. Do you have a simple argument here? –  Norbert Dec 5 '13 at 16:50
Firstly, Hausdorff's proof is nice: math.stackexchange.com/questions/83526/… I used to know even simpler proof of the inequality $|\beta \mathbb{N}|>\mathfrak{c}$ (which is what we need). Give me some time to remind that myself. –  Tomek Kania Dec 5 '13 at 16:55

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