Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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.

share|improve this answer
    
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  
+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).

share|improve this answer
    
this is simple modulo very strong results –  no identity 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? –  no identity 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.