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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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    $\begingroup$ How are pure states defined in your textbook? $\endgroup$ Nov 1, 2011 at 11:20
  • $\begingroup$ It is a extreme point of state space. $\endgroup$
    – Strongart
    Nov 2, 2011 at 10:54

3 Answers 3

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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 $F_x$.

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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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  • $\begingroup$ Nice answer!The matrix units is good! $\endgroup$
    – Strongart
    Feb 12, 2012 at 10:47
  • $\begingroup$ 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? $\endgroup$ Mar 28, 2012 at 13:25
  • $\begingroup$ Looks like you are right, Matthew, thank you. I'll try to think about it later today. $\endgroup$ Mar 28, 2012 at 17:09
  • $\begingroup$ It's fixed now. Thanks again, Matthew! $\endgroup$ Mar 28, 2012 at 18:28
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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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    $\begingroup$ this is simple modulo very strong results $\endgroup$
    – Norbert
    Dec 5, 2013 at 15:00
  • $\begingroup$ 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 $\Gamma$. $\endgroup$ Dec 5, 2013 at 16:49
  • $\begingroup$ and what about $\ell_\infty=C(\beta\Gamma)$. Do you have a simple argument here? $\endgroup$
    – Norbert
    Dec 5, 2013 at 16:50
  • $\begingroup$ 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. $\endgroup$ Dec 5, 2013 at 16:55

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