34.10 States

Proof. By the Cauchy-Schwarz inequality.$\square$

Proof. Since the evaluation map is weak* continuous and

the state space is an intersection of convex and weak*-closed sets, so it is closed and convex.

By Theorem 34.9.2, $S(A) \subset \ol{B_{A^*}(0, 1)}$, which is weak* compact by the Banach-Alaoglu Theorem. Therefore $S(A)$ is compact by Proposition 5.16.3.

By the Krein-Milman Theorem, $S(A)$ is the weak*-closed convex hull of $P(A)$.$\square$

Proof. (1): Let $\phi \in \Omega(A)$. By Proposition 33.7.2, $\norm{\phi}_{A^*}= \dpn{1, \phi}{A}= 1$. Thus $\phi$ is a state by Theorem 34.9.2, and $\Omega(A) \subset S(A)$.

Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^{*}x \in \ker(\phi)$ as well. As $t \ne 0$, $x^{*}x \in \ker(\psi)$ and $x^{*}x \in \ker(\rho)$. By the Cauchy-Schwarz inequality,

Likewise, $\dpn{x, \rho}{A}= 0$ as well. Hence $\ker(\psi), \ker(\rho) \supset \ker(\phi)$. Thus there exist scalars $\alpha, \beta \in \complex$ such that $\phi = \alpha \psi = \beta \rho$. However, since $\phi, \psi, \rho \in S(A)$, $\alpha = \beta = 1$, and $\phi = \psi = \rho$. Therefore $\phi$ is a pure state.

(2): Using the Gelfand-Naimark Theorem, identify $A$ with $C(\Omega(A); \complex)$ and $S(A)$ as Radon probability measures on $\Omega(A)$.

Let $\cm = \bracs{t\mu|\mu \in S(A), t \in [0, 1]}$. By Proposition 23.7.4, the extreme points of $\cm$ are the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, and possibly $0$. For any $\mu \in S(A)$, $\nu, \rho \in \cm$, and $t \in (0, 1)$, $\mu = (1 - t)\nu + t\rho$ implies that $\nu(\Omega(A)) = \rho(\Omega(A)) = 1$, and $\nu, \rho \in S(A)$ as well. Thus the extreme points of $S(A)$ are exactly the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, which correspond to $\Omega(A)$ itself.$\square$

Proof. (1): By Theorem 34.9.2, $\norm{\phi}_{B^*}= \dpn{1_A, \phi}{B}$. By the Hahn-Banach Theorem, there exists $\Phi \in A^{*}$ such that $\Phi|_{B} = \phi$ and $\norm{\Phi}_{A^*}= \norm{\phi}_{B^*}= \dpn{1_A, \Phi}{A}$. Thus Theorem 34.9.2 implies that $\Phi \in S(A)$.

(2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the Krein-Milman Theorem.

Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_{B} + t\rho|_{B}$. Since $\phi \in P(B)$, $\phi = \psi|_{B} = \rho|_{B}$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$.$\square$

Proof. Let $\lambda \in \sigma_{A}(x)$. By Proposition 33.8.2, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]}= \lambda$. By Proposition 34.10.5, $\phi \in P(A[x])$. The pure state extension theorem implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]}= \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A}= \lambda$, and $\sigma_{A}(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$.

Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the Gelfand-Naimark Theorem, the Spectral Theorem, and the Riesz Representation Theorem, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_{A}(x)$. In which case,

Finally, since $S(A)$ is compact and convex by Proposition 34.10.4,

by Proposition 5.16.3 and Proposition 5.5.3.$\square$

Proof, [Theorem 13.7, Zhu93]. Let $B = \text{span}\bracs{x, 1}$. For each $\alpha x + \beta \in B$, let $\dpn{\alpha x + \beta, \phi_0}{B}= \alpha \lambda + \beta$. Since $\sigma_{A}(1) = \bracs{1}$, $\phi_{0} \in B^{*}$ is a well-defined linear functional with $\dpn{x, \phi_0}{B}= \lambda$ and $\dpn{1, \phi_0}{B}= 1$.

In addition, for each $\alpha x + \beta \in B$, $\alpha \lambda + \beta \in \sigma_{A}(\alpha x + \beta)$ by Proposition 33.5.11, and

Thus $\norm{\phi_0}_{B^*}= \dpn{1, \phi_0}{B}= 1$. By the Hahn-Banach Theorem, there exists $\phi \in A^{*}$ such that $\phi|_{B} = \phi_{0}$ and $\norm{\phi}_{A^*}= \norm{\phi_0}_{B^*}$. In which case, $\dpn{x, \phi}{A}= \lambda$ and $\dpn{1, \phi}{A}= \norm{\phi}_{A^*}= 1$. By Theorem 34.9.2, $\phi$ is positive and hence a state.$\square$

Proof, [Theorem 13.9, Zhu93]. (1): Let $x \in A$ such that $\dpn{x, \phi}{A}= 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By Theorem 34.10.8, $\sigma_{A}(x) = \bracs{0}$, and $\norm{x}_{A} = [x]_{sp}= 0$ by Theorem 34.4.3.

Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$,

because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A}= 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well.

(2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^{*}$ by Lemma 17.1.5.

(3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A}\in \real$.

On the other hand, if $\dpn{x, \phi}{A}\in \real$, then $\dpn{x, \phi}{A}= \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A}=0$. If this holds for all $\phi \in P(A)$, then $x - x^{*} = 0$ by (1), and $x$ is self-adjoint.

(4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A}\ge 0$.

On the other hand, if $\dpn{x, \phi}{A}\ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By Corollary 34.10.7, $\sigma_{A}(x) \subset [0, \infty)$. As such, $x$ is positive by Corollary 34.6.3.$\square$