[proofplan] We build the [Hilbert space](/page/Hilbert%20Space) by turning the state into a positive semidefinite sesquilinear form on $\mathcal A$, then quotienting by the null space of that form. The [Cauchy-Schwarz inequality](/theorems/432) for this positive semidefinite form identifies the null space as the radical and shows that it is stable under left multiplication. Left multiplication therefore descends to the quotient, giving a representation on a dense domain in the Hilbert space completion. The unit class is cyclic, and evaluating the representation against this unit class recovers the original state. [/proofplan] [step:Define the positive semidefinite form induced by the state] Define a sesquilinear form \begin{align*} \langle \cdot,\cdot\rangle_\varphi: \mathcal A \times \mathcal A \to \mathbb C \end{align*} by \begin{align*} \langle a,b\rangle_\varphi := \varphi(b^*a) \end{align*} for $a,b \in \mathcal A$. This form is linear in the first variable because $\varphi$ is linear and the involution is conjugate-linear in the second input after applying $b^*a$. It is conjugate-linear in the second variable because, for $\lambda,\mu \in \mathbb C$ and $b,c \in \mathcal A$, \begin{align*} \langle a,\lambda b+\mu c\rangle_\varphi = \varphi((\lambda b+\mu c)^*a) = \overline{\lambda}\varphi(b^*a)+\overline{\mu}\varphi(c^*a). \end{align*} It is Hermitian. First we verify the standard consequence of positivity that $\varphi(x^*)=\overline{\varphi(x)}$ for every $x \in \mathcal A$. Positivity of $(1_{\mathcal A}+t x)^*(1_{\mathcal A}+t x)$ for every $t \in \mathbb R$ implies that $\varphi(x)+\varphi(x^*) \in \mathbb R$. Positivity of $(1_{\mathcal A}+i t x)^*(1_{\mathcal A}+i t x)$ for every $t \in \mathbb R$ implies that $i\varphi(x)-i\varphi(x^*) \in \mathbb R$. These two realness conditions give $\varphi(x^*)=\overline{\varphi(x)}$. Applying this with $x=b^*a$ gives \begin{align*} \langle b,a\rangle_\varphi = \varphi(a^*b) = \overline{\varphi(b^*a)} = \overline{\langle a,b\rangle_\varphi}. \end{align*} Finally, \begin{align*} \langle a,a\rangle_\varphi = \varphi(a^*a) \geq 0 \end{align*} for every $a \in \mathcal A$, so $\langle \cdot,\cdot\rangle_\varphi$ is positive semidefinite. [guided] The state $\varphi$ is meant to become a vector state, so the first task is to manufacture an [inner product](/page/Inner%20Product) from it. The only expression built from $\varphi$ and two algebra elements $a,b \in \mathcal A$ that has the correct adjoint symmetry is $\varphi(b^*a)$, so we define \begin{align*} \langle \cdot,\cdot\rangle_\varphi: \mathcal A \times \mathcal A \to \mathbb C \end{align*} by \begin{align*} \langle a,b\rangle_\varphi := \varphi(b^*a). \end{align*} This is linear in the first variable: for $\lambda,\mu \in \mathbb C$ and $a,c,b \in \mathcal A$, \begin{align*} \langle \lambda a+\mu c,b\rangle_\varphi = \varphi(b^*(\lambda a+\mu c)) = \lambda\varphi(b^*a)+\mu\varphi(b^*c) = \lambda\langle a,b\rangle_\varphi+\mu\langle c,b\rangle_\varphi. \end{align*} It is conjugate-linear in the second variable because the involution is conjugate-linear: \begin{align*} \langle a,\lambda b+\mu c\rangle_\varphi = \varphi((\lambda b+\mu c)^*a) = \overline{\lambda}\varphi(b^*a)+\overline{\mu}\varphi(c^*a). \end{align*} The form is Hermitian. Positivity of $\varphi$ implies $\varphi(x^*)=\overline{\varphi(x)}$ for every $x \in \mathcal A$, which follows by applying positivity to $(1_{\mathcal A}+tx)^*(1_{\mathcal A}+tx)$ for real and imaginary scalar parameters $t$ and comparing the resulting real quadratic polynomials. Taking $x=b^*a$ gives \begin{align*} \langle b,a\rangle_\varphi = \varphi(a^*b) = \overline{\varphi(b^*a)} = \overline{\langle a,b\rangle_\varphi}. \end{align*} Finally, the defining positivity of a state gives \begin{align*} \langle a,a\rangle_\varphi = \varphi(a^*a) \geq 0. \end{align*} Thus $\langle \cdot,\cdot\rangle_\varphi$ is a positive semidefinite sesquilinear form. It may fail to be an inner product only because nonzero elements can have zero length; the quotient in the next step removes exactly those elements. [/guided] [/step] [step:Identify the null space as a left ideal] Define the null space \begin{align*} N_\varphi := \{a \in \mathcal A : \langle a,a\rangle_\varphi = 0\} = \{a \in \mathcal A : \varphi(a^*a)=0\}. \end{align*} For the positive semidefinite form $\langle \cdot,\cdot\rangle_\varphi$, we prove the Cauchy-Schwarz inequality: \begin{align*} |\langle x,y\rangle_\varphi|^2 \leq \langle x,x\rangle_\varphi \langle y,y\rangle_\varphi \end{align*} for all $x,y \in \mathcal A$. If $\langle y,y\rangle_\varphi=0$, positivity of $\langle x+\lambda y,x+\lambda y\rangle_\varphi$ for all $\lambda \in \mathbb C$ forces $\langle x,y\rangle_\varphi=0$; otherwise minimizing \begin{align*} \langle x+\lambda y,x+\lambda y\rangle_\varphi \end{align*} over $\lambda \in \mathbb C$ gives the displayed inequality. It follows that \begin{align*} N_\varphi = \{n \in \mathcal A : \langle n,b\rangle_\varphi = 0 \text{ for every } b \in \mathcal A\}. \end{align*} Therefore $N_\varphi$ is a complex linear subspace of $\mathcal A$. Moreover, if $n \in N_\varphi$ and $a \in \mathcal A$, then for every $b \in \mathcal A$, \begin{align*} \langle an,b\rangle_\varphi = \varphi(b^*an) = \langle n,a^*b\rangle_\varphi = 0. \end{align*} Thus $an \in N_\varphi$, so $N_\varphi$ is a left ideal in $\mathcal A$. [/step] [step:Complete the quotient into a Hilbert space] Let $\mathcal A/N_\varphi$ denote the complex quotient [vector space](/page/Vector%20Space), and write $[a]$ for the equivalence class of $a \in \mathcal A$. Define \begin{align*} ([\cdot],[\cdot])_\varphi: (\mathcal A/N_\varphi) \times (\mathcal A/N_\varphi) \to \mathbb C \end{align*} by \begin{align*} ([a],[b])_\varphi := \langle a,b\rangle_\varphi = \varphi(b^*a). \end{align*} This is well-defined because $N_\varphi$ is the radical of $\langle \cdot,\cdot\rangle_\varphi$: if $a-a' \in N_\varphi$ and $b-b' \in N_\varphi$, then all mixed differences have zero pairing with every element of $\mathcal A$. The form $([\cdot],[\cdot])_\varphi$ is positive definite on $\mathcal A/N_\varphi$ because $([a],[a])_\varphi=0$ implies $a \in N_\varphi$, hence $[a]=0$. Let $\mathcal D_\varphi := \mathcal A/N_\varphi$ equipped with this inner product, viewed as a pre-Hilbert space. Let $H_\varphi$ be the Hilbert space completion of $\mathcal D_\varphi$, and regard $\mathcal D_\varphi$ as its canonical dense subspace. [/step] [step:Define left multiplication on the quotient] For each $a \in \mathcal A$, define a [linear map](/page/Linear%20Map) \begin{align*} \pi_\varphi(a): \mathcal D_\varphi \to \mathcal D_\varphi \end{align*} by \begin{align*} \pi_\varphi(a)[b] := [ab] \end{align*} for $b \in \mathcal A$. This is well-defined: if $[b]=[c]$, then $b-c \in N_\varphi$, and since $N_\varphi$ is a left ideal, $a(b-c) \in N_\varphi$, so $[ab]=[ac]$. The map $a \mapsto \pi_\varphi(a)$ is linear because multiplication in $\mathcal A$ is bilinear. It is unital since \begin{align*} \pi_\varphi(1_{\mathcal A})[b]=[1_{\mathcal A}b]=[b] \end{align*} for every $b \in \mathcal A$. It is multiplicative because, for $a,c,b \in \mathcal A$, \begin{align*} \pi_\varphi(a)\pi_\varphi(c)[b] = \pi_\varphi(a)[cb] = [acb] = \pi_\varphi(ac)[b]. \end{align*} Finally, the $*$-relation holds on the common dense domain $\mathcal D_\varphi$: for $a,b,c \in \mathcal A$, \begin{align*} (\pi_\varphi(a)[b],[c])_\varphi = ([ab],[c])_\varphi = \varphi(c^*ab), \end{align*} while \begin{align*} ([b],\pi_\varphi(a^*)[c])_\varphi = ([b],[a^*c])_\varphi = \varphi((a^*c)^*b) = \varphi(c^*ab). \end{align*} Hence \begin{align*} (\pi_\varphi(a)u,v)_\varphi = (u,\pi_\varphi(a^*)v)_\varphi \end{align*} for all $u,v \in \mathcal D_\varphi$. Thus $\pi_\varphi$ is a unital $*$-representation of $\mathcal A$ on the dense domain $\mathcal D_\varphi$. [/step] [step:Recover the state from the cyclic vector] Define \begin{align*} \xi_\varphi := [1_{\mathcal A}] \in \mathcal D_\varphi. \end{align*} For every $a \in \mathcal A$, \begin{align*} (\pi_\varphi(a)\xi_\varphi,\xi_\varphi)_{H_\varphi} = ([a1_{\mathcal A}],[1_{\mathcal A}])_\varphi = ([a],[1_{\mathcal A}])_\varphi = \varphi(1_{\mathcal A}^*a) = \varphi(a). \end{align*} Also, \begin{align*} \pi_\varphi(\mathcal A)\xi_\varphi = \{[a] : a \in \mathcal A\} = \mathcal D_\varphi. \end{align*} Since $\mathcal D_\varphi$ is dense in $H_\varphi$ by construction, the vector $\xi_\varphi$ is cyclic. [/step] [step:Extend the representation when the left multiplication operators are bounded] Assume now that for every $a \in \mathcal A$ there exists a constant $C_a \geq 0$ such that \begin{align*} \|\pi_\varphi(a)u\|_{H_\varphi} \leq C_a\|u\|_{H_\varphi} \end{align*} for every $u \in \mathcal D_\varphi$. Since $\mathcal D_\varphi$ is dense in $H_\varphi$, each $\pi_\varphi(a):\mathcal D_\varphi \to \mathcal D_\varphi$ extends uniquely to a [bounded linear operator](/page/Bounded%20Linear%20Operator) \begin{align*} \overline{\pi_\varphi(a)} \in \mathcal L(H_\varphi). \end{align*} The identities \begin{align*} \pi_\varphi(1_{\mathcal A})=\operatorname{id}_{\mathcal D_\varphi} \end{align*} and \begin{align*} \pi_\varphi(a)\pi_\varphi(c)=\pi_\varphi(ac) \end{align*} hold on the dense subspace $\mathcal D_\varphi$, hence their bounded extensions agree on all of $H_\varphi$. The adjoint relation likewise extends from $\mathcal D_\varphi$ to $H_\varphi$, so \begin{align*} \overline{\pi_\varphi(a)}^*=\overline{\pi_\varphi(a^*)}. \end{align*} Thus, under the stated boundedness hypothesis, $\pi_\varphi$ extends to a unital $*$-representation of $\mathcal A$ in $\mathcal L(H_\varphi)$. This completes the construction. [/step]