[proofplan] We use the trace-class predual identification of $\mathcal{L}(H)$: normal linear functionals are exactly trace pairings against trace-class operators. Starting from a normal state, this gives a trace-class representative $\rho$; positivity of the state is tested on rank-one projections to force $\rho\geq 0$, and normalization gives $\operatorname{Tr}(\rho)=1$. Conversely, a positive trace-class operator of trace one defines a positive normalized functional by the trace pairing, and normality follows from the same predual identification. [/proofplan]

[step:Represent the normal functional by a trace-class operator] Assume first that $\phi:\mathcal{L}(H)\to\mathbb C$ is a normal state. By the trace-class predual identification in [Trace Class Predual Of B H][citetheorem:9274], applied to the [Hilbert space](/page/Hilbert%20Space) $H$, there exists a trace-class operator $\rho\in\mathcal T(H)$ such that \begin{align*} \phi(a)=\operatorname{Tr}(\rho a) \end{align*} for every $a\in\mathcal{L}(H)$. Here the hypothesis needed for the cited result is exactly that $H$ is a Hilbert space and that $\phi$ is a normal linear functional on $\mathcal{L}(H)$; both are in force because a normal state is, by definition, a positive normalized normal linear functional. [/step]

[step:Test positivity on rank-one projections to prove $\rho$ is positive]Let $\xi\in H$ be a unit vector. Define the rank-one projection $p_\xi\in\mathcal{L}(H)$ by \begin{align*} p_\xi:H&\to H \end{align*} \begin{align*} \eta&\mapsto (\eta,\xi)_H\xi. \end{align*} Then $p_\xi$ is a positive operator. Since $\phi$ is positive, we have \begin{align*} 0\leq \phi(p_\xi)=\operatorname{Tr}(\rho p_\xi). \end{align*} For the rank-one projection $p_\xi$, the trace pairing satisfies \begin{align*} \operatorname{Tr}(\rho p_\xi)=(\rho\xi,\xi)_H. \end{align*} Hence \begin{align*} (\rho\xi,\xi)_H\geq 0 \end{align*} for every unit vector $\xi\in H$. If $\eta\in H$ is arbitrary and $\eta\neq 0$, applying this to $\xi=\eta/\|\eta\|_H$ gives \begin{align*} (\rho\eta,\eta)_H=\|\eta\|_H^2\left(\rho\frac{\eta}{\|\eta\|_H},\frac{\eta}{\|\eta\|_H}\right)_H\geq 0. \end{align*} For $\eta=0$ the same inequality is immediate. Thus the quadratic form of $\rho$ is nonnegative on all of $H$, so $\rho$ is a positive operator.[/step]

[guided]The only point that is not automatic from the predual identification is positivity of the representing trace-class operator. The functional $\phi$ is positive, meaning that $\phi(a)\geq 0$ whenever $a\in\mathcal{L}(H)$ is positive. To convert this into positivity of $\rho$, we test $\phi$ on the simplest positive operators: rank-one projections. Let $\xi\in H$ be a unit vector. Define \begin{align*} p_\xi:H&\to H \end{align*} \begin{align*} \eta&\mapsto (\eta,\xi)_H\xi. \end{align*} This map is bounded, self-adjoint, idempotent, and has range $\mathbb C\xi$, so it is the [orthogonal projection](/theorems/437) onto $\mathbb C\xi$. In particular $p_\xi\geq 0$. Since $\phi$ is positive, \begin{align*} 0\leq \phi(p_\xi). \end{align*} The trace-class representation from the previous step gives \begin{align*} \phi(p_\xi)=\operatorname{Tr}(\rho p_\xi). \end{align*} For this rank-one projection, the trace is computed by evaluating on the vector spanning its range: \begin{align*} \operatorname{Tr}(\rho p_\xi)=(\rho\xi,\xi)_H. \end{align*} Therefore \begin{align*} (\rho\xi,\xi)_H\geq 0 \end{align*} for every unit vector $\xi\in H$. Now take an arbitrary vector $\eta\in H$. If $\eta\neq 0$, define $\xi=\eta/\|\eta\|_H$. Then $\xi$ is a unit vector, and the previous inequality gives \begin{align*} 0\leq (\rho\xi,\xi)_H =\left(\rho\frac{\eta}{\|\eta\|_H},\frac{\eta}{\|\eta\|_H}\right)_H. \end{align*} Multiplying by $\|\eta\|_H^2$ yields \begin{align*} (\rho\eta,\eta)_H\geq 0. \end{align*} If $\eta=0$, this inequality is also true. Hence the quadratic form associated to $\rho$ is nonnegative on every vector in $H$, which is precisely the operator positivity condition $\rho\geq 0$.[/guided]

[step:Use normalization to compute the trace of the representative] Let $I_H\in\mathcal{L}(H)$ denote the identity operator on $H$. Since $\phi$ is a state, $\phi(I_H)=1$. Using the trace representation with $a=I_H$, we obtain \begin{align*} 1=\phi(I_H)=\operatorname{Tr}(\rho I_H)=\operatorname{Tr}(\rho). \end{align*} Thus the representing trace-class operator satisfies $\rho\geq 0$ and $\operatorname{Tr}(\rho)=1$. [/step]

[step:Build a normal state from a positive trace-class operator] Conversely, let $\rho\in\mathcal T(H)$ be a positive trace-class operator with $\operatorname{Tr}(\rho)=1$. Define \begin{align*} \phi:\mathcal{L}(H)&\to\mathbb C \end{align*} \begin{align*} a&\mapsto \operatorname{Tr}(\rho a). \end{align*} The trace pairing is linear in $a$, so $\phi$ is a linear functional. If $a\in\mathcal{L}(H)$ is positive, then $\rho^{1/2}a\rho^{1/2}$ is a positive trace-class operator, and cyclicity of the trace for products of bounded and trace-class operators gives \begin{align*} \phi(a)=\operatorname{Tr}(\rho a)=\operatorname{Tr}(\rho^{1/2}a\rho^{1/2}). \end{align*} The trace of a positive trace-class operator is nonnegative, hence $\phi(a)\geq 0$. Therefore $\phi$ is positive. Also, \begin{align*} \phi(I_H)=\operatorname{Tr}(\rho I_H)=\operatorname{Tr}(\rho)=1. \end{align*} Thus $\phi$ is normalized. Finally, by the trace-class predual identification in [Trace Class Predual Of B H][citetheorem:9274], every functional of the form $a\mapsto \operatorname{Tr}(\rho a)$ with $\rho\in\mathcal T(H)$ is normal on $\mathcal{L}(H)$. Hence $\phi$ is a normal state. [/step]

[step:Conclude the equivalence] The first direction produced, from an arbitrary normal state $\phi$, a positive trace-class operator $\rho$ with $\operatorname{Tr}(\rho)=1$ and $\phi(a)=\operatorname{Tr}(\rho a)$ for every $a\in\mathcal{L}(H)$. The converse direction showed that every such positive trace-class operator defines a normal state by the same formula. This proves the stated equivalence. [/step]