[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.
[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]
[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]
