[proofplan] Let $\tau$ and $\sigma$ be faithful normal tracial states on $M$. We compare $\sigma$ to $\tau$ using the [Radon-Nikodym theorem](/theorems/1247) for normal weights relative to the faithful normal finite trace $\tau$, obtaining a positive self-adjoint affiliated density $h$ such that $\sigma(x)=\tau(hx)$ for positive $x$. The tracial identity for $\sigma$ forces this density to be central, and in a factor every central affiliated positive operator is scalar. Finally, the normalization $\tau(1)=\sigma(1)=1$ forces that scalar to be $1$, so $\sigma=\tau$. [/proofplan]

[step:Represent one trace by a Radon-Nikodym density relative to the other]Let $\tau:M\to\mathbb C$ and $\sigma:M\to\mathbb C$ be faithful normal tracial states. Since $\tau$ is positive, faithful, normal, tracial, and satisfies $\tau(1)=1<\infty$, it is a faithful normal semifinite trace on $M$. Since $\sigma$ is positive and normal with $\sigma(1)=1$, it is a finite normal weight on $M$. We use the Pedersen-Takesaki [Radon-Nikodym theorem](/theorems/2640) for normal weights relative to a faithful normal semifinite trace: there exists a positive self-adjoint operator $h$ affiliated with $M$ such that, for every $x\in M_+$, \begin{align*} \sigma(x)=\tau_h(x). \end{align*} The functional $\tau_h:M_+\to[0,\infty]$ is defined as follows. Let $E_h:\mathcal B([0,\infty))\to\mathcal P(M)$ denote the spectral measure of $h$; the projections belong to $M$ by the [Spectral Projection Criterion for Affiliation]([citetheorem:9305]). For each $m\in\mathbb N$, define the bounded positive operator $h_m\in M$ by \begin{align*} h_m=hE_h([0,m]). \end{align*} Then \begin{align*} \tau_h(x)=\sup_{m\in\mathbb N}\tau(x^{1/2}h_mx^{1/2}). \end{align*} This number is finite for every $x\in M_+$ because it equals $\sigma(x)$ and $\sigma$ is a finite normal weight. This definition is the extended trace pairing; it avoids treating the product $hx$ as an ordinary bounded operator.[/step]

[guided]We need a way to compare two normal traces on the same von Neumann algebra. The correct comparison object is a Radon-Nikodym density. The reference trace is $\tau:M\to\mathbb C$. Because $\tau$ is a faithful normal tracial state, it is positive, faithful, normal, tracial, and finite with \begin{align*} \tau(1)=1. \end{align*} A finite trace is automatically semifinite, so $\tau$ satisfies the hypotheses of the Radon-Nikodym theorem for normal weights relative to a faithful normal semifinite trace. The second trace is $\sigma:M\to\mathbb C$. Since $\sigma$ is a normal state, it is a positive normal linear functional and satisfies \begin{align*} \sigma(1)=1. \end{align*} Thus $\sigma$ is a finite normal weight on $M$. Applying the Pedersen-Takesaki Radon-Nikodym theorem to $\sigma$ relative to $\tau$ gives a positive self-adjoint operator $h$ affiliated with $M$ such that, for each positive bounded operator $x\in M_+$, \begin{align*} \sigma(x)=\tau_h(x). \end{align*} To define $\tau_h$ without pretending that $hx$ is bounded, let $E_h:\mathcal B([0,\infty))\to\mathcal P(M)$ be the spectral measure of $h$. Since $h$ is affiliated with $M$, the [Spectral Projection Criterion for Affiliation]([citetheorem:9305]) gives $E_h(B)\in M$ for every Borel set $B\subseteq[0,\infty)$. For each $m\in\mathbb N$, define \begin{align*} h_m=hE_h([0,m])\in M_+. \end{align*} Then the extended trace pairing is \begin{align*} \tau_h(x)=\sup_{m\in\mathbb N}\tau(x^{1/2}h_mx^{1/2}). \end{align*} The supremum is finite here because the Radon-Nikodym theorem identifies it with $\sigma(x)$ and $\sigma(1)=1$. This is exactly why the Radon-Nikodym theorem is the right tool: it packages the comparison of normal weights into a single positive affiliated density while keeping unbounded products inside bounded spectral approximants.[/guided]

[step:Use the tracial Radon-Nikodym theorem to place the density in the center]We claim that the affiliated operator $h$ is central. We use the tracial part of the Pedersen-Takesaki Radon-Nikodym theorem: if $\tau$ is a faithful normal semifinite trace on a von Neumann algebra $M$, $h$ is a positive self-adjoint operator affiliated with $M$, and the normal weight $\tau_h$ defined by the bounded spectral cutdowns of $h$ is tracial, then $h$ is affiliated with $Z(M)$. The hypotheses match the present situation. The reference weight $\tau$ is a faithful normal semifinite trace, as verified in the preceding step. The operator $h$ is positive self-adjoint and affiliated with $M$ by the Radon-Nikodym theorem. The weight $\tau_h$ equals $\sigma$ on $M_+$, and $\sigma$ is tracial by hypothesis. Hence the tracial Radon-Nikodym theorem gives that $h$ is affiliated with $Z(M)$.[/step]

[guided]The delicate point is that $h$ may be unbounded, so expressions such as $ha^*xa$ are not ordinary products in $M$. We therefore do not use cyclicity of $\tau$ on unbounded products. Instead we invoke the precise theorem designed for this situation: the tracial part of the Pedersen-Takesaki Radon-Nikodym theorem says that, for a faithful normal semifinite trace $\tau$, a positive affiliated density $h$ gives a tracial normal weight $\tau_h$ exactly when the density $h$ is affiliated with the center $Z(M)$. We verify its hypotheses one by one. First, $\tau$ is a faithful normal semifinite trace because it is a faithful normal tracial state. Second, the preceding Radon-Nikodym step produced a positive self-adjoint operator $h$ affiliated with $M$. Third, the normal weight associated to this density is precisely $\tau_h$, defined for $x\in M_+$ by \begin{align*} \tau_h(x)=\sup_{m\in\mathbb N}\tau(x^{1/2}h_mx^{1/2}), \end{align*} where $h_m=hE_h([0,m])\in M_+$ is the bounded spectral cutoff of $h$. Fourth, the Radon-Nikodym formula gives \begin{align*} \tau_h(x)=\sigma(x) \end{align*} for every $x\in M_+$. Since $\sigma$ is tracial, $\tau_h$ is tracial. The theorem therefore applies and yields that $h$ is affiliated with $Z(M)$. This is the step where traciality is used. The conclusion is not obtained by formally commuting an unbounded operator through bounded elements; it is exactly the centrality conclusion in the Radon-Nikodym theorem for tracial weights.[/guided]

[step:Reduce the central affiliated density to a scalar] Because $M$ is a factor, the [Factor Criterion by Central Projections]([citetheorem:9284]) implies that the only central projections in $M$ are $0$ and $1$. Since $h$ is affiliated with $Z(M)$, every spectral projection $E_h(B)$ belongs to $Z(M)$ for every Borel set $B\subseteq[0,\infty)$, by the [Spectral Projection Criterion for Affiliation]([citetheorem:9305]). Hence every spectral projection of $h$ is either $0$ or $1$. The [spectral theorem for self-adjoint operators](/theorems/6911) then gives a scalar $\lambda\in[0,\infty)$ such that \begin{align*} h=\lambda 1. \end{align*} Indeed, the spectral measure $E_h$ is a projection-valued probability measure on $[0,\infty)$ whose values are only $0$ and $1$, so it is concentrated at a single point $\lambda$; since $h$ is a densely defined self-adjoint operator, this point is finite. The normalization determines the scalar: \begin{align*} 1=\sigma(1)=\tau_h(1)=\tau(\lambda 1)=\lambda\tau(1)=\lambda. \end{align*} Thus $h=1$. [/step]

[step:Conclude that the two normalized traces are equal] For every $x\in M_+$, the Radon-Nikodym formula and the identity $h=1$ give \begin{align*} \sigma(x)=\tau_h(x)=\tau(x). \end{align*} Every element $y\in M$ is a complex linear combination of four positive elements, obtained by decomposing the self-adjoint and skew-adjoint parts into positive and negative parts. Since $\sigma$ and $\tau$ are linear, equality on $M_+$ implies \begin{align*} \sigma(y)=\tau(y) \end{align*} for every $y\in M$. Hence any two faithful normal tracial states on $M$ are equal, so the faithful normal tracial state on $M$ is unique. [/step]