[proofplan] We compute the free cumulants of the normalized sums $S_n$ directly from multilinearity, freeness, and identical distribution. The first cumulant is zero, the second cumulant is one, and each cumulant of order at least three is multiplied by a factor $n^{1-m/2}$, hence tends to zero. The free moment-cumulant formula then converts cumulant convergence into moment convergence term by term over the finite lattice of noncrossing partitions. In the limit, only noncrossing pairings survive, and these are counted by the Catalan number $C_r$. [/proofplan]

[step:Compute the cumulants of the normalized sums]For each $m\in\mathbb N$, let $\kappa_m:\mathcal A^m\to\mathbb C$ denote the $m$th free cumulant functional associated to $\varphi$. For $n\in\mathbb N$, multilinearity of $\kappa_m$ gives \begin{align*} \kappa_m(S_n,\dots,S_n)=n^{-m/2}\sum_{i_1=1}^n\cdots\sum_{i_m=1}^n \kappa_m(a_{i_1},\dots,a_{i_m}). \end{align*} By freeness, all mixed free cumulants vanish (citing a result not yet in the wiki: vanishing of mixed free cumulants for free subalgebras). Thus the only nonzero summands are those with $i_1=\cdots=i_m$. Since the variables are identically distributed, each remaining summand equals $\kappa_m(a_1,\dots,a_1)$. Therefore \begin{align*} \kappa_m(S_n,\dots,S_n)=n^{1-m/2}\kappa_m(a_1,\dots,a_1). \end{align*}[/step]

[guided]Fix a moment order $m\in\mathbb N$. The object we need to understand is the $m$th free cumulant of $S_n$, where \begin{align*} S_n := n^{-1/2}\sum_{i=1}^n a_i. \end{align*} Because $\kappa_m:\mathcal A^m\to\mathbb C$ is multilinear in its $m$ arguments, expanding each copy of $S_n$ gives \begin{align*} \kappa_m(S_n,\dots,S_n)=n^{-m/2}\sum_{i_1=1}^n\cdots\sum_{i_m=1}^n \kappa_m(a_{i_1},\dots,a_{i_m}). \end{align*} The key free-probability input is that mixed cumulants vanish for freely independent families (citing a result not yet in the wiki: vanishing of mixed free cumulants for free subalgebras). Here “mixed” means that the entries of the cumulant come from at least two different free subalgebras. Since the elements $a_i$ are freely independent, the summand \begin{align*} \kappa_m(a_{i_1},\dots,a_{i_m}) \end{align*} is zero unless all indices $i_1,\dots,i_m$ are equal. Thus the full $m$-fold sum collapses to the $n$ diagonal choices \begin{align*} (i_1,\dots,i_m)=(j,\dots,j)\quad\text{with }1\leq j\leq n. \end{align*} For each such $j$, identical distribution gives \begin{align*} \kappa_m(a_j,\dots,a_j)=\kappa_m(a_1,\dots,a_1). \end{align*} Consequently, \begin{align*} \kappa_m(S_n,\dots,S_n)=n^{-m/2}\sum_{j=1}^n \kappa_m(a_j,\dots,a_j)=n^{1-m/2}\kappa_m(a_1,\dots,a_1). \end{align*} This is the scaling mechanism behind the theorem: order $m$ cumulants scale like $n^{1-m/2}$ after normalization by $\sqrt n$.[/guided]

[step:Take the cumulant limits] For $m=1$, the formula gives \begin{align*} \kappa_1(S_n)=n^{1/2}\kappa_1(a_1). \end{align*} Since $\kappa_1(a_1)=\varphi(a_1)=0$, we have $\kappa_1(S_n)=0$ for every $n$. For $m=2$, the second free cumulant satisfies \begin{align*} \kappa_2(a_1,a_1)=\varphi(a_1^2)-\varphi(a_1)^2=1. \end{align*} Hence \begin{align*} \kappa_2(S_n,S_n)=\kappa_2(a_1,a_1)=1 \end{align*} for every $n$. For every fixed $m\geq 3$, the scalar $\kappa_m(a_1,\dots,a_1)$ is independent of $n$, and $1-m/2<0$. Therefore \begin{align*} \lim_{n\to\infty}\kappa_m(S_n,\dots,S_n)=\lim_{n\to\infty} n^{1-m/2}\kappa_m(a_1,\dots,a_1)=0. \end{align*} Thus the limiting cumulants are precisely \begin{align*} c_1=0,\qquad c_2=1,\qquad c_m=0\quad\text{for every }m\geq 3, \end{align*} where $c_m:=\lim_{n\to\infty}\kappa_m(S_n,\dots,S_n)$. [/step]

[step:Convert cumulant convergence into moment convergence] Fix $m\in\mathbb N$. Let $NC(m)$ denote the finite set of noncrossing partitions of $\{1,\dots,m\}$. For a partition $\pi\in NC(m)$ and a block $V=\{j_1<\cdots<j_\ell\}\subset\{1,\dots,m\}$, define the block cumulant factor of $S_n$ by \begin{align*} \kappa_V(S_n):=\kappa_\ell(S_n,\dots,S_n). \end{align*} Define \begin{align*} \kappa_\pi(S_n):=\prod_{V\in\pi}\kappa_V(S_n). \end{align*} By the free moment-cumulant formula (citing a result not yet in the wiki: free moment-cumulant formula over noncrossing partitions), \begin{align*} \varphi(S_n^m)=\sum_{\pi\in NC(m)}\kappa_\pi(S_n). \end{align*} Since $NC(m)$ is finite, the limit may be passed through the sum and through each finite product: \begin{align*} \lim_{n\to\infty}\varphi(S_n^m)=\sum_{\pi\in NC(m)}\prod_{V\in\pi} c_{|V|}. \end{align*} Here $|V|$ denotes the cardinality of the block $V$. [/step]

[step:Identify the surviving noncrossing partitions] Because $c_1=0$, any partition with a singleton block contributes zero to the limiting sum. Because $c_\ell=0$ for every $\ell\geq 3$, any partition with a block of size at least three also contributes zero. Since $c_2=1$, a partition contributes one precisely when every block has size two. Therefore \begin{align*} \lim_{n\to\infty}\varphi(S_n^m)=\#\{\pi\in NC(m):\text{ every block of }\pi\text{ has size }2\}. \end{align*} If $m$ is odd, no partition of $\{1,\dots,m\}$ can have all blocks of size two, so \begin{align*} \lim_{n\to\infty}\varphi(S_n^m)=0. \end{align*} If $m=2r$, the surviving partitions are exactly the noncrossing pairings of $\{1,\dots,2r\}$. The standard Catalan enumeration of noncrossing pairings gives \begin{align*} \#\{\pi\in NC(2r):\text{ every block of }\pi\text{ has size }2\}=C_r=\frac{1}{r+1}\binom{2r}{r}. \end{align*} Thus \begin{align*} \lim_{n\to\infty}\varphi(S_n^{2r})=C_r. \end{align*} [/step]

[step:Conclude convergence to the standard semicircular distribution] The standard semicircular element $s$ is characterized by free cumulants \begin{align*} \kappa_1(s)=0,\qquad \kappa_2(s,s)=1,\qquad \kappa_m(s,\dots,s)=0\quad\text{for every }m\geq 3. \end{align*} The preceding steps show that for every fixed $m\in\mathbb N$, \begin{align*} \lim_{n\to\infty}\kappa_m(S_n,\dots,S_n)=\kappa_m(s,\dots,s). \end{align*} Applying the free moment-cumulant formula to $s$ gives the same limiting moments computed above: odd moments vanish, and the $2r$th moment equals $C_r$. Hence, for every $m\in\mathbb N$, \begin{align*} \lim_{n\to\infty}\varphi(S_n^m)=\varphi(s^m). \end{align*} This is convergence in noncommutative distribution, so $S_n$ converges in distribution to the standard semicircular element $s$. [/step]