[proofplan] We compute all mixed free cumulants of the normalized sums. Multilinearity expands each cumulant as a sum over tuple labels, and freeness of the subalgebras generated by the tuples forces every surviving summand to have one common label. Identical distribution then gives the scaling factor $n^{1-m/2}$, so only the second-order cumulants survive in the limit. The joint [moment-cumulant formula for free cumulants](/theorems/7107) then identifies the limiting moments with those of the semicircular system of covariance $C$. [/proofplan]

[step:Compute the mixed free cumulants of the normalized sums]Fix $m \in \mathbb N$ and indices $i_1,\dots,i_m \in \{1,\dots,d\}$. For $r \in \{1,\dots,m\}$, the element $S_{n,i_r} \in \mathcal A$ is \begin{align*} S_{n,i_r}=\frac{1}{\sqrt n}\sum_{k=1}^{n} a_{k,i_r}. \end{align*} Let $\kappa_m:\mathcal A^m \to \mathbb C$ denote the $m$-th free cumulant functional associated to $(\mathcal A,\varphi)$. By multilinearity of $\kappa_m$ in its $m$ arguments, \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m}) = n^{-m/2} \sum_{k_1=1}^{n}\cdots\sum_{k_m=1}^{n} \kappa_m(a_{k_1,i_1},\dots,a_{k_m,i_m}). \end{align*}[/step]

[guided]We begin with cumulants because free independence is expressed most directly through the vanishing of mixed free cumulants. Fix the word of coordinate indices $i_1,\dots,i_m$. For each position $r$, the variable in that position is the normalized sum \begin{align*} S_{n,i_r}=\frac{1}{\sqrt n}\sum_{k=1}^{n} a_{k,i_r}. \end{align*} The $m$-th free cumulant $\kappa_m:\mathcal A^m \to \mathbb C$ is multilinear, so we may expand it one argument at a time. Pulling out the scalar factor $n^{-1/2}$ from each of the $m$ arguments gives the total factor $n^{-m/2}$. The remaining choices are the tuple labels $k_1,\dots,k_m$, one label for each position in the word. Therefore \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m}) = n^{-m/2} \sum_{k_1=1}^{n}\cdots\sum_{k_m=1}^{n} \kappa_m(a_{k_1,i_1},\dots,a_{k_m,i_m}). \end{align*} This expansion is the place where the normalization by $\sqrt n$ enters the proof. The rest of the argument determines how many of the $n^m$ summands actually survive.[/guided]

[step:Use freeness to keep only diagonal label choices]For each $k \in \mathbb N$, the element $a_{k,i}$ belongs to the unital $*$-subalgebra $\mathcal A_k$ for every $1 \le i \le d$. By the free-cumulant characterization of free independence, since the subalgebras $(\mathcal A_k)_{k \ge 1}$ are freely independent, every mixed free cumulant involving elements from at least two distinct subalgebras vanishes. Hence \begin{align*} \kappa_m(a_{k_1,i_1},\dots,a_{k_m,i_m})=0 \end{align*} unless $k_1=\cdots=k_m$. Therefore \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m}) = n^{-m/2} \sum_{k=1}^{n} \kappa_m(a_{k,i_1},\dots,a_{k,i_m}). \end{align*} Because the tuples $a_k$ are identically distributed, their joint free cumulants are the same. Thus \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m}) = n^{1-m/2}\kappa_m(a_{1,i_1},\dots,a_{1,i_m}). \end{align*}[/step]

[guided]Now we use the part of the hypothesis that is specific to free probability. For every $k$, all coordinates $a_{k,1},\dots,a_{k,d}$ lie in the same unital $*$-subalgebra \begin{align*} \mathcal A_k=\operatorname{alg}(1,a_{k,1},\dots,a_{k,d}). \end{align*} The free-cumulant characterization of free independence says that a free cumulant vanishes whenever its arguments come from more than one member of a freely independent family of subalgebras. Its hypotheses apply here because the subalgebras $(\mathcal A_k)_{k \ge 1}$ are freely independent. In the summand \begin{align*} \kappa_m(a_{k_1,i_1},\dots,a_{k_m,i_m}), \end{align*} the $r$-th argument lies in $\mathcal A_{k_r}$. If the labels $k_1,\dots,k_m$ are not all equal, then the cumulant is mixed across at least two freely independent subalgebras, so it is zero. Consequently the only surviving terms are the diagonal choices \begin{align*} k_1=\cdots=k_m=k. \end{align*} The expanded cumulant therefore reduces to \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m}) = n^{-m/2} \sum_{k=1}^{n} \kappa_m(a_{k,i_1},\dots,a_{k,i_m}). \end{align*} The tuples are identically distributed, and free cumulants are polynomial functions of the joint moments through the moment-cumulant relations. Therefore the joint free cumulants of the tuple $a_k$ agree with those of $a_1$: \begin{align*} \kappa_m(a_{k,i_1},\dots,a_{k,i_m}) = \kappa_m(a_{1,i_1},\dots,a_{1,i_m}) \end{align*} for every $k$. The sum contains $n$ equal terms, giving \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m}) = n^{1-m/2}\kappa_m(a_{1,i_1},\dots,a_{1,i_m}). \end{align*} This is the central scaling identity of the proof.[/guided]

[step:Identify the limiting free cumulants]For $m=1$, the centering hypothesis gives \begin{align*} \kappa_1(S_{n,i_1})=\varphi(S_{n,i_1})=0. \end{align*} For $m=2$, the second free cumulant satisfies \begin{align*} \kappa_2(x,y)=\varphi(xy)-\varphi(x)\varphi(y) \end{align*} for all $x,y \in \mathcal A$. Since $\varphi(a_{1,i})=0$ for every $i$, \begin{align*} \kappa_2(S_{n,i_1},S_{n,i_2})=\kappa_2(a_{1,i_1},a_{1,i_2})=\varphi(a_{1,i_1}a_{1,i_2})=c_{i_1 i_2}. \end{align*} For $m \ge 3$, the scaling identity gives \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m})=n^{1-m/2}\kappa_m(a_{1,i_1},\dots,a_{1,i_m}), \end{align*} and $1-m/2<0$, so \begin{align*} \lim_{n \to \infty}\kappa_m(S_{n,i_1},\dots,S_{n,i_m})=0. \end{align*} Thus the limiting cumulants are zero in order $1$, equal to $c_{ij}$ in order $2$, and zero in every order at least $3$.[/step]

[guided]We now extract the actual limits from the scaling identity. In order $1$, the first free cumulant is the state itself, so the centering assumption gives \begin{align*} \kappa_1(S_{n,i_1})=\varphi(S_{n,i_1})=0. \end{align*} In order $2$, the second free cumulant is covariance with respect to the state: \begin{align*} \kappa_2(x,y)=\varphi(xy)-\varphi(x)\varphi(y) \end{align*} for all $x,y \in \mathcal A$. Since each coordinate of $a_1$ is centered, the correction term vanishes, and the scaling identity with $m=2$ gives \begin{align*} \kappa_2(S_{n,i_1},S_{n,i_2})=\kappa_2(a_{1,i_1},a_{1,i_2})=\varphi(a_{1,i_1}a_{1,i_2})=c_{i_1 i_2}. \end{align*} For every order $m \ge 3$, the same scaling identity reads \begin{align*} \kappa_m(S_{n,i_1},\dots,S_{n,i_m})=n^{1-m/2}\kappa_m(a_{1,i_1},\dots,a_{1,i_m}). \end{align*} The exponent satisfies $1-m/2<0$, while the fixed cumulant $\kappa_m(a_{1,i_1},\dots,a_{1,i_m})$ is a complex number independent of $n$. Hence \begin{align*} \lim_{n \to \infty}\kappa_m(S_{n,i_1},\dots,S_{n,i_m})=0. \end{align*} This shows that the limiting cumulant data are exactly the semicircular cumulant data: no first cumulants, covariance matrix $C$ in order $2$, and no cumulants in order at least $3$.[/guided]

[step:Convert cumulant convergence into joint moment convergence]Let $NC(m)$ denote the set of noncrossing partitions of $\{1,\dots,m\}$. For a partition $\pi \in NC(m)$, define the block-cumulant functional $\kappa_\pi:\mathcal A^m \to \mathbb C$ by \begin{align*} \kappa_\pi(x_1,\dots,x_m):= \prod_{V \in \pi} \kappa_{|V|}(x_{v_1},\dots,x_{v_{|V|}}) \end{align*} for elements $x_1,\dots,x_m \in \mathcal A$, where $V=\{v_1<\cdots<v_{|V|}\}$ is a block of $\pi$. The free moment-cumulant formula for free cumulants gives \begin{align*} \varphi(S_{n,i_1}\cdots S_{n,i_m}) = \sum_{\pi \in NC(m)} \kappa_\pi(S_{n,i_1},\dots,S_{n,i_m}). \end{align*} Since $NC(m)$ is finite and each block cumulant has the limit computed above, we may pass the limit through the finite sum. Any partition with a singleton block contributes $0$ because all first cumulants vanish. Any partition with a block of size at least $3$ contributes $0$ in the limit because all cumulants of order at least $3$ vanish in the limit. Hence only noncrossing pair partitions survive, and for each $\pi \in NC_2(m)$, \begin{align*} \lim_{n \to \infty} \kappa_\pi(S_{n,i_1},\dots,S_{n,i_m}) = \prod_{\{p,q\}\in \pi} c_{i_p i_q}. \end{align*} Therefore \begin{align*} \lim_{n \to \infty}\varphi(S_{n,i_1}\cdots S_{n,i_m}) = \sum_{\pi \in NC_2(m)} \prod_{\{p,q\}\in \pi} c_{i_p i_q}. \end{align*}[/step]

[guided]The cumulant computation gives the limiting building blocks. To obtain convergence in joint distribution, we must translate those building blocks back into moments of words. The translation is the free moment-cumulant formula, which states that the moment of a word is the sum over noncrossing partitions of products of block cumulants. For the fixed word $S_{n,i_1}\cdots S_{n,i_m}$, let $NC(m)$ be the finite set of noncrossing partitions of $\{1,\dots,m\}$. If $\pi \in NC(m)$ and $V=\{v_1<\cdots<v_{|V|}\}$ is a block of $\pi$, the block contributes the cumulant \begin{align*} \kappa_{|V|}(S_{n,i_{v_1}},\dots,S_{n,i_{v_{|V|}}}). \end{align*} Thus, define the block-cumulant functional $\kappa_\pi:\mathcal A^m \to \mathbb C$ by \begin{align*} \kappa_\pi(x_1,\dots,x_m) := \prod_{V \in \pi} \kappa_{|V|}(x_{v_1},\dots,x_{v_{|V|}}) \end{align*} for elements $x_1,\dots,x_m \in \mathcal A$, where each block is written as $V=\{v_1<\cdots<v_{|V|}\}$. Applied to the word $S_{n,i_1},\dots,S_{n,i_m}$, the moment-cumulant formula gives \begin{align*} \varphi(S_{n,i_1}\cdots S_{n,i_m}) = \sum_{\pi \in NC(m)} \kappa_\pi(S_{n,i_1},\dots,S_{n,i_m}). \end{align*} The set $NC(m)$ is finite, so there is no issue of interchanging a limit with an infinite series. We pass the limit through this finite sum and inspect which partitions can survive. If a partition has a singleton block, then that block contributes a first cumulant, and every first cumulant is zero because the variables are centered. If a partition has a block of size at least $3$, then the cumulant attached to that block tends to zero by the higher-order part of the cumulant computation. Therefore the only partitions that can contribute in the limit are those whose every block has size $2$, namely the noncrossing pair partitions $NC_2(m)$. For a pair block $\{p,q\}$, the limiting second cumulant is exactly \begin{align*} c_{i_p i_q}. \end{align*} Hence each noncrossing pair partition $\pi$ contributes \begin{align*} \prod_{\{p,q\}\in \pi} c_{i_p i_q}, \end{align*} and summing over all such partitions gives \begin{align*} \lim_{n \to \infty}\varphi(S_{n,i_1}\cdots S_{n,i_m}) = \sum_{\pi \in NC_2(m)} \prod_{\{p,q\}\in \pi} c_{i_p i_q}. \end{align*} This formula is exactly the joint moment formula for a semicircular system with covariance matrix $C$.[/guided]

[step:Identify the limiting distribution as the semicircular system with covariance $C$] First verify that $C$ is an admissible covariance matrix. Since $\varphi$ is tracial and each $a_{1,i}$ is self-adjoint, for all $1 \le i,j \le d$ we have \begin{align*} \overline{c_{ij}}=\overline{\varphi(a_{1,i}a_{1,j})}=\varphi((a_{1,i}a_{1,j})^*)=\varphi(a_{1,j}a_{1,i})=c_{ji}. \end{align*} Thus $C$ is Hermitian. Since $\varphi$ is tracial, $c_{ij}=c_{ji}$, and the Hermitian identity above gives $\overline{c_{ij}}=c_{ij}$; hence $C$ is real symmetric. For any vector $\lambda=(\lambda_1,\dots,\lambda_d) \in \mathbb C^d$, define the element $x_\lambda \in \mathcal A$ by \begin{align*} x_\lambda:=\sum_{i=1}^{d}\lambda_i a_{1,i}. \end{align*} Since each $a_{1,i}$ is self-adjoint and $\varphi$ is positive, \begin{align*} \sum_{i=1}^{d}\sum_{j=1}^{d}\overline{\lambda_i}c_{ij}\lambda_j=\varphi(x_\lambda^*x_\lambda)\ge 0. \end{align*} Thus $C$ is real symmetric positive semidefinite. By the construction theorem for semicircular systems with prescribed real symmetric positive semidefinite covariance, there exists a tracial noncommutative $*$-probability space $(\mathcal B,\psi)$ and a self-adjoint semicircular system $(s_1,\dots,s_d)$ in $\mathcal B$ with covariance matrix $C$. By definition, this semicircular system has free cumulants determined by \begin{align*} \kappa_1(s_i)=0, \end{align*} \begin{align*} \kappa_2(s_i,s_j)=c_{ij}, \end{align*} and \begin{align*} \kappa_m(s_{i_1},\dots,s_{i_m})=0 \end{align*} for every $m \ge 3$ and every $i_1,\dots,i_m \in \{1,\dots,d\}$. Applying the free moment-cumulant formula to $(s_1,\dots,s_d)$ in the noncommutative probability space $(\mathcal B,\psi)$ gives \begin{align*} \psi(s_{i_1}\cdots s_{i_m}) = \sum_{\pi \in NC_2(m)} \prod_{\{p,q\}\in \pi} c_{i_p i_q}. \end{align*} This agrees with the limit of $\varphi(S_{n,i_1}\cdots S_{n,i_m})$ for every word. Therefore $(S_{n,1},\dots,S_{n,d})$ converges in joint distribution to the semicircular system $(s_1,\dots,s_d)$ with covariance matrix $C$. [/step]