[proofplan]
We compare the moments of $s$ with its free cumulants using the free moment-cumulant formula over noncrossing partitions. If all free cumulants except the second vanish and $\kappa_2(s,s)=1$, then only noncrossing pairings contribute to each moment, giving precisely the Catalan moment sequence. Conversely, if $s$ has the Catalan moments of the standard semicircle law, we isolate the one-block cumulant in the moment-cumulant formula and induct on the order of the cumulant. The lower-order cumulants account exactly for the noncrossing pairing contribution, so every higher cumulant must vanish.
[/proofplan]
[/proofplan]
[step:Introduce the cumulant notation and the noncrossing partition expansion]
For each $m\in\mathbb N$, let $NC(m)$ denote the set of noncrossing partitions of $\{1,\dots,m\}$.
Let $NC_2(m)\subset NC(m)$ denote the subset consisting of those partitions whose blocks all have cardinality $2$.
Define
\begin{align*}
c_m:=\kappa_m(s,\dots,s).
\end{align*}
For a noncrossing partition $\pi\in NC(m)$, define its block cumulant contribution by
\begin{align*}
\kappa_\pi[s,\dots,s]:=\prod_{V\in\pi} c_{|V|},
\end{align*}
where $V$ ranges over the blocks of $\pi$ and $|V|$ denotes the number of elements of $V$.
The free moment-cumulant formula gives, for every $m\in\mathbb N$,
\begin{align*}
\varphi(s^m)=\sum_{\pi\in NC(m)} \kappa_\pi[s,\dots,s].
\end{align*}
[/step]
[step:Derive the Catalan moments from the vanishing of all nonquadratic cumulants]
Assume
\begin{align*}
c_1=0,\qquad c_2=1,\qquad c_m=0 \quad \text{for every } m\in\mathbb N \text{ with } m\ne 2.
\end{align*}
The hypothesis $s=s^*$ supplies the self-adjointness part of the definition of a standard semicircular element. It remains to identify the moment sequence. Let $m\in\mathbb N$. For a partition $\pi\in NC(m)$, the product
\begin{align*}
\kappa_\pi[s,\dots,s]=\prod_{V\in\pi} c_{|V|}
\end{align*}
is nonzero only when every block $V$ of $\pi$ satisfies $|V|=2$. Thus the only contributing partitions are the noncrossing pairings of $\{1,\dots,m\}$.
If $m$ is odd, no partition of $\{1,\dots,m\}$ can have all blocks of size $2$, so the moment-cumulant formula gives
\begin{align*}
\varphi(s^m)=0.
\end{align*}
If $m=2r$ is even, every noncrossing pairing contributes
\begin{align*}
\prod_{V\in\pi} c_{|V|}=\prod_{V\in\pi} c_2=1.
\end{align*}
Therefore
\begin{align*}
\varphi(s^{2r})=|NC_2(2r)|=C_r,
\end{align*}
where $C_r$ is the number of noncrossing pairings of $\{1,\dots,2r\}$. For $r=0$, the statement is
\begin{align*}
\varphi(s^0)=\varphi(1_{\mathcal A})=1=C_0,
\end{align*}
by the normalization of the noncommutative probability space and the convention that the empty set has one noncrossing pairing. Hence $s$ has the standard semicircular moment sequence.
[guided]
Assume that the cumulants satisfy
\begin{align*}
c_1=0,\qquad c_2=1,\qquad c_m=0 \quad \text{for every } m\in\mathbb N \text{ with } m\ne 2.
\end{align*}
Here $c_m:=\kappa_m(s,\dots,s)$, and $NC(m)$ denotes the set of noncrossing partitions of $\{1,\dots,m\}$, with $NC_2(m)\subset NC(m)$ the subset of pairings. The theorem already assumes $s=s^*$, so the self-adjointness required in the definition of a standard semicircular element is in force. We want to compute each moment $\varphi(s^m)$. The moment-cumulant formula says that this moment is a sum over all noncrossing partitions:
\begin{align*}
\varphi(s^m)=\sum_{\pi\in NC(m)} \prod_{V\in\pi} c_{|V|}.
\end{align*}
The key question is: which partitions can contribute a nonzero term? A block $V$ with $|V|\ne 2$ contributes a factor $c_{|V|}=0$. Therefore the whole product attached to $\pi$ is nonzero only if every block of $\pi$ has size $2$. Such partitions are exactly the noncrossing pairings.
If $m$ is odd, the set $\{1,\dots,m\}$ cannot be decomposed into two-element blocks. Hence there are no noncrossing pairings, and every term in the moment-cumulant sum is zero. Therefore
\begin{align*}
\varphi(s^m)=0.
\end{align*}
If $m=2r$, the contributing partitions are precisely the elements of $NC_2(2r)$. For each such partition $\pi$, every block has size $2$, so each block contributes $c_2=1$. Hence
\begin{align*}
\kappa_\pi[s,\dots,s]=\prod_{V\in\pi} c_2=1.
\end{align*}
Summing one contribution for each noncrossing pairing gives
\begin{align*}
\varphi(s^{2r})=|NC_2(2r)|=C_r.
\end{align*}
This proves the positive even moments. The remaining even case is $r=0$: since $s^0=1_{\mathcal A}$ and the noncommutative probability space is normalized,
\begin{align*}
\varphi(s^0)=\varphi(1_{\mathcal A})=1=C_0.
\end{align*}
Thus the moments of $s$ are exactly the standard semicircular moments.
[/guided]
[/step]
[step:Recover the first two cumulants from the standard semicircular moments]
Conversely, assume $s$ is standard semicircular. Then
\begin{align*}
\varphi(s)=0,\qquad \varphi(s^2)=1.
\end{align*}
For $m=1$, the moment-cumulant formula gives
\begin{align*}
0=\varphi(s)=c_1.
\end{align*}
For $m=2$, the noncrossing partitions of $\{1,2\}$ are the one-block partition and the two singleton partition, so
\begin{align*}
1=\varphi(s^2)=c_2+c_1^2.
\end{align*}
Since $c_1=0$, this gives
\begin{align*}
c_2=1.
\end{align*}
[/step]
[step:Use induction to force every higher cumulant to vanish]
We prove by induction on $m\ge 3$ that $c_m=0$. Fix $m\ge 3$ and assume that for every $j<m$,
\begin{align*}
c_j=0 \quad \text{unless } j=2,
\end{align*}
and that $c_2=1$.
Let $1_m\in NC(m)$ denote the one-block partition of $\{1,\dots,m\}$. Separating the $1_m$ term in the moment-cumulant formula gives
\begin{align*}
\varphi(s^m)=c_m+\sum_{\substack{\pi\in NC(m), \pi\ne 1_m}}\prod_{V\in\pi} c_{|V|}.
\end{align*}
If $\pi\ne 1_m$, then every block $V$ of $\pi$ satisfies $|V|<m$. By the induction hypothesis, the product attached to $\pi$ is nonzero exactly when every block of $\pi$ has size $2$, and in that case the product equals $1$. Therefore
\begin{align*}
\sum_{\substack{\pi\in NC(m), \pi\ne 1_m}}\prod_{V\in\pi} c_{|V|}=|NC_2(m)|.
\end{align*}
Since $s$ is standard semicircular, its $m$-th moment is also $|NC_2(m)|$: this number is $0$ when $m$ is odd and is $C_r$ when $m=2r$. Hence
\begin{align*}
|NC_2(m)|=\varphi(s^m)=c_m+|NC_2(m)|.
\end{align*}
Subtracting $|NC_2(m)|$ gives
\begin{align*}
c_m=0.
\end{align*}
By induction, $c_m=0$ for every $m\ge 3$. Together with $c_1=0$ and $c_2=1$, this is exactly the claimed free cumulant characterization.
[/step]
