[proofplan] We expand each power of $s(f)=\ell(f)+\ell(f)^*$ as a sum over words in one creation operator and one annihilation operator. A word contributes to the vacuum expectation exactly when, as it acts on $\Omega$ from right to left, it never attempts to annihilate the vacuum and ends back in the vacuum sector. These admissible words are in bijection with Dyck words, hence are absent in odd length and are counted by the Catalan number in even length. Finally, the normalization $\|f\|_H=1$ makes every admissible word contribute exactly $1$. [/proofplan]

[step:Expand the moment as a sum over creation and annihilation words] Set \begin{align*} C := \ell(f) \in \mathcal{L}(\mathcal{F}(H)) \end{align*} and \begin{align*} A := \ell(f)^* \in \mathcal{L}(\mathcal{F}(H)). \end{align*} For an integer $n \geq 0$, let $\mathcal{W}_n$ denote the set of all words $\varepsilon=(\varepsilon_1,\dots,\varepsilon_n)$ with each $\varepsilon_i \in \{c,a\}$. For $\varepsilon \in \mathcal{W}_n$, define the operator \begin{align*} T_\varepsilon: \mathcal{F}(H) \to \mathcal{F}(H) \end{align*} by \begin{align*} T_\varepsilon := T_{\varepsilon_1}\cdots T_{\varepsilon_n}, \end{align*} where $T_c := C$ and $T_a := A$. For $n=0$, $\mathcal{W}_0$ consists of the empty word and $T_\varepsilon$ is the identity operator on $\mathcal{F}(H)$. By distributivity in the noncommutative algebra $\mathcal{L}(\mathcal{F}(H))$, \begin{align*} s(f)^n = (C+A)^n = \sum_{\varepsilon \in \mathcal{W}_n} T_\varepsilon. \end{align*} Applying the linear functional $\tau$ gives \begin{align*} \tau(s(f)^n) = \sum_{\varepsilon \in \mathcal{W}_n} (T_\varepsilon \Omega,\Omega)_{\mathcal{F}(H)}. \end{align*} [/step]

[step:Characterize the words that survive the vacuum expectation]For $\varepsilon=(\varepsilon_1,\dots,\varepsilon_n) \in \mathcal{W}_n$ and $0 \leq k \leq n$, define the right-to-left height \begin{align*} h_k(\varepsilon) := \#\{j \in \{n-k+1,\dots,n\} : \varepsilon_j=c\} - \#\{j \in \{n-k+1,\dots,n\} : \varepsilon_j=a\}. \end{align*} The word $\varepsilon$ is called admissible if $h_k(\varepsilon) \geq 0$ for every $0 \leq k \leq n$ and $h_n(\varepsilon)=0$. We claim that \begin{align*} (T_\varepsilon\Omega,\Omega)_{\mathcal{F}(H)} = \begin{cases} 1, & \text{if } \varepsilon \text{ is admissible}, \end{align*} \begin{align*} 0, & \text{if } \varepsilon \text{ is not admissible}. \end{cases} \end{align*} Indeed, the operators act on $\Omega$ from right to left. The left creation operator satisfies $C\Omega=f$ and sends a tensor $h_1\otimes\cdots\otimes h_r$ to $f\otimes h_1\otimes\cdots\otimes h_r$. The annihilation operator satisfies $A\Omega=0$ and \begin{align*} A(h_1\otimes\cdots\otimes h_r) = (h_1,f)_H h_2\otimes\cdots\otimes h_r \end{align*} for $r \geq 1$. Thus a nonzero intermediate vector can occur only if, after every suffix of the word has acted, the number of creations in that suffix is at least the number of annihilations in that suffix. This is precisely the condition $h_k(\varepsilon)\geq 0$ for all $k$. If this condition fails, some annihilation operator is applied to $\Omega$, and the resulting vector is $0$. If the nonnegativity condition holds but $h_n(\varepsilon)>0$, then $T_\varepsilon\Omega$ lies in the $h_n(\varepsilon)$-particle sector, which is orthogonal to the vacuum sector $\mathbb{C}\Omega$, so $(T_\varepsilon\Omega,\Omega)_{\mathcal{F}(H)}=0$. If $h_k(\varepsilon)\geq 0$ for all $k$ and $h_n(\varepsilon)=0$, then each annihilation removes a leading copy of $f$. Since $\|f\|_H=1$, \begin{align*} (f,f)_H = 1. \end{align*} Repeatedly applying the displayed formula for $A$ therefore returns exactly $\Omega$, so $(T_\varepsilon\Omega,\Omega)_{\mathcal{F}(H)}=1$.[/step]

[guided]We now determine exactly which words in the expansion can have nonzero vacuum expectation. The important point is the direction of action: although the word is written as $T_{\varepsilon_1}\cdots T_{\varepsilon_n}$, the operator $T_{\varepsilon_n}$ acts on $\Omega$ first. For $\varepsilon=(\varepsilon_1,\dots,\varepsilon_n)$ and $0 \leq k \leq n$, define \begin{align*} h_k(\varepsilon) := \#\{j \in \{n-k+1,\dots,n\} : \varepsilon_j=c\} - \#\{j \in \{n-k+1,\dots,n\} : \varepsilon_j=a\}. \end{align*} This number records the particle count after the last $k$ letters of the word have acted, provided no annihilation has already killed the vector. A creation letter increases this count by $1$, and an annihilation letter decreases it by $1$. The full Fock space action gives \begin{align*} C\Omega = f \end{align*} and, for $r \geq 1$, \begin{align*} C(h_1\otimes\cdots\otimes h_r)=f\otimes h_1\otimes\cdots\otimes h_r. \end{align*} The adjoint annihilation operator satisfies \begin{align*} A\Omega=0 \end{align*} and \begin{align*} A(h_1\otimes\cdots\otimes h_r)=(h_1,f)_Hh_2\otimes\cdots\otimes h_r. \end{align*} Why does this produce the height condition? If at some stage more annihilations than creations have occurred in the suffix already acting, then an annihilation is being applied to the vacuum vector. Since $A\Omega=0$, the whole word then contributes zero. This is exactly the failure of $h_k(\varepsilon)\geq 0$ for some $k$. Suppose instead that $h_k(\varepsilon)\geq 0$ for every $k$. Then the word never tries to annihilate the vacuum. If $h_n(\varepsilon)>0$, the final vector lies in a positive-particle tensor sector of $\mathcal{F}(H)$, while $\Omega$ lies in the zero-particle sector. Distinct tensor sectors in the full Fock space are orthogonal, so \begin{align*} (T_\varepsilon\Omega,\Omega)_{\mathcal{F}(H)}=0. \end{align*} It remains to consider the case $h_k(\varepsilon)\geq 0$ for every $k$ and $h_n(\varepsilon)=0$. Then every created tensor is a copy of $f$, and every annihilation removes a leading copy of $f$. Each such removal contributes the scalar \begin{align*} (f,f)_H=\|f\|_H^2=1. \end{align*} After all letters have acted, the particle count is again zero, so the final vector is precisely $\Omega$. Hence \begin{align*} (T_\varepsilon\Omega,\Omega)_{\mathcal{F}(H)}=(\Omega,\Omega)_{\mathcal{F}(H)}=1. \end{align*} Therefore the vacuum expectation of a word is $1$ exactly for admissible words and $0$ for all other words.[/guided]

[step:Identify admissible words with Dyck words and count them] If $n$ is odd, no admissible word exists, because $h_n(\varepsilon)=0$ requires the total number of creation letters and annihilation letters to be equal. Hence, for every $m \geq 0$, \begin{align*} \tau(s(f)^{2m+1})=0. \end{align*} Now let $n=2m$. Define \begin{align*} R:\mathcal{W}_{2m}\to\mathcal{W}_{2m} \end{align*} by reversing words: \begin{align*} R(\varepsilon_1,\dots,\varepsilon_{2m}) := (\varepsilon_{2m},\dots,\varepsilon_1). \end{align*} A word $\varepsilon$ is admissible exactly when $R(\varepsilon)$ has $m$ creation letters, $m$ annihilation letters, and every initial segment has at least as many creation letters as annihilation letters. Thus $R$ is a bijection from admissible words of length $2m$ to Dyck words with $m$ up-steps and $m$ down-steps, where $c$ is read as an up-step and $a$ as a down-step. The standard Catalan enumeration of Dyck words gives the number of such words as \begin{align*} C_m=\frac{1}{m+1}\binom{2m}{m}. \end{align*} Equivalently, this is the classical count of balanced length-$2m$ paths whose partial sums never become negative. [/step]

[step:Sum the surviving word contributions] Combining the expansion with the word characterization, for $m \geq 0$ we have \begin{align*} \tau(s(f)^{2m}) = \sum_{\varepsilon\in\mathcal{W}_{2m}}(T_\varepsilon\Omega,\Omega)_{\mathcal{F}(H)}. \end{align*} All non-admissible words contribute $0$, and each admissible word contributes $1$. Since the number of admissible words is $C_m$, it follows that \begin{align*} \tau(s(f)^{2m})=C_m. \end{align*} Together with the odd-moment computation, this proves the claimed semicircular moment formula. [/step]