[proofplan] We prove the result directly from the left-creation model on full Fock space. First, for each unit vector $e_i$, the operator $s_i = \ell(e_i) + \ell(e_i)^*$ acts on the one-vector Fock subspace as the adjacency operator of the half-line, so its vacuum moments count Dyck paths and hence are the Catalan moments of the centered variance-one semicircle law. Second, orthogonality of the vectors $e_i$ implies that annihilation by $\ell(e_i)^*$ cannot remove a tensor whose first letter lies in a different one-dimensional subspace. This gives the standard alternating centered-word criterion for free independence. [/proofplan] [step:Record the creation and annihilation rules on tensor words] For $h \in H$, the adjoint $\ell(h)^*: \mathcal{F}(H) \to \mathcal{F}(H)$ is determined on elementary tensors by \begin{align*} \ell(h)^*\Omega = 0 \end{align*} and, for $m \geq 1$ and $h_1,\dots,h_m \in H$, \begin{align*} \ell(h)^*(h_1 \otimes \cdots \otimes h_m) = (h_1,h)_H h_2 \otimes \cdots \otimes h_m, \end{align*} where for $m=1$ the tensor $h_2 \otimes \cdots \otimes h_m$ is interpreted as $\Omega$. Indeed, for every $\eta \in H^{\otimes (m-1)}$, \begin{align*} (\ell(h)\eta,h_1 \otimes \cdots \otimes h_m)_{\mathcal{F}(H)} = (h,h_1)_H(\eta,h_2 \otimes \cdots \otimes h_m)_{\mathcal{F}(H)}. \end{align*} Since the Hilbert-space [inner product](/page/Inner%20Product) is linear in the first variable, this is exactly the adjoint relation for the displayed formula. In particular, for each $i \in I$, the operator $s_i = \ell(e_i)+\ell(e_i)^*$ is self-adjoint. [/step] [step:Compute the vacuum moments of one Fock semicircular operator] Fix $i \in I$ and define $K_i := \mathbb{C}e_i \subset H$. Let \begin{align*} \mathcal{F}(K_i) := \mathbb{C}\Omega \oplus \bigoplus_{m=1}^{\infty} K_i^{\otimes m} \subset \mathcal{F}(H). \end{align*} The subspace $\mathcal{F}(K_i)$ is invariant under $s_i$. For $m \geq 0$, define $u_m \in \mathcal{F}(K_i)$ by $u_0 := \Omega$ and $u_m := e_i^{\otimes m}$ for $m \geq 1$. Since $\|e_i\|_H=1$, the family $(u_m)_{m \geq 0}$ is orthonormal, and the creation-annihilation rules give \begin{align*} s_i u_0 = u_1 \end{align*} and, for $m \geq 1$, \begin{align*} s_i u_m = u_{m+1}+u_{m-1}. \end{align*} For $n \in \mathbb{N}$, the moment $\varphi(s_i^n)$ is the coefficient of $u_0$ in $s_i^n u_0$. Expanding the action by the preceding recurrence, each application of $\ell(e_i)$ raises the index $m$ by $1$, and each application of $\ell(e_i)^*$ lowers the index $m$ by $1$, with the convention that a lowering step from $0$ contributes $0$. Hence $\varphi(s_i^n)$ is the number of lattice paths of length $n$ starting at $0$, ending at $0$, using steps $+1$ and $-1$, and never going below $0$. If $n=2r+1$ is odd, no such path exists because a path returning to $0$ has the same number of upward and downward steps. If $n=2r$, these paths are the Dyck paths of length $2r$, whose number is the Catalan number \begin{align*} C_r := \frac{1}{r+1}\binom{2r}{r}. \end{align*} Therefore \begin{align*} \varphi(s_i^{2r+1}) = 0 \end{align*} and \begin{align*} \varphi(s_i^{2r}) = C_r. \end{align*} These are precisely the moments of the centered semicircular distribution of variance $1$. Thus each $s_i$ is centered semicircular of variance $1$ with respect to $\varphi$. [guided] Fix one index $i \in I$. The point of passing to $K_i := \mathbb{C}e_i$ is that the operator $s_i$ only creates and annihilates copies of the single vector $e_i$. Thus all vacuum moments of $s_i$ are already computed inside the one-dimensional Fock space \begin{align*} \mathcal{F}(K_i) := \mathbb{C}\Omega \oplus \bigoplus_{m=1}^{\infty} K_i^{\otimes m}. \end{align*} Define $u_0 := \Omega$ and $u_m := e_i^{\otimes m}$ for $m \geq 1$. Since $e_i$ is a unit vector, the tensors $u_m$ form an [orthonormal basis](/page/Orthonormal%20Basis) of $\mathcal{F}(K_i)$. The creation operator adds one copy of $e_i$ on the left, so $\ell(e_i)u_m = u_{m+1}$ for all $m \geq 0$. The annihilation operator removes the first copy of $e_i$ when one is present, and kills the vacuum, so \begin{align*} \ell(e_i)^*u_0 = 0 \end{align*} and, for $m \geq 1$, \begin{align*} \ell(e_i)^*u_m = u_{m-1}. \end{align*} Adding the two formulas gives \begin{align*} s_i u_0 = u_1 \end{align*} and, for $m \geq 1$, \begin{align*} s_i u_m = u_{m+1}+u_{m-1}. \end{align*} Now compute $\varphi(s_i^n)$. By definition of the vacuum state, \begin{align*} \varphi(s_i^n) = (s_i^n\Omega,\Omega)_{\mathcal{F}(H)} = (s_i^n u_0,u_0)_{\mathcal{F}(H)}. \end{align*} Thus $\varphi(s_i^n)$ is the coefficient of $u_0$ in $s_i^n u_0$. Each factor of $s_i$ gives a choice: create, which moves $u_m$ to $u_{m+1}$, or annihilate, which moves $u_m$ to $u_{m-1}$ if $m \geq 1$ and gives $0$ if $m=0$. Therefore the surviving terms are exactly the paths that start at height $0$, move by steps $+1$ and $-1$, never cross below $0$, and return to height $0$ after $n$ steps. Such a path has the same number of upward and downward steps. Hence no path exists when $n$ is odd, so $\varphi(s_i^{2r+1})=0$. When $n=2r$, these paths are the Dyck paths of length $2r$, and the standard Catalan count gives \begin{align*} \#\{\text{Dyck paths of length }2r\} = C_r := \frac{1}{r+1}\binom{2r}{r}. \end{align*} Consequently \begin{align*} \varphi(s_i^{2r}) = C_r. \end{align*} The centered semicircular law of variance $1$ is characterized by the moment sequence with odd moments equal to $0$ and even moments equal to the Catalan numbers $C_r$. Therefore $s_i$ has centered semicircular distribution of variance $1$ with respect to the vacuum state. [/guided] [/step] [step:Show that centered polynomials in one generator create words beginning in its subspace] For $i \in I$, let $\mathcal{A}_i$ denote the unital algebra generated by $s_i$ inside $\mathcal{L}(\mathcal{F}(H))$. Define \begin{align*} \mathcal{F}^+(K_i) := \bigoplus_{m=1}^{\infty} K_i^{\otimes m}. \end{align*} We prove the following word property. [claim:Centered words force the first tensor letter into the corresponding one-dimensional subspace] Let $i \in I$ and let $a \in \mathcal{A}_i$ satisfy $\varphi(a)=0$. If $\xi \in \mathcal{F}(H)$ is either $\Omega$ or an elementary tensor whose first tensor factor is orthogonal to $e_i$, then $a\xi$ is a finite linear combination of elementary tensors whose first tensor factor lies in $K_i$. [/claim] [proof] Since $\mathcal{A}_i$ is generated by the single self-adjoint operator $s_i$, there is a polynomial $P \in \mathbb{C}[t]$ such that $a=P(s_i)$. Write \begin{align*} P(t)=\sum_{r=0}^{N} c_r t^r \end{align*} with $N \in \mathbb{N}$ and coefficients $c_0,\dots,c_N \in \mathbb{C}$. First take $\xi=\Omega$. The vector $P(s_i)\Omega$ lies in $\mathcal{F}(K_i)$ because both $\ell(e_i)$ and $\ell(e_i)^*$ preserve $\mathcal{F}(K_i)$. Its vacuum coefficient is \begin{align*} (P(s_i)\Omega,\Omega)_{\mathcal{F}(H)} = \varphi(P(s_i)) = \varphi(a)=0. \end{align*} Hence $P(s_i)\Omega \in \mathcal{F}^+(K_i)$, so every nonzero elementary tensor in its expansion has first tensor factor in $K_i$. Now let $\xi=h_1 \otimes \cdots \otimes h_m$ with $m \geq 1$ and $(h_1,e_i)_H=0$. Consider a monomial $s_i^r$. Expanding $s_i^r$ as a sum of words in $\ell(e_i)$ and $\ell(e_i)^*$, any nonzero word applied to $\xi$ has one of two forms. Either every created copy of $e_i$ is eventually annihilated before the operators reach the original first tensor factor $h_1$, in which case the output is exactly $\xi$; or at least one created copy of $e_i$ remains at the left, in which case the output has first tensor factor in $K_i$. A word whose annihilation operator attempts to remove the original first tensor factor gives zero, because \begin{align*} \ell(e_i)^*(h_1 \otimes \cdots \otimes h_m) = (h_1,e_i)_H h_2 \otimes \cdots \otimes h_m = 0. \end{align*} The coefficient of the unchanged vector $\xi$ in $s_i^r\xi$ is exactly the coefficient of $\Omega$ in $s_i^r\Omega$, namely $\varphi(s_i^r)$, because in both cases it is counted by the same balanced creation-annihilation words that never try to annihilate below the created stack. Therefore the coefficient of $\xi$ in $P(s_i)\xi$ is \begin{align*} \sum_{r=0}^{N} c_r \varphi(s_i^r) = \varphi(P(s_i)) = \varphi(a)=0. \end{align*} All remaining nonzero elementary tensors have first tensor factor in $K_i$. This proves the claim. [/proof] [/step] [step:Apply the centered alternating word criterion for free independence] We verify free independence of the family $(\mathcal{A}_i)_{i \in I}$. Let $n \in \mathbb{N}$, let $i_1,\dots,i_n \in I$ satisfy $i_k \neq i_{k+1}$ for $1 \leq k < n$, and choose elements $a_k \in \mathcal{A}_{i_k}$ such that $\varphi(a_k)=0$ for every $1 \leq k \leq n$. We must prove \begin{align*} \varphi(a_1a_2\cdots a_n)=0. \end{align*} Apply the claim from right to left. Since $a_n$ is centered, $a_n\Omega$ is a finite linear combination of non-vacuum tensors whose first tensor factor lies in $K_{i_n}$. Because $i_{n-1}\neq i_n$, every vector in $K_{i_n}$ is orthogonal to $e_{i_{n-1}}$. Applying the claim to $a_{n-1}$ shows that $a_{n-1}a_n\Omega$ is a finite linear combination of non-vacuum tensors whose first tensor factor lies in $K_{i_{n-1}}$. Repeating this argument inductively, after applying $a_k$ the resulting vector is a finite linear combination of non-vacuum tensors whose first tensor factor lies in $K_{i_k}$. In particular, $a_1a_2\cdots a_n\Omega$ is orthogonal to $\Omega$. Therefore \begin{align*} \varphi(a_1a_2\cdots a_n) = (a_1a_2\cdots a_n\Omega,\Omega)_{\mathcal{F}(H)} = 0. \end{align*} This is exactly the defining centered alternating-word condition for free independence, so the algebras $(\mathcal{A}_i)_{i \in I}$ are freely independent with respect to $\varphi$. [guided] We now prove freeness using the definition. For each $i \in I$, the algebra $\mathcal{A}_i$ is the unital algebra generated by $s_i$. Free independence requires the following test: whenever the indices alternate and each chosen element has vacuum expectation zero, the vacuum expectation of the product must vanish. So let $n \in \mathbb{N}$, let $i_1,\dots,i_n \in I$ satisfy $i_k \neq i_{k+1}$ for $1 \leq k < n$, and choose $a_k \in \mathcal{A}_{i_k}$ with $\varphi(a_k)=0$. We need to show \begin{align*} \varphi(a_1a_2\cdots a_n)=0. \end{align*} The reason the proof works is that a centered polynomial in $s_i$ cannot leave a vector unchanged when the first tensor letter is orthogonal to $e_i$. The unchanged part is exactly its vacuum expectation, and the centered hypothesis removes that scalar part. What remains must begin with a newly created copy of $e_i$. Start on the right with $\Omega$. Since $a_n$ is centered and belongs to $\mathcal{A}_{i_n}$, the claim gives that $a_n\Omega$ is a finite linear combination of non-vacuum tensors whose first tensor factor lies in \begin{align*} K_{i_n}=\mathbb{C}e_{i_n}. \end{align*} Now move one factor left. Since $i_{n-1}\neq i_n$ and the family $(e_i)_{i\in I}$ is orthonormal, every vector in $K_{i_n}$ is orthogonal to $e_{i_{n-1}}$. Therefore the claim applies to $a_{n-1}$ acting on each elementary tensor appearing in $a_n\Omega$. It follows that $a_{n-1}a_n\Omega$ is a finite linear combination of non-vacuum tensors whose first tensor factor lies in $K_{i_{n-1}}$. The same argument repeats for $a_{n-2},a_{n-3},\dots,a_1$. At the $k$-th stage from the left, the vector already constructed has first tensor factor in $K_{i_{k+1}}$, and the alternating-index condition gives $i_k \neq i_{k+1}$. Orthogonality of the one-dimensional subspaces $K_{i_k}$ and $K_{i_{k+1}}$ lets us apply the claim to $a_k$. Hence the new vector has first tensor factor in $K_{i_k}$ and is still orthogonal to the vacuum vector. After all factors are applied, $a_1a_2\cdots a_n\Omega$ is a finite linear combination of non-vacuum tensors. The vacuum vector $\Omega$ is orthogonal to every tensor in $\bigoplus_{m=1}^{\infty}H^{\otimes m}$, so \begin{align*} (a_1a_2\cdots a_n\Omega,\Omega)_{\mathcal{F}(H)}=0. \end{align*} By definition of the vacuum state, \begin{align*} \varphi(a_1a_2\cdots a_n)=0. \end{align*} This proves the centered alternating-word condition, and therefore the algebras $\mathcal{A}_i$ are freely independent. [/guided] [/step] [step:Conclude that the operators form a semicircular family] For each $i \in I$, the preceding moment computation shows that $s_i$ is centered semicircular of variance $1$ with respect to $\varphi$. The preceding free-independence argument shows that the unital algebras generated by the operators $s_i$ are freely independent. Therefore the family $(s_i)_{i \in I}$ is a freely independent family of centered semicircular elements of variance $1$ with respect to the vacuum expectation $\varphi$. [/step]