Proof of Calculus of the Free Fock Space: Orthonormal Monomials, Adjoints, Commutation Relations and the Vacuum

Each result cited is universally quantified over the data in its own statement.

Throughout, $\langle\cdot,\cdot\rangle$ and $\lVert\cdot\rVert$ are the inner product and norm of the complex Hilbert space $\mathcal{F}_{n}$ of The Free Fock Space: Creation, Annihilation and Semicircular Operators §space, and $\Omega=\Omega_{\mathcal{F}}$. The map $p\mapsto[p]$ is the canonical map of the completion of $(\mathcal{P}_{n},h_{n})$, so by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §isometry it is complex-linear (in particular $[0]=0$) and $\langle[p],[q]\rangle=h_{n}(p,q)$ for all $p,q\in\mathcal{P}_{n}$. By The Free Fock Space: Creation, Annihilation and Semicircular Operators §operators, for $j\in[n]$ and $w\in W_{n}$ we have $l_{j}[x_{w}]=[x_{(j)w}]$ and $r_{j}[x_{w}]=[x_{w(j)}]$; $\bar{l}_{j}[x_{v}]$ is $[x_{w}]$ if $v=(j)w$ for a word $w$, and is $[0]=0$ if $v$ is empty or its first letter is not $j$; and $\bar{r}_{j}[x_{v}]$ is $[x_{w}]$ if $v=w(j)$, and $0$ if $v$ is empty or its last letter is not $j$. A nonempty word $v$ of length $m$ with first letter $v_{1}=j$ is of the form $(j)w$ (with $w=\varnothing$ if $m=1$, and otherwise $w$ the word of length $m-1$ with $w_{i}=v_{i+1}$), by the definition of concatenation; likewise, by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §last-letter, a nonempty word $v$ with last letter $j$ is of the form $w(j)$. Sums, scalar multiples and composites of elements of $\mathcal{L}(\mathcal{F}_{n})$ lie in $\mathcal{L}(\mathcal{F}_{n})$, which is a complex vector space in which composition distributes over sums, by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations; every $A\in\mathcal{L}(\mathcal{F}_{n})$ has an adjoint $A^{*}\in\mathcal{L}(\mathcal{F}_{n})$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint, and by Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint and conjugate symmetry

$$\langle A^{*}\eta,\xi\rangle=\langle\eta,A\xi\rangle\quad\text{and}\quad\langle A\xi,\eta\rangle=\langle\xi,A^{*}\eta\rangle\qquad(\xi,\eta\in\mathcal{F}_{n}).$$

Three preliminary facts. (F1) If $\varphi:\mathcal{P}_{n}\to\mathbb{C}$ is linear and $\varphi(x_{w})=0$ for every $w\in W_{n}$, then $\varphi=0$: the zero map is linear and has the same values on monomials, so this is the uniqueness in Noncommutative Polynomials Form a Unital Complex Algebra with Involution: Linear Extension from Monomials, Products, Adjoints and Self-Adjoint Parts §linear-extension (a).

(F2) If $A,B\in\mathcal{L}(\mathcal{F}_{n})$ and $A[x_{w}]=B[x_{w}]$ for every $w\in W_{n}$, then $A=B$. Indeed, let $C=A-B\in\mathcal{L}(\mathcal{F}_{n})$. For fixed $\eta\in\mathcal{F}_{n}$ the map $p\mapsto\langle\eta,C[p]\rangle$ is linear, since $[\cdot]$ and $C$ are linear and the inner product is linear in its second argument, and it vanishes on monomials; so it is $0$ by (F1). Taking $\eta=C[p]$ gives $\lVert C[p]\rVert^{2}=0$, so $C[p]=0$ for every $p\in\mathcal{P}_{n}$. Now let $\zeta\in\mathcal{F}_{n}$ and let $\varepsilon>0$ be real. By The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §dense there is $p\in\mathcal{P}_{n}$ with $\lVert\zeta-[p]\rVert<\varepsilon$, and since $\lVert C\rVert_{\mathrm{op}}$ is a bound for $C$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound,

$$\lVert C\zeta\rVert=\lVert C(\zeta-[p])\rVert\le\lVert C\rVert_{\mathrm{op}}\,\varepsilon .$$

As $\varepsilon>0$ is arbitrary, $C\zeta=0$. Hence $A=B$.

(F3) If $A,B\in\mathcal{L}(\mathcal{F}_{n})$ and $\langle A[x_{u}],[x_{v}]\rangle=\langle[x_{u}],B[x_{v}]\rangle$ for all $u,v\in W_{n}$, then $A^{*}=B$. Indeed, for fixed $u$ the map $q\mapsto\langle A[x_{u}],[q]\rangle-\langle[x_{u}],B[q]\rangle$ is linear and vanishes on monomials, hence is $0$ by (F1). Then for fixed $q$ the map $p\mapsto\langle[q],A[p]\rangle-\langle B[q],[p]\rangle$ is linear, and it is the complex conjugate of $p\mapsto\langle A[p],[q]\rangle-\langle[p],B[q]\rangle$, which vanishes on monomials by what was just shown; so it is $0$ by (F1). Thus

$$\langle A[p],[q]\rangle=\langle[p],B[q]\rangle\qquad(p,q\in\mathcal{P}_{n}).\tag{$*$}$$

Fix $q$ and let $\zeta=A^{*}[q]-B[q]$. For every $p$, $\langle\zeta,[p]\rangle=\langle[q],A[p]\rangle-\langle B[q],[p]\rangle=0$, by the displayed property of $A^{*}$ and the complex conjugate of $(*)$. Let $\varepsilon>0$ be real, and by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §dense choose $p$ with $\lVert\zeta-[p]\rVert<\varepsilon$. Then, by Cauchy-Schwarz Inequality in a Complex Inner Product Space,

$$\lVert\zeta\rVert^{2}=\langle\zeta,\zeta-[p]\rangle\le\lVert\zeta\rVert\,\varepsilon ,$$

so $\lVert\zeta\rVert\le\varepsilon$ (trivially if $\zeta=0$). Hence $\zeta=0$, that is $A^{*}[q]=B[q]$ for every $q$, and $A^{*}=B$ by (F2).

Clause 1 (Orthonormal monomials). Let $u,v\in W_{n}$. By the definition of monomials, $x_{u}$ has support $\{u\}$ and $x_{u}(u)=1$, so by the isometry property recalled above and the definition of $h_{n}$ in The Free Fock Space: Creation, Annihilation and Semicircular Operators,

$$\langle[x_{u}],[x_{v}]\rangle=h_{n}(x_{u},x_{v})=\overline{x_{u}(u)}\,x_{v}(u)=x_{v}(u),$$

which is $1$ if $u=v$ and $0$ otherwise. Since $\Omega=[1]=[x_{\varnothing}]$ by The Algebra of Noncommutative Polynomials in Finitely Many Self-Adjoint Variables §monomials, $\lVert\Omega\rVert^{2}=\langle[x_{\varnothing}],[x_{\varnothing}]\rangle=1$, so $\lVert\Omega\rVert=1$. The map $P$ is linear because the inner product is linear in its second argument, and by Cauchy-Schwarz Inequality in a Complex Inner Product Space, $\lVert P\zeta\rVert=|\langle\Omega,\zeta\rangle|\,\lVert\Omega\rVert\le\lVert\zeta\rVert$ for every $\zeta$; so $1$ is a bound for $P$ and $P\in\mathcal{L}(\mathcal{F}_{n})$. For later use, $P$ is its own adjoint: for $\zeta,\eta\in\mathcal{F}_{n}$, by conjugate-linearity in the first argument and conjugate symmetry,

$$\langle P\zeta,\eta\rangle=\overline{\langle\Omega,\zeta\rangle}\,\langle\Omega,\eta\rangle=\langle\zeta,\Omega\rangle\langle\Omega,\eta\rangle=\langle\zeta,P\eta\rangle ,$$

so $P$ is an adjoint of $P$ and $P^{*}=P$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique.

Clause 2 (Adjoints). Let $j\in[n]$ and $u,v\in W_{n}$. By Clause 1, $\langle l_{j}[x_{u}],[x_{v}]\rangle=\langle[x_{(j)u}],[x_{v}]\rangle$ is $1$ if $v=(j)u$ and $0$ otherwise. If $v=(j)w$ for a word $w$, then $\langle[x_{u}],\bar{l}_{j}[x_{v}]\rangle=\langle[x_{u}],[x_{w}]\rangle$ is $1$ if $u=w$ and $0$ otherwise, and $u=w$ holds if and only if $v=(j)u$, because the word $w$ is determined by $(j)w$ (The Free Fock Space: Creation, Annihilation and Semicircular Operators §operators). Otherwise $v$ is empty or has first letter different from $j$, so $\bar{l}_{j}[x_{v}]=0$, while $v\neq(j)u$ because $(j)u$ has first letter $j$; both sides are then $0$. Hence $\langle l_{j}[x_{u}],[x_{v}]\rangle=\langle[x_{u}],\bar{l}_{j}[x_{v}]\rangle$ for all $u,v$, and $l_{j}^{*}=\bar{l}_{j}$ by (F3). The same argument with $r_{j}[x_{u}]=[x_{u(j)}]$, $\bar{r}_{j}$, $u(j)$ and "last letter" in place of $l_{j}[x_{u}]=[x_{(j)u}]$, $\bar{l}_{j}$, $(j)u$ and "first letter" gives $r_{j}^{*}=\bar{r}_{j}$.

Let also $k\in[n]$ and $w\in W_{n}$. Then $l_{j}^{*}l_{k}[x_{w}]=\bar{l}_{j}[x_{(k)w}]$, which is $[x_{w}]$ if $k=j$ and $0$ if $k\neq j$ (the first letter of $(k)w$ being $k$); so $l_{j}^{*}l_{k}[x_{w}]=(\delta_{jk}I)[x_{w}]$, and $l_{j}^{*}l_{k}=\delta_{jk}I$ by (F2). Likewise $r_{j}^{*}r_{k}[x_{w}]=\bar{r}_{j}[x_{w(k)}]=\delta_{jk}[x_{w}]$, so $r_{j}^{*}r_{k}=\delta_{jk}I$ by (F2). Finally $l_{j}^{*}\Omega=\bar{l}_{j}[x_{\varnothing}]=0$ and $r_{j}^{*}\Omega=\bar{r}_{j}[x_{\varnothing}]=0$, the word $\varnothing$ being empty.

Clause 3 (Commuting creation operators). Let $j,k\in[n]$ and $w\in W_{n}$. Then $l_{j}r_{k}[x_{w}]=[x_{(j)(w(k))}]$ and $r_{k}l_{j}[x_{w}]=[x_{((j)w)(k)}]$, and these are equal since $(j)(w(k))=((j)w)(k)$ by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §monoid. By (F2), $l_{j}r_{k}=r_{k}l_{j}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus the adjoint of $l_{j}r_{k}$ is $r_{k}^{*}l_{j}^{*}$ and the adjoint of $r_{k}l_{j}$ is $l_{j}^{*}r_{k}^{*}$; since these are adjoints of the same operator, $l_{j}^{*}r_{k}^{*}=r_{k}^{*}l_{j}^{*}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique.

Clause 4 (Vacuum commutators). Let $j,k\in[n]$; both $r_{j}^{*}l_{k}-l_{k}r_{j}^{*}$ and $\delta_{jk}P$ lie in $\mathcal{L}(\mathcal{F}_{n})$ (Clause 1), so by (F2) it suffices to compare them on $[x_{w}]$, $w\in W_{n}$, using Clause 2 ($r_{j}^{*}=\bar{r}_{j}$).

If $w=\varnothing$: since $(k)=\varnothing(k)$, $r_{j}^{*}l_{k}[x_{\varnothing}]=\bar{r}_{j}[x_{(k)}]$ is $[x_{\varnothing}]=\Omega$ if $j=k$ and $0$ otherwise, that is $\delta_{jk}\Omega$; $l_{k}r_{j}^{*}[x_{\varnothing}]=l_{k}0=0$ by Clause 2; and $\delta_{jk}P[x_{\varnothing}]=\delta_{jk}\langle\Omega,\Omega\rangle\Omega=\delta_{jk}\Omega$ by Clause 1.

If $w$ has length $m\in\mathbb{N}$: then $\delta_{jk}P[x_{w}]=\delta_{jk}\langle[x_{\varnothing}],[x_{w}]\rangle\Omega=0$ by Clause 1, as $w\neq\varnothing$. Write $w=v(i)$ with $v\in W_{n}$ and $i=w_{m}$ its last letter, as recalled above; then $(k)w=((k)v)(i)$ by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §monoid. If $i=j$, then

$$r_{j}^{*}l_{k}[x_{w}]=\bar{r}_{j}[x_{((k)v)(j)}]=[x_{(k)v}]=l_{k}[x_{v}]=l_{k}\bar{r}_{j}[x_{v(j)}]=l_{k}r_{j}^{*}[x_{w}].$$

If $i\neq j$, then $w$ and $(k)w$ both have last letter $i\neq j$, so $r_{j}^{*}l_{k}[x_{w}]=\bar{r}_{j}[x_{(k)w}]=0$ and $l_{k}r_{j}^{*}[x_{w}]=l_{k}0=0$. In both cases $(r_{j}^{*}l_{k}-l_{k}r_{j}^{*})[x_{w}]=0=\delta_{jk}P[x_{w}]$. This proves $r_{j}^{*}l_{k}-l_{k}r_{j}^{*}=\delta_{jk}P$ for all $j,k$.

For the second relation, apply the first with $j$ and $k$ interchanged: $r_{k}^{*}l_{j}-l_{j}r_{k}^{*}=\delta_{kj}P$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus (each operator is the adjoint of its adjoint, and adjoints of composites, sums and real multiples are formed as stated there), the left side has adjoint $l_{j}^{*}r_{k}-r_{k}l_{j}^{*}$, and the right side has adjoint $\delta_{jk}P^{*}=\delta_{jk}P$ by Clause 1. Equal operators have equal adjoints (Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique), so $l_{j}^{*}r_{k}-r_{k}l_{j}^{*}=\delta_{jk}P$.

Clause 5 (Semicircular operators). Let $j\in[n]$. By Clause 2, $S_{j}=l_{j}+l_{j}^{*}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $l_{j}$ is the adjoint of $l_{j}^{*}$, so $S_{j}$ has the adjoint $l_{j}^{*}+l_{j}=S_{j}$; thus $S_{j}$ is an adjoint of itself, hence self-adjoint by the same clause. The same argument with $D_{j}=r_{j}+r_{j}^{*}$ shows that $D_{j}$ is self-adjoint. By The Free Fock Space: Creation, Annihilation and Semicircular Operators §operators, each of $l_{j},\bar{l}_{j},r_{j},\bar{r}_{j}$ is the unique operator in $\mathcal{L}(\mathcal{F}_{n})$ with the stated values on classes, and the operator provided by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §extension-linear with $C=1$ is such an operator and has bound $1$; so $1$ is a bound for each of them, and their operator norms are at most $1$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. By the triangle inequality in Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations, $\lVert S_{j}\rVert_{\mathrm{op}}\le\lVert l_{j}\rVert_{\mathrm{op}}+\lVert\bar{l}_{j}\rVert_{\mathrm{op}}\le2$ and $\lVert D_{j}\rVert_{\mathrm{op}}\le2$.

Clause 6 (Left and right commute). Let $j,k\in[n]$. Using $S_{j}=l_{j}+l_{j}^{*}$, $D_{k}=r_{k}+r_{k}^{*}$ (Clause 2) and distributivity of composition over sums,

$$S_{j}D_{k}-D_{k}S_{j}=(l_{j}r_{k}-r_{k}l_{j})-(r_{k}^{*}l_{j}-l_{j}r_{k}^{*})+(l_{j}^{*}r_{k}-r_{k}l_{j}^{*})+(l_{j}^{*}r_{k}^{*}-r_{k}^{*}l_{j}^{*}).$$

The first and last brackets vanish by Clause 3; by Clause 4 (the first relation with $j,k$ interchanged, and the second relation) the middle terms are $-\delta_{kj}P+\delta_{jk}P=0$. Hence $S_{j}D_{k}=D_{k}S_{j}$.

Clause 7 (Reversal). First, an auxiliary fact (C): $S_{u}D_{k}=D_{k}S_{u}$ for every $u\in W_{n}$ and $k\in[n]$. For $u=\varnothing$, $S_{\varnothing}=I$ by Evaluation of Noncommutative Polynomials at a Tuple of Bounded Operators §word-products. For words of positive length we induct on the length $m\in\mathbb{N}$, the statement for $m$ being that $S_{u}D_{k}=D_{k}S_{u}$ for every $k$ and every $u$ of length $m$. For $m=1$, $u=(i)$ is a letter by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §last-letter, $S_{u}=S_{i}$ by Evaluation of Noncommutative Polynomials at a Tuple of Bounded Operators §word-products, and Clause 6 applies. If the statement holds for $m$ and $u$ has length $m+1$, then $u=v(i)$ with $v$ of length $m$ by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §last-letter, and $S_{u}=S_{v}S_{i}$ by Evaluation of Noncommutative Polynomials at Bounded Operators: a Unital Homomorphism Compatible with Adjoints, Substitution, Representations and the GNS Multiplication Operators §values; so $S_{u}D_{k}=S_{v}D_{k}S_{i}=D_{k}S_{v}S_{i}=D_{k}S_{u}$ by Clause 6 and the induction hypothesis.

Now we prove $S_{w}\Omega=D_{w^{\mathrm{rev}}}\Omega$. For $w=\varnothing$ both sides are $I\Omega=\Omega$, as $\varnothing^{\mathrm{rev}}=\varnothing$ (Words over a Finite Alphabet: the Empty Word, Concatenation and Reversal §reversal). For words of positive length we induct on the length $m\in\mathbb{N}$, the statement for $m$ being that $S_{w}\Omega=D_{w^{\mathrm{rev}}}\Omega$ for every $w$ of length $m$. For $m=1$, $w=(j)$ is a letter and $w^{\mathrm{rev}}=(j)$ by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §reversal; since $\Omega=[x_{\varnothing}]$ and $(j)\varnothing=\varnothing(j)=(j)$, Clause 2 gives

$$S_{j}\Omega=l_{j}[x_{\varnothing}]+l_{j}^{*}\Omega=[x_{j}]=r_{j}[x_{\varnothing}]+r_{j}^{*}\Omega=D_{j}\Omega .$$

If the statement holds for $m$ and $w$ has length $m+1$, write $w=v(j)$ with $v$ of length $m$ (Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §last-letter); then $w^{\mathrm{rev}}=(j)v^{\mathrm{rev}}$ by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §reversal, and by Evaluation of Noncommutative Polynomials at Bounded Operators: a Unital Homomorphism Compatible with Adjoints, Substitution, Representations and the GNS Multiplication Operators §values, the case $m=1$, (C) and the induction hypothesis,

$$S_{w}\Omega=S_{v}S_{j}\Omega=S_{v}D_{j}\Omega=D_{j}S_{v}\Omega=D_{j}D_{v^{\mathrm{rev}}}\Omega=D_{(j)v^{\mathrm{rev}}}\Omega=D_{w^{\mathrm{rev}}}\Omega .$$

Clause 8 (Right annihilation of the vacuum orbit). Fix $j\in[n]$. For $a\in[n]$, expanding $S_{a}=l_{a}+l_{a}^{*}$ and using Clause 4 (first relation, with $k=a$) and Clause 3 (with $a,j$ in place of $j,k$),

$$r_{j}^{*}S_{a}-S_{a}r_{j}^{*}=(r_{j}^{*}l_{a}-l_{a}r_{j}^{*})+(r_{j}^{*}l_{a}^{*}-l_{a}^{*}r_{j}^{*})=\delta_{ja}P .\tag{$\dagger$}$$

We induct on the length $k\in\mathbb{N}$ of $w$, the statement for $k$ being the displayed formula of Clause 8 for every word $w$ of length $k$.

For $k=1$, $w=(a)$ is a letter, $S_{w}\Omega=S_{a}\Omega=[x_{a}]$ (Clause 7, case $m=1$), and since $(a)=\varnothing(a)$, $r_{j}^{*}[x_{(a)}]=\bar{r}_{j}[x_{\varnothing(a)}]=\delta_{ja}\Omega$. On the right side, $L_{j}(w)$ is $\{1\}$ if $a=j$ and empty otherwise; for $s=1$, $w_{<1}=w_{>1}=\varnothing$, so the term is $\langle\Omega,\Omega\rangle\Omega=\Omega$. Both sides equal $\delta_{ja}\Omega$.

Suppose the statement holds for $k$ and let $w$ have length $k+1$. Let $a=w_{1}$ and let $v$ be the word of length $k$ with $v_{t}=w_{t+1}$ for $t\in[k]$; then $w=(a)v$ by Words over a Finite Alphabet: the Empty Word, Concatenation and Reversal §concatenation, and $S_{w}=S_{a}S_{v}$ by Evaluation of Noncommutative Polynomials at Bounded Operators: a Unital Homomorphism Compatible with Adjoints, Substitution, Representations and the GNS Multiplication Operators §values. By $(\dagger)$, the induction hypothesis for $v$ and linearity of $S_{a}$,

$$r_{j}^{*}S_{w}\Omega=S_{a}r_{j}^{*}S_{v}\Omega+\delta_{ja}P S_{v}\Omega=\sum_{t\in L_{j}(v)}\langle\Omega,S_{v_{>t}}\Omega\rangle\,S_{a}S_{v_{<t}}\Omega+\delta_{ja}\langle\Omega,S_{v}\Omega\rangle\Omega .$$

For $t\in[k]$ put $s=t+1\in[k+1]$. Then $w_{s}=v_{t}$, so $s\in L_{j}(w)$ if and only if $t\in L_{j}(v)$; moreover $w_{<s}=(w_{1},\dots,w_{t})=(a)v_{<t}$ and $w_{>s}=(w_{t+2},\dots,w_{k+1})=v_{>t}$, so $S_{a}S_{v_{<t}}=S_{w_{<s}}$ by Evaluation of Noncommutative Polynomials at Bounded Operators: a Unital Homomorphism Compatible with Adjoints, Substitution, Representations and the GNS Multiplication Operators §values. The index $s=1$ lies in $L_{j}(w)$ if and only if $a=j$, and $w_{<1}=\varnothing$, $w_{>1}=v$, so its term is $\langle\Omega,S_{v}\Omega\rangle S_{\varnothing}\Omega=\langle\Omega,S_{v}\Omega\rangle\Omega$. Since $L_{j}(w)$ is the disjoint union of $\{t+1:\ t\in L_{j}(v)\}$ and of $\{1\}$ or the empty set according as $a=j$ or not, reindexing the finite sum by the bijection $t\mapsto t+1$ gives

$$r_{j}^{*}S_{w}\Omega=\sum_{s\in L_{j}(w)}\langle\Omega,S_{w_{>s}}\Omega\rangle\,S_{w_{<s}}\Omega ,$$

which is the statement for $k+1$. This completes the induction and the proof.