[proofplan]
The proof is a direct calculation in the [tensor product](/page/Tensor%20Product) model supplied by tensor independence. We pass from the generated algebra $B$ to $A_1 \otimes A_2$ using the inverse of the multiplication identification, where each $a_i$ lives in the first tensor factor and each $b_i$ lives in the second tensor factor. Multiplication in the tensor product gathers the $A_1$ letters into $a_1\cdots a_n$ and the $A_2$ letters into $b_1\cdots b_n$. Finally, tensor independence says that the state on this tensor product is the product of the two restricted states.
[/proofplan]
[step:Pass to the tensor product model supplied by tensor independence]
Let $B \subset A$ denote the unital subalgebra generated by $A_1 \cup A_2$. Define the multiplication map $m: A_1 \otimes A_2 \to B$ by $m(x \otimes y)=xy$ for $x \in A_1$ and $y \in A_2$, extended linearly. By tensor independence, $m$ is a unital algebra isomorphism and the state $\varphi|_B$ corresponds to the product state $\varphi_1 \otimes \varphi_2$, where $\varphi_1: A_1 \to \mathbb{C}$ and $\varphi_2: A_2 \to \mathbb{C}$ are the restrictions
\begin{align*}
\varphi_1=\varphi|_{A_1}.
\end{align*}
\begin{align*}
\varphi_2=\varphi|_{A_2}.
\end{align*}
Let $\Psi: B \to A_1 \otimes A_2$ denote the inverse algebra isomorphism $\Psi=m^{-1}$. Since $m(a \otimes 1)=a$ for $a \in A_1$ and $m(1 \otimes b)=b$ for $b \in A_2$, we have
\begin{align*}
\Psi(a)=a\otimes 1
\end{align*}
for every $a \in A_1$, and
\begin{align*}
\Psi(b)=1\otimes b
\end{align*}
for every $b \in A_2$.
[/step]
[step:Gather the alternating word into the two tensor factors]
Fix $n \in \mathbb{N}$, elements $a_1,\dots,a_n \in A_1$, and elements $b_1,\dots,b_n \in A_2$. Since $\Psi$ is an algebra homomorphism, applying $\Psi$ to the alternating word gives
\begin{align*}
\Psi(a_1b_1a_2b_2\cdots a_nb_n)=\prod_{i=1}^{n}\bigl((a_i\otimes 1)(1\otimes b_i)\bigr).
\end{align*}
In the algebraic tensor product, multiplication is defined by
\begin{align*}
(x\otimes y)(x'\otimes y')=(xx')\otimes(yy')
\end{align*}
for $x,x' \in A_1$ and $y,y' \in A_2$. Hence
\begin{align*}
(a_i\otimes 1)(1\otimes b_i)=a_i\otimes b_i
\end{align*}
for each $i \in \{1,\dots,n\}$, and repeated use of the same multiplication rule yields
\begin{align*}
\Psi(a_1b_1a_2b_2\cdots a_nb_n)=(a_1a_2\cdots a_n)\otimes(b_1b_2\cdots b_n).
\end{align*}
[guided]
We want to compute the mixed word in a setting where the two independent algebras occupy separate coordinates. Tensor independence gives exactly that setting: the generated algebra $B$ is identified with $A_1\otimes A_2$, and the inverse identification $\Psi=m^{-1}:B\to A_1\otimes A_2$ is an algebra homomorphism.
Because $m(a\otimes 1)=a$ for $a\in A_1$ and $m(1\otimes b)=b$ for $b\in A_2$, the inverse map satisfies
\begin{align*}
\Psi(a_i)=a_i\otimes 1
\end{align*}
and
\begin{align*}
\Psi(b_i)=1\otimes b_i
\end{align*}
for every $i\in\{1,\dots,n\}$. Since $\Psi$ preserves multiplication, the image of the full alternating product is the product of the images:
\begin{align*}
\Psi(a_1b_1a_2b_2\cdots a_nb_n)=\prod_{i=1}^{n}\bigl((a_i\otimes 1)(1\otimes b_i)\bigr).
\end{align*}
Now the multiplication rule in the algebraic tensor product is
\begin{align*}
(x\otimes y)(x'\otimes y')=(xx')\otimes(yy')
\end{align*}
for $x,x'\in A_1$ and $y,y'\in A_2$. Applying this first to each adjacent pair gives
\begin{align*}
(a_i\otimes 1)(1\otimes b_i)=a_i\otimes b_i.
\end{align*}
Applying the same rule repeatedly then gathers all first tensor-factor entries together and all second tensor-factor entries together:
\begin{align*}
(a_1\otimes b_1)(a_2\otimes b_2)\cdots(a_n\otimes b_n)=(a_1a_2\cdots a_n)\otimes(b_1b_2\cdots b_n).
\end{align*}
Therefore
\begin{align*}
\Psi(a_1b_1a_2b_2\cdots a_nb_n)=(a_1a_2\cdots a_n)\otimes(b_1b_2\cdots b_n).
\end{align*}
[/guided]
[/step]
[step:Apply the product state to obtain the mixed moment factorization]
By tensor independence, the equality $\varphi|_B=(\varphi_1\otimes\varphi_2)\circ\Psi$ holds on $B$. Therefore,
\begin{align*}
\varphi(a_1b_1a_2b_2\cdots a_nb_n)=(\varphi_1\otimes\varphi_2)\bigl((a_1a_2\cdots a_n)\otimes(b_1b_2\cdots b_n)\bigr).
\end{align*}
By the definition of the product state on $A_1\otimes A_2$,
\begin{align*}
(\varphi_1\otimes\varphi_2)(x\otimes y)=\varphi_1(x)\varphi_2(y)
\end{align*}
for all $x\in A_1$ and $y\in A_2$. Taking $x=a_1a_2\cdots a_n$ and $y=b_1b_2\cdots b_n$ gives
\begin{align*}
\varphi(a_1b_1a_2b_2\cdots a_nb_n)=\varphi_1(a_1a_2\cdots a_n)\varphi_2(b_1b_2\cdots b_n).
\end{align*}
Since $\varphi_1=\varphi|_{A_1}$ and $\varphi_2=\varphi|_{A_2}$, this is exactly
\begin{align*}
\varphi(a_1b_1a_2b_2\cdots a_nb_n)=\varphi(a_1a_2\cdots a_n)\varphi(b_1b_2\cdots b_n).
\end{align*}
This proves the asserted factorization.
[/step]
