
> 
> For $A= (A\_1,\cdots,A\_d)\in {\cal L}(E)^d$ such that $A\_iA\_j=A\_jA\_i$ for all $i,j$. Why
>  $$\sum\_{f\in F(n,d)} A\_{f}^\*A\_{f}=\displaystyle\sum\_{|\alpha|=n}\frac{n!}{\alpha!}{A^\*}^{\alpha}A^{\alpha}\,?$$
>  Note that $F(n,d)$ denotes the set of all functions from $\{1,\cdots,n\}$ into $\{1,\cdots,d\}$ and $A\_f:=A\_{f(1)}\cdots A\_{f(n)}$, for $f\in F(n,d)$. Also $\alpha = (\alpha\_1, \alpha\_2,...,\alpha\_d) \in \mathbb{Z}\_+^d;\;\alpha!: =\alpha\_1!\cdots\alpha\_d!,\;|\alpha|:=\displaystyle\sum\_{j=1}^d|\alpha\_j|$; $A^\*=(A\_1^\*,\cdots,A\_d^\*)$ and $A^\alpha:=A\_1^{\alpha\_1} A\_2^{\alpha\_2}\cdots A\_d^{\alpha\_d}$.
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The above formula figures in Remarks. 1. of this paper ([1](http://matwbn.icm.edu.pl/ksiazki/sm/sm103/sm103310.pdf)).



Why $\sum_{f\in F(n,d)} A_{f}^*A_{f}=\sum_{|\alpha|=n}\frac{n!}{\alpha!}{A^*}^{\alpha}A^{\alpha}?$
More generally for any $L\in\mathcal{L}(H)$ 
$$\sum\_{f\in [d]^n} A\_{f}^\*LA\_{f}=\displaystyle\sum\_{|\alpha|=n}{n\choose \alpha}\ {A^\*}^{\alpha}LA^{\alpha}\ .$$
It is just an instance of the expansion of the $n$-power of the sum of $d$ commuting objects in a ring, $$(X\_1+\dots +X\_d)^n=\sum\_{\alpha\in\mathbb{N}^d\atop |a|=n} {n\choose \alpha}X^\alpha, $$ with $X:=X\_1^{\alpha\_1}X\_2^{\alpha\_2}\dots X\_d^{\alpha\_d}.$
In your case $X\_j$ is the linear operator on $\mathcal{L}(H)$ defined by $L\mapsto A\_j^\*LA\_j.$ 


