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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 $A^\alpha:=A_1^{\alpha_1} A_2^{\alpha_2}\cdots A_d^{\alpha_d}$.

The above formula figures in Remarks. 1. of this paper (1).

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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 $L\mapsto A_j^*LA_j.$

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  • $\begingroup$ Thank you for your answer. But I don't understand why $$\sum_{f\in [d]^n} A_{f}^*LA_{f}=\displaystyle\sum_{|\alpha|=n}{n\choose \alpha}\ {A^*}^{\alpha}LA^{\alpha}\ .$$ Also what do you mean by $ [d]^n$? Thank you for your help. $\endgroup$
    – Student
    Commented Feb 6, 2018 at 12:33
  • $\begingroup$ 1) Because it is the operator $(X_1+\dots+X_d)^n$ applied to $L$. 2) Yes, $[d]:=\{1,2,\dots, d\}$, and $[d]^n$ is the same as $F(n,d)$ $\endgroup$
    – Pietro Majer
    Commented Feb 6, 2018 at 14:35
  • $\begingroup$ If $X=(A_1^*LA_1,\cdots,A_d^*LA_d)$, why $X^\alpha={A^*}^\alpha LA^\alpha$? Thank you very much for your help $\endgroup$
    – Student
    Commented Feb 6, 2018 at 15:37
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    $\begingroup$ If $X_j$ is the operator that takes $L$ to $A_j^*L A_j$ then e.g. $X_2X_1$ takes $L$ to $A_2^*A_1^*L A_1 A_2$, and $X_d^{\alpha_d}X_{d-1}^{\alpha_{d-1}}\dots X_2^{\alpha_2}X_1^{\alpha_1}$ takes $L$ to ${A_d^*}^{\alpha_d}\dots {A_2^*}^{\alpha_2}{A_1^*}^{\alpha_1}L A_1^{\alpha_1}A_2^{\alpha_2}\dots A_d^{\alpha_d}$ $\endgroup$
    – Pietro Majer
    Commented Feb 6, 2018 at 17:27
  • $\begingroup$ Are you clear with the expansion of $\big(X_1+X_2+\dots+X_d\big)^n$ in a commuting ring as I wrote? (The multinomial coefficient ${n\choose \alpha}$ is ${n!\over\alpha!}:={n!\over\alpha_1!\alpha_2!\dots\alpha_d!}$). $\endgroup$
    – Pietro Majer
    Commented Feb 17, 2018 at 8:20

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