Is $Q_T^n(X)=[Q_T(X)]^n$?

Question

Let ${\bf T}=(T_1,T_2,\cdots,T_d) \in \mathcal{B}(\mathcal{H})^d$ be a commuting $d$-tuple of bounded linear operators on a Hilbert space $\mathcal{H}$ (i.e. $T_iT_j=T_jT_i$, for all $i,j\in\{1,\cdots,d\}$). Let us consider the operator \begin{eqnarray*} Q_T:\mathcal{B}(\mathcal{H})&\longrightarrow& \mathcal{B}(\mathcal{H})\\ X&\longmapsto& Q_T(X):=\sum_{j=1}^d T_j^*XT_j. \end{eqnarray*} I see this result in this paper (1) that $$Q_T^n(X)=\sum_{|\alpha|=n}\frac{n!}{\alpha!}{{\bf T}^{*}}^{\alpha}X{\bf T}^\alpha$$

Is $Q_T^n(X)=[Q_T(X)]^n$?

Answer 1

We have $Q_T^2=Q_T \circ Q_T$, $Q_T^3=Q_T^2 \circ Q_T$, and so on.

If we denote $Q_T^n$ by $A_n$, then

$Q_T^n(X) =A_n(X)$.

With the notation from above we have $[Q_T(X)]^n=(A_1(X))^n$, so in general:

$$Q_T^n(X) \ne [Q_T(X)]^n.$$