Let $K$ and $H$ be Hilbert spaces. Let $\{e_i:i\in I\}$ be an orthogonal basis of $H$. Define $$ U_i:K\to K\overset{.}{\otimes} H: x\mapsto x\overset{.}{\otimes} e_i $$ Assume $T\in\mathcal{B}(K\overset{.}{\otimes} H)$ commutes with all of $U_i U_j^*$.
Then how do I show that $U_i^*TU_j=0$ for different $i,j\in I$ and $U_i^* T U_i=U_j^*TU_j$?
Since $\langle U_i(x),U_j(y)\rangle=\delta_{i,j}\langle x,y\rangle$ for all $x,y\in K$, then $$ U_i^* U_j=\delta_{ij}1_K\tag{1} $$ By assumption $TU_iU_j^*=U_iU_j^*T$. Multiply it from the right by $U_j$ and use $(1)$ to get $$ TU_i=U_iU_j^*TU_j\tag{2} $$ Then multiply it from the left by $U_i^*$ and use $(1)$ to get $$ U_i^*TU_i=U_j^*T U_j\tag{3} $$ What was desired.
From $(1)$ and $(2)$ for $i\neq j$ we also get $$ U^*_j T U_i=U^*_j (T U_i)=U^*_j (U_iU_j^*TU_j)=(U^*_j U_i)U_j^*TU_j=0\cdot U_j^*TU_j=0\tag{4} $$ What was desired.