Let $A$ be a C$^*$-algebra. A pre-Hilbert $A$-module $H$ is a right $A$ module with a $A$-valued inner product (which is linear in the second variable and conjugate linear in the first variable) such that

1.$\langle \xi, \beta \cdot T \rangle_{A} = \langle \xi, \beta\rangle_{A}T$,

2.$\langle \xi, \xi\rangle_{A} \geq 0$, and $\langle \xi, \xi\rangle_{A}=0$ implies $\xi=0$,

where $\xi, \beta \in H$ and $T \in A$. If $H$ is complete with respect to the norm $\|\langle \xi, \xi\rangle_{A}\|^{1/2}$, then $H$ is called a Hilbert $A$-module.

If $H$ is a pre-Hilbert $A$-module, let $\overline{H}$ be the completion of $H$ with respect to the norm $\|\langle \xi, \xi\rangle_{A}\|^{1/2}$. Let $B(H)$ be the set of bounded adjointable mapping from $H$ to $H$. It is clear that each $T \in B(H)$ extends to a element in $B(\overline{H})$ (the bounded adjointable mapping from $\overline{H}$ to $\overline{H}$). Thus, we can identify $B(H)$ as a subalgebra in $B(\overline{H})$. My question is if $B(H)$ is dense in $B(\overline{H})$?