Unitary operator on dense set, Unique extension? given two Hilberspace $H_1$ and $H_2$. Let $V\subset H_1$ and $W\subset H_2$ be dense subspaces. 
Furthermore let $U: V \rightarrow W$ be an unitary operator.
I just want to know whether there is a unique extension $\tilde U$ of $U$ from $H_1$ to $H_2$.
 A: First of all, if a unitary extension exists, it has to be unique due to continuity and $V$ being dense. 
For each $x \in H_1$ and any sequence $(a_n(x))_n$ in $V$ converging to $x$ define $\widetilde{U} x = \lim\limits_{n \rightarrow \infty} U a_{n}(x)$. We have to check that this is well-defined. In fact, $(U a_{n}(x))_n$ is a Cauchy-sequence as $U$ is unitary (it suffices to assume $U$ to be bounded) and convergent as $H_2$ is complete. Moreover, as $U$ is unitary the definition is independent of the choice of $(a_n(x))_n$ (Here again, boundedness is sufficient). Now it is easy to show that $\widetilde{U}$ is indeed a linear operator $H_1 \rightarrow H_2$ extending $U$. 
Finally, we have to check that $\widetilde{U}$ is unitary.Note that the adjoint $\widetilde{U}^*$ of $\widetilde{U}$ extends the adjoint $U^*$ of $U$ and 
the compositions $\widetilde{U}^* \widetilde{U}$ and $\widetilde{U}\widetilde{U}^*$ are identities when restricted to $V$ and $W$, respectively. So we are done by continuity and $V$ and $W$ being dense.
A: Assuming you require $\tilde{U}$ to be continuous, yes, such an extension exists and is unique.  This is a consequence of the so-called BLT theorem.  $\tilde{U}$ will even be unitary.
If you don't require $\tilde{U}$ to be continuous the answer is no; there will in general be many linear extensions which are not continuous, but you will need the axiom of choice to "construct" them.
