Proposition: Symmetry of an operator is not integration

The symmetric operator

has deficiency indices $(1,0)$. It has no self-adjoint extension and cannot be the restriction of the generator of a unitary one-parameter group.

Proof. The deficiency equation $(S^{\ast }-\mathrm{i})f=0$ has the square-integrable solution $f(x)=C{e}^{-x}$, giving index one. The equation $(S^{\ast }+\mathrm{i})f=0$ has only $f(x)=C{e}^x$, which is not square-integrable unless $C=0$. The unequal deficiency indices exclude a self-adjoint extension. A unitary one-parameter group has a self-adjoint generator, so such a group is impossible.