Question about the kinetic energy operator The Kinetic Energy Operator is essentially self-adjoint. Under what circumstances does it have a unique extension?
 A: The "true" kinetic energy is the self-adjoint extension you are referring to.  As you know, the domain of an operator is integral to its definition.  The "formula" we want for the kinetic energy operator is (ignoring some constants) given by $-\nabla^2$, but we need to decide which domain it should act on so as to be self-adjoint.
A reasonable start is to define the operator $T_0:\psi\mapsto -\nabla^2\psi$ whose domain is $C^\infty_c(\mathbb R^n)$ - the compactly-supported smooth functions on $\mathbb R^n$.  This is a very nice domain to work with, but one can show that $T_0$ is merely essentially self-adjoint.  To get a self-adjoint operator, which must consider its closure $T:=T_0^{**}$; one can show that $T:\psi\mapsto -\nabla^2\psi$ with domain
$$H^2(\mathbb R^n)=\left\{ \psi \in L^2(\mathbb R^n)\ \bigg| \ \psi\text{ is twice weakly-differentiable and }\nabla^2\psi\in L^2(\mathbb R^n)\right\}$$
More generally, it's much easier to determine whether an operator is essentially self-adjoint than it is to determine whether or not it is self-adjoint.  Working in practice with an essentially self-adjoint operator is perfectly fine, as long as you bear in mind that its true self-adjoint extension has a larger domain.
