Domain of the quantum free Hamiltonian in 1D

Consider the quantum free Hamiltonian $H_0 =-\frac{d^2}{dx^2}$ (the Laplacian on the real line). I want to show that it is (essentially) self-adjoint in its domain of definition. The usual approach works with the Fourier transform and proceeds to show that $H_0$ is unitarily equivalent to a multiplication operator. The domain of definition is taken to be those $\psi\in L^2(\mathbb{R})$ such that $-\psi''\in L^2(\mathbb{R})$ in the weak (distributional) sense. My question is: is it possible to prove self-adjointness without resorting to Fourier and distributions, just by using Hilbert space techniques?

• Using the Cayley transform is one possible way to show that a symmetric operator is essentially self-adjoint. See Wikipedia and many books on functional analysis. – Urgje Mar 31 '14 at 8:36
• To apply results related to the Cayley transform of symmetric operators you need to show first that the domain is dense (I think this is included in the definition of `symmetric'). In fact, this is almost what I am asking for: once we get a densely defined domain for $H_0$ it is (relatively) easy to show self-adjointness. What I want to know is if it is possible to show this without using distributions... – Grimolatto Mar 31 '14 at 8:41
• Yes, a dense domain is required. This is not necessarily the case for a general symmetric operator. Your H_0 is symmetric with domain C_0, the infinitely differential compactly supported functions. Now note that C_0 is dense in L^2(R). – Urgje Mar 31 '14 at 9:09

Let's try to show that the closure of a "smaller" symmetric operator is the desired self-adjoint operator. (A self-adjoint operator is always closed.) Take $\Delta = H_0$ just on the domain of the test-functions $D(\Delta) = \mathcal{C}_0^\infty$ (dense in $L^2$) as a starting point. It is clearly symmetric as twice partial integration shows. Now every symmetric operator is closable, i.e. it has a smallest closed extension $\overline{\Delta}$ on some larger domain $D(\overline{\Delta}) \supset D(\Delta)$.
To construct the closure operator take all $x \in L^2$ s.t. a sequence $\{x_n\} \subset D(\Delta)$ that converges to $x$ (in $L^2$-norm) also has $\{\Delta x_n\}$ converging to some $y \in L^2$ (in $L^2$-norm). We set $\overline{\Delta}x=y$ and thus define the extension. (This is unique, i.e. does not depend on the selected sequence, because the operator is symmetric.)
To get the domain of the closure operator we need to collect all these $x$. Putting the two convergences together it says we take all $x$ that have correponding sequences $\{x_n\} \subset D(\Delta)$ converging to $x$ but w.r.t to the stronger norm $\|x\|_\Delta = \|x\|_2 + \|\Delta x\|_2$. (This is called the graph norm of $\Delta$.) But the closure of the test functions w.r.t. to this graph norm is just the Sobolev space $H^2$ of twice weakly (not distributionally) differentiable functions. So this is our new domain $D(\overline\Delta) = H^2$ and we suspect this is the right domain for a self-adjoint operator.
For self-adjointness we need $D(\overline\Delta) = D(\overline\Delta^*)$ so the domains of the operator and its adjoint must coincide. The domain of the adjoint is defined as $$D(\overline\Delta^*) = \{ x\in L^2 | \exists y \in L^2 \,\forall z \in D(\overline\Delta) : \langle \overline\Delta z,x \rangle = \langle z,y \rangle \}.$$
But this is just the definition for twice weakly differentiation with the additional condition that the second (weak) derivative $y$ is still in $L^2$ (because we are not using test functions $z$). So $x \in L^2$ and $y = \overline\Delta x \in L^2$ and thus $$D(\overline\Delta^*) = H^2 = D(\overline\Delta)$$ $\Box$