The derivative operator in an interval is given in many expositions of self-adjoint extensions of symmetric operators as an illustration. At first, the solution of making a self-adjoint extension is usually given by changing the domain of the symmetric operator from $\phi(0) = \phi(1) = 0$ to $\phi(0) = \phi(1)$, so that the domain of the operator and its adjoint are made identical.
Usually what follows is that this same symmetric operator is discussed where deficiency subspaces are computed, and an isometry is used upon them, which extends the Cayley transform of the original symmetric operator to a unitary operator whose inverse Cayley transform is self adjoint.
(Wikipedia touches on both. See the included specifics from Wikipedia, below.)
It seems to me then, that in the former approach, the operator is unchanged, but the domain of the operator and its adjoint are changed to be identical. In the latter case, though, the operator is changed too: The operator is no longer $A$, but rather $A \oplus A_α$ (in the notation of the second Wikipedia paragraph.)
First of all, I can't see how the operator $A \oplus A_α$ relates to the basic differentiation operator. I don't think the operator $A \oplus A_α$ can even be expressed in "closed-form." I think they are not really related, which brings me to the question:
- Are these two solutions to self-adjoint extension related in any way other than being ways to extend the original symmetric operator?
- Does either solution offer a larger domain than the other approach? (It seems that the latter approach, with the Cayley transform has a smaller domain.)
- The Cayley transform method is somewhat procedural, where one can simply apply the algorithm and get the result, but is there any procedural method that alters the domain $\phi(0) = \phi(1) = 0$ to $\phi(0) = \phi(1)$, or is that purely from a stroke of genius?
A symmetric operator that is not essentially self-adjoint We first consider the Hilbert space $L^2[0, 1]$ and the differential operator
$$ \phi \mapsto \frac{1}{i} \phi'$$
defined on the space of continuously differentiable complex-valued functions on [0,1], satisfying the boundary conditions $$\phi(0) = \phi(1) = 0$$
Then $D$ is a symmetric operator as can be shown by integration by parts. The spaces $N_+$, $N_-$ (defined below) are given respectively by the distributional solutions to the equation
$$-i u' = i u $$
$$-i u' = -i u$$
which are in $L^2[0, 1]$. One can show that each one of these solution spaces is 1-dimensional, generated by the functions $x →e^{−x}$ and $x → e^x$ respectively. This shows that $D$ is not essentially self-adjoint, but does have self-adjoint extensions. These self-adjoint extensions are parametrized by the space of unitary mappings $N_+ → N_−$, which in this case happens to be the unit circle $\mathbb T$.
In this case, the failure of essential self-adjointness is due to an "incorrect" choice of boundary conditions in the definition of the domain of $D$. Since $D$ is a first-order operator, only one boundary condition is needed to ensure that $D$ is symmetric. If we replaced the boundary conditions given above by the single boundary condition $$\phi(0) = \phi(1),$$
then $D$ would still be symmetric and would now, in fact, be essentially self-adjoint. This change of boundary conditions gives one particular essentially self-adjoint extension of $D$. Other essentially self-adjoint extensions come from imposing boundary conditions of the form $\phi(1) = e^{i\theta}\phi(0)$.
This simple example illustrates a general fact about self-adjoint extensions of symmetric differential operators $P$ on an open set $M$. They are determined by the unitary maps between the eigenvalue spaces
$$ N_± = \{ u \in L^2(M): P_{dist} u = ± i u\}$$where $P_{dist}$ is the distributional extension of $P$.
In other places, this same problem is discussed where deficiency subspaces are computed, and an isometry is used upon them to extends the Cayley transform of the operator. For example, https://en.wikipedia.org/wiki/Extensions_of_symmetric_operators#An_example
Consider the Hilbert space $L^2[0,1]$. On the subspace of absolutely continuous function that vanish on the boundary, define the operator $A$ by
$$A f = i \frac{d}{dx} f.$$
Integration by parts shows $A$ is symmetric. Its adjoint $A^*$ is the same operator with Dom($A^*$) being the absolutely continuous functions with no boundary condition. We will see that extending $A$ amounts to modifying the boundary conditions, thereby enlarging Dom($A$) and reducing Dom($A^*$), until the two coincide.
Direct calculation shows that $K_+$ and $K_-$ are one-dimensional subspaces given by
$$K_+ = \operatorname{span}\{\phi_+ = a \cdot e^x \}$$
and
$$K_- = \operatorname{span}\{\phi_- = a \cdot e^{-x} \}$$
where $a$ is a normalizing constant. So the self-adjoint extensions of $A$ are parametrized by the unit circle in the complex plane, $ \{{ \lvert a \rvert =1 \}} $. For each unitary $U_α$ : $K_- → K_+$, defined by $U_α(φ_-) = αφ_+$, there corresponds an extension $A_α$ with domain
$$\operatorname{Dom}(A_{\alpha}) = \{ f + \beta (\alpha \phi_{-} - \phi_+) | f \in \operatorname{Dom}(A) , \; \beta \in \mathbb{C} \}.$$
If $f ∈ \operatorname{Dom}(A_α)$, then $f$ is absolutely continuous and
$$\lvert\frac{f(0)}{f(1)}\rvert = \left|\frac{e\alpha -1}{\alpha - e}\right| = 1.$$