What is self-adjoint anyway? The adjoint $A^*$ of an operator $A$ (always densely defined on a Hilbert space $\mathcal{H}$) is given by all pairs $(y,z)$ (the graph of $A^*$) that obey \[ \langle y,Ax \rangle = \langle z,x \rangle \quad\forall x \in D(A) \] and thus it holds $z = A^*y$ and $y \in D(A^*)$. If $A=A^*$ on $D(A)$ the operator is called symmetric and one easily gets $D(A) \subseteq D(A^*)$. If further the domains coincide $D(A) = D(A^*)$ the operator is self-adjoint. For bounded operators we always have $D(A)=\mathcal{H}$ so the two notions are equivalent in that case.

Example. $i\partial_x$ with its natural domain $H^1([0,1]) = \{ f \in L^2([0,1]) \mid \partial_x f \in L^2([0,1]) \}$ (Sobolev_space#The_case_p_.3D_2) with all derivatives assumed weak. We study: \[ \langle f,i\partial_x g \rangle = i\int_0^1 \bar{f}\partial_x g \,dx = i \bar{f}(1)g(1)-i \bar{f}(0)g(0) - i\int_0^1 \partial_x\bar{f} g \,dx = i \bar{f}(1)g(1)-i \bar{f}(0)g(0) + \langle i\partial_x f,g \rangle. \]

The operator thus fails to be symmetric. On $H_0^1([0,1])$ instead it is symmetric, but having $g \in H_0^1([0,1])$ is enough for the boundary terms to vanish, so $f \in D((i\partial_x)^*) \supsetneq D(i\partial_x)$. To get a self-adjoint operator we must make $D(i\partial_x)$ larger which lets $D((i\partial_x)^*)$ shrink, until they match. A boundary condition $f(1)=\alpha f(0), \alpha \in \mathbb{C}$ yields \[ \bar{f}(1)g(1)-\bar{f}(0)g(0)=|\alpha|^2\bar{f}(0)g(0)-\bar{f}(0)g(0) \] that vanishes if $|\alpha|^2 = 1$. Any such choice of $\alpha$ gives a self-adjoint extension of the original operator but there is no unique such self-adjoint extension.

Physical note: Clearly different $\alpha$ describe different “phase jumps” at the boundary that is now actually the joint where the two endpoints of the interval $[0,1]$ are identified. Without periodicity (or an infinitely large configuration space $\mathbb{R}$) there cannot be a momentum eigenstate (it would hit the wall), which somehow is supposed to be connected to self-adjoint operators.