Hermiticity/self-adjointness of the Laplacian operator I'm trying to understand the concept of hermicity/self-adjointness but even after reading through some threads here, I just don't get it (first problem being that I don't understand the difference between the two, if there is any). But anyway, let's look e. g. at the one-dimensional eigenvalue problem of the Laplace operator $$\Delta f(x)=\frac{d^2}{dx^2} f(x)=cf(x)$$ where $c$ is a constant and the domain $\Omega$ for $x$ is e. g. $[x_0;x_1]$. From what I've read, the Laplace operator is self-adjoint/hermitian for boundary constraints $f(x_0)=f(x_1)=0$ (though I don't know why that is). Could someone expain why? (And maybe also showcase a set of BCs for which the Laplacian operator is not self-adjoint/hermitian?)
 A: The Laplacian is only partially defined on $L^2$ and it is unbounded. Usually, the Laplacian is defined on the dense subspace of compactly supported smooth functions and is extended to a self-adjoint closure (still partially defined only).
For a partially defined operator $A: D(A)\subseteq L^2 \to L^2$, it is self-adjoint by definition if and only if $D(A) = D(A^*)$ and for every $f,g\in D(A)$ it follows $$\int f Ag = \int Af g. $$
I will explain the symmetry for functions in one variable for sake of simplicity. Consider $\Delta: C_c^\infty(a, b) \subseteq L^2(a, b) \to L^2(a, b)$ defined by $\Delta f = f''$ and $f,g\in C_c^\infty(a, b)$. By integration by parts, it follows
$$ \langle f, \Delta g \rangle = \int_a^b f g'' = \underbrace{[f g']_a^b}_{=0} - \int_a^b f' g' = -\underbrace{[f' g]_a^b}_{=0} + \int_a^b f'' g = \langle \Delta f, g \rangle. $$
That is, $\Delta$ is symmetric, and thus, we have $D(\Delta) \subseteq D(\Delta^*)$. But, for example $D(\Delta^*)$ also contains $C^2(a,b)$ and is larger than $D(\Delta)$. Thus, $\Delta$ is not self-adjoint. However, as $-\Delta$ is a non-negative (or positive semidefinite), there are many ways to extend $-\Delta$ to a self-adjoint operator.
A: If $f,g$ are twice absolutely continuous on $[x_0,x_1]$ with $f,g,f'',g''\in L^2[x_0,x_1]$, then
$$
       \langle f'',g\rangle-\langle f,g''\rangle=\int_{x_0}^{x_1}f''\overline{g}-f\overline{g''}dx \\
   = \int_{x_0}^{x_1}\frac{d}{dx}(f'\overline{g}-f\overline{g'})dx \\
   = \left.\det\left(\begin{array}{cc}\overline{g(x)} & f(x) \\ \overline{g'(x)} & f'(x)\end{array}\right)\right|_{x_0}^{x_1}.
$$
$\Delta$ is symmetric if the determinants given above at $x_1,x_2$ are $0$ for all $f,g$ in the domain. This typically involves imposing endpoint conditions on $f,g$ such as
$$
                    \cos\alpha f(x_0)+\sin\alpha f'(x_0) = 0, \\
                    \cos\beta f(x_1)+\sin\beta f'(x_1) = 0.
$$
For $\alpha = 0 = \beta$, this amounts to imposing $f(x_0)=f(x_1)=0$. If $f,g$ satisfy these conditions, then $\langle f'',g\rangle = \langle f,g''\rangle$ for all $f,g$ satisfying $0$ endpoint conditions at $x=x_0,x_1$. But, as you can see, many other conditions give rise to a symmetric operator $\frac{d^2}{dx^2}$.
Symmetry does not imply self-adjointness of $Lf=f''$. However, symmetry is necessary. It turns out that $L_{\alpha,\beta}$ given by $Lf=f''$ with $f$ being twice absolutely continuous with $f''\in L^2$ is self-adjoint. There are periodic conditions that can imposed that will also result in self-adjoint operators. Examples of non-selfadjoint versions of $L$ include conditions
$$
               f(x_0)=f'(x_0)=f(x_1)=f'(x_1)=0.
$$
Then $L^*$ is the unrestricted operator $L$ (i.e., one with no endpoint conditions.) In particular $\langle Lf,g\rangle = \langle f,L^*g\rangle$ for all $f$ satisfying $f(x_0)=f'(x_0)=f(x_1)=f'(x_1)=0$ and $g$ unrestricted.
If $A : \mathcal{D}(A)\rightarrow\mathcal{H}$ is a linear operator on a dense subspace of a Hilbert space $\mathcal{H}$, then one way to define the adjoint $A^*$ is that the domain $\mathcal{D}(A^*)$ consists of all $y\in\mathcal{H}$ such that
$$
      |\langle Ax,y\rangle | \le M \|x\|,\;\;\; x\in\mathcal{H}.
$$
Then, $x\mapsto \langle Ax,y\rangle$ extends by continuity to a continuous linear function on $\mathcal{H}$ and, so, there exists a unique $z\in\mathcal{H}$ such that $\langle Ax,y\rangle = \langle x,z\rangle$ for all $x\in\mathcal{H}$. Define the adjoint operator $A^*$ on $y$ by $A^*y=z$. Then,
$$
              \langle Ax,y\rangle = \langle x,A^*y\rangle,\;\; x\in\mathcal{D}(A),\;y\in\mathcal{D}(A^*).
$$
$0$ is always in the domain of the adjoint with $A^*0=0$. In order to prove more about the domain of $A^*$, you can write the adjoint relation as an inner product condition in $\mathcal{H}\times\mathcal{H}$:
$$
         \left\langle \left[\begin{array}{c}Ax \\ x\end{array}\right],\left[\begin{array}{c}y\\-A^*y\end{array}\right]\right\rangle = 0
$$
Using this observation, it can be shown that, if $A$ is densely defined and closed (which means its graph is closed), then $A^*$ is closed and densely-defined. This is done by looking at orthogonal complements in $\mathcal{H}\times\mathcal{H}$.
