Examples of skew adjoint differential operators

Question

I just need some references which studies examples of skew adjoint differential operators generating unitary strongly continuous groups of operators, and its applications to partial differential equations.

The example I know is the differential operator defined on the hilbert space $H=L^2(\mathbb{R})$ by $$Af=f'$$, which has as domain $$D(A)=\{f \in L^2(\mathbb{R}), absolutely \ continuous, \ with \ f'\in L^2(\mathbb{R}) \}.$$

This operator generates a strongly continuous unitary group: $$(U(t)f)=f(t+s).$$ By unitary I mean $U(t)^{-1}=U(t)^*$. By a Stone's Theorem, this implies that $A$ must be skew adjoint.

Answer 1

You can take any self-adjoint operator and multiply it by $i$. Example: $i\Delta$ generates the Schrödinger equation for a free particle (the potential $V$ is identically zero). The wave equation $u_{tt}=c^2u_{xx}$ can also be interpreted in this way, by considering it as evolution of $(u,cu_x)$ in phase space: the generating operator is $\begin{pmatrix} 0 & c\frac{d}{dx} \\ c\frac{d}{dx} & 0\end{pmatrix}$, which is skew-adjoint.

Reference: Mathematical Methods in Quantum Mechanics by Gerald Teschl: very readable and free to download. Or pretty much any PDE book with "Hamiltonian" or "quantum mechanics" in it.

Answer 2

If you have any densely-defined selfadjoint linear operator $A$ on a complex Hilbert space $X$, then $e^{itA}$ is a unitary semigroup, which is really $e^{tB}$ where $B^{\star}=-B$.

More generally, suppose that $U : [0,\infty)\rightarrow \mathscr{L}(X)$ is an isometric $C^{0}$ semigroup, meaning that $$ \begin{align} & 1.\;\; U(t)U(t')=U(t+t'),\;\;\; t, t' \ge 0;\\ & 2.\;\; U(0)=I;\\ & 3.\;\; \|U(t)x\|=\|x\|,\;\; t \ge 0;\\ & 4.\;\; \lim_{t\downarrow 0}U(t)x=x,\;\; x \in X; \end{align} $$ It turns out that these assumptions are enough to guarantee that the following is a dense linear subspace of $X$: $$ \mathcal{D}(A)=\left\{ x \in X\; :\; Ax=\lim_{h\downarrow 0}\frac{1}{h}(U(h)-I)x\mbox{ exists}\right\} $$ Automatically, $U(t)\mathcal{D}(A)\subseteq\mathcal{D}(A)$ for all $t \ge 0$, and the right derivative of $U(t)x$ is $\frac{d}{dt}U(t)x= AU(t)x=U(t)Ax$ for all $x \in\mathcal{D}(A)$. Note that property (3) implies that the right derivative also satisfies the following for all $t \ge 0$ and $x \in \mathcal{D}(A)$: $$ \frac{d}{dt}\|U(t)x\|^{2}=\frac{d}{dt}(U(t)x,U(t)x)=(U(t)Ax,U(t)x)+(U(t)x,U(t)Ax)=0. $$ In particular, setting $t=0$ and applying $U(0)=I$ gives $$ (Ax,x)+(x,Ax)=0,\;\;\; x \in \mathcal{D}(A). $$ By the polarization identity for complex Hilbert spaces, the above gives $$ (Ax,y)=-(x,Ay),\;\;\; x,y \in \mathcal{D}(A). $$ So $A$ is antisymmetric because of property (3). In fact, property (3) is equivalent to the antisymmetry of $A$. Equivalently, $iA$ is symmetric, i.e., $((iA)x,y)=(x,(iA)y)$ for all $x,y \in X$. If $A$ is bounded, then $iA$ will be self-adjoint. If $A$ is not bounded, it is possible that $(iA)^{\star} \ne (iA)$ because the adjoint $(iA)^{\star}$ may have a larger domain, even though the two agree on the smaller domain $\mathcal{D}(A)$. This is sometimes written as $(iA)\prec (iA)^{\star}$ to mean that the operator on the right is a proper extension of the one on the left.

If you assume that $U(t)$ is unitary for all $t \ge 0$ (which is (3)+surjective), then $(iA)=(iA)^{\star}$. Conversely, if $(iA)=(iA)^{\star}$, then $U(t)$ is unitary for all $t \ge 0$.