Suppose $\mathcal{H}$ is a Hilbert space, $A\in \mathcal{L}(\mathcal{H})$ is self adjoint, and define $U(t)=e^{itA}$. Show that $U(t)-I$ is compact for all $t\in \mathbb{R}$ if and only if $A$ is compact.
One direction is relatively straightforward: if $A$ is compact, write
$$
U(t)-I=A\big(itI+\frac{(it)^2}{2!}A+\frac{(it)^3}{3!}A^2+\dots\big),
$$
from which it is apparent that $U(t)-I$ is itself compact. However, I am having trouble with the other direction. One notes that as $A$ is self adjoint, $U(t)$ is unitary, but I am not sure where to head from here. I have also attempted to apply the spectral theorem, but to no avail. A hint would be appreciated!
Best Answer
Define the function $U:\mathbb{R}\to L(H)$ by $U(t)=e^{itA}$. Note that $U$ is differentiable, in the sense that the limit $\lim_{h\to 0}\frac{U(t+h)-U(t)}{h}$ exists in $L(H)$ for all $t$. Denote the derivative by $U'(t)$ as usual and note that $U'(t)=iAU(t)$ for all $t$. If any of the claims here are unclear to you, check out Murphy's book, p.11-12.
You have already handled one direction. For the other, assume that $U(t)-I$ is compact for all $t$. Let $t\in\mathbb{R}$ and $h>0$. Then $$\frac{U(t+h)-U(t)}{h}=\frac{(U(t+h)-I)-(U(t)-I)}{h}$$ which is a compact operator, as a linear combination of compacts. Since the compact operators form a closed set, we can take the limit as $h\to0^+$ and deduce that $U'(t)$ is also compact. This is true for all $t\in\mathbb{R}$.
But $U'(0)=iAU(0)=iA$, so $iA$ is compact and therefore $A$ is compact.