When is $\exp(-iHt)$ well-defined?

Question

If $H$ is a linear operator, what restrictions should be put on $H$ in order for $\exp(-iHt)$ to be well defined?

How do you define $\exp(-iHt)$ when $H$ is infinite-dimensional? (If it is possible)

Answer 1

There is a generation theorem for a $C^0$ semigroup that answers this question if you want $t \ge 0$.

You can make sense of $e^{tA}$ for $t \ge 0$ if $A : \mathcal{D}(A)\subseteq \mathcal{H}\rightarrow\mathcal{H}$ is a closed, densely-defined linear operator whose spectrum is contained in the closed left half-plane and if $$ \|(A-\lambda I)^{-1}\| \le \frac{1}{\Re\lambda},\;\; \Re\lambda > 0. $$ This can be weakened to only require that $\|(A-\lambda I)^{-n}\| \le M/(\Re\lambda)^n$ for all $n=1,2,3,\cdots$ and for some constant $M$. These conditions are essentially necessary in order to generate a $C^0$ semigroup $e^{tA}$ such that $\|e^{tA}\| \le M$ for all $t \ge 0$. (The $C^0$ requirement is that $\|e^{tA}x-x\|\rightarrow 0$ as $t\downarrow 0$ for all $x\in\mathcal{H}$.

You can recover the resolvent from the $C^0$ semigroup by the Laplace transform $$ \int_{0}^{\infty}e^{tA} e^{-st}x dt = (sI-A)^{-1}x,\;\; \Re\lambda > 0. $$ You can obtain the $C^0$ semigroup from the inverse Laplace transform of the resolvent $(A-sI)^{-1}$: $$ e^{tA}x = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}(sI-A)^{-1}xds,\;\; t > \gamma > 0. $$ The construction is not simple, but it does work.