This is a generation theorem in the setting of Hilbert spaces that links self-adjoint operators to unitary semigroups. It simply states that the Cauchy problem \[ i\frac{d}{dt}\psi(t) = H\psi(t) \] with self-adjoint but possibly unbounded Hamiltonian $H$ and initial state $\psi(0)=\psi_0\in \mathcal{H}$ has a (semi)group solution for all $t\in\mathbb{R}$ like given in the proof. We thus have another reason for self-adjointness, at least for the Hamiltonian of a (closed) system. The evolution semigroup is now written as $U(t)$ and defines the exponential map $\exp(-iHt):=U(t)$.

Proof. Take the so-called Yosida approximation $H_\lambda = -i\lambda - \lambda^2 (i \lambda - H)^{-1}$ ($\lambda>0$ for $t>0$) is a bounded operator by the resolvent lemma from before which yields $\|(i\lambda - H)^{-1}\| \leq |\lambda|^{-1}$. We show that $H_\lambda \rightarrow H$ strongly on $D(H)$ for $|\lambda|\rightarrow \infty$. A simple calculation shows $(i \lambda - H) H_\lambda = i\lambda H$ and thus $H_\lambda = i\lambda (i\lambda - H)^{-1} H$ on $D(H)$. Take $\psi \in D(H)$ then \begin{align*} \|(i\lambda (i\lambda - H)^{-1} - \mathrm{id})\psi\| &= \|i\lambda^{-1}(\lambda^2(i\lambda - H)^{-1} +i\lambda)\psi\| \\ &= |\lambda|^{-1} \|H_\lambda \psi\| \\ &\leq |\lambda|^{-1} \|i\lambda (i\lambda - H)^{-1}\| \cdot \|H \psi\| \\ &\leq |\lambda|^{-1} \|H\psi\| \longrightarrow 0, \end{align*} when $|\lambda| \rightarrow \infty$ and therefore $i\lambda (i\lambda - H)^{-1} \rightarrow \mathrm{id}$ strongly on $D(H)$. Because $(i\lambda - H)^{-1}$ is uniformly bounded with respect to $\lambda$ and $D(H)$ dense in $\mathcal{H}$ we get convergence of the whole Hilbert space which establishes $H_\lambda \rightarrow H$ strongly on $D(H)$.

Take now $\exp(-iH_\lambda t)$ as the ordinary exponential map (infinite sum) which converges because of boundedness of $H_\lambda$. Taking the original definition of $H_\lambda$ one shows that it is bounded by 1. It remains to show that there is a strong limit $\exp(-iH_\lambda t) \rightarrow \exp(-iHt)$ for $|\lambda| \rightarrow \infty$. This is achieved by showing that the sequence has the Cauchy property. To make use of the usual properties of the exponential map, the elements of the family $\{\exp(-i H_\lambda t)\}_{\lambda>0}$ have to commutate. This follows directly from the fact that $\{H_\lambda\}_{\lambda>0}$ is a commuting family. We form \[ (\exp(-i H_\lambda t) - \exp(-i H_\mu t)) \psi = \int_0^t \frac{d}{ds} ( \exp(-i H_\lambda s) \exp(-i H_\mu (t-s)) \psi ) \,ds. \] and estimate its norm by \[ \int_0^t \| \exp(-i H_\lambda s) \exp(-i H_\mu (t-s)) \| \cdot \|(H_\lambda-H_\mu)\psi\| \,ds \leq |t| \|(H_\lambda-H_\mu)\psi\|. \] For $|\lambda|,|\mu| \rightarrow \infty$ this converges to 0 for all $\psi \in D(H)$ and the thus formed Cauchy sequence defines $\exp(-i H t)$ uniquely on $D(H)$ which extends again to the whole $\mathcal{H}$ because of uniform boundedness. For initial values in $D(H)$ it can be shown that $\exp(-i H t)\psi_0$ is indeed a solution to the Cauchy problem.

Finally we show that the new-formed evolution operator is unitary, which also yields uniqueness of the so-formed solution as well as conservation of propabilities. $H_\lambda$ is not even self-adjoint but the related $H_{(\lambda)} = \frac{1}{2}(H_\lambda + H_{-\lambda})$ that converges to $H$ just as well is. This means \[ \langle \exp(-i H t)\varphi,\exp(-i H t)\psi \rangle = \lim_{|\lambda| \rightarrow \infty} \langle \exp(-i H_{(\lambda)} t)\varphi,\exp(-i H_{(\lambda)} t)\psi \rangle = \langle \varphi,\psi \rangle. \Box \]