The theory as I know it
Let $\mathcal{H}$ be a Hilbert space and $(A, D(A))$ a self-adjoint operator acting on it. The Spectral Theorem (cfr. Reed & Simon Methods of modern mathematical physics, vol.I, §VIII.6) asserts the existence of a unique projection-valued measure (PVM) $P$ s.t.
$$A= \int_{-\infty}^\infty \lambda dP_{\lambda},$$
thus allowing us to define functions of $A$ and especially, for real $t$,
$$e^{i t A}=\int_{-\infty}^{\infty}e^{i t \lambda} dP_{\lambda}.$$
Turns out (Reed & Simon, §VIII.7) that $(e^{itA})_{t \in \mathbb{R}}$ is a strongly continuous unitary group with generator $(A, D(A))$.
In particular, since the operator $-\Delta$ is self-adjoint on $H^2(\mathbb{R}^d)$, we can define solution of the free Schrödinger equation with inital datum $u_0 \in H^2(\mathbb{R}^d)$
$$\text{(S)}\ \begin{cases} i u_t= \Delta u \\ u(0, x)=u_0(x) \end{cases}$$
the function $u(t,x)=e^{i t \Delta}u_0(x)$.
Problem
Is the requirement $u_0 \in H^2(\mathbb{R}^d)$ really necessary? Would $u_0 \in H^1(\mathbb{R}^d)$ suffice? In fact I read in some course notes: "since $-\Delta$ is self-adjoint on $L^2(\mathbb{R}^d)$ with form domain $H^1(\mathbb{R}^d)$, we can define the Schrödinger group $e^{i t \Delta}$ [...] and so $u(t, x)=e^{i t \Delta}u_0(x)$ is solution of (S) for $u_0 \in H^1(\mathbb{R}^d)$."
The author then stops explaining. How could he weaken the regularity request on $u_0$ so much?
Well, this is sloppy language. You can solve the equation only for $H^2$. However, you can define the exponential function and the unitary group as you do, and call $u(t,\cdot) = e^{it\Delta}u_0(\cdot)$ the generalized (mild) solution for all $u_0\in L^2$. Why not?
Now if multiply the original equation by a nice function with compact support, then you can integrate by part once, and what you get you can write down also for $H^1$ functions. They remain invariant under the group, and are usually called the weak solutions.