[proofplan] Use the [Spectral Theorem for Self-Adjoint Operators](/theorems/6911) in its projection-valued-measure form, together with its bounded [Borel functional calculus](/theorems/2696), to define $e^{-it\mathcal H}$. The functions $\lambda\mapsto e^{-it\lambda}$ have modulus one and multiply according to addition of times, so the resulting operators form a unitary group. Strong continuity and differentiability on $D(\mathcal H)$ are proved by reducing norm estimates to scalar spectral integrals, using the spectral domain characterization of $D(\mathcal H)$, and applying the [Dominated Convergence Theorem](/theorems/4). Finally, any strong solution becomes constant after multiplication by the inverse unitary group, which gives uniqueness; the generator-domain part of the [Stone Theorem For Unitary Groups](/theorems/6975) then identifies this group as the unique strongly continuous unitary group with generator $-i\mathcal H$. [/proofplan]

[step:Construct the unitary group by spectral functional calculus] Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, and let $\mathcal L(H)$ denote the Banach algebra of bounded linear operators $H\to H$. By the Spectral Theorem for [Self-Adjoint Operators](/page/Self-Adjoint%20Operators), let $E:\mathcal B(\mathbb R)\to\mathcal L(H)$ denote the spectral projection-valued measure associated to $\mathcal H$. Let $I_H\in\mathcal L(H)$ denote the identity operator on $H$. For each $t\in\mathbb R$, define the bounded Borel map \begin{align*} m_t:\mathbb R\to\mathbb C,\qquad \lambda\mapsto e^{-it\lambda}. \end{align*} Since $|m_t(\lambda)|=1$ for every $\lambda\in\mathbb R$, the bounded Borel functional calculus supplied by the spectral theorem defines a bounded operator \begin{align*} U(t):=m_t(\mathcal H)=\int_{\mathbb R} e^{-it\lambda}\,dE(\lambda)\in\mathcal L(H). \end{align*} The functional calculus is multiplicative on bounded Borel functions. Since $m_{t+s}=m_t m_s$ and $m_0=1$ for all $s,t\in\mathbb R$, we obtain \begin{align*} U(t+s)=U(t)U(s) \end{align*} and \begin{align*} U(0)=I_H. \end{align*} Also $\overline{m_t}=m_{-t}$, so the adjoint relation in the functional calculus gives \begin{align*} U(t)^*=U(-t). \end{align*} Therefore \begin{align*} U(t)^*U(t)=U(-t)U(t)=U(0)=I_H. \end{align*} Similarly, \begin{align*} U(t)U(t)^*=U(t)U(-t)=U(0)=I_H. \end{align*} Thus $U(t)$ is unitary for every $t\in\mathbb R$, and $(U(t))_{t\in\mathbb R}$ is a one-parameter unitary group. [/step]

[step:Prove strong continuity from scalar spectral measures]Fix $\psi\in H$. Define the finite positive Borel measure \begin{align*} \mu_\psi:\mathcal B(\mathbb R)\to[0,\infty),\qquad A\mapsto \|E(A)\psi\|_H^2. \end{align*} By the isometry identity in the spectral theorem, \begin{align*} \|U(t)\psi-\psi\|_H^2=\int_{\mathbb R}|e^{-it\lambda}-1|^2\,d\mu_\psi(\lambda). \end{align*} For each fixed $\lambda\in\mathbb R$, the integrand satisfies $|e^{-it\lambda}-1|^2\to 0$ as $t\to 0$. Also \begin{align*} |e^{-it\lambda}-1|^2\le 4 \end{align*} for every $t,\lambda\in\mathbb R$, and $4$ is integrable with respect to $\mu_\psi$ because \begin{align*} \int_{\mathbb R}4\,d\mu_\psi(\lambda)=4\|\psi\|_H^2<\infty. \end{align*} By the [Dominated Convergence Theorem](/theorems/4), \begin{align*} \lim_{t\to 0}\|U(t)\psi-\psi\|_H^2=0. \end{align*} Hence $U(t)\psi\to\psi$ in $H$ as $t\to0$. Using the group identity, \begin{align*} \|U(t)\psi-U(t_0)\psi\|_H=\|U(t_0)(U(t-t_0)\psi-\psi)\|_H. \end{align*} Since $U(t_0)$ is unitary, \begin{align*} \|U(t_0)(U(t-t_0)\psi-\psi)\|_H=\|U(t-t_0)\psi-\psi\|_H. \end{align*} Therefore $t\mapsto U(t)\psi$ is continuous at every $t_0\in\mathbb R$. Thus $(U(t))_{t\in\mathbb R}$ is strongly continuous.[/step]

[guided]Fix a vector $\psi\in H$. To prove strong continuity, we must prove convergence in the Hilbert norm: \begin{align*} \|U(t)\psi-\psi\|_H\to0 \end{align*} as $t\to0$. The spectral theorem converts this Hilbert-space norm into a scalar integral. Define the finite positive Borel measure \begin{align*} \mu_\psi:\mathcal B(\mathbb R)\to[0,\infty),\qquad A\mapsto \|E(A)\psi\|_H^2. \end{align*} This measure is finite because $\mu_\psi(\mathbb R)=\|E(\mathbb R)\psi\|_H^2=\|\psi\|_H^2$. The spectral isometry identity gives \begin{align*} \|U(t)\psi-\psi\|_H^2=\int_{\mathbb R}|e^{-it\lambda}-1|^2\,d\mu_\psi(\lambda). \end{align*} Now the problem is reduced to a scalar dominated convergence argument. For each fixed spectral parameter $\lambda\in\mathbb R$, continuity of the exponential function gives \begin{align*} |e^{-it\lambda}-1|^2\to0 \end{align*} as $t\to0$. We also need an integrable dominating function independent of $t$. Since both $e^{-it\lambda}$ and $1$ have modulus one, \begin{align*} |e^{-it\lambda}-1|^2\le4. \end{align*} The constant function $4$ is integrable with respect to $\mu_\psi$ because \begin{align*} \int_{\mathbb R}4\,d\mu_\psi(\lambda)=4\|\psi\|_H^2<\infty. \end{align*} Therefore the [Dominated Convergence Theorem](/theorems/4) applies and yields \begin{align*} \lim_{t\to0}\|U(t)\psi-\psi\|_H^2=0. \end{align*} This proves continuity at $0$. Continuity at an arbitrary time $t_0\in\mathbb R$ follows from the group law and unitarity. First, the group law gives \begin{align*} \|U(t)\psi-U(t_0)\psi\|_H=\|U(t_0)(U(t-t_0)\psi-\psi)\|_H. \end{align*} Because $U(t_0)$ is unitary, it preserves the Hilbert norm, so \begin{align*} \|U(t_0)(U(t-t_0)\psi-\psi)\|_H=\|U(t-t_0)\psi-\psi\|_H. \end{align*} The final expression tends to $0$ as $t\to t_0$, so $t\mapsto U(t)\psi$ is continuous everywhere.[/guided]

[step:Differentiate the orbit for initial data in the operator domain]Assume $\psi_0\in D(\mathcal H)$. The spectral domain characterization for self-adjoint operators, obtained from the unbounded functional calculus in the Spectral Theorem for Self-Adjoint Operators applied to the coordinate function $\lambda\mapsto\lambda$, gives \begin{align*} \psi_0\in D(\mathcal H)\iff \int_{\mathbb R}\lambda^2\,d\mu_{\psi_0}(\lambda)<\infty. \end{align*} For $h\ne0$, the spectral isometry identity gives \begin{align*} \left\|\frac{U(h)\psi_0-\psi_0}{h}+i\mathcal H\psi_0\right\|_H^2=\int_{\mathbb R}\left|\frac{e^{-ih\lambda}-1}{h}+i\lambda\right|^2\,d\mu_{\psi_0}(\lambda). \end{align*} For each fixed $\lambda\in\mathbb R$, \begin{align*} \frac{e^{-ih\lambda}-1}{h}\to -i\lambda \end{align*} as $h\to0$. Also the mean value estimate for the scalar function $r\mapsto e^{-ir\lambda}$ gives \begin{align*} \left|\frac{e^{-ih\lambda}-1}{h}\right|\le |\lambda|, \end{align*} and hence \begin{align*} \left|\frac{e^{-ih\lambda}-1}{h}+i\lambda\right|^2\le4\lambda^2. \end{align*} Since $\lambda^2$ is integrable with respect to $\mu_{\psi_0}$, the [Dominated Convergence Theorem](/theorems/4) yields \begin{align*} \lim_{h\to0}\left\|\frac{U(h)\psi_0-\psi_0}{h}+i\mathcal H\psi_0\right\|_H=0. \end{align*} Therefore the orbit $\psi:\mathbb R\to H$, $t\mapsto U(t)\psi_0$, is differentiable at $0$ and \begin{align*} \psi'(0)=-i\mathcal H\psi_0. \end{align*} We next verify domain invariance, because $\mathcal H$ is generally unbounded. For $\psi_0\in D(\mathcal H)$, the scalar spectral measure of $U(t)\psi_0$ satisfies \begin{align*} \mu_{U(t)\psi_0}(A)=\int_A |e^{-it\lambda}|^2\,d\mu_{\psi_0}(\lambda)=\mu_{\psi_0}(A) \end{align*} for every Borel set $A\subset\mathbb R$. Therefore \begin{align*} \int_{\mathbb R}\lambda^2\,d\mu_{U(t)\psi_0}(\lambda)=\int_{\mathbb R}\lambda^2\,d\mu_{\psi_0}(\lambda)<\infty. \end{align*} By the spectral domain characterization, $U(t)\psi_0\in D(\mathcal H)$. On $D(\mathcal H)$, the unbounded functional calculus gives \begin{align*} \mathcal H U(t)\psi_0=U(t)\mathcal H\psi_0 \end{align*} because multiplication by $\lambda$ and multiplication by $e^{-it\lambda}$ commute pointwise and the preceding integrability estimate verifies that both sides are defined. Applying the differentiability at $0$ to the vector $\psi_0\in D(\mathcal H)$, and using the group law, gives \begin{align*} \psi'(t)=\lim_{h\to0}\frac{U(t+h)\psi_0-U(t)\psi_0}{h}. \end{align*} By the group law and boundedness of $U(t)$, \begin{align*} \lim_{h\to0}\frac{U(t+h)\psi_0-U(t)\psi_0}{h}=U(t)\lim_{h\to0}\frac{U(h)\psi_0-\psi_0}{h}. \end{align*} Using the derivative at $0$ and the commutation relation on $D(\mathcal H)$, \begin{align*} U(t)\lim_{h\to0}\frac{U(h)\psi_0-\psi_0}{h}=-iU(t)\mathcal H\psi_0=-i\mathcal H U(t)\psi_0. \end{align*} Thus \begin{align*} i\psi'(t)=\mathcal H\psi(t) \end{align*} for every $t\in\mathbb R$, with $\psi(0)=\psi_0$.[/step]

[guided]The delicate point is that $\mathcal H$ may be unbounded, so differentiability cannot be proved by treating $\mathcal H$ as an ordinary bounded matrix. We use the Spectral Theorem for Self-Adjoint Operators to reduce the difference quotient to scalar multiplication. Since $\psi_0\in D(\mathcal H)$, the spectral domain characterization obtained from the unbounded functional calculus for $\lambda\mapsto\lambda$ gives \begin{align*} \int_{\mathbb R}\lambda^2\,d\mu_{\psi_0}(\lambda)<\infty. \end{align*} For $h\ne0$, the spectral isometry identity gives \begin{align*} \left\|\frac{U(h)\psi_0-\psi_0}{h}+i\mathcal H\psi_0\right\|_H^2=\int_{\mathbb R}\left|\frac{e^{-ih\lambda}-1}{h}+i\lambda\right|^2\,d\mu_{\psi_0}(\lambda). \end{align*} For each fixed $\lambda\in\mathbb R$, the scalar derivative of $h\mapsto e^{-ih\lambda}$ at $0$ is $-i\lambda$, so \begin{align*} \frac{e^{-ih\lambda}-1}{h}\to -i\lambda \end{align*} as $h\to0$. To justify passing the limit through the spectral integral, we need a dominating function independent of $h$. The mean value estimate for $r\mapsto e^{-ir\lambda}$ gives \begin{align*} \left|\frac{e^{-ih\lambda}-1}{h}\right|\le |\lambda|. \end{align*} Hence \begin{align*} \left|\frac{e^{-ih\lambda}-1}{h}+i\lambda\right|^2\le4\lambda^2. \end{align*} The function $\lambda\mapsto4\lambda^2$ is integrable with respect to $\mu_{\psi_0}$ by the domain characterization, so the [Dominated Convergence Theorem](/theorems/4) gives \begin{align*} \lim_{h\to0}\left\|\frac{U(h)\psi_0-\psi_0}{h}+i\mathcal H\psi_0\right\|_H=0. \end{align*} Thus the orbit is differentiable at $0$ with derivative $-i\mathcal H\psi_0$. We also need to know that the orbit remains inside $D(\mathcal H)$. This is where the modulus-one property of the multiplier is used a second time. For every Borel set $A\subset\mathbb R$, the projection $E(A)$ is the functional-calculus operator associated to the indicator function $\mathbb{1}_A:\mathbb R\to\{0,1\}$. Since the functional calculus is multiplicative, $E(A)U(t)$ is the functional-calculus operator associated to $\mathbb{1}_A(\lambda)e^{-it\lambda}$. Applying the spectral isometry identity to this bounded Borel multiplier gives \begin{align*} \mu_{U(t)\psi_0}(A)=\|E(A)U(t)\psi_0\|_H^2=\int_{\mathbb R}\mathbb{1}_A(\lambda)|e^{-it\lambda}|^2\,d\mu_{\psi_0}(\lambda). \end{align*} Because $|e^{-it\lambda}|=1$ for every $\lambda\in\mathbb R$, this becomes \begin{align*} \mu_{U(t)\psi_0}(A)=\int_A 1\,d\mu_{\psi_0}(\lambda)=\mu_{\psi_0}(A). \end{align*} Therefore \begin{align*} \int_{\mathbb R}\lambda^2\,d\mu_{U(t)\psi_0}(\lambda)=\int_{\mathbb R}\lambda^2\,d\mu_{\psi_0}(\lambda)<\infty, \end{align*} so $U(t)\psi_0\in D(\mathcal H)$. On this domain, multiplication by $\lambda$ and by $e^{-it\lambda}$ commute in the spectral representation, giving \begin{align*} \mathcal H U(t)\psi_0=U(t)\mathcal H\psi_0. \end{align*} Finally, the group law gives \begin{align*} \frac{U(t+h)\psi_0-U(t)\psi_0}{h}=U(t)\frac{U(h)\psi_0-\psi_0}{h}. \end{align*} Since $U(t)$ is bounded, it may be passed through the norm limit. Hence \begin{align*} \psi'(t)=U(t)(-i\mathcal H\psi_0)=-i\mathcal H U(t)\psi_0. \end{align*} Equivalently, $i\psi'(t)=\mathcal H\psi(t)$ for every $t\in\mathbb R$, and $\psi(0)=\psi_0$ follows from $U(0)=I_H$.[/guided]

[step:Show uniqueness by conjugating any strong solution to a constant path] Let $\varphi:\mathbb R\to H$ be a strong solution with $\varphi(0)=\psi_0$. Define \begin{align*} \eta:\mathbb R\to H,\qquad t\mapsto U(-t)\varphi(t). \end{align*} Fix $t\in\mathbb R$. Since $\varphi(t)\in D(\mathcal H)$, the differentiability result already proved gives \begin{align*} \lim_{h\to0}\frac{U(-(t+h))\varphi(t)-U(-t)\varphi(t)}{h}=i\mathcal H U(-t)\varphi(t). \end{align*} We justify the derivative of $\eta$ by a difference quotient. For $h\ne0$, \begin{align*} \frac{\eta(t+h)-\eta(t)}{h}=\frac{U(-(t+h))\varphi(t+h)-U(-t)\varphi(t)}{h}. \end{align*} Insert and subtract $U(-(t+h))\varphi(t)$ to obtain \begin{align*} \frac{\eta(t+h)-\eta(t)}{h}=U(-(t+h))\frac{\varphi(t+h)-\varphi(t)}{h}+\frac{U(-(t+h))\varphi(t)-U(-t)\varphi(t)}{h}. \end{align*} Because $U(-(t+h))\to U(-t)$ strongly and $\varphi$ is differentiable in $H$, the first term converges in $H$ to $U(-t)\varphi'(t)$. The second term converges in $H$ to $i\mathcal H U(-t)\varphi(t)$ by the differentiability of the unitary orbit through the fixed vector $\varphi(t)$. Hence \begin{align*} \eta'(t)=i\mathcal H U(-t)\varphi(t)+U(-t)\varphi'(t). \end{align*} The commutation relation gives $\mathcal H U(-t)\varphi(t)=U(-t)\mathcal H\varphi(t)$, and the differential equation gives $\varphi'(t)=-i\mathcal H\varphi(t)$. Hence \begin{align*} \eta'(t)=iU(-t)\mathcal H\varphi(t)-iU(-t)\mathcal H\varphi(t)=0. \end{align*} Since $\eta:\mathbb R\to H$ is differentiable and $\eta'(t)=0$ for every $t\in\mathbb R$, the Banach-space zero-derivative-implies-constant theorem, equivalently the [fundamental theorem of calculus](/theorems/632) for $H$-valued $C^1$ maps on intervals, gives that $\eta$ is constant on $\mathbb R$. Since $\eta(0)=U(0)\varphi(0)=\psi_0$, we have \begin{align*} U(-t)\varphi(t)=\psi_0 \end{align*} for every $t\in\mathbb R$. Applying $U(t)$ to both sides gives \begin{align*} \varphi(t)=U(t)\psi_0=e^{-it\mathcal H}\psi_0. \end{align*} Thus the strong solution with initial value $\psi_0\in D(\mathcal H)$ is unique. It remains only to make explicit the uniqueness of the generated unitary group. Let $(V(t))_{t\in\mathbb R}$ be a strongly continuous unitary group on $H$ whose infinitesimal generator is $-i\mathcal H$. We invoke the generator-domain theorem for strongly continuous groups, as stated in the Stone Theorem For Unitary Groups, in the following standard form: if $(V(t))_{t\in\mathbb R}$ is a strongly continuous group with generator $A$, then every $\xi\in D(A)$ satisfies $V(t)\xi\in D(A)$ for every $t\in\mathbb R$, and the orbit $t\mapsto V(t)\xi$ is differentiable with derivative $V(t)A\xi=AV(t)\xi$. Applying this theorem with $A=-i\mathcal H$, whose domain is $D(A)=D(\mathcal H)$, gives that if $\xi\in D(\mathcal H)$, then $V(t)\xi\in D(\mathcal H)$ for every $t\in\mathbb R$ and \begin{align*} \frac{d}{dt}V(t)\xi=V(t)(-i\mathcal H\xi)=(-i\mathcal H)V(t)\xi. \end{align*} Define the comparison orbit $\phi:\mathbb R\to H$ by $\phi(t):=V(t)\xi$. Multiplying this identity by $i$ gives that this explicitly defined map $\phi$ is a strong solution of $i\phi'(t)=\mathcal H\phi(t)$ with initial value $\xi$. The uniqueness just proved gives $V(t)\xi=U(t)\xi$ for every $t\in\mathbb R$ and every $\xi\in D(\mathcal H)$. Since $\mathcal H$ is self-adjoint, $D(\mathcal H)$ is dense in $H$; since $U(t)$ and $V(t)$ are bounded unitary operators, equality on the dense subspace $D(\mathcal H)$ extends to equality on all of $H$. Therefore $V(t)=U(t)$ for every $t\in\mathbb R$. For arbitrary $\psi_0\in H$, the formula $t\mapsto U(t)\psi_0=e^{-it\mathcal H}\psi_0$ is therefore the unique orbit through $\psi_0$ of the strongly continuous unitary group generated by $\mathcal H$. [/step]