[proofplan] We define the infinitesimal generator $A$ by differentiating the orbit $t \mapsto U(t)\psi$ at $t=0$. Smooth time-averages of orbit vectors give a dense supply of differentiable vectors, so $A$ is densely defined. Laplace-transform resolvent formulas show that $\lambda I_H-A$ is onto for every real $\lambda \neq 0$, which upgrades skew-symmetry to skew-adjointness. Then $\mathcal H:=iA$ is self-adjoint, and the [spectral theorem for self-adjoint operators](/theorems/6911) identifies the original group with $e^{-it\mathcal H}$. [/proofplan]

[step:Define the infinitesimal generator on differentiable orbit vectors] Define the domain \begin{align*} D(A):=\left\{\psi \in H:\lim_{t \to 0}\frac{U(t)\psi-\psi}{t}\text{ exists in }H\right\}. \end{align*} Define the operator $A:D(A)\to H$ by \begin{align*} A\psi:=\lim_{t \to 0}\frac{U(t)\psi-\psi}{t}. \end{align*} For $\psi \in D(A)$ and $s \in \mathbb{R}$, the group law gives \begin{align*} \frac{U(t)U(s)\psi-U(s)\psi}{t} = U(s)\frac{U(t)\psi-\psi}{t}. \end{align*} Since $U(s):H \to H$ is bounded and linear, taking $t \to 0$ gives $U(s)\psi \in D(A)$ and \begin{align*} A U(s)\psi=U(s)A\psi. \end{align*} Thus $D(A)$ is invariant under every $U(s)$, and $A$ commutes with the group on its domain. [/step]

[step:Construct a dense subspace of differentiable vectors by time averaging]Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $\varphi \in C_c^\infty(\mathbb{R})$ and $\psi \in H$. Define the averaged vector \begin{align*} \psi_\varphi:=\int_{\mathbb{R}}\varphi(t)U(t)\psi\,d\mathcal{L}^1(t), \end{align*} where the integral is the Bochner integral in $H$. The map $t \mapsto \varphi(t)U(t)\psi$ is continuous with compact support, hence Bochner integrable. For $h \neq 0$, using the group law and the substitution $r=t+h$, which preserves one-dimensional Lebesgue measure $d\mathcal{L}^1(r)=d\mathcal{L}^1(t)$, we obtain \begin{align*} \frac{U(h)\psi_\varphi-\psi_\varphi}{h} = \int_{\mathbb{R}}\frac{\varphi(t-h)-\varphi(t)}{h}U(t)\psi\,d\mathcal{L}^1(t). \end{align*} Since $\varphi \in C_c^\infty(\mathbb{R})$, the difference quotients converge uniformly to $-\varphi'$ and are supported in a fixed compact set for all sufficiently small $h$. Therefore the norm convergence of the Bochner integrals follows from [uniform convergence](/page/Uniform%20Convergence) of the scalar coefficient and boundedness of $\|U(t)\psi\|_H=\|\psi\|_H$. Hence $\psi_\varphi \in D(A)$ and \begin{align*} A\psi_\varphi = -\int_{\mathbb{R}}\varphi'(t)U(t)\psi\,d\mathcal{L}^1(t). \end{align*} It remains to see that these vectors are dense. Choose $\eta \in C_c^\infty(\mathbb{R})$ with $\eta \geq 0$ and \begin{align*} \int_{\mathbb{R}}\eta(t)\,d\mathcal{L}^1(t)=1. \end{align*} For $\varepsilon>0$, define $\eta_\varepsilon:\mathbb{R}\to\mathbb{R}$ by \begin{align*} \eta_\varepsilon(t):=\varepsilon^{-1}\eta(t/\varepsilon). \end{align*} Then \begin{align*} \psi_{\eta_\varepsilon}-\psi = \int_{\mathbb{R}}\eta_\varepsilon(t)(U(t)\psi-\psi)\,d\mathcal{L}^1(t). \end{align*} The strong continuity of $U$ at $0$ implies that this norm tends to $0$ as $\varepsilon \downarrow 0$. Hence $D(A)$ is dense in $H$.[/step]

[guided]The point of the averaging construction is to manufacture differentiability in the group parameter. A general orbit $t \mapsto U(t)\psi$ is only assumed continuous, so differentiating it directly may fail. But after averaging against a smooth compactly supported scalar function, the differentiability can be transferred onto that scalar function. Let $\varphi \in C_c^\infty(\mathbb{R})$ and $\psi \in H$, and define \begin{align*} \psi_\varphi:=\int_{\mathbb{R}}\varphi(t)U(t)\psi\,d\mathcal{L}^1(t). \end{align*} This Bochner integral is well-defined because $t \mapsto U(t)\psi$ is continuous, $\varphi$ is continuous with compact support, and $\|U(t)\psi\|_H=\|\psi\|_H$ for every $t \in \mathbb{R}$. We now compute the difference quotient. For $h \neq 0$, \begin{align*} U(h)\psi_\varphi = \int_{\mathbb{R}}\varphi(t)U(h+t)\psi\,d\mathcal{L}^1(t). \end{align*} Apply the substitution $r=t+h$. The transformation $t \mapsto r=t+h$ preserves one-dimensional Lebesgue measure, so $d\mathcal{L}^1(r)=d\mathcal{L}^1(t)$, and the integration domain remains all of $\mathbb{R}$. Thus \begin{align*} U(h)\psi_\varphi = \int_{\mathbb{R}}\varphi(r-h)U(r)\psi\,d\mathcal{L}^1(r). \end{align*} Renaming $r$ as $t$ gives \begin{align*} \frac{U(h)\psi_\varphi-\psi_\varphi}{h} = \int_{\mathbb{R}}\frac{\varphi(t-h)-\varphi(t)}{h}U(t)\psi\,d\mathcal{L}^1(t). \end{align*} Because $\varphi$ is smooth with compact support, the scalar difference quotients \begin{align*} t \mapsto \frac{\varphi(t-h)-\varphi(t)}{h} \end{align*} converge uniformly to $-\varphi'(t)$ as $h \to 0$, and for $|h|$ sufficiently small their supports lie inside one fixed compact subset of $\mathbb{R}$. Since $\|U(t)\psi\|_H=\|\psi\|_H$, the norm of the error is bounded by the uniform scalar error times $\|\psi\|_H$ times the finite Lebesgue measure of that compact set. Therefore the Bochner integrals converge in $H$, and \begin{align*} A\psi_\varphi = -\int_{\mathbb{R}}\varphi'(t)U(t)\psi\,d\mathcal{L}^1(t). \end{align*} So every such averaged vector is in $D(A)$. It remains to explain why these differentiable averaged vectors are dense. Choose $\eta \in C_c^\infty(\mathbb{R})$ with $\eta \geq 0$ and \begin{align*} \int_{\mathbb{R}}\eta(t)\,d\mathcal{L}^1(t)=1. \end{align*} For $\varepsilon>0$, define the rescaled smooth function $\eta_\varepsilon:\mathbb{R}\to\mathbb{R}$ by \begin{align*} \eta_\varepsilon(t):=\varepsilon^{-1}\eta(t/\varepsilon). \end{align*} Then $\eta_\varepsilon$ is supported in a shrinking neighbourhood of $0$ as $\varepsilon \downarrow 0$, and \begin{align*} \psi_{\eta_\varepsilon}-\psi= \int_{\mathbb{R}}\eta_\varepsilon(t)(U(t)\psi-\psi)\,d\mathcal{L}^1(t). \end{align*} To prove convergence in $H$, fix $\delta>0$. Strong continuity at $0$ gives $\rho>0$ such that $\|U(t)\psi-\psi\|_H<\delta$ whenever $|t|<\rho$. For all sufficiently small $\varepsilon$, the support of $\eta_\varepsilon$ lies in $(-\rho,\rho)$, and therefore \begin{align*} \|\psi_{\eta_\varepsilon}-\psi\|_H\leq \int_{\mathbb{R}}\eta_\varepsilon(t)\|U(t)\psi-\psi\|_H\,d\mathcal{L}^1(t) \leq \delta\int_{\mathbb{R}}\eta_\varepsilon(t)\,d\mathcal{L}^1(t)=\delta. \end{align*} Since $\delta>0$ was arbitrary, $\psi_{\eta_\varepsilon}\to\psi$ in $H$. Each $\psi_{\eta_\varepsilon}$ belongs to $D(A)$ by the differentiability computation above, so $D(A)$ is dense in $H$.[/guided]

[step:Prove that the generator is skew-symmetric] Throughout the proof, the [Hilbert space](/page/Hilbert%20Space) [inner product](/page/Inner%20Product) $(\cdot,\cdot)_H$ is linear in the first variable and conjugate-linear in the second variable. Let $\psi,\phi \in D(A)$. Since each $U(t)$ is unitary, \begin{align*} (U(t)\psi,U(t)\phi)_H=(\psi,\phi)_H \end{align*} for every $t \in \mathbb{R}$. Hence \begin{align*} \left(\frac{U(t)\psi-\psi}{t},U(t)\phi\right)_H + \left(\psi,\frac{U(t)\phi-\phi}{t}\right)_H = 0. \end{align*} Taking $t \to 0$, using strong continuity of $U(t)\phi \to \phi$ and the defining limits for $A\psi$ and $A\phi$, gives \begin{align*} (A\psi,\phi)_H+(\psi,A\phi)_H=0. \end{align*} Thus $A$ is skew-symmetric on the dense domain $D(A)$. [/step]

[step:Identify the real resolvents by Laplace transform formulas] Let $\mathcal{L}(H)$ denote the [Banach space](/page/Banach%20Space) of bounded linear operators from $H$ to $H$. Let $\lambda>0$. Define the [bounded linear operator](/page/Bounded%20Linear%20Operator) $R_\lambda \in \mathcal{L}(H)$ by \begin{align*} R_\lambda\psi:=\int_0^\infty e^{-\lambda t}U(t)\psi\,d\mathcal{L}^1(t). \end{align*} The integral is Bochner integrable because \begin{align*} \int_0^\infty e^{-\lambda t}\|U(t)\psi\|_H\,d\mathcal{L}^1(t) = \|\psi\|_H\int_0^\infty e^{-\lambda t}\,d\mathcal{L}^1(t) = \frac{1}{\lambda}\|\psi\|_H. \end{align*} Thus $\|R_\lambda\|_{\mathcal{L}(H)}\leq \lambda^{-1}$. We will repeatedly use the following endpoint-average estimate for Bochner integrals. If $g:[0,a]\to H$ is continuous at $0$, then \begin{align*} \left\|\frac{1}{h}\int_0^h g(r)\,d\mathcal{L}^1(r)-g(0)\right\|_H \leq \sup_{0\leq r\leq h}\|g(r)-g(0)\|_H \end{align*} for $0<h\leq a$, and the right-hand side tends to $0$ as $h\downarrow 0$. The identical estimate on $[-a,0]$ follows by replacing $[0,h]$ with $[-h,0]$. We also use that if $g:\mathbb{R}\to H$ is Bochner integrable, then the absolute continuity of the Bochner integral gives $\int_E g\,d\mathcal{L}^1\to 0$ in $H$ whenever $\mathcal{L}^1(E)\to 0$ along measurable sets contained in a fixed finite-measure region. For $h>0$, the group law and the substitution $r=t+h$, which preserves $\mathcal{L}^1$, give \begin{align*} \frac{U(h)R_\lambda\psi-R_\lambda\psi}{h} = \frac{e^{\lambda h}-1}{h}\int_h^\infty e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r) - \frac{1}{h}\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} As $h \downarrow 0$, the first term converges to $\lambda R_\lambda\psi$: the scalar factor converges to $\lambda$, and the missing interval contribution $\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r)$ has norm at most $h\|\psi\|_H$. The second term converges to $\psi$ by the continuity of averages at the endpoint. Indeed, \begin{align*} \left\|\frac{1}{h}\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r)-\psi\right\|_H \leq \sup_{0\leq r\leq h}\|e^{-\lambda r}U(r)\psi-\psi\|_H. \end{align*} The right-hand side tends to $0$ as $h\downarrow 0$ because $r\mapsto e^{-\lambda r}U(r)\psi$ is continuous at $0$. For negative increments, let $k=-h>0$. Using the substitution $r=t-k$, which again preserves $\mathcal{L}^1$, we obtain \begin{align*} \frac{U(-k)R_\lambda\psi-R_\lambda\psi}{-k} = \frac{1-e^{-\lambda k}}{k}R_\lambda\psi - \frac{e^{-\lambda k}}{k}\int_{-k}^0 e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} As $k \downarrow 0$, the first term converges to $\lambda R_\lambda\psi$. The second term converges to $\psi$ by the same endpoint-average argument: the map $r\mapsto e^{-\lambda r}U(r)\psi$ is continuous at $0$, and therefore \begin{align*} \left\|\frac{1}{k}\int_{-k}^0 e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r)-\psi\right\|_H \leq \sup_{-k\leq r\leq 0}\|e^{-\lambda r}U(r)\psi-\psi\|_H \end{align*} tends to $0$. Therefore the two-sided limit exists, so $R_\lambda\psi \in D(A)$ and \begin{align*} A R_\lambda\psi=\lambda R_\lambda\psi-\psi. \end{align*} Therefore \begin{align*} (\lambda I_H-A)R_\lambda\psi=\psi. \end{align*} So $\lambda I_H-A$ is surjective. For $\lambda<0$, define \begin{align*} R_\lambda\psi:=-\int_{-\infty}^0 e^{-\lambda t}U(t)\psi\,d\mathcal{L}^1(t). \end{align*} The integral is Bochner integrable because $\lambda<0$ makes $e^{-\lambda t}$ integrable on $(-\infty,0]$ and $\|U(t)\psi\|_H=\|\psi\|_H$. For $h>0$, using the substitution $r=t+h$, which preserves one-dimensional Lebesgue measure and changes the interval $(-\infty,0]$ into $(-\infty,h]$, we obtain \begin{align*} \frac{U(h)R_\lambda\psi-R_\lambda\psi}{h} = \frac{e^{\lambda h}-1}{h}R_\lambda\psi - \frac{e^{\lambda h}}{h}\int_0^h e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} As $h\downarrow 0$, the first term converges to $\lambda R_\lambda\psi$, and the second term converges to $\psi$ by the endpoint-average argument applied to the continuous map $r\mapsto e^{-\lambda r}U(r)\psi$ at $0$. For negative increments, let $k=-h>0$. Using the substitution $r=t-k$, which preserves one-dimensional Lebesgue measure and changes $(-\infty,0]$ into $(-\infty,-k]$, we get \begin{align*} \frac{U(-k)R_\lambda\psi-R_\lambda\psi}{-k} = \frac{e^{-\lambda k}-1}{k}\int_{-\infty}^{-k} e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r) - \frac{1}{k}\int_{-k}^0 e^{-\lambda r}U(r)\psi\,d\mathcal{L}^1(r). \end{align*} The first term converges to $\lambda R_\lambda\psi$, because the truncated integral converges in $H$ to $-R_\lambda\psi$ and the scalar factor converges to $-\lambda$. The second term converges to $\psi$ by the same endpoint-average argument. Hence $R_\lambda\psi\in D(A)$ and \begin{align*} A R_\lambda\psi=\lambda R_\lambda\psi-\psi. \end{align*} Therefore \begin{align*} (\lambda I_H-A)R_\lambda\psi=\psi. \end{align*} Thus $\lambda I_H-A$ is surjective for every real $\lambda \neq 0$. [/step]

[step:Show that the skew-symmetric generator is skew-adjoint] We prove that $A^*=-A$. Since $A$ is densely defined and skew-symmetric, we have $A \subset -A^*$. Let $\eta \in D(A^*)$. We show $\eta \in D(A)$ and $A^*\eta=-A\eta$. Fix $\lambda>0$. Since $\lambda I_H+A$ is surjective by the already proved surjectivity of $\mu I_H-A$ with $\mu=-\lambda$, choose $\xi \in D(A)$ such that \begin{align*} (\lambda I_H+A)\xi=(\lambda I_H-A^*)\eta. \end{align*} For every $\phi \in D(A)$, the definition of the adjoint gives \begin{align*} ((\lambda I_H-A)\phi,\eta-\xi)_H = (\phi,(\lambda I_H-A^*)\eta)_H-(\phi,(\lambda I_H-A^*)\xi)_H. \end{align*} Because $\xi\in D(A)$ and $A\subset -A^*$, we have $A^*\xi=-A\xi$, so \begin{align*} (\lambda I_H-A^*)\xi=(\lambda I_H+A)\xi. \end{align*} The defining choice of $\xi$ therefore gives \begin{align*} ((\lambda I_H-A)\phi,\eta-\xi)_H=0 \end{align*} for every $\phi\in D(A)$. Because $\lambda I_H-A$ is surjective, its range is all of $H$, so $\eta-\xi=0$. Hence $\eta=\xi \in D(A)$. Substituting $\eta=\xi$ into \begin{align*} (\lambda I_H+A)\xi=(\lambda I_H-A^*)\eta \end{align*} gives $A\eta=-A^*\eta$, or equivalently \begin{align*} A^*\eta=-A\eta. \end{align*} Thus $D(A^*)\subset D(A)$ and $A^*=-A$. Therefore $A$ is skew-adjoint. [/step]

[step:Convert the skew-adjoint generator into a self-adjoint operator] Define the operator $\mathcal H:D(A)\to H$ by \begin{align*} \mathcal H\psi:=iA\psi. \end{align*} Since $A^*=-A$, we have \begin{align*} \mathcal H^*=(iA)^*=-iA^*=-i(-A)=iA=\mathcal H. \end{align*} Thus $\mathcal H$ is self-adjoint, with $D(\mathcal H)=D(A)$. By the definition of $A$, \begin{align*} \lim_{t \to 0}\frac{U(t)\psi-\psi}{t}=A\psi=-i\mathcal H\psi \end{align*} for every $\psi \in D(\mathcal H)$. [/step]

[step:Recover the original unitary group from the self-adjoint generator] By the [Spectral Theorem for Normal Operators](/theorems/2695), applied in its self-adjoint projection-valued-measure form to the [self-adjoint operator](/page/Self-Adjoint%20Operators) $\mathcal H$, there is a projection-valued measure $E$ on the Borel subsets of $\mathbb{R}$ such that \begin{align*} \mathcal H=\int_{\mathbb{R}}\lambda\,dE(\lambda), \end{align*} and the functional calculus for self-adjoint operators defines a strongly continuous unitary group $(V(t))_{t \in \mathbb{R}}$ by \begin{align*} V(t):=e^{-it\mathcal H} = \int_{\mathbb{R}}e^{-it\lambda}\,dE(\lambda). \end{align*} For $\psi \in D(\mathcal H)$, the functional calculus gives $V(t)\psi \in D(\mathcal H)$ for every $t \in \mathbb{R}$, because multiplication by $e^{-it\lambda}$ preserves the integrability condition defining $D(\mathcal H)$. It also gives \begin{align*} \lim_{t \to 0}\frac{V(t)\psi-\psi}{t}=-i\mathcal H\psi=A\psi. \end{align*} Indeed, for each $\psi\in H$, define the finite positive Borel measure $\mu_\psi$ on $\mathbb{R}$ by $\mu_\psi(B):=(E(B)\psi,\psi)_H$ for every Borel set $B\subset\mathbb{R}$. In the spectral representation, the difference quotient is the scalar multiplier \begin{align*} \lambda \mapsto \frac{e^{-it\lambda}-1}{t}, \end{align*} which converges pointwise to $-i\lambda$. If $\psi\in D(\mathcal H)$, then $\int_{\mathbb{R}}\lambda^2\,d\mu_\psi(\lambda)<\infty$, and the elementary bound \begin{align*} \left|\frac{e^{-it\lambda}-1}{t}\right|\leq |\lambda| \end{align*} gives convergence in $L^2(\mathbb{R},\mu_\psi)$ by the [Dominated Convergence Theorem](/theorems/4), applied to the squared absolute values and dominated by $\lambda^2$. Thus $U(t)$ and $V(t)$ are strongly continuous unitary groups with the same generator $A$. To prove equality, fix $\psi \in D(A)$ and $t \in \mathbb{R}$. Define the map $F:[0,t]\to H$ by \begin{align*} F(s):=V(t-s)U(s)\psi \end{align*} if $t\geq 0$; for $t<0$ the same argument is applied on the interval $[t,0]$. Since $D(A)=D(\mathcal H)$ is invariant under both groups, $A$ commutes with $U(s)$ on $D(A)$, and the functional calculus gives $AV(r)\zeta=V(r)A\zeta$ for every $\zeta \in D(A)$ and $r \in \mathbb{R}$. We now verify the differentiability used in the product rule. For $\zeta\in D(A)$ and $s\in\mathbb{R}$, \begin{align*} \lim_{h\to 0}\frac{U(s+h)\zeta-U(s)\zeta}{h}=U(s)A\zeta=A U(s)\zeta. \end{align*} The same computation for $V$ gives the derivative $\frac{d}{dr}V(r)\zeta=A V(r)\zeta$ on $D(A)$. Therefore $F$ is differentiable by the product rule for strongly differentiable Hilbert-space-valued maps and \begin{align*} F'(s)=V(t-s)A U(s)\psi - V(t-s)A U(s)\psi=0. \end{align*} Hence $F$ is constant. Evaluating at $s=0$ and $s=t$ gives \begin{align*} V(t)\psi=U(t)\psi. \end{align*} Since $D(A)$ is dense in $H$ and $U(t),V(t):H\to H$ are bounded linear operators, equality extends to all $\psi \in H$. Therefore \begin{align*} U(t)=e^{-it\mathcal H} \end{align*} for every $t \in \mathbb{R}$. [/step]

[step:Prove uniqueness of the self-adjoint generator] Suppose $\mathcal K:D(\mathcal K)\subset H\to H$ is another self-adjoint operator satisfying \begin{align*} U(t)=e^{-it\mathcal K} \end{align*} for every $t \in \mathbb{R}$. By the functional calculus for self-adjoint operators, for every $\psi \in D(\mathcal K)$, \begin{align*} \lim_{t \to 0}\frac{U(t)\psi-\psi}{t} = \lim_{t \to 0}\frac{e^{-it\mathcal K}\psi-\psi}{t} = -i\mathcal K\psi. \end{align*} Hence $D(\mathcal K)\subset D(A)$ and $A\psi=-i\mathcal K\psi$ on $D(\mathcal K)$, so $\mathcal K\subset iA=\mathcal H$. Both $\mathcal K$ and $\mathcal H$ are self-adjoint. A self-adjoint operator has no proper self-adjoint extension, because equality with its adjoint fixes both the action and the domain. Therefore $\mathcal K=\mathcal H$. This proves existence, the stated domain characterization, the derivative identity, and uniqueness. [/step]