[proofplan] We use the [spectral theorem for self-adjoint operators](/theorems/6911) to represent $e^{-it\mathcal H}$ as multiplication by the scalar function $\lambda\mapsto e^{-it\lambda}$ in the spectral calculus. Since this multiplier has absolute value $1$ at every spectral parameter $\lambda\in\mathbb R$, the spectral integral computing the squared norm is unchanged. Taking square roots gives conservation of the Hilbert-space norm for every initial vector $\psi_0\in H$, without requiring $\psi_0\in D(\mathcal H)$. [/proofplan] [step:Represent the evolution through the spectral measure of $\mathcal H$] Let $(\cdot,\cdot)_H:H\times H\to \mathbb C$ denote the [inner product](/page/Inner%20Product) on the [Hilbert space](/page/Hilbert%20Space) $H$, linear in the first argument. Since $\mathcal H:D(\mathcal H)\subset H\to H$ is [self-adjoint](/page/Self-Adjoint%20Operators), the spectral theorem for self-adjoint operators, together with its [Borel functional calculus](/theorems/2696), applies. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, let $\mathcal L(H)$ denote the [Banach space](/page/Banach%20Space) of bounded linear operators from $H$ to $H$, and let $I_H:H\to H$ denote the identity operator on $H$. Let \begin{align*} E:\mathcal B(\mathbb R)\to \mathcal L(H) \end{align*} denote the projection-valued spectral measure of $\mathcal H$. For the fixed vector $\psi_0\in H$, define the finite positive Borel measure \begin{align*} \mu_{\psi_0}:\mathcal B(\mathbb R)\to [0,\infty),\qquad B\mapsto (E(B)\psi_0,\psi_0)_H. \end{align*} The spectral functional calculus defines, for each $t\in\mathbb R$, the bounded operator \begin{align*} U(t):H\to H \end{align*} by $U(t)(\eta)=e^{-it\mathcal H}\eta$ for $\eta\in H$, as the spectral multiplier associated with the bounded Borel function \begin{align*} m_t:\mathbb R\to \mathbb C \end{align*} given by $m_t(\lambda)=e^{-it\lambda}$ for $\lambda\in\mathbb R$. Thus $\psi(t)=U(t)\psi_0$. [guided] Let $(\cdot,\cdot)_H:H\times H\to \mathbb C$ denote the inner product on the Hilbert space $H$, linear in the first argument. The point of using the spectral theorem for self-adjoint operators is that a [self-adjoint](/page/Self-Adjoint%20Operators) operator can be treated as multiplication by the real spectral variable after passing to its spectral resolution, and its Borel functional calculus gives the corresponding operators $m(\mathcal H)$ for bounded Borel multipliers $m:\mathbb R\to\mathbb C$. The hypothesis needed for this representation is exactly self-adjointness of $\mathcal H$ on the complex Hilbert space $H$. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, let $\mathcal L(H)$ denote the Banach space of bounded linear operators from $H$ to $H$, and let $I_H:H\to H$ denote the identity operator on $H$. Let \begin{align*} E:\mathcal B(\mathbb R)\to \mathcal L(H) \end{align*} be the projection-valued spectral measure of $\mathcal H$ supplied by the spectral theorem for self-adjoint operators. Fix the initial vector $\psi_0\in H$. From $E$ and $\psi_0$ we form the scalar spectral measure \begin{align*} \mu_{\psi_0}:\mathcal B(\mathbb R)\to [0,\infty),\qquad B\mapsto (E(B)\psi_0,\psi_0)_H. \end{align*} This measure is finite because $E(\mathbb R)=I_H$, so \begin{align*} \mu_{\psi_0}(\mathbb R)=(E(\mathbb R)\psi_0,\psi_0)_H=(\psi_0,\psi_0)_H=\|\psi_0\|_H^2. \end{align*} Now fix a time $t\in\mathbb R$. The relevant scalar multiplier is the bounded Borel function \begin{align*} m_t:\mathbb R\to \mathbb C \end{align*} given by $m_t(\lambda)=e^{-it\lambda}$ for $\lambda\in\mathbb R$. The spectral functional calculus defines the bounded operator \begin{align*} U(t):H\to H \end{align*} by $U(t)(\eta)=e^{-it\mathcal H}\eta$ for $\eta\in H$. The solution in the theorem is therefore $\psi(t)=U(t)\psi_0$. This formulation is important because it makes sense for every $\psi_0\in H$, even when $\psi_0$ is not in the operator domain $D(\mathcal H)$. Why does this representation prove conservation of norm? The spectral functional calculus converts the Hilbert-space norm of $U(t)\psi_0$ into an integral against the scalar spectral measure $\mu_{\psi_0}$. Applying the norm identity in the spectral functional calculus to the bounded Borel function $m_t$ gives \begin{align*} \|\psi(t)\|_H^2=\|U(t)\psi_0\|_H^2=\int_{\mathbb R}|m_t(\lambda)|^2\,d\mu_{\psi_0}(\lambda). \end{align*} For each $\lambda\in\mathbb R$, the number $e^{-it\lambda}$ lies on the complex unit circle, and therefore \begin{align*} |m_t(\lambda)|^2=|e^{-it\lambda}|^2=1. \end{align*} Substituting this pointwise identity into the spectral integral yields \begin{align*} \|\psi(t)\|_H^2=\int_{\mathbb R}1\,d\mu_{\psi_0}(\lambda)=\mu_{\psi_0}(\mathbb R). \end{align*} Using the already computed value of the total mass of $\mu_{\psi_0}$, we obtain \begin{align*} \|\psi(t)\|_H^2=\|\psi_0\|_H^2. \end{align*} Both sides are squares of non-negative [real numbers](/page/Real%20Numbers), so taking the non-negative square root gives \begin{align*} \|\psi(t)\|_H=\|\psi_0\|_H. \end{align*} Since $t\in\mathbb R$ was arbitrary, this proves norm conservation for every time $t\in\mathbb R$. [/guided] [/step] [step:Compute the norm using the modulus of the spectral multiplier] Fix $t\in\mathbb R$. By the norm identity in the spectral functional calculus from the spectral theorem for self-adjoint operators, applied to the bounded Borel function $m_t$, we have \begin{align*} \|\psi(t)\|_H^2=\|U(t)\psi_0\|_H^2=\int_{\mathbb R}|m_t(\lambda)|^2\,d\mu_{\psi_0}(\lambda). \end{align*} For every $\lambda\in\mathbb R$, \begin{align*} |m_t(\lambda)|^2=|e^{-it\lambda}|^2=1. \end{align*} Therefore \begin{align*} \|\psi(t)\|_H^2=\int_{\mathbb R}1\,d\mu_{\psi_0}(\lambda)=\mu_{\psi_0}(\mathbb R). \end{align*} Using $E(\mathbb R)=I_H$, we get \begin{align*} \mu_{\psi_0}(\mathbb R)=(I_H\psi_0,\psi_0)_H=\|\psi_0\|_H^2. \end{align*} Hence \begin{align*} \|\psi(t)\|_H^2=\|\psi_0\|_H^2. \end{align*} [/step] [step:Take square roots to obtain conservation for every time] Both $\|\psi(t)\|_H$ and $\|\psi_0\|_H$ are non-negative real numbers. Taking square roots in \begin{align*} \|\psi(t)\|_H^2=\|\psi_0\|_H^2 \end{align*} gives \begin{align*} \|\psi(t)\|_H=\|\psi_0\|_H. \end{align*} Since $t\in\mathbb R$ was arbitrary, the identity holds for every $t\in\mathbb R$. [/step]