[proofplan] We pass to the spectral representation of the self-adjoint operator $\mathcal H$. In that representation, $\mathcal H$ is multiplication by a real measurable function and the unitary evolution $e^{-it\mathcal H}$ is multiplication by a complex phase of modulus $1$. The modulus-one phase preserves the weighted $L^2$ condition defining the operator domain, and it cancels from the quadratic form representing the energy. [/proofplan]

[step:Represent the Hamiltonian and the propagator as multiplication operators] By the [spectral theorem for self-adjoint operators](/theorems/6911) (citing a result not yet in the wiki: Spectral Theorem for [Self-Adjoint Operators](/page/Self-Adjoint%20Operators)), there exist a [measure space](/page/Measure%20Space) $(X,\mathcal A,\mu)$, a real-valued measurable function $h:X\to\mathbb R$, and a unitary map \begin{align*} U:H\to L^2(X,\mathcal A,\mu) \end{align*} such that $\mathcal H=U^{-1}M_hU$, where the multiplication operator \begin{align*} M_h:D(M_h)\subset L^2(X,\mathcal A,\mu)\to L^2(X,\mathcal A,\mu) \end{align*} has domain \begin{align*} D(M_h)=\{f\in L^2(X,\mathcal A,\mu):hf\in L^2(X,\mathcal A,\mu)\} \end{align*} and satisfies $M_h f=hf$ for $f\in D(M_h)$. The same functional calculus gives, for each $t\in\mathbb R$, \begin{align*} Ue^{-it\mathcal H}U^{-1}=M_{\theta_t}, \end{align*} where $\theta_t:X\to\mathbb C$ is the measurable function $\theta_t(x)=e^{-ith(x)}$. Since $h$ is real-valued, $|\theta_t(x)|=1$ for every $x\in X$. [/step]

[step:Use the modulus-one phase to preserve the operator domain]Fix $t\in\mathbb R$. Define $f_0\in L^2(X,\mathcal A,\mu)$ by $f_0=U\psi_0$. Since $\psi_0\in D(\mathcal H)$ and $\mathcal H=U^{-1}M_hU$, we have $f_0\in D(M_h)$, equivalently $hf_0\in L^2(X,\mathcal A,\mu)$. Now \begin{align*} U\psi(t)=Ue^{-it\mathcal H}\psi_0=M_{\theta_t}f_0=\theta_t f_0. \end{align*} Because $|\theta_t|=1$, we have \begin{align*} |h\theta_t f_0|^2=|hf_0|^2 \end{align*} pointwise on $X$. Therefore $h\theta_t f_0\in L^2(X,\mathcal A,\mu)$, so $\theta_t f_0\in D(M_h)$. Since $D(\mathcal H)=U^{-1}D(M_h)$, it follows that $\psi(t)\in D(\mathcal H)$.[/step]

[guided]Fix a time $t\in\mathbb R$. The domain question becomes transparent after applying the unitary spectral transform. Define \begin{align*} f_0=U\psi_0. \end{align*} This is an element of $L^2(X,\mathcal A,\mu)$ because $U:H\to L^2(X,\mathcal A,\mu)$ is unitary. More is true: since $\psi_0\in D(\mathcal H)$ and $\mathcal H=U^{-1}M_hU$, the vector $f_0$ belongs to the domain of $M_h$. By the definition of $D(M_h)$, this means exactly that \begin{align*} hf_0\in L^2(X,\mathcal A,\mu). \end{align*} The evolved vector is obtained by multiplying $f_0$ by the phase $\theta_t$. Indeed, the functional calculus identity $Ue^{-it\mathcal H}U^{-1}=M_{\theta_t}$ gives \begin{align*} U\psi(t)=Ue^{-it\mathcal H}\psi_0=M_{\theta_t}f_0=\theta_t f_0. \end{align*} To prove that $\psi(t)\in D(\mathcal H)$, it is enough to prove that $U\psi(t)\in D(M_h)$, because $D(\mathcal H)=U^{-1}D(M_h)$. Thus we must check that $h\theta_t f_0\in L^2(X,\mathcal A,\mu)$. Since $h$ is real-valued, the complex exponential $\theta_t(x)=e^{-ith(x)}$ has modulus $1$ for every $x\in X$. Hence \begin{align*} |h(x)\theta_t(x)f_0(x)|^2=|h(x)f_0(x)|^2 \end{align*} for every $x\in X$ where the functions are represented. Therefore \begin{align*} \int_X |h\theta_t f_0|^2\,d\mu(x)=\int_X |hf_0|^2\,d\mu(x)<\infty. \end{align*} So $\theta_t f_0\in D(M_h)$, and consequently $\psi(t)\in D(\mathcal H)$.[/guided]

[step:Compute the energy in the spectral representation and cancel the phase] Since $\psi(t)\in D(\mathcal H)$, the energy [inner product](/page/Inner%20Product) is well-defined. Using $\mathcal H=U^{-1}M_hU$, unitarity of $U$, and $U\psi(t)=\theta_t f_0$, we obtain \begin{align*} (\mathcal H\psi(t),\psi(t))_H=(M_h\theta_t f_0,\theta_t f_0)_{L^2(X,\mathcal A,\mu)}. \end{align*} By the definition of the $L^2$ inner product, linear in the first argument, \begin{align*} (M_h\theta_t f_0,\theta_t f_0)_{L^2(X,\mathcal A,\mu)}=\int_X h\theta_t f_0\,\overline{\theta_t f_0}\,d\mu(x). \end{align*} Since $|\theta_t|^2=1$, this becomes \begin{align*} \int_X h\theta_t f_0\,\overline{\theta_t f_0}\,d\mu(x)=\int_X h|f_0|^2\,d\mu(x). \end{align*} Again using unitarity of $U$ and $\mathcal H=U^{-1}M_hU$, \begin{align*} \int_X h|f_0|^2\,d\mu(x)=(M_hf_0,f_0)_{L^2(X,\mathcal A,\mu)}=(\mathcal H\psi_0,\psi_0)_H. \end{align*} Thus \begin{align*} (\mathcal H\psi(t),\psi(t))_H=(\mathcal H\psi_0,\psi_0)_H. \end{align*} Since $t\in\mathbb R$ was arbitrary, the equality holds for every $t\in\mathbb R$. [/step]