Spectral Theorem applied to $\frac{d^2}{dx^2}$

Question

By the spectral theorem we have for a self adjoint densely defined operator $A \colon D(A) \subset \mathcal{H} \to \mathcal{H}$ that there exists a uniquely determined projection valued measure $P_A$ so that
$$f(A) = \int_{\sigma(A)} f(\lambda) dP_A(\lambda)$$
for every Borel function $f$. Let $\mathcal{F}$ denote the Fourier transform.
We consider the operator $H_0$ with domain $D(H_0) = H^2(\mathbb{R}^n)$ and $H_0\psi = -\Delta \psi$, where $\Delta$ is the Laplace operator and $H^2(\mathbb{R}^n)$ is the Sobolev space defined by $\{f \in L^2(\mathbb{R}^n) \colon |p|^2 \mathcal{F}f(p) \in L^2(\mathbb{R}^n)\}$. It can be shown, see for example page 198 Theorem 7.17 in the linked book below, that $H_0$ is self adjoint. Hence an application of the spectral theorem to $H_0$ is justified and one can apparently verify
$$e^{-itH_0} \psi(x) = \mathcal{F}^{-1} \Big(e^{-itp^2}\mathcal{F} \psi(p)\Big)(x) \qquad (\star)$$
(this is from Gerald Teschl's book "Mathematical Methods in Quantum Mechanics" page 199 equation 7.38: https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe2.pdf). Does anyone know how formula $(\star)$ is derived?

Answer 1

Let $\mathcal{H}$ and $\tilde{\mathcal{H}}$ be two Hilbert spaces. Recall that a map $\mathcal{U} \colon \mathcal{H} \to \tilde{\mathcal{H}}$ is called a unitary isomorphism between $\mathcal{H}$ and $\tilde{\mathcal{H}}$ if $\mathcal{U}$ is linear, bijective and $\|\mathcal{U}\psi\| = \|\psi\|$ for all $\psi \in \mathcal{H}$. If $A \colon D(A) \subset \mathcal{H} \to \mathcal{H}$ and $\tilde{A} \colon D(\tilde{A}) \subset \tilde{\mathcal{H}} \to \tilde{\mathcal{H}}$ are two densely defined operators, then we say $A$ and $\tilde{A}$ are unitarily equivalent if there exists a unitary isomorphism $\mathcal{U} \colon \mathcal{H} \to \tilde{\mathcal{H}}$ so that $$\mathcal{U}A = \tilde{A}\mathcal{U} \qquad \mathcal{U}D(A) = D(\tilde{A})$$ Now note that if $A$ and $\tilde{A}$ are unitarily equivalent (by means of $\mathcal{U}$) as above, then we have $$d\mu_{\varphi,\psi} = d\tilde{\mu}_{\mathcal{U}\varphi,\mathcal{U}\psi}$$ for all $\varphi,\psi \in \mathcal{H}$, where $d\mu_{\varphi,\psi}$ is the corresponding spectral measure for $A$ with respect to the vectors $\varphi,\psi$ and $d\tilde{\mu}_{\mathcal{U}\varphi,\mathcal{U}\psi}$ is the corresponding spectral measure for $\tilde{A}$ with respect to the vectors $\mathcal{U}\varphi,\mathcal{U}\psi$. This can be seen as follows: Let $R_A, R_{\tilde{A}}$ be the respective resolvents, then $\mathcal{U}R_A(z)\mathcal{U}^{-1} = R_{\tilde{A}}(z)$ is easily seen. But then $$\int \frac{1}{\lambda-z} d\mu_{\varphi,\psi}(\lambda) = \langle \varphi \mid R_A(z) \psi\rangle = \langle \mathcal{U}\varphi \mid R_{\tilde{A}}(z)\mathcal{U}\psi\rangle = \int \frac{1}{\lambda-z} d\tilde{\mu}_{\mathcal{U}\varphi,\mathcal{U}\psi}(\lambda)$$ But the Borel transform of a Borel measure is uniquely determined and hence $d\mu_{\varphi,\psi} = d\tilde{\mu}_{\mathcal{U}\varphi,\mathcal{U}\psi}$, as wanted. But then we also $$\mathcal{U}f(A) = f(\tilde{A})\mathcal{U}$$ for every (bounded) Borel function $f$, since $$\langle \varphi \mid f(A)\psi\rangle = \int f(\lambda) d\mu_{\varphi, \psi}(\lambda) = \int f(\lambda)d\mu_{\mathcal{U}\varphi, \mathcal{U}\psi}(\lambda)= \langle \mathcal{U}\varphi \mid f(\tilde{A})\mathcal{U}\psi\rangle = \langle \varphi \mid \mathcal{U}^{-1}f(\tilde{A})\mathcal{U}\psi\rangle$$ holds for all $\varphi, \psi \in \mathcal{H}$.

Now the last ingredient in all this is that for $H_0 = -\Delta$ with $D(H_0) = H^2(\mathbb{R}^n)$ we have the unitary equivalence $$\mathcal{F}H_0\mathcal{F}^{-1} = |p|^2$$ where $|p|^2$ is the maximally defined multiplication operator with domain $\{\varphi \in L^2(\mathbb{R}^n) \mid |p|^2\varphi(p) \in L^2(\mathbb{R}^n)\}$. With that in our toolbox equation $(\star)$ is obvious, since for $f(z) = e^{-izt}$ we have $$e^{-itH_0} = f(H_0) = \mathcal{F}^{-1}f(|p|^2)\mathcal{F} = \mathcal{F}^{-1}e^{-it|p|^2}\mathcal{F}$$ as wanted.