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

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?

• I couldn't find your formula $(\star)$ on p.$210$ of the cited reference. Assuming $\ \hat{\psi}\$ is synonymous with $\ \mathcal{F}\psi\$, it would appear to be equation $(7.38)$ on p.$199$. Nov 25, 2021 at 2:04
• @lonza leggiera Yes it's page 199! Sorry for that, corrected that typo in my question! Nov 25, 2021 at 6:35
• The spectral theorem does not simply hold for any densely defined operator, but only for normal ones (I think Teschl only deals with self-adjoint ones). That's a big difference. So as a first step, you will have to define $H_0$ properly, including its domain. Nov 25, 2021 at 9:16
• @MaoWao Thank you, I fixed the post accordingly. Nov 25, 2021 at 9:30
• @h3fr43nd Yes, that's my reading of what equation $(7.35)$ is too. It's justification might be obvious to someone who's more familiar with spectral measures than I now am, but I certainly wouldn't say it's obvious to me (although it does seem very plausible). Nov 29, 2021 at 4:56

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.