1. Let $A$ be a self-adjoint (possibly unbounded) operator on $H$. Then corresponding to the purely continuous spectrum, there are no eigenvectors of $A$ in $H$, but there exist generalized eigenvectors of $A$ in the topological dual of the dense subspace.
2. the spectral theorem in this nuclear setting (theorem 1 in the link), or in more concrete terms: Let $A$ be a self-adjoint operator on an infinite dimensional separable Hilbert space $H$ with simple absolutely continuous spectrum $\sigma(A)=\mathbb{R}^{+}$. Then, there exists a Rigged Hilbert space $\phi \subset H \subset \phi^{\times}$, such that

a) $A \phi \subset \phi$ and $A$ is continuous on $\phi$. Therefore it may be continuously extented to $\phi^{\times}$.

b) For almost all $\omega \in \mathbb{R}^{+}$, with respect to the Lebesgue measure, there exists $| \omega \rangle \in \phi^{\times}$ so that $A | \omega \rangle=\omega | \omega \rangle$.

c) For any pair of vectors $\varphi,\psi \in \phi$, and any measurable function $f:\mathbb{R}^{+} \to \mathbb{C}$, we have that $$\langle \varphi| f(A) \psi \rangle= \int_{0}^{\infty} f(\omega) \langle \varphi | \omega \rangle \langle \omega| \psi \rangle d \omega,$$ with $\langle \omega|\psi \rangle^{*}=\langle \psi | \omega \rangle$.

d) The above spectral decomposition is implemented by a unitary operator $U:H \to L^2(\mathbb{R}^{+})$, with $U (\psi)=\langle \omega| \psi \rangle=\psi(\omega)$ and $(U A U)^{-1} \psi(\omega)=\omega \psi(\omega)=\omega \langle \omega| \psi \rangle$ for any $\psi \in \phi$. This means that $UAU^{-1}$ is the multiplication operator on $U \psi$.

e) For any pre existent Rigged Hilbert space $\phi \subset H \subset \phi^{\times}$, such that $A \phi \subset \phi$ with continuity and $A$ is satisfying our hypotheses, then items $b)-d)$ hold.

1. Let $A$ be a self-adjoint (possibly unbounded) operator on $H$. Then corresponding to the purely continuous spectrum, there are no eigenvectors of $A$ in $H$, but there exist generalized eigenvectors of $A$ in the topological dual of the dense subspace.

2. the spectral theorem in this nuclear setting (theorem 1 in the link), or in more concrete terms: Let $A$ be a self-adjoint operator on an infinite dimensional separable Hilbert space $H$ with simple absolutely continuous spectrum $\sigma(A)=\mathbb{R}^{+}$. Then, there exists a Rigged Hilbert space $\phi \subset H \subset \phi^{\times}$, such that

• a) $A \phi \subset \phi$ and $A$ is continuous on $\phi$. Therefore it may be continuously extented to $\phi^{\times}$.

• b) For almost all $\omega \in \mathbb{R}^{+}$, with respect to the Lebesgue measure, there exists $| \omega \rangle \in \phi^{\times}$ so that $A | \omega \rangle=\omega | \omega \rangle$.

• c) For any pair of vectors $\varphi,\psi \in \phi$, and any measurable function $f:\mathbb{R}^{+} \to \mathbb{C}$, we have that $$\langle \varphi| f(A) \psi \rangle= \int_{0}^{\infty} f(\omega) \langle \varphi | \omega \rangle \langle \omega| \psi \rangle d \omega,$$ with $\langle \omega|\psi \rangle^{*}=\langle \psi | \omega \rangle$.

• d) The above spectral decomposition is implemented by a unitary operator $U:H \to L^2(\mathbb{R}^{+})$, with $U (\psi)=\langle \omega| \psi \rangle=\psi(\omega)$ and $(U A U)^{-1} \psi(\omega)=\omega \psi(\omega)=\omega \langle \omega| \psi \rangle$ for any $\psi \in \phi$. This means that $UAU^{-1}$ is the multiplication operator on $U \psi$.

• e) For any pre existent Rigged Hilbert space $\phi \subset H \subset \phi^{\times}$, such that $A \phi \subset \phi$ with continuity and $A$ is satisfying our hypotheses, then items $b)-d)$ hold.

In physics, we very often say: Let $|x \rangle$ be the position basis, let $| p \rangle$ be the momentum basis. We then proceed to say $\langle x| x' \rangle=i \hbar \delta(x-x')$ and similarly up to some factor for the momentum. They are related by the Fourier transform $\langle x |p \rangle=e^{ipx}$. Strictly speaking, this cannot be correct, as the "eigenfunctions" of the position and momentum operator are not elements of the Hilbert space, they are not square integrable. One could say, the set of eigenvectors in $H$ is empty.

In physics, we very often say: Let $|x \rangle$ be the position basis, let $| p \rangle$ be the momentum basis. We then proceed to say $\langle x| x' \rangle= \delta(x-x')$ and similarly up to some factor for the momentum. They are related by the Fourier transform $\langle x |p \rangle=e^{ipx}$. Strictly speaking, this cannot be correct, as the "eigenfunctions" of the position and momentum operator are not elements of the Hilbert space, they are not square integrable. One could say, the set of eigenvectors in $H$ is empty.