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--- hamburgminicourse2017:the_spectrum_of_operators [2017/04/03 16:10] – markus
+++ hamburgminicourse2017:the_spectrum_of_operators [2017/04/04 17:38] (current) – admin
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 We start with two examples of the most familiar Hamiltonians in QM:
-**Example.** The free Hamiltonian $T = -\Delta$ is symmetric on the space of all infinitely-differentiable functions with compact support $\mathcal{C}^\infty_0(\Omega)$ and self-adjoint on either $H^2(\Omega)\cap H^1_0(\Omega)$ (zero boundary conditions), $H^2_p(\Omega)$ (periodic boundary conditions) or $H^2(\mathbb{R}^n)$ (no boundary). The spectrum of this operator is the continuum $[0,\infty)$ but it is not possible to find any eigenstates in the Hilbert space $L^2(\mathbb{R}^n)$. One could also assign a bilinear form $\langle \varphi,-\Delta \psi \rangle = \langle \nabla \varphi,\nabla \psi \rangle$ defined on $H^1 \times H^1$ to the Hamiltonian which defines the so-called [[wp>Energetic_space#Energetic_extension|energetic extension]] of $T$ as the representing operator of this bilinear form.
+**Example.** The free Hamiltonian $T = -\Delta$ is symmetric on the space of all infinitely-differentiable functions with compact support $\mathcal{C}^\infty_0(\Omega)$ and self-adjoint on either $H^2(\Omega)\cap H^1_0(\Omega)$ (zero boundary conditions), $H^2_p(\Omega)$ (periodic boundary conditions) or $H^2(\mathbb{R}^n)$ (no boundary). The spectrum of this operator on $H^2(\mathbb{R}^n)$ is the continuum $[0,\infty)$ but it is not possible to find any eigenstates in the Hilbert space $L^2(\mathbb{R}^n)$. One could also assign a bilinear form $\langle \varphi,-\Delta \psi \rangle = \langle \nabla \varphi,\nabla \psi \rangle$ defined on $H^1 \times H^1$ to the Hamiltonian which defines the so-called [[wp>Energetic_space#Energetic_extension|energetic extension]] of $T$ as the representing operator of this bilinear form.
 **Example.** Now take the hydrogen Hamiltonian $H = -\Delta -1/|x|$ on $L^2(\mathbb{R^3})$ with its well known spectrum that is discrete for $E<0$ and continuous above. The domain for the Hamiltonian to be self-adjoint will be discussed later. We already have an infinite sequence of eigenstates assigned to the negative eigenvalues, so do we need the //full// spectrum to represent an arbitrary state? Yes! Take a little gaussian blob $\psi$ far away from the origin, then the potential energy is clearly $\approx 0$ while the kinetic energy can be made arbirarily large. So we definitely need parts of the Hilbert space with positive energy expectation value to represent this state.
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   - Finally the last case might hold with $A-\lambda\mathrm{id}$ having a non-dense range, so the $\varepsilon_i$ are limited to certain subspaces and $\lambda$ is called part of the //residual spectrum//. This is possible for bounded operators but not for self-adjoint ones.
-**Note.** The [[wp>Lebesgue decomposition theorem]] allows a different [[wp>Decomposition_of_spectrum_(functional_analysis)#Decomposing_the_spectrum|partitioning of the spectrum]] into an //absolutely continuous//, //singular continuous//, and //pure point// part.
+**Note.** The spectral theorem and the [[wp>Lebesgue decomposition theorem]] allow a different [[wp>Decomposition_of_spectrum_(functional_analysis)#Decomposing_the_spectrum|partitioning of the spectrum]] for normal operators into an //absolutely continuous//, //singular continuous//, and //pure point// part.
 For a self-adjoint operator it holds that all $z \in \mathbb{C}$ with $\Im z \neq 0$ are in the resolvent set and the bounded operator $(A-z\mathrm{id})^{-1}$ is called a //[[wp>Resolvent formalism|resolvent]]// that establishes a very useful link to compex analysis. The norm of this operator obeys the following estimate: