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--- hamburgminicourse2017:the_spectrum_of_operators [2017/04/02 13:28] – markus
+++ hamburgminicourse2017:the_spectrum_of_operators [2017/04/04 17:38] (current) – admin
@@ Line 7: / Line 7: @@
 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:
@@ Line 39: / Line 39: @@
 **Lemma.** $z \in \mathbb{C}, \Im z \neq 0$, and $A$ self-adjoint, then $\|(A-z\mathrm{id})^{-1}\| \leq |\Im z|^{-1}$.
-**Proof.** Take $\varphi \in D(A)$, the domain of the oprator. Then
+**Proof.** Take $\varphi \in D(A)$, the domain of the operator. Then
-\[
+\begin{align*}
-\|(A-z\mathrm{id})\varphi\|^2 = |z|^2\|\varphi\|^2 + \|A\varphi\|^2 - 2(\Re z) \langle \varphi,A\varphi \rangle.
+\|(A-z\mathrm{id})\varphi\|^2 &= |z|^2\|\varphi\|^2 + \|A\varphi\|^2 - 2(\Re z) \langle \varphi,A\varphi \rangle\\
-\]
+&\geq |z|^2\|\varphi\|^2 + \|A\varphi\|^2 - 2|\Re z| \cdot |\langle \varphi,A\varphi \rangle|.
-By the Cauchy-Schwarz inequality $\langle \varphi,A\varphi \rangle \leq \|\varphi\| \cdot \|A\varphi\|$, so
+\end{align*}
+By the Cauchy--Schwarz inequality $|\langle \varphi,A\varphi \rangle| \leq \|\varphi\| \cdot \|A\varphi\|$, so
 \begin{align*}
-\|(A-z\mathrm{id})\varphi\|^2 &\geq |z|^2\|\varphi\|^2 + \|A\varphi\|^2 - 2(\Re z) \|\varphi\| \cdot \|A\varphi\| \\
+\|(A-z\mathrm{id})\varphi\|^2 &\geq |z|^2\|\varphi\|^2 + \|A\varphi\|^2 - 2|\Re z| \cdot \|\varphi\| \cdot \|A\varphi\| \\
-&= (\|A\varphi\| - (\Re z) \|\varphi\|)^2 + (\Im z)^2 \|\varphi\|^2 \geq (\Im z)^2 \|\varphi\|^2.
+&= (\|A\varphi\| - |\Re z| \cdot \|\varphi\|)^2 + |\Im z|^2 \|\varphi\|^2 \geq |\Im z|^2 \|\varphi\|^2.
 \end{align*}
-Take $(A-z\mathrm{id})\varphi = \psi \in \mathcal{H}$, then
+It holds that an inverse operator $(A-z\mathrm{id})^{-1}$ exists with dense domain (proof omitted here), thus taking $(A-z\mathrm{id})\varphi = \psi$ we can argue for a bounded inverse that fulfills
 \[
-|\Im z|^{-1} \|\psi\| \geq \|(A-z\mathrm{id})^{-1}\psi\|.
+|\Im z|^{-1} \|\psi\| \geq \|(A-z\mathrm{id})^{-1}\psi\|. \Box
 \]
  >> [[The abstract Cauchy problem in Banach space]]