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--- hamburgminicourse2017:why_self-adjoint_operators [2017/03/31 15:01] – markus
+++ hamburgminicourse2017:why_self-adjoint_operators [2017/04/03 16:12] (current) – markus
@@ Line 7: / Line 7: @@
 and thus it holds $z = A^*y$ and $y \in D(A^*)$. If $A=A^*$ on $D(A)$ the operator is called //symmetric// and one easily gets $D(A) \subseteq D(A^*)$. If further the domains coincide $D(A) = D(A^*)$ the operator is //self-adjoint//. For bounded operators we always have $D(A)=\mathcal{H}$ so the two notions are equivalent in that case.
-**Example.** $i\partial_x$ with its natural domain $H^1([0,1]) = \{ f \in L^2([0,1]) \mid \partial_x f \in L^2([0,1]) \}$ ([[wp>Sobolev_space#The_case_p_.3D_2]]) with all derivatives assumed weak. We study:
+**Example.** $i\partial_x$ with its natural domain $H^1([0,1]) = \{ f \in L^2([0,1]) \mid \partial_x f \in L^2([0,1]) \}$ (see [[wp>Sobolev_space#The_case_p_.3D_2|Sobolev space]]) with all derivatives assumed weak. We study:
 \[
 \langle f,i\partial_x g \rangle = i\int_0^1 \bar{f}\partial_x g \,dx = i \bar{f}(1)g(1)-i \bar{f}(0)g(0) - i\int_0^1 \partial_x\bar{f} g \,dx = i \bar{f}(1)g(1)-i \bar{f}(0)g(0) + \langle i\partial_x f,g \rangle.
 \]
-The operator thus fails to be symmetric. On $H_0^1([0,1])$ instead it //is// symmetric, but having $g \in H_0^1([0,1])$ is enough for the boundary terms to vanish, so $f \in D((i\partial_x)^*) \supsetneq D(i\partial_x)$. To get a self-adjoint operator we must make $D(i\partial_x)$ larger which lets $D((i\partial_x)^*)$ shrink, until they match. A boundary condition $f(1)=\alpha f(0), \alpha \in \mathbb{C}$ yields
+The operator thus fails to be symmetric. On $H_0^1([0,1])$ instead it //is// symmetric, but having $g \in H_0^1([0,1])$ is enough for the boundary terms to vanish, so $f \in D((i\partial_x)^*) \supsetneq D(i\partial_x)$. To get a self-adjoint operator we must make $D(i\partial_x)$ larger which lets $D((i\partial_x)^*)$ shrink, until they match. A boundary condition $f(1)=\alpha f(0)$, $\alpha \in \mathbb{C}$, yields
 \[
 \bar{f}(1)g(1)-\bar{f}(0)g(0)=|\alpha|^2\bar{f}(0)g(0)-\bar{f}(0)g(0)
@@ Line 18: / Line 18: @@
 that vanishes if $|\alpha|^2 = 1$. Any such choice of $\alpha$ gives a self-adjoint extension of the original operator but there is no unique such self-adjoint extension.
-**Physical note:** Clearly different $\alpha$ describe different "phase jumps" at the boundary that is now actually the joint where the two endpoints of the interval $[0,1]$ are identified. Without periodicity (or an infinitely large configuration space $\mathbb{R}$) there cannot be a momentum eigenstate (it would hit the wall), which somehow is supposed to be connected to self-adjoint operators.
+**Physical note.** Clearly different $\alpha$ describe different "phase jumps" at the boundary that is now actually the joint where the two endpoints of the interval $[0,1]$ are identified. Without periodicity (or an infinitely large configuration space $\mathbb{R}$) there cannot be a momentum eigenstate (it would hit the wall), which somehow is supposed to be connected to self-adjoint operators.
 **Why self-adjoint operators?** Sometimes it is argued that observables are represented by self-adjoint operators because they have //real// eigenvalues exclusively that are identified as the possible outcomes of a measurement. But why should a $z\in\mathbb{C}$ not be the outcome of a measurement, e.g. the pointer position of a clock? On the other side a non-self-adjoint operator may have only real eigenvalues. So this cannot really be the reason.
-Looking back at the old matrix mechanics of Heisenberg one identifies the central method of diagonalization. For an arbitrary linear operator $A$ on a Hilbert space it is asked, whether there is a unitary $U$ such that $UAU^*$ is diagonal, i.e. a multiplication operator (a function $x \mapsto f(x)$, after all it's //functional analysis//) on $L^2(\Omega,\mathcal{A},\mu)$ on some measure space $(\Omega,\mathcal{A},\mu)$. In the case of $A=i\partial_x$ this can be achieved through Fourier transform. In general the diagonalization is achieved through a **resolution of identity** (the generalization of an orthonormal basis), i.e. //every Hilbert space state can be assigned to one or more specific outcomes//.
+Looking back at the old matrix mechanics of Heisenberg one identifies the central method of diagonalization. For an arbitrary linear operator $A$ on a Hilbert space it is asked, whether there is a unitary $U$ such that $UAU^*$ is diagonal, i.e. a multiplication operator (a function $x \mapsto f(x)$, after all it's //functional analysis//) on $L^2(\Omega,\mathcal{A},\mu)$ on some measure space $(\Omega,\mathcal{A},\mu)$. In the case of $A=i\partial_x$ this can be achieved through Fourier transform. In general the diagonalization is achieved through a **resolution of identity** (the generalization of an orthonormal eigenbasis), i.e. //every Hilbert space state can be assigned to one or more specific outcomes//.
 The mathematical result telling us which operators can be diagonalized is called the **spectral theorem**. Needless to say self-adjoint operators allow for such a diagonalization but also the more general //normal// operators do. They are to self-adjoint operators what complex numbers are to reals: